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# What is the percent difference between 3 and 39 relative percent difference Using this tool you can find the percent difference between any two values. So, we think you reached us looking for answers to questions like: 1) What is the percentage difference between 3 and 39? 2) What is the absolute difference between 3 and 39? Or may be: What is the percent difference between 3 and 39 relative percent difference See the solution to these problems just after the Percent Difference Calculator below. ### Percent Difference Calculator Please fill in (or change values of) the two first boxes below and get answers to any combination of values: What is the percent difference between and ? Answer: % ### How to work out percentage differences - Step by Step To find the percent difference between two values x and y, use this formula: % difference = \|x - y\| Ã— 100% (\|x\| + \|y\|)/2 Where: x and y are two values of a give variable. Note that percent difference is not equivalent or equal to percent change. Percent change is used when comparing a new value to an old value where the old value is used as a reference. Percent difference is used just to compare two values without considering any order. It doesn't matter which of the two values is written first. That is why the absolute value() is often used. Percent difference is also called relative percent difference. () Note: absolute value means that we must consider always a positive value as for example: \|1\|=1 (absolute value of 1 is 1) and \|-1\|= 1 (absolute value of minus 1 is also 1). See more about percent difference here. Here are the solutions to the questions stated above: ### 1) What is the percentage difference between 3 and 39? Use the above formula to find the percent difference between two numbers. So, replacing the given values, we have the percent difference equation % Difference = [(\|39\| - \|3\|) / (\|3\| + \|39\|)/2] x 100 = [(39 - 3) / (3 + 39) / 2] x 100 = [36 / 42 / 2] x 100 = (36 / 21) x 100 = 171.42857142857 Where: y = 3 and x = 39, so \|y\| = 3 and \|x\| = 39 ### 2) What is the absolute difference between 3 and 39? This problem is not about percent difference, but about absolute difference: The absolute difference of two real numbers x, y is given by \|x âˆ’ y\|, the absolute value of their difference. The minus sign denotes subtraction and \|z\| means absolute value. Absolute difference describes the distance between the points corresponding to x and y on the real line. Now that you know what is absolute difference, The solution is very simple: absolute difference = \|Y - X\| = \|39 - 3\| = \|39 - 3\| = 36 ### Percent Difference Calculator Please link to this page! Just right click on the above image, choose copy link address, then past it in your HTML. ## Contact Us! Please get in touch with us if you: 1. Have any suggestions 2. Have any questions 3. Have found an error/bug 4. Anything else ... To contact us, please . ### Disclaimer While every effort is made to ensure the accuracy of the information provided on this website, we offer no warranties in relation to these informations.
# What are the number of necklaces made from 7 beads of different color? Contents ## How many necklaces can be formed with a different Coloured beads? 2520 Ways 8 beads of different colours be strung as a necklace if can be wear from both side. ## How many necklaces can you make with 10 beads of colors? This is easy: count all permutations of 10 beads, 10!, then divide by 20 because we counted each permutation 10 times due to rotation, and counted each of these twice because you can flip the necklace over. Thus the answer is 10!/20 = 181440. ## How many different necklaces can be formed using 9 different Coloured beads? This leaves us with 18,150 – 6 = 18,144 strings. The total number of necklaces we can form with these strings is 18,144 ÷ 9 = 2016. ## What are the number of ways in which 10 beads can be arranged to form a necklace? Answer: This is called a cyclic permutation. The formula for this is simply (n-1)!/2, since all the beads are identical. Hence, the answer is 9!/2 = 362880/2 = 181440. THIS IS AMAZING: How many commandments are in Mosaic law? ## How many necklace of 12 beads each can be made from 18 beads of different Colours? Correct Option: C First, we can select 12 beads out of 18 beads in 18C12 ways. Now, these 12 beads can make a necklace in 11! / 2 ways as clockwise and anti-clockwise arrangements are same. So, required number of ways = [ 18C12 . 11! ] / 2! ## How many ways can 12 beads be arranged on a bracket? 12 different beads can be arranged among themselves in a circular order in (12-1)!= 11! Ways. Now, in the case of necklace, there is not distinction between clockwise and anti-clockwise arrangements. ## How many necklaces can be formed with 6 white and 5 red beads if each necklace is unique how many can be formed? 5! but correct answer is 21. 6P6 = 720 or 6! ## How many different bangles can be formed from 8 different colored beads? How many different bangles can be formed from 8 different colored beads? Answer: 5,040 bangles . ## How many ways can you make a bracelet with 5 different beads? Thus for n=5, there are possible 4!/2=12 different bracelets.
# 2018 AMC 10A Problems/Problem 12 ## Problem How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? $$x+3y=3$$ $$\big\|\|x\|-\|y\|\big\|=1$$ $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$ ## Solution The graph looks something like this: $[asy] draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3,0)); dot((0,1)); [/asy]$ Now it's clear that there are $\boxed{3}$ intersection points. (pinetree1) x+3y=3 can be rewritten to x=3-3y. Substituting 3-3y for x in the second equation will give \|\|3-3y\|-y\|=1. Splitting this question into casework for the ranges of y will give us the total number of solutions. Case 1: y>1 3-3y will be negative so \|3-3y\| = 3y-3. \|3y-3-y\| = \|2y-3\| = 1 Subcase 1: y>3/2 2y-3 is positive so 2y-3 = 1 and y = 2 and x = 3-3(2) = -3 Subcase 2: 1<y<3/2 2y-3 is negative so \|2y-3\| = 3-2y = 1. 2y = 2 and so there are no solutions (y can't equal to 1) Case 2: y = 1 x = 0 Case 3: y<1 3-3y will be positive so \|3-3y-y\| = \|3-4y\| = 1 Subcase 1: y>4/3 3-4y will be negative so 4y-3 = 1 --> 4y = 4. There are no solutions (again, y can't equal to 1) Subcase 2: y<4/3 3-4y will be positive so 3-4y = 1 --> 4y = 2. y = 1/2 and x = 3/2 NOTE: Please fix this up using latex I have no idea how Solution by Danny Li JHS
# Bus308 2018 homework week 2 Week 2: Identifying Significant Differences – part 1 To Ensure full credit for each question, you need to show how you got your results. This involves either showing where the data you used is located or showing the excel formula in each cell. Be sure to copy the appropriate data columns from the data tab to the right for your use this week. As with our examination of compa-ratio in the lecture, the first question we have about salary between the genders involves equality – are they the same or different? What we do, depends upon our findings. 1 As with the compa-ratio lecture example, we want to examine salary variation within the groups – are they equal? Use Cell K10 for the Excel test outcome location. a What is the data input ranged used for this question: b Which is needed for this question: a one- or two-tail hypothesis statement and test ? Why: c. Step 1: Ho: Ha: Step 2: Significance (Alpha): Step 3: Test Statistic and test: Why this test? Step 4: Decision rule: Step 5: Conduct the test – place test function in cell k10 Step 6: Conclusion and Interpretation What is the p-value: What is your decision: REJ or NOT reject the null? Why? What is your conclusion about the variance in the population for male and female salaries? 2 Once we know about variance quality, we can move on to means: Are male and female average salaries equal? Use Cell K35 for the Excel test outcome location. (Regardless of the outcome of the above F-test, assume equal variances for this test.) a What is the data input ranged used for this question: b Does this question need a one or two-tail hypothesis statement and test? Why: c. Step 1: Ho: Ha: Step 2: Significance (Alpha): Step 3: Test Statistic and test: Why this test? Step 4: Decision rule: Step 5: Conduct the test – place test function in cell K35 Step 6: Conclusion and Interpretation What is the p-value: What is your decision: REJ or NOT reject the null? Why? What is your conclusion about the means in the population for male and female salaries? 3 Education is often a factor in pay differences. Do employees with an advanced degree (degree = 1) have higher average salaries? Use Cell K60 for the Excel test outcome location. Note: assume equal variance for the salaries in each degree for this question. a What is the data input ranged used for this question: b Does this question need a one or two-tail hypothesis statement and test? Why: c. Step 1: Ho: Ha: Step 2: Significance (Alpha): Step 3: Test Statistic and test: Why this test? Step 4: Decision rule: Step 5: Conduct the test – place test function in cell K60 Step 6: Conclusion and Interpretation What is the p-value: Is the t value in the t-distribution tail indicated by the arrow in the Ha claim? What is your decision: REJ or NOT reject the null? Why? What is your conclusion about the impact of education on average salaries? 4 Considering both the compa-ratio information from the lectures and your salary information, what conclusions can you reach about equal pay for equal work? Why – what statistical results support this conclusion? We offer the best custom writing services. We have done this question before; we can also do it for you… ## Calculate the price of your order 550 words We'll send you the first draft for approval by September 11, 2018 at 10:52 AM Total price: \$26 The price is based on these factors: Number of pages Urgency Basic features • Free title page and bibliography • Unlimited revisions • Plagiarism-free guarantee • Money-back guarantee On-demand options • Writer’s samples • Part-by-part delivery • Overnight delivery • Copies of used sources Paper format • 275 words per page • 12 pt Arial/Times New Roman • Double line spacing • Any citation style (APA, MLA, Chicago/Turabian, Harvard) # Our guarantees Delivering a high-quality product at a reasonable price is not enough anymore. That’s why we have developed 5 beneficial guarantees that will make your experience with our service enjoyable, easy, and safe. ### Money-back guarantee You have to be 100% sure of the quality of your product to give a money-back guarantee. This describes us perfectly. Make sure that this guarantee is totally transparent. ### Zero-plagiarism guarantee Each paper is composed from scratch, according to your instructions. It is then checked by our plagiarism-detection software. There is no gap where plagiarism could squeeze in. ### Free-revision policy Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result.
# From Ratios to Rates Videos and solutions to help Grade 6 students learn that ratios can be transformed to rates and unit rates. Common Core Standards: 5.NF.4a Related Topics: Lesson Plans and Worksheets for Grade 6 Lesson Plans and Worksheets for all Grades More Lessons for Grade 6 Common Core For Grade 6 ### New York State Common Core Math Module 1, Grade 6, Lesson 16 Lesson 16 Student Outcomes • Students recognize that they can associate a ratio of two quantities, such as the ratio of miles per hour is 5:2, to another quantity called the rate. • Given a ratio, students precisely identify the associated rate. They identify the unit rate and the rate unit. Lesson 16 Summary • A ratio of two quantities, such as 5 miles per 2 hours, can be written as another quantity called a rate. • The numerical part of the rate is called the unit rate and is simply the value of the ratio, in this case 2:5. This means that in 1 hour, the car travels 2. 5 miles. The unit for the rate is miles/hour, read “miles per hour”. NYS Math Module 1 Grade 6 Lesson 16 Classwork Example 1: Introduction to Rates and Unit Rates Diet cola was on sale last week; it cost \$10 for every 4 packs of diet cola. a. How much do 2 packs of diet cola cost? b. How much does 1 pack of diet cola cost Exploratory Challenge 1. Teagan went to Gamer Realm to buy new video games. Gamer Realm was having a sale: \$65 for 4 video games. He bought 3 games for himself and one game for his friend, Diego, but Teagan does not know how much Diego owes him for the one game. What is the unit price of the video games? What is the rate unit? 2. Four football fans took turns driving the distance from New York to Oklahoma to see a big game. Each driver set the cruise control during his leg of the trip, enabling him to travel at a constant speed. The men changed drivers each time they stopped for gas and recorded their driving times and distances in the table below. Use the given data to answer the following questions. a. What two quantities are being compared? b. What is the ratio of the two quantities for Andre’s car? What is the associated rate? c. Answer the same two questions in part (b) for the other three drivers. d. For each driver, circle the unit rate and put a box around the rate unit. 3. A publishing company is looking for new employees to type novels that will soon be published. The publishing company wants to find someone who can type at least 45 words per minute. Dominique discovered she can type at a constant rate of words in minutes. Does Dominique type at a fast enough rate to qualify for the job? Explain why or why not. Problem Set The Scott family is trying to save as much money as possible. One way to cut back on the money they spend is by finding deals while grocery shopping; however, the Scott family needs help determining which stores have the better deals. 1. At Grocery Mart, strawberries cost \$2.99 for 2lbs., and at Baldwin Hills Market strawberries are \$3.99 for 3 lbs. a. What is the unit price of strawberries at each grocery store? If necessary, round to the nearest penny. b. If the Scott family wanted to save money, where should they go to buy strawberries? Why? 2. Potatoes are on sale at both Grocery Mart and Baldwin Hills Market. At Grocery Mart, 5 lb. a bag of potatoes cost \$2.85 and at Baldwin Hills Market a 7 lb. bag of potatoes costs \$4.20. Which store offers the best deal on potatoes? How do you know? How much better is the deal? This video covers lessons 16 and 17 Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. [?] Subscribe To This Site
# § 5.5 Factoring Special Forms. Blitzer, Intermediate Algebra, 4e – Slide #67 The Difference of Two Squares If A and B are real numbers, variables, or. ## Presentation on theme: "§ 5.5 Factoring Special Forms. Blitzer, Intermediate Algebra, 4e – Slide #67 The Difference of Two Squares If A and B are real numbers, variables, or."— Presentation transcript: § 5.5 Factoring Special Forms Blitzer, Intermediate Algebra, 4e – Slide #67 The Difference of Two Squares If A and B are real numbers, variables, or algebraic expressions, then In words: The difference of the squares of two terms, factors as the product of a sum and a difference of those terms. Blitzer, Intermediate Algebra, 4e – Slide #68 The Difference of Two SquaresEXAMPLE SOLUTION Factor: We must express each term as the square of some monomial. Then we use the formula for factoring Express as the difference of two squares Factor using the Difference of Two Squares method Blitzer, Intermediate Algebra, 4e – Slide #69 The Difference of Two SquaresEXAMPLE SOLUTION Factor: The GCF of the two terms of the polynomial is 6. We begin by factoring out 6. Factor the GCF out of both terms Factor using the Difference of Two Squares method Blitzer, Intermediate Algebra, 4e – Slide #70 The Difference of Two SquaresEXAMPLE SOLUTION Factor completely: Express as the difference of two squares The factors are the sum and difference of the expressions being squared The factor is the difference of two squares and can be factored Blitzer, Intermediate Algebra, 4e – Slide #71 The Difference of Two Squares The factors of are the sum and difference of the expressions being squared CONTINUED Thus, Blitzer, Intermediate Algebra, 4e – Slide #72 Factoring CompletelyEXAMPLE SOLUTION Factor completely: Group terms with common factors Factor out the common factor from each group Factor out x + 3 from both terms Factor as the difference of two squares Blitzer, Intermediate Algebra, 4e – Slide #73 Factoring Special Forms Factoring Perfect Square Trinomials Let A and B be real numbers, variables, or algebraic expressions. Blitzer, Intermediate Algebra, 4e – Slide #74 Factoring Perfect Square TrinomialsEXAMPLE SOLUTION Factor: We suspect that is a perfect square trinomial because. The middle term can be expressed as twice the product of 4x and -5y. Express in form Factor Blitzer, Intermediate Algebra, 4e – Slide #75 Grouping & Difference of Two SquaresEXAMPLE SOLUTION Factor: Group as minus a perfect square trinomial to obtain a difference of two squares Factor the perfect square trinomial Rewrite as the difference of two squares Blitzer, Intermediate Algebra, 4e – Slide #76 Grouping & Difference of Two Squares Factor the difference of two squares. The factors are the sum and difference of the expressions being squared. Simplify CONTINUED Thus, Blitzer, Intermediate Algebra, 4e – Slide #77 The Sum & Difference of Two Cubes Factoring the Sum & Difference of Two Cubes 1)Factoring the Sum of Two Cubes: Same Signs Opposite Signs 2) Factoring the Difference of Two Cubes: Same Signs Opposite Signs Blitzer, Intermediate Algebra, 4e – Slide #78 The Sum & Difference of Two CubesEXAMPLE SOLUTION Factor: Rewrite as the Sum of Two Cubes Factor the Sum of Two Cubes Simplify Thus, Blitzer, Intermediate Algebra, 4e – Slide #79 The Sum & Difference of Two CubesEXAMPLE SOLUTION Factor: Rewrite as the Difference of Two Cubes Factor the Difference of Two Cubes Simplify Thus, Download ppt "§ 5.5 Factoring Special Forms. Blitzer, Intermediate Algebra, 4e – Slide #67 The Difference of Two Squares If A and B are real numbers, variables, or." Similar presentations
# Definition and Properties of Logarithms Please provide a rating, it takes seconds and helps us to keep this resource free for all to use The following math revision questions are provided in support of the math tutorial on Definition and Properties of Logarithms. In addition to this tutorial, we also provide revision notes, a video tutorial, revision questions on this page (which allow you to check your understanding of the topic) and calculators which provide full, step by step calculations for each of the formula in the Definition and Properties of Logarithms tutorials. The Definition and Properties of Logarithms calculators are particularly useful for ensuring your step-by-step calculations are correct as well as ensuring your final result is accurate. Not sure on some or part of the Definition and Properties of Logarithms questions? Review the tutorials and learning material for Definition and Properties of Logarithms Logarithms Learning Material Tutorial IDTitleTutorialVideo Tutorial Revision Notes Revision Questions 13.1Definition and Properties of Logarithms ## Definition and Properties of Logarithms Revision Questions 1. . Which of the following is equivalent to 35 = 243? 1. log3 5 = 243 2. log5 3 = 243 3. log3 243 = 5 4. log5 243 = 3 2. . The expression log 62 is equivalent to 1. 6 log 2 2. 2 log 6 3. 2 log 10 4. log6 36 3. . The number 32 in the expression log2 32 = 5 represents the 1. Base 2. Argument 3. Logarithm 4. Exponent 4. . The expression log16 4 is equal to 1. 2 2. 4 3. 64 4. 1/2 5. . What is the result of the operation 4 log2 8 + log3 9? 1. 5 2. 9 3. 12 4. 14 6. . What is the result of the operations log2 16 ∙ log4 64 1. 12 2. 24 3. 80 4. 1024 7. . What is the result of the operation log4 80 - log4 5 1. 2 2. 75/4 3. 16 4. 100 8. . What is the value of the expression log4 8 - log8 4 1. 5/6 2. 6/5 3. 1/2 4. -1/2 9. . The expression 2 log4 16/log7 49 - log 10 1. 2 2. 1 3. 0 4. -8 10. . Which of the following options is equivalent to log35x7 1. 1/2 log3 5 + 7 log3 x 2. 1/2 (log3 5 + 7 log3 x ) 3. 7 log3 5x 4. 7/2 log3 5x 11. . Which of the following options is equivalent to 40 log3 a - (8 log3 b + 16 log3 c) 1. log3 40a/148bc 2. log3 (40a - 8b - 16c) 3. log3 a40/b8 c16 4. 48 log3 a/bc 12. . What is the value of the expression log1/4 32 + 5 log3 (1/3)2 1. 41/4 2. 15/2 3. -25/2 4. -15/2 13. . What is the value of the expression log3 1/3 + log2 1/2 - 2 log5 1/5 1. -4 2. -2 3. 0 4. 4 14. . What is log 1 + 3 log5 25 1. 4 2. 6 3. 7 4. 26 15. . What is log 1/100 + log 1/10 - log 1 + log 10 - log 100 1. -1 2. 0 3. 1 4. 2 ## Whats next? Enjoy the "Definition and Properties of Logarithms" practice questions? People who liked the "Definition and Properties of Logarithms" practice questions found the following resources useful: 1. Practice Questions Feedback. Helps other - Leave a rating for this practice questions (see below) 2. Logarithms Math tutorial: Definition and Properties of Logarithms. Read the Definition and Properties of Logarithms math tutorial and build your math knowledge of Logarithms 3. Logarithms Video tutorial: Definition and Properties of Logarithms. Watch or listen to the Definition and Properties of Logarithms video tutorial, a useful way to help you revise when travelling to and from school/college 4. Logarithms Revision Notes: Definition and Properties of Logarithms. Print the notes so you can revise the key points covered in the math tutorial for Definition and Properties of Logarithms 5. Check your calculations for Logarithms questions with our excellent Logarithms calculators which contain full equations and calculations clearly displayed line by line. See the Logarithms Calculators by iCalculator™ below. 6. Continuing learning logarithms - read our next math tutorial: Exponential and Logarithmic Equations ## Help others Learning Math just like you Please provide a rating, it takes seconds and helps us to keep this resource free for all to use
## Precalculus (6th Edition) Blitzer The inverse is $A=\left[ \begin{matrix} 1 & 0 & 1 \\ 2 & 1 & 3 \\ -1 & 1 & 1 \\ \end{matrix} \right]$. Consider the given matrix, $A=\left[ \begin{matrix} -2 & 1 & -1 \\ -5 & 2 & -1 \\ 3 & -1 & 1 \\ \end{matrix} \right]$ We have to find the inverse of matrix $A$ by using a calculator. We will follow the steps given below: (1) We select the matrix $A$ from the matrix edit menu and then enter or accept the dimensions. (2) Use ${{2}^{nd}}$ and press the matrix button. (3) Select the matrix $A$ and use the function ${{X}^{-1}}$ for the inverse of matrix $A$ and press enter. (4) We find the ${{A}^{-1}}$ Therefore, the inverse of the matrix $A$ is $A=\left[ \begin{matrix} 1 & 0 & 1 \\ 2 & 1 & 3 \\ -1 & 1 & 1 \\ \end{matrix} \right]$ Now, check the result: show that we can multiply matrix $A$ with ${{A}^{-1}}$ to obtain the identity matrix. Hence, the inverse of the given matrix is correct.
# Distance - Rate - Time – Different Directions06:36 minutes Video Transcript ## TranscriptDistance - Rate - Time – Different Directions Tim Carcrashian, true to his name, works for a vehicle crash testing company. His job is to set up the slow-motion-camera used to document car crashes. The set-up of the camera is very expensive and time-consuming. To do the job just right, Tim needs to know exactly when and where the cars will crash. How can we help him? ### The distance-rate-time pyramid This is a distance-rate-time problem - with cars coming from different directions. For the test crash, Car 1 and Car 2 will drive towards each other. Car 1 will pass Point A going 50 miles per hour, and Car 2 will pass Point B going 70 miles per hour. The distance from Point A to Point B is 1.2 miles. Exactly where and when will the cars crash? Let’s figure out how to solve this problem. Distance is equal to rate times time. This pyramid can help us remember the formula. A shorter, more common, way to say this d = rt. ### Write equations Ok, let's get started...It's a good idea to use a table to organize the data. First, let's fill in the values with what we already know, the rate for each car. We don’t know the value for time yet, just that it's the same for both cars, so for now, we can represent this with the variable, 't'. We can write two equations to describe what's going on here: d = 50t. And...d = 70t. Let's write that information into the table. Since we know the total distance the two cars will a travel together is 1.2 miles, we can set up an equation and solve for the unknown time. 1.2 = 50t + 70t. Evaluating this we get: t = 1/100 of an hour. Let's fill in the table... Now, we can use the value we just calculated for the time to determine the actual distances both cars will drive before crashing and fill in the table. d = 50(0.01), which is equal to 0.5 d = 70(0.01), giving us 0.7 ### The distance-rate-time formula This is just great! Tim is so happy! Using the distance-rate-time formula, coming from different directions, now, Tim knows Car 1 will drive exactly 1/2 of a mile and Car 2 will drive 7/10 of a mile, then crash, bang, boom! As a bonus, Tim also knows exactly when this will happen. The cars will crash after 36 seconds, which is equal to 1/100 of an hour. He sets up the Slow-Motion-Camera Unit just right. He’s so good at his job, maybe he'll get a promotion and a big raise! Oh no. One week later, Tim learns the test track will be under construction in order to improve the testing conditions. During construction, the right side of the track will be shortened. Instead of the track having a distance of 1/2 of a mile on the left side and 7/10 of a mile on the right side... ### Calculate a new rate The right side will be only 4/10 of a mile long. Tim doesn't want to change the set-up of the camera. What can he do so the cars will still crash at the same place and time? He can't change the distance or the time, but he can change the speed of Car 2. Let's calculate the new rate using the distance-rate time formula. As before, we have d = rt. To isolate the rate, divide distance by time. r = 0.4 / 0.01. The rate is 40 miles per hour. So, if Car 2 travels on the track at a rate of 40 miles per hour for 4/10 of a mile, after 1/100 of an hour, the two cars will crash in just the right spot. ### Change the distance Another week goes by... Tim gets an email... The track improvements are all done, and the new plan is to crash test one of the cars driving at a very high speed. Car 2 will be tested driving 90 miles per hour. Tim doesn't have time to change the camera. What should he do? He can't change the speed or the time, but he can change the distance Car 2 will drive. If Car 2 drives 90 miles per hour for 1/100 of an hour, use the distance-rate-time formula to calculate the new distance it must drive so the camera can capture the crash. Let's reuse our formula from before: distance is equal to rate times time. 90(0.01) = 0.9 miles. So Tim can get the perfect camera shot, Car 2 must drive 9/10 of a mile at 90 miles per hour for 1/100 of an hour . ### Summary Let's summarize: The distance-rate-time pyramid is a handy graphic organizer to help you remember and understand how to solve these kind of problems. t = d/r, r = d/t, d = rt. After all his hard work, Tim can finally watch the crash test. Oh no, that's not test Car 2...That looks a lot like Tim's boss' car... ## Videos in this Topic Equations (7 Videos) ## Distance - Rate - Time – Different Directions Exercise ### Would you like to practice what you’ve just learned? Practice problems for this video Distance - Rate - Time – Different Directions help you practice and recap your knowledge. • #### Determine when and where the car will crash. ##### Hints Fill in the table: $\begin{array}{\|l\|c\|c\|c\|} \hline &r&t&d\\ \hline \text{car 1}&50&t&?\\ \hline \text{car 2}&70&t&?\\ \hline \end{array}$ Don't forget the mnemonic device. Determine the unknown quantity, and use the appropriate equation: • $d=r\times t$ • $r=\frac dt$ • $t=\frac dr$ The sum of the distances is equal to $1.2$ (miles). ##### Solution The red car travels at $50$ mph, while the blue car travels in the opposite direction at $70$ mph. This is a distance-rate-time-problem with different directions. $\begin{array}{\|l\|c\|c\|c\|} \hline &r&t&d\\ \hline \text{car 1}&50&t&50t\\ \hline \text{car 2}&70&t&70t\\ \hline \end{array}$ Use the equation $d=r\times t$. Since the car cannot travel a distance greater than $1.2$ miles, we write the equation: $\begin{array}{rclcl} 50t+70t & = & 1.2 &&\|\text{Combine Like Terms} \\ 120t&=&1.2 \\ \color{#669900}{\div120} && \color{#669900}{\div120} &&\|\text{Opposite Operations} \\ t &=& 0.01 \end{array}$ $t=0.01$ is the time in hours. If we want to know how many seconds this is, we can calculate the conversion. There are $60$ minutes in an hour and $60$ seconds for every minute. Multiply... there are $3,600$ seconds in one hour. $\begin{array}{rcll} \dfrac{60 ~\text{min.}}{1 ~\text{hr.}} \times \dfrac{60 ~\text{sec.}}{1 ~\text{min.}} &=& \dfrac{3,600 ~\text{sec.}}{1 ~\text{hr.}} \end{array}$ $~$ $\dfrac{3,600 ~\text{ sec.}}{1 ~\text{ hr.}}\times 0.01 ~\text{ hr.} = 36 ~\text{ sec.}$ Now we know that the crash will happen after $36$ seconds. To find out where the crash will occur, we use the equation $d=r\times t$. Plugging in the known values gives us: $d = \dfrac{50 ~\text{mi.}} {\text{hr.}} \times 0.01 ~\text{hr.} = 0.5~\text{mi.}$ for the red car $d = \dfrac{70 ~\text{mi.}} {\text{hr.}} \times 0.01 ~\text{hr.} = 0.7~\text{mi.}$ for the blue car To document the crash, Johnny has to set up the camera $0.5$ miles far away from Point A. • #### Calculate the rate at which the car travels. ##### Hints Keep this triangle in mind - it's magic. If it takes $0.5$ hours to travel a distance of $30$ miles, the rate is $60$ mph. ##### Solution You can remember the formulas by using the Distance Rate Time Triangle. Here we know • Time: $t=0.01$ hours • Distance: $0.4$ miles We need to know the rate of the blue car, so we choose the formula in the middle $r=\frac dt$. By plugging in the known values, we get: $r=\frac{0.4~\text{mi.}}{0.01~\text{hr.}}=40 ~\text{mph.}$ • #### Evaluate the time, if the rate changes. ##### Hints First gather all given information: • $r=80$ • $t=0.01$ Decide which formula to use. If you already know the rate and the time, use $d=r\times t$ to find the distance. For example: The given information: • $r=90$ • $t=0.01$ To calculate the distance, we can calculate: $d = rt$ $d=90\times 0.01=0.9$ miles ##### Solution The speed of the blue car is increased to $80$ mph. Johnny's a bit lazy and doesn't want to change the camera position. He can't change the rate or the time, but he can change the distance: To figure this out, use the equation: $d=r\times t$. Given: • Rate: $80$ mph • Time: $0.01$ hrs. $d=80\times 0.01=0.8$ So the blue car must be a distance of $0.8$ miles from the crash point. The distance between the two cars at the start is $0.5+0.8=1.3$ miles. • #### Find out when and where the two brothers will meet. ##### Hints This is a distance, rate, and time problem with different directions. Use the formula $d=r\times t$. Keep in mind that the sum of the distances traveled is $3$ miles. Use the opposite operations: • Multiplication ($\times~\longleftrightarrow~\div$) • Division ($\div~\longleftrightarrow~\times$) • Addition ($+~\longleftrightarrow~-$) • Subtraction ($-~\longleftrightarrow~+$) You have to calculate the time first, and then you can calculate the distance. ##### Solution Eugene is riding his bike at a rate of $10$ mph, while Shane is running at a rate of $2$ mph. Here we have a distance rate time problem involving different directions. First let's consider the distance. We know that $d=r\times t$, and the time $t$ is unknown. • $d=10t$ for Eugene • $d=2t$ for Shane But, we do know the total distance, so we can write the equation: $10t+2t=3$ Combine the like terms: $12t=3$. Finally, divide by $12$: $t=\frac3{12}=\frac14=0.25$ hours. So the brothers will meet after $0.25$ hours or $15$ minutes. To determine the distance each boy will travel, we use the formula $d=r\times t$ once again: • Eugene: $d=10\times 0.25=2.5$ miles • Shane: $d=2\times 0.25=0.5$ miles In summary: • We know that the brothers will meet after 15 minutes. • Eugene starts $2.5$ miles away from the meeting point. • Shane starts $0.5$ miles away from the meeting point. • #### Summarize the distance, rate, and time problem. ##### Hints Use the magic triangle: • $d$ over $r$ as well as $d$ over $t$ • $r$ and $t$ on the same level Remember what the variables stand for: • Distance • Rate • Time ##### Solution To solve problems, use the Distance Rate Time triangle. • Distance • Rate • Time Starting at the top, if we want to calculate distance, $d$, we can see that $r$ and $t$ are on the same level. So our equation is modified: $d=r\times t$. To calculate the rate, $r$, we look at the other two variables, $d$ and $t$. $d$ stands over $t$ in the triangle, so we have $r=\frac dt$. To find time, $t$, we notice that $d$ is over $r$, so our final equation is: $t=\frac dr$. • #### Determine when Jim has to leave his home to make it to his appointments on time. ##### Hints First determine the time, $t$, Jim needs to travel each distance. You should use the following equation: $t=\frac dr$ Next, calculate the time in hours. Convert to minutes if necessary. Subtract your answer from the time Jim plans to arrive at each destination. Remember, there are $60$ minutes in an hour. ##### Solution What do we know? • The distance from Jim's house to each destination • Jim's walking rate is $2$ mph To calculate the time it takes Jim to travel to each of these destinations, we use the formula $t=\frac dr$. School Distance = $0.8$ miles $t=\frac{0.8}{2}=0.4$ hours. Multiply by 60: $t=0.4\times 60=24$ minutes. Subtract $24$ minutes from the arrival time: $8:00-0:24=7:36$ a.m. To get to school on time, Jim has to leave his house by $7:36$ a.m. Grandma's House Distance = $1$ mile $t=\frac{1}{2}=0.5$ hours. Again, multiply by $60$: $t=0.5\times 60=30$ minutes. Subtracting $30$ minutes from the arrival time gives us: $10:00-0:30=9:30$ a.m. Jim has to leave by $9:30$ a.m. to get to his grandma's house by $10:00$ a.m. Jack Distance = $3$ miles $t=\frac{3}{2}=1.5$ hours or, $1$ hour and $30$ minutes. Subtract this time from the arrival time (We used the $24$-hour clock for convenience.): $14:00-1:30=12:30$, or 12:30 pm. Jim has to leave by 12:30 p.m. to meet Jack at $2$ p.m. Cinema Distance = $4$ miles Oh, what a day. Jim deserves a little bit of time to kick back and relax. To get to the movies on time, he has to start $t=\frac42=2$ hours before the film starts playing. So he should leave his house at $6$ p.m. He walks $2$ hours to watch a movie!? I hope it's a good movie!
# How many terms of the A.P. are needed to give the sum –25? Question: How many terms of the A.P. $-6,-\frac{11}{2},-5, \ldots$ are needed to give the sum $-25 ?$ Solution: Let the sum of n terms of the given A.P. be –25. It is known that, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, where $n=$ number of terms, $a=$ first term, and $d=$ common difference Here, $a=-6$ $d=-\frac{11}{2}+6=\frac{-11+12}{2}=\frac{1}{2}$ Therefore, we obtain $-25=\frac{n}{2}\left[2 \times(-6)+(n-1)\left(\frac{1}{2}\right)\right]$ $\Rightarrow-50=n\left[-12+\frac{n}{2}-\frac{1}{2}\right]$ $\Rightarrow-50=n\left[-\frac{25}{2}+\frac{n}{2}\right]$ $\Rightarrow-100=n(-25+n)$ $\Rightarrow n^{2}-25 n+100=0$ $\Rightarrow n^{2}-5 n-20 n+100=0$ $\Rightarrow n(n-5)-20(n-5)=0$ $\Rightarrow n=20$ or 5
Question Video: Solving a Matrix Equation by Finding the Inverse \| Nagwa Question Video: Solving a Matrix Equation by Finding the Inverse \| Nagwa # Question Video: Solving a Matrix Equation by Finding the Inverse Mathematics • Third Year of Secondary School ## Join Nagwa Classes Given that [−1, −6, −6 and 0, −2, −4 and 2, 4, 7] [𝑥, 𝑦, 𝑧] = [−8, 2, −9], find the values of 𝑥, 𝑦, and 𝑧. 07:50 ### Video Transcript Given that the product of a three-by-three matrix with elements negative one, negative six, negative six, zero, negative two, negative four, two, four, and seven and a vector containing elements 𝑥, 𝑦, and 𝑧 is equal to the vector containing elements negative eight, two, and negative nine, find the values of 𝑥, 𝑦, and 𝑧. We know that if matrix 𝐴 is multiplied by vector 𝐗 and it equals vector 𝐁, that vector 𝐗 equals the inverse matrix of 𝐴 multiplied by vector 𝐁. We should also remember that order matters. We cannot multiply 𝐁 by the inverse matrix of 𝐴. We must multiply it in this order, the inverse of 𝐴 times 𝐁. This means that our first step is to find the inverse of matrix 𝐴. You could use a graphing calculator to help you do this step. But we will take a look at how to do the steps by hand without a graphing calculator. This is a multistep process and will be the most complex part of solving this problem. The first step in finding the inverse of matrix 𝐴, if the inverse exists, is to find the cofactor matrix, 𝐶, equal to the three-by-three matrix 𝐶 sub 𝑖𝑗, whose entries are the determinants of the corresponding matrix minors multiplied by the alternating sign, negative one, to the power of 𝑖 plus 𝑗. Then, we transpose the cofactor matrix to find the adjugate matrix, sometimes referred to as the classical adjoint matrix. Then, we will calculate the determinant of 𝐴 and multiply the adjugate matrix by the reciprocal of the determinant. Since 𝐴 is a three-by-three matrix, we must use the following formula to find the three-by-three cofactor matrix. Notice the pattern of alternating positive and negative signs in front of each determinant. If we assign the wrong sign to one element, it can completely change our final result. To begin, we find the determinant in the first row, first column. We look to matrix 𝐴 to find the four elements in the 𝑒-, 𝑓-, ℎ-, and 𝑖-positions. These are negative two, negative four, four, and seven. To calculate the determinant, we multiply the top-left element by the bottom-right element. And then we subtract the bottom-left element multiplied by the top-right element. And so the first element of our cofactor matrix is found by negative two times seven minus four times negative four, which is negative 14 minus negative 16, which equals positive two. So the top-left element in our new matrix is going to be positive two. The determinant from the first row, second column would then look like this: zero times seven minus two times negative four, which is zero minus negative eight, which equals positive eight. Notice, however, that because of its position in the cofactor matrix, we need to apply a negative sign. So that element in the first row, second column is negative eight. Now we’re looking for the element in the first row, third column: zero times four minus two times negative two. This element comes out to positive four. Let’s repeat this process for the second row of cofactors. The signs for this row are negative, positive, negative. The first determinant in this row comes out to negative 18. But we need to take the negative of negative 18. And so we have positive 18 in the second row, first column position. In the second row, second column, we have positive five. And the determinant in the second row, third column comes out to positive eight. But we need to take the negative of that value. So this element will actually be negative eight. And finally we’ll calculate the determinants in the third row using first a positive sign, then negative, then positive again. For the third row, first column, we get positive 12. In the third row, second column, we get four. But we need to multiply that by negative one, which gives us negative four. And in the final position, we get positive two. We now have our complete cofactor matrix. Our next step is to find the adjugate matrix. We do this by transposing the rows and the columns. And that looks like a reflection across the diagonal. The elements along the diagonal remain the same. But we’ll switch the negative eight and positive 18, the four with the 12, and the negative eight with the negative four. So the adjugate of matrix 𝐴 is two, 18, 12, negative eight, five, negative four, four, negative eight, two. Our final step is to then multiply the adjugate matrix by the reciprocal of the determinant of the original matrix 𝐴. To calculate the determinant, we can use this formula. We’ll multiply each element in the top row by the matching cofactors we found in the previous step. The determinant of the first row, first column was two. The first row, second column was eight. And the first row, third column was four. The expression simplifies to negative two plus 48 minus 24. So the determinant of 𝐴 is 22. So the reciprocal of the determinant of matrix 𝐴 is then one over 22. Altogether, the inverse of matrix 𝐴 is one over 22 times the adjugate. After multiplying each element by one over 22, the product looks like this. As we originally said, to solve this system, we need to calculate the inverse of matrix 𝐴 times 𝐁. The resulting array will contain three elements: a value for 𝑥, a value for 𝑦, and a value for 𝑧. To find 𝑥, we multiply negative eight by one over 11, two by nine over 11, and negative nine by six over 11, then find the sum of these products. This expression simplifies to negative 44 over 11. So the value of 𝑥 is negative four. The value of 𝑦 is found by calculating negative eight times negative four over 11 plus two times five over 22 plus negative nine times negative two over 11, which simplifies to 55 over 11. So 𝑦 is equal to positive five. And finally to solve for 𝑧, we must calculate negative eight times two over 11 plus two times negative four over 11 plus negative nine times one over 11. This expression simplifies to negative 33 over 11. So 𝑧 is equal to negative three. Therefore, under these conditions, 𝑥 is equal to negative four, 𝑦 is equal to five, and 𝑧 is equal to negative three. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
Simplifying Radicals Topic Index \| Algebra Index \| Regents Exam Prep Center (For this lesson, the term "radical" will refer only to "square root".) When working with the simplification of radicals you must remember some basic information about perfect square numbers. You need to remember: Perfect Squares 4 = 2 x 2 9 = 3 x 3 16 = 4 x 4 25 = 5 x 5 36 = 6 x 6 49 = 7 x 7 64 = 8 x 8 81 = 9 x 9 100 = 10 x 10 Radicals (square roots) = 2 = 3 = 4 = 5 = 6 = 7 = 8 = 9 = 10 While there are certainly many more perfect squares, the ones appearing in the charts above are the ones most commonly used. To simplify means to find another expression with the same value. It does not mean to find a decimal approximation. To simplify (or reduce) a radical: 1. Find the largest perfect square which will divide evenly into the number under your radical sign. This means that when you divide, you get no remainders, no decimals, no fractions. Reduce: the largest perfect square that divides evenly into 48 is 16. If the number under your radical cannot be divided evenly by any of the perfect squares, your radical is already in simplest form and cannot be reduced further. 2. Write the number appearing under your radical as the product (multiplication) 3. Give each number in the product its own radical sign. 4. Reduce the "perfect" radical which you have now created. What happens if I do not choose the largest perfect square to start the process? If instead of choosing 16 as the largest perfect square to start this process, you choose 4, look what happens..... Unfortunately, this answer is not in simplest form. The 12 can also be divided by the perfect square (4). If you do not choose the largest perfect square to start the process, you will have to repeat the process. Example: Reduce: Don't let the number in front of the radical distract you. It is simply "along for the ride" and will be multiplied times our final answer. The largest perfect square dividing evenly into 50 is 25. Reduce the "perfect" radical and multiply times the 3 (who is "along for the ride") Note: The examples shown in these lessons on radicals show ALL of the steps in the process. It may NOT be necessary for you to list EVERY step. As long as you understand the process and can arrive at the correct answer, you are ALL SET!! Topic Index \| Algebra Index \| Regents Exam Prep Center Created by Donna Roberts
Home Practice For learners and parents For teachers and schools Textbooks Full catalogue Pricing Support We think you are located in United States. Is this correct? # Working with formulas ## 22.4 Working with formulas In science and in many other fields of work, we need to be able to work with formulas. For example, if you know the area of a square and need to find its side length, you can rearrange the formula to make the side length the subject of the formula. In other words, you solve the equation for the variable that you need. The formula for the area of a square is: $A = s^{2}$ where $$A$$ is the area of the square and $$s$$ is the side of the square. We need to solve this equation for $$s$$. To find $$s$$, we use the inverse operation, which is the square root: \begin{align} \sqrt{A} &= \sqrt{s^{2}} \\ &= s \\ \therefore s &= \sqrt{A} \end{align} So if the area of a square is $$25 \text{ cm}^2$$, then the side length is $$s = \sqrt{25} = 5 \text{ cm}$$. ## Worked Example 22.4 Working with formulas Use this equation to find a formula for $$a$$. $F = ma$ We need to rearrange the given equation in order to have just $$a$$ on its own. ### Use inverse operations to rearrange the equation. Divide both sides by $$m$$. $\frac{F}{m} = \frac{ma}{m}$ ### Simplify. $\frac{F}{m} = a$ ### Switch the two sides around. $a = \frac{F}{m}$ You now have a formula that you can solve to find values for $$a$$. ## Worked Example 22.5 Working with formulas Find a formula for $$V$$. $C = \frac{n}{V}$ We need to rearrange the given equation in order to have $$V$$ on its own. ### Use inverse operations to rearrange the equation. Multiply both sides by $$V$$ and simplify. \begin{align} CV &= \left( \frac{n}{V} \right)V \\ &= n \end{align} ### Divide both sides by $$C$$ and simplify. \begin{align} \frac{CV}{C} &= \frac{n}{C} \\ V &= \frac{n}{C} \end{align} You now have a formula for $$V$$. temp text
# Rational Inequality How to solve a rational inequality: 2 examples and their solutions. ## Example 1 ### Solution Before solving the inequality, first set (denominator) ≠ 0. Excluded Value See 1/x. The denominator is x. So x ≠ 0. Next, see 1/2x. The denominator is 2x. So 2x ≠ 0. Then x ≠ 0. From the denominators, you found that x ≠ 0. The x values should satisfy this condition. Next, solve the rational inequality. First, find the least common multiple, LCM, of the denominators. The denominators are x and 2x. So the LCM is 2x. Change the denominators of the rational expression to the LCM: 2x. See 1/x. The denominator is x. The factor 2 is missing. So multiply 2 to both of the numerator and the denominator. 1/x = [1/x]⋅[2/2] See -1/2x. The denominator is 2x: the LCM. So you don't have to change -1/2x. Write the inequality sign ≥. See the right side 3. The denominator is 1. 2x is missing. So multiply 2x to both of the numerator and the denominator. 3 = 3⋅[2x/2x] So 1/x - 1/2x ≥ 3 becomes [1/x]⋅[2/2] - 1/2x ≥ 3⋅[2x/2x]. [1/x]⋅[2/2] = 2/2x 3⋅[2x/2x] = 6x/2x 2/2x - 1/2x = 1/2x Move 6x/2x to the right side. Multiply -1 to both sides. Then (6x - 1)/2x ≤ 0. Multiplying (-) on both sides changes the order of the inequality sign: ≥ → ≤. Linear Inequality (One Variable) Divide both sides by 2. Then (6x - 1)/x ≤ 0. Move the denominator x to the numerator. (6x - 1)/x → x(6x - 1) This is multiplying the square of the denominator, x2, to both sides. x ≠ 0 So you can multiply x2 on both sides. And x2 is (+). So the order of the inequality sign doesn't change. And write ≠ 0 under x. Find the zeros of x(6x - 1). Case 1) x = 0 (We know that x ≠ 0. This is the zero you're going to use when graphing the inequality on the x-axis.) Case 2) 6x - 1 = 0 Then x = 1/6. Case 1) x = 0 Case 2) x = 1/6 So the zeros are x = 0, 1/6. Draw y = x(6x - 1) on the x-axis. First point the zeros x = 0 and 1/6. And draw a parabola that passes through x = 0 and 1/6. x ≠ 0 So draw an empty circle on x = 0. See x(6x - 1) ≤ 0. The left side is less than or equal to 0. So color the region where the graph is below the x-axis (y = 0). So draw a full circle on x = 1/6. (Don't fill the empty circle on x = 0. This is the excluded value.) The colored region is 0 < x ≤ 1/6. So 0 < x ≤ 1/6 ## Example 2 ### Solution Before solving the inequality, first set (denominator) ≠ 0. See 4/(x - 1). The denominator is (x - 1). So x - 1 ≠ 0. Then x ≠ 1. Next, see 1/x. The denominator is x. So x ≠ 0. From the denominators, you found that x - 1 ≠ 0, x ≠ 0. The x values should satisfy these conditions. Next, solve the rational inequality. First, find the least common multiple, LCM, of the denominators. The denominators are (x - 1) and x. So the LCM is x(x - 1). Change the denominators of the rational expression to the LCM: x(x - 1). See 4/(x - 1). The denominator is (x - 1). The factor x is missing. So multiply x to both of the numerator and the denominator. 4/(x - 1) = [4/(x - 1)]⋅[x/x] See the next term +1. The denominator is 1. x(x - 1) is missing. So multiply x(x - 1) to both of the numerator and the denominator. +1 = +1⋅[[x(x - 1)]/[x(x - 1)]] Write the inequality sign ≤. See the right side 1/x. The denominator is x. The factor (x - 1) is missing. So multiply (x - 1) to both of the numerator and the denominator. 1/x = [1/x]⋅[(x - 1)/(x - 1)] So 4/(x - 1) + 1 ≤ 1/x becomes [4/(x - 1)]⋅[x/x] + 1⋅[[x(x - 1)]/[[x(x - 1)]] ≤ [1/x]⋅[(x - 1)/(x - 1)]. 4⋅x = 4x +1⋅x(x - 1) = +x2 - x Multiply a Monomial and a Polynomial 1⋅(x - 1) = x - 1 4x + x2 - x = x2 + 3x Move (x - 1)/[x(x - 1)] to the left side. Then (x2 + 3x - x + 1)/[x(x - 1)]. +3x - x = +2x x2 + 2x + 1 = x2 + 2⋅x⋅1 + 12 = (x + 1)2 Factor a Perfect Square Trinomial Move the denominator x(x - 1) to the numerator. (x + 1)2/[x(x - 1)] → (x + 1)2x(x - 1) This is multiplying the square of the denominator, [x(x - 1)]2, to both sides. (x - 1) ≠ 0, x ≠ 0 So x(x - 1) ≠ 0. So you can multiply [x(x - 1)]2 on both sides. And [x(x - 1)]2 is (+). So the order of the inequality sign doesn't change. And write ≠ 0 under x and (x - 1). Write the zeros. x = -1, 0, 1 (We know that x ≠ 0, 1. These are the zeros you're going to use when graphing the inequality on the x-axis.) Draw the x-axis. Point the zeros x = -1, 0, 1. Draw y = (x + 1)2x(x - 1) on the x-axis. Polynomial Inequality The highest degree term of y = (x + 1)2x(x - 1) is x4. The coefficient is (+). So starting from the top right of the x-axis, draw the graph that goes toward the nearest zero: x = 1. See the factor (x - 1). (x - 1) = (x - 1)1 The exponent is 1. It's odd. Then draw the graph that passes through the x-axis at x = 1. Next, see the factor x. x = x1 The exponent is 1. It's odd. Then draw the graph that passes through the x-axis at x = 0. See the factor (x + 1)2. The exponent is 2. It's even. Then draw the graph that bounces off the x-axis at x = -1. So this is the graph of the polynomial y = (x + 1)2x(x - 1) on the x-axis. x ≠ 0, (x - 1) ≠ 0 So draw empty circles on x = 0 and x = 1. See (x + 1)2x(x - 1) ≤ 0. The left side is less than 0. So color the region where the graph is below the x-axis (y = 0). The inequality sign does not include equal to [=]. So draw a full circle on x = -1. (Don't fill the empty circles on x = 0 and x = 1. These are the excluded values.) The colored regions are x = -1, 0 < x < 1. So x = -1, 0 < x < 1
# At a Glance - Rolle's Theorem Rolle's Theorem says: Let f be a function that • is continuous on the closed interval [ab] • is differentiable on the open interval (ab), and • has (a) = (b). Then there is some c in the open interval (ab) with f ' (c) = 0. Sometimes the third condition is stated as (a) = (b) = 0, but for the proof, it doesn't matter. In pictures, we're saying suppose f is a nice smooth function with the same starting and ending height: If f increases or decreases from its starting height, it needs to turn around and come back in order to end at the same height it started at: Since f is a nice smooth differentiable function, its derivative at that turn-around point must be 0: If f doesn't go up or down from its starting point, then f is constant: In this case, ' (c) is 0 for every value of c in the interval (ab). Rolle's Theorem is reminiscent of the Intermediate Value Theorem. Rolle's Theorem says if f satisfies some assumptions (more mathematically known as hypotheses) then f ' will be zero at some point in (ab). We could have a constant function, in which case f ' will be 0 infinitely many times: We could have a function that turns around once: Or we could have a function that turns around many times: Rolle's Theorem doesn't tell us where or how many times f ' will be zero; it tells us f ' must be zero at least once if the hypotheses are all satisfied. ### Sample Problem Suppose f is not continuous on [ab]. Then there doesn't need to be any c in (ab) with ' (c) = 0. Here's an example: This function is not continuous. At the point of discontinuity, f ' doesn't exist. At all other points in the interval, f ' is positive: There is no point c in (a, b) where ' (c) = 0. We found earlier that the derivative of the absolute value function doesn't exist at 0. When x is negative the slope of the absolute value function is -1; when x is positive the slope of the absolute value function is 1: There is no value of c anywhere, in any interval (a, b), with ' (c) = 0. The derivative of the absolute value function isn't 0 anywhere. If a function fails any of the hypotheses, we aren't allowed to use Rolle's Theorem. #### Example 1 Let f(x) = x2. Prove that there is some c in (-2, 2) with f ' (c) = 0. #### Example 2 For the function f shown below, determine if we're allowed to use Rolle's Theorem to guarantee the existence of some c in (a, b) with f ' (c) = 0. If not, explain why not. #### Example 3 Let . Determine if Rolle's Theorem guarantees the existence of some c in (-1, 1) with f ' (c) = 0. If not, explain why not. #### Example 4 Let f(x) = x2 – x. Does Rolle's Theorem guarantees the existence of some c in (0, 1) with f ' (c) = 0? If not, explain why not. #### Exercise 1 Let f(x) = sin(x). This function is differentiable everywhere. Prove that there is some c in (0, 2π) with f ' (c) = 0. By looking at the graph of f, determine how many such values of c there are in (0, 2π). #### Exercise 2 For the function f(x) = 2x, determine whether we're allowed to use Rolle's Theorem to guarantee the existence of some c in (0, 1) with f ' (c) = 0. If not, explain why not. #### Exercise 3 For the function f shown below, determine whether we're allowed to use Rolle's Theorem to guarantee the existence of some c in (-1 ,1) with f ' (c) = 0. If not, explain why not. (Insert graph of f(x) = 3 for x ≤ -1, f(x) = x2 for -1 < x < 1 and f(x) = 3 for x ≥ 1) #### Exercise 4 For the function f shown below, determine we're allowed to use Rolle's Theorem to guarantee the existence of some c in (a, b) with f ' (c) = 0. If not, explain why not. (Insert graph of the function f(x) = -2(x-a) for x ≤ a, f(x) = 0 for a < x < b and f(x) = 2(x-b) for x ≥ b) #### Exercise 5 For the function f shown below, determine we're allowed to use Rolle's Theorem to guarantee the existence of some c in (a, b) with f ' (c) = 0. If not, explain why not. (Insert graph of f(x) = sin(x) on the interval (0, 2π) On the x-axis, label the origin as a, and then label x = 3π/2 as b.) #### Exercise 6 For each given function f and interval, determine if Rolle's Theorem guarantees the existence of some c in that interval with f ' (c) = 0. If not, explain why not. • f(x) = x3 on the interval (-2, 2) #### Exercise 7 For the function f and interval, determine if Rolle's Theorem guarantees the existence of some c in that interval with f ' (c) = 0. If not, explain why not. • f(x) = cos(x) on the interval (-π, 3π) (yes, cos(x) is differentiable) #### Exercise 8 For the given function f and interval, determine if Rolle's Theorem guarantees the existence of some c in that interval with f ' (c) = 0. If not, explain why not. • f(x) = (x – 2)2 + 4 on the interval (-2, 2) #### Exercise 9 For the given function f and interval, determine if Rolle's Theorem guarantees the existence of some c in that interval with f ' (c) = 0. If not, explain why not. • on the interval (-1, 1).
Lesson 18: Introduction to Algebra: Expressions and Variables Size: px Start display at page: Transcription 2 Lesson 18 Warm-up: Solve the movie ticket problem Time: 10 Minutes Write on the board: A movie theatre owner wants to know if there is a relationship between the number of tickets sold and how long the movie has been playing. Here is how many tickets have been sold so far: 3,000 tickets sold Week 1; 2,750 tickets sold Week 2; 2,500 tickets sold Week 3; and 2,250 tickets sold Week 4. Basic Questions: Make a graph with the above information. o Give students time to make their graphs. They may make a line graph or a bar graph. The line graph should look like the example below. You can connect the dots to make a line of best fit or you can leave it as a scatter plot. Either way there is a negative correlation. What can the owner predict might happen on Week 5? (Attendance will be lower, there will be 2,000 tickets sold) Extension Questions: What is the average (mean) attendance for the first 4 weeks? (Total of 10,500 divided by 4 = 2,625 tickets sold per week) Is the correlation positive or negative? (It is negative because the more time goes by, the lower the ticket sales. Some students may say the slope is negative, which is true. Slope will be taught in a later lesson.) Lesson 18 Activity 1: Vocabulary Time: 10 Minutes 1) This activity can be projected on the board and done as a whole class. Have students volunteer to fill in the blank after everyone has a chance to do the activity. 2) Explain variable as a letter that takes the place of a number until we figure out what that number is. 3) The first example for 3 and 4 is numeric because there are no variables. 4) The second example for 3 and for 4 is algebraic because there is a variable. D. Legault, Minnesota Literacy Council, 3 5) Solicit more examples of each from students. Answers: 1. A variable is a letter or symbol used to write a value (number) that can change. 2. A constant is a value that does not change. 3. An equation is a math statement that shows the two sides are equal or balanced. 4. An expression is a combination of numbers and operations. constant expression variable equation 1. A is a letter or symbol used to write a value (a number) that can change. Some examples are x, y, a, c. 2. A is a value that does not change. Some examples are 56, 47, ½, An is a math statement that shows the two sides are equal. Some example are = 9 and 4 + x = 9 4. An is a combination of numbers and operations. Some examples are and 4 + x. Lesson 18 Activity 2: Write Expressions Time: Minutes 1) Write on the board: a b 5 x D. Legault, Minnesota Literacy Council, 4 y 16 2) Ask students how they would read aloud each of the expressions. a. For the first one, they may say: a plus 4, 4 added to a, the sum of 4 and a, 4 more than a, etc. b. For the second one, they may say 17 minus b, b subtracted from 17, the difference of 17 and b, b less than 17. (Note: This last one is tricky for students so add more examples if necessary. This example can be you have \$17 and I have less money and it s b fewer dollars). c. For the third one, they may say 5 times x, the product of 5 and x, etc. This expression can also be written as 5x, which is the case in algebra. Make sure to point this out. d. For the fourth one, they may say y divided by 16, the quotient of y and 16. 3) Do Worksheet Do the first few together on the board. Circulate to help. Have volunteers write answers on the board. 4) Do pages in the student book together. Page 50 has more examples of the above and on page 51 students also choose which expression best matches the problem. Lesson 18 Activity 3: Word Problems Time: Minutes 1) After you have done the problems in the student book together, students can work independently in the workbook pages ) You may need to review operations with integers from a previous lesson because there are some negative numbers in some of the problems. 3) Circulate to help. Discuss and explain the questions that students found challenging. Validate correct reasoning and redirect/explain if reasoning is not correct. Lesson 18 Activity 4: Evaluate Expressions Time: 20 Minutes 1) Tell the students: You have just done many problems in which you had to match an expression to a word problem. Now we will learn to evaluate expressions for different variables. That means we will find the answer for different values. 2) Example A: You work h hours per week and make \$15 an hour. a. Write an expression to represent how much you get paid. (15h or 15 x h) b. Evaluate the expression for 10 hours in one week (15 x 10 = \$150) c. Evaluate the expression for 25 hours in one week (15 x 25 = \$375) D. Legault, Minnesota Literacy Council, 5 3) Example B: In the warm up activity about the movie ticket sales, the owner wants to know how much money she will make if the tickets are sold at \$11.25 each. a. Write an expression with t for tickets. (\$11.25t ) b. Evaluate the expression for the first week when 3,000 tickets were sold. (3,000 x \$11.25 = \$33,750) c. Evaluate the expression for the second week when 2,750 tickets were sold. (2,750 x \$11.25 = \$30,937.50) 4) Practice evaluating expressions with Worksheet Do the first few together. Application Option A: Make Algebra Concrete Time: 15 Minutes This activity may be useful to make algebraic more concrete for students. Use the following as a guideline. I wasn t able to change the words (document is a pdf) so choose items in the classroom and ignore the reference to child. Also, you can make the activity more interesting by saying a few examples yourself and then having students work in pairs to create an expression and dictate it to the rest of the class. D. Legault, Minnesota Literacy Council, 6 D. Legault, Minnesota Literacy Council, 7 Application Option B: Write Expressions from a Menu Time: 10 Minutes (See printable handout on the next page.) Use the menu from this restaurant to write an algebraic expression for each question. The variable is the first letter of the menu item. Once you write the variable, evaluate the expression. French Fries \$2 Pizza Slice \$3 Hotdog \$3 Ice Cream Sundae \$3 Shake \$2 Taco \$2 Coke \$1 Milk \$1 1. Mike ordered 3 hotdogs for his family. How much was the meal? 2. Stanley wants to eat an ice cream sundae after he finishes his meal of coke and a slice of pizza. How much was his bill? 3. There are four people in the Cullwell family. Two people order French fries and two people order tacos. Each one of them orders milk with their meal. How much do the drinks cost for the Cullwell family? 4. If the whole class decided to eat lunch, and the teacher bought everyone a taco and a shake, how much would the teacher have to pay? Answers: 1. 3h, 3(\$3)= \$9 2. c + p + i, \$1+\$3+\$3=\$7 3. 2f + 2t + 4m, 2(\$2) + 2(\$2) + 4(\$1) = \$12 4. Answer will vary D. Legault, Minnesota Literacy Council, 8 Writing Expressions from a Menu Use the menu from this restaurant to write an algebraic expression for each question. The variable is the first letter of the menu item. Once you write the variable, evaluate the expression. French Fries \$2 Pizza Slice \$3 Hotdog \$3 Ice Cream Sundae \$3 Shake \$2 Taco \$2 Coke \$1 Milk \$1 2. Mike ordered 3 hotdogs for his family. How much was the meal? 4. Stanley wants to eat an ice cream sundae after he finishes his meal of coke and a slice of pizza. How much was his bill? 5. There are four people in the Cullwell family. Two people order French fries and two people order tacos. Each one of them orders milk with their meal. How much do the drinks cost for the Cullwell family? 5. If the whole class decided to eat lunch, and the teacher bought everyone a taco and a shake, how much would the teacher have to pay? D. Legault, Minnesota Literacy Council, 9 Lesson 18: Group Exit Ticket Time: 5 Minutes Do expression dictation. For example, say 15 less than n. Have students write the answer on a piece of paper (n 15). You may continue with more dictations yourself or students can take turns saying the dictation problems for the whole class to write down. As each problem is said, students should write down the expression individually. D. Legault, Minnesota Literacy Council, 10 Kuta Software Lesson - Infinite 18: Algebra Introduction 1 to Algebra: Expressions Name and Variables Variable and Verbal Expressions Worksheet 18.1 Write Expressions Write each as an algebraic expression. 1) the difference of 10 and 5 2) the quotient of 14 and 7 Date Period 3) u decreased by 17 4) half of 14 5) x increased by 6 6) the product of x and 7 7) the sum of q and 8 8) 6 squared 9) twice q 10) the product of 8 and 12 11) the quotient of 18 and n 12) n cubed Write each as a verbal expression. 13) x 2 14) a ) ) 5n D. Legault, Minnesota Literacy Council, 11 Kuta Software Lesson - Infinite 18: Algebra Introduction 1 to Algebra: Expressions Name and Variables Variable and Verbal Expressions Worksheet 18.1 Answers Write each as an algebraic expression. Date Period 1) the difference of 10 and ) u decreased by 17 u ) x increased by 6 x + 6 2) the quotient of 14 and ) half of ) the product of x and 7 x 7 7) the sum of q and 8 q + 8 8) 6 squared 6 2 9) twice q 2q 10) the product of 8 and ) the quotient of 18 and n 18 n 12) n cubed n 3 Write each as a verbal expression. 13) x 2 half of x 14) a + 9 a increased by 9 15) 19-3 the difference of 19 and 3 16) 5n 5 times a number D. Legault, Minnesota Literacy Council, 12 Kuta Software Lesson - Infinite 18: Algebra Introduction 1 to Algebra: Expressions Name and Variables Evaluating Expressions Worksheet 18.2 Evaluate Expressions Evaluate each using the values given. 1) y 2 + x; use x = 1, and y = 2 2) a b; use a = 10, and b = 4 Date Period 3) p 2 + m; use m = 1, and p = 5 4) y x; use x = 1, and y = 3 5) m + p 5; use m = 1, and p = 5 6) y 2 - x; use x = 7, and y = 7 7) z(x + y); use x = 6, y = 8, and z = 6 8) x + y + y; use x = 9, and y = 10 9) p m; use m = 9, and p = 3 10) 6q + m - m; use m = 8, and q = 3 11) p 2 m 4; use m = 4, and p = 7 12) y - (z + z 2 ); use y = 10, and z = 2 13) z - (y 3-1); use y = 3, and z = 7 14) (y + x) 2 + x; use x = 1, and y = 1 D. Legault, Minnesota Literacy Council, 13 Kuta Software - Infinite Algebra 1 Name Evaluating Expressions Worksheet 18.2 Answers Evaluate each using the values given. Date Period 1) y 2 + x; use x = 1, and y = 2 2 2) a b; use a = 10, and b = 4 1 3) p 2 + m; use m = 1, and p = ) y x; use x = 1, and y = ) m + p 5; use m = 1, and p = 5 2 6) y 2 - x; use x = 7, and y = ) z(x + y); use x = 6, y = 8, and z = ) x + y + y; use x = 9, and y = ) p m; use m = 9, and p = ) 6q + m - m; use m = 8, and q = ) p 2 m 4; use m = 4, and p = ) y - (z + z 2 ); use y = 10, and z = ) z - (y 3-1); use y = 3, and z = ) (y + x) 2 + x; use x = 1, and y = 1 2 D. Legault, Minnesota Literacy Council, 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 Lesson 4: Convert Fractions, Review Order of Operations Lesson 4: Convert Fractions, Review Order of Operations LESSON 4: Convert Fractions, Do Order of Operations Weekly Focus: fractions, decimals, percent, order of operations Weekly Skill: convert, compute Direct Translation is the process of translating English words and phrases into numbers, mathematical symbols, expressions, and equations. Section 1 Mathematics has a language all its own. In order to be able to solve many types of word problems, we need to be able to translate the English Language into Math Language. is the process of translating Current California Math Standards Balanced Equations Balanced Equations Current California Math Standards Balanced Equations Grade Three Number Sense 1.0 Students understand the place value of whole numbers: 1.1 Count, read, and write whole numbers to 10,000. 4 Mathematics Curriculum New York State Common Core 4 Mathematics Curriculum G R A D E GRADE 4 MODULE 1 Topic F Addition and Subtraction Word Problems 4.OA.3, 4.NBT.1, 4.NBT.2, 4.NBT.4 Focus Standard: 4.OA.3 Solve multistep word Curriculum Alignment Project Curriculum Alignment Project Math Unit Date: Unit Details Title: Solving Linear Equations Level: Developmental Algebra Team Members: Michael Guy Mathematics, Queensborough Community College, CUNY Jonathan 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 Lesson 3: Using Inequalities to Problem Solve Lesson 3: Using Inequalities to Problem Solve Selected Content Standards Benchmarks Addressed: N-1-M Demonstrating that a rational number can be expressed in many forms, and selecting an appropriate form Name of Lesson: Properties of Equality A Review. Mathematical Topic: The Four Properties of Equality. Course: Algebra I Name of Lesson: Properties of Equality A Review Mathematical Topic: The Four Properties of Equality Course: Algebra I Time Allocation: One (1) 56 minute period Pre-requisite Knowledge: The students will 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 Using Proportions to Solve Percent Problems I RP7-1 Using Proportions to Solve Percent Problems I Pages 46 48 Standards: 7.RP.A. Goals: Students will write equivalent statements for proportions by keeping track of the part and the whole, and by solving Counting Change and Changing Coins Grade Two Counting Change and Changing Coins Content Standards Overview Students share the book The Penny Pot, by Stuart J. Murphy, to learn about choices, producers and consumers, and counting money. Kindergarten Math Curriculum Course Description and Philosophy Text Reference: Kindergarten Math Curriculum Course Description and Philosophy How do numbers and math affect our daily lives? What types of problems can be solved by understanding numbers and math concepts? Through inquiry, Materials: Student-paper, pencil, circle manipulatives, white boards, markers Teacher- paper, pencil, circle manipulatives, worksheet Topic: Creating Mixed Numbers and Improper Fractions SOL: 5.7- The student will add and subtract with fractions and mixed numbers, with and without regrouping, and express answers in simplest form. Problems Understanding Income and Expenses EPISODE # 123 Understanding Income and Expenses EPISODE # 123 LESSON LEVEL Grades 4-6 KEY TOPICS Entrepreneurship Income and expenses Cash flow LEARNING OBJECTIVES 1. Understand what your income and expenses are. 2. Algebra Unit Plans. Grade 7. April 2012. Created By: Danielle Brown; Rosanna Gaudio; Lori Marano; Melissa Pino; Beth Orlando & Sherri Viotto Algebra Unit Plans Grade 7 April 2012 Created By: Danielle Brown; Rosanna Gaudio; Lori Marano; Melissa Pino; Beth Orlando & Sherri Viotto Unit Planning Sheet for Algebra Big Ideas for Algebra (Dr. Small) OA3-10 Patterns in Addition Tables Pages 60 63 Standards: 3.OA.D.9 Goals: Students will identify and describe various patterns in addition tables. Prior Knowledge Required: Can add two numbers within 20 HFCC Math Lab Beginning Algebra 13 TRANSLATING ENGLISH INTO ALGEBRA: WORDS, PHRASE, SENTENCES HFCC Math Lab Beginning Algebra 1 TRANSLATING ENGLISH INTO ALGEBRA: WORDS, PHRASE, SENTENCES Before being able to solve word problems in algebra, you must be able to change words, phrases, and sentences 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, Verbal Phrases to Algebraic Expressions Student Name: Date: Contact Person Name: Phone Number: Lesson 13 Verbal Phrases to s Objectives Translate verbal phrases into algebraic expressions Solve word problems by translating sentences into equations .2 Methods of Addition Objectives Relate addition stories to number bonds. Write two addition facts for a given number bond. Solve picture problems using addition. Learn addition facts through, and the Solving Rational Equations Lesson M Lesson : Student Outcomes Students solve rational equations, monitoring for the creation of extraneous solutions. Lesson Notes In the preceding lessons, students learned to add, subtract, multiply, Tickets, Please Using Substitution to Solve a Linear System, Part 2 Problem 1 Students write linear systems of equations representing several situations. Using substitution, they will solve each linear system and interpret the solution of the system in terms of each problem Summer Assignment for incoming Fairhope Middle School 7 th grade Advanced Math Students Summer Assignment for incoming Fairhope Middle School 7 th grade Advanced Math Students Studies show that most students lose about two months of math abilities over the summer when they do not engage in Money: Week 1 of 2. Beginning Level (CASAS reading scores of 181-200) The Minnesota Literacy Council created this curriculum with funding from the MN Department of Education. We invite you to adapt it for your own classrooms. Beginning Level (CASAS reading scores of 181-200) Contents. Sample worksheet from www.mathmammoth.com Contents Introduction... 4 Warmup: Mental Math 1... 8 Warmup: Mental Math 2... 10 Review: Addition and Subtraction... 12 Review: Multiplication and Division... 15 Balance Problems and Equations... 19 More Wants and Needs. Grade One. Overview. Prerequisite Skills. Lesson Objectives. Materials List Grade One Wants and Needs Overview Students share the book Something Good, by Robert Munsch, to learn about unlimited wants, limited resources, choice, and counting money. They complete worksheets on determining FINAL SIOP LESSON PLAN. Preparation Name: Stephanie Hart DOMINICAN UNIVERSITY OF CALIFORNIA FINAL SIOP LESSON PLAN Content Area: Mathematics, Solving Inequalities Grade Level: 8 English Learners: This is a sheltered class within a mainstream Saving and Creating a Personal Budget Grade Five Saving and Creating a Personal Budget Overview Students share several chapters of the book From the Mixed-Up Files of Mrs. Basil E. Frankweiler, by E.L. Konigsburg, to learn about the role of Comparing Simple and Compound Interest Comparing Simple and Compound Interest GRADE 11 In this lesson, students compare various savings and investment vehicles by calculating simple and compound interest. Prerequisite knowledge: Students should The Crescent Primary School Calculation Policy The Crescent Primary School Calculation Policy Examples of calculation methods for each year group and the progression between each method. January 2015 Our Calculation Policy This calculation policy has Performance Assessment Task Bikes and Trikes Grade 4. Common Core State Standards Math - Content Standards Performance Assessment Task Bikes and Trikes Grade 4 The task challenges a student to demonstrate understanding of concepts involved in multiplication. A student must make sense of equal sized groups of Lesson 1.1: Earth and Space Science - Introduction Weekly Focus: Main Idea Weekly Skill: Introduction Lesson Summary: This week students will take a pre- self-evaluation to determine their background knowledge in Earth and space science. They will also Using Algebra Tiles for Adding/Subtracting Integers and to Solve 2-step Equations Grade 7 By Rich Butera Using Algebra Tiles for Adding/Subtracting Integers and to Solve 2-step Equations Grade 7 By Rich Butera 1 Overall Unit Objective I am currently student teaching Seventh grade at Springville Griffith Middle Unit 7 The Number System: Multiplying and Dividing Integers Unit 7 The Number System: Multiplying and Dividing Integers Introduction In this unit, students will multiply and divide integers, and multiply positive and negative fractions by integers. Students will Clifton High School Mathematics Summer Workbook Algebra 1 1 Clifton High School Mathematics Summer Workbook Algebra 1 Completion of this summer work is required on the first day of the school year. Date Received: Date Completed: Student Signature: Parent Signature: Ohio Standards Connection Patterns, Functions and Algebra Benchmark E Solve open sentences and explain strategies. Indicator 4 Solve open sentences by representing an expression in more than one way using A Concrete Introduction. to the Abstract Concepts. of Integers and Algebra using Algebra Tiles A Concrete Introduction to the Abstract Concepts of Integers and Algebra using Algebra Tiles Table of Contents Introduction... 1 page Integers 1: Introduction to Integers... 3 2: Working with Algebra Tiles... Unit 1 Equations, Inequalities, Functions Unit 1 Equations, Inequalities, Functions Algebra 2, Pages 1-100 Overview: This unit models real-world situations by using one- and two-variable linear equations. This unit will further expand upon pervious CAHSEE on Target UC Davis, School and University Partnerships UC Davis, School and University Partnerships CAHSEE on Target Mathematics Curriculum Published by The University of California, Davis, School/University Partnerships Program 006 Director Sarah R. Martinez, All the examples in this worksheet and all the answers to questions are available as answer sheets or videos. BIRKBECK MATHS SUPPORT www.mathsupport.wordpress.com Numbers 3 In this section we will look at - improper fractions and mixed fractions - multiplying and dividing fractions - what decimals mean and exponents Counting Money and Making Change Grade Two Ohio Standards Connection Number, Number Sense and Operations Benchmark D Determine the value of a collection of coins and dollar bills. Indicator 4 Represent and write the value of money using the sign EXTRA ACTIVITy pages EXTRA FUN ACTIVITIES This booklet contains extra activity pages for the student as well as the tests. See the next page for information about the activity pages. Go to page 7 to find the Alpha tests. EXTRA F.IF.7b: Graph Root, Piecewise, Step, & Absolute Value Functions F.IF.7b: Graph Root, Piecewise, Step, & Absolute Value Functions F.IF.7b: Graph Root, Piecewise, Step, & Absolute Value Functions Analyze functions using different representations. 7. Graph functions expressed Discovering Math: Data and Graphs Teacher s Guide Teacher s Guide Grade Level: K 2 Curriculum Focus: Mathematics Lesson Duration: Two class periods Program Description Discovering Math: Data and Graphs From simple graphs to sampling to determining what Rational Number Project Rational Number Project Fraction Operations and Initial Decimal Ideas Lesson : Overview Students estimate sums and differences using mental images of the 0 x 0 grid. Students develop strategies for adding Sample Fraction Addition and Subtraction Concepts Activities 1 3 Sample Fraction Addition and Subtraction Concepts Activities 1 3 College- and Career-Ready Standard Addressed: Build fractions from unit fractions by applying and extending previous understandings of operations Sneaking up on Slope Sneaking up on Slope After the warm- up, the lesson starts with a handout where the students will be graphing points and looking at the pattern it creates to find the next point in the line. The handout 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 POLYNOMIAL FUNCTIONS POLYNOMIAL FUNCTIONS Polynomial Division.. 314 The Rational Zero Test.....317 Descarte s Rule of Signs... 319 The Remainder Theorem.....31 Finding all Zeros of a Polynomial Function.......33 Writing a Unit 1 Number Sense. In this unit, students will study repeating decimals, percents, fractions, decimals, and proportions. Unit 1 Number Sense In this unit, students will study repeating decimals, percents, fractions, decimals, and proportions. BLM Three Types of Percent Problems (p L-34) is a summary BLM for the material NS6-50 Dividing Whole Numbers by Unit Fractions Pages 16 17 NS6-0 Dividing Whole Numbers by Unit Fractions Pages 6 STANDARDS 6.NS.A. Goals Students will divide whole numbers by unit fractions. Vocabulary division fraction unit fraction whole number PRIOR KNOWLEDGE 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 Lesson 2.15: Physical Science Speed, Velocity & Acceleration Weekly Focus: Reading for Comprehension Weekly Skill: Numeracy Skills in Science Lesson Summary: This week students will continue reading for comprehension with reading passages on speed, velocity, and Barter vs. Money. Grade One. Overview. Prerequisite Skills. Lesson Objectives. Materials List Grade One Barter vs. Money Overview Students share the book Sheep in a Shop, by Nancy Shaw, to learn about choice, making decisions, trade, and the barter system. They complete worksheets on comparing Transportation: Week 2 of 2 The Minnesota Literacy Council created this curriculum with funding from the MN Department of Education. We invite you to adapt it for your own classrooms. Beginning Level (CASAS reading scores of 181-200) Problem of the Month: Perfect Pair Problem of the Month: The Problems of the Month (POM) are used in a variety of ways to promote problem solving and to foster the first standard of mathematical practice from the Common Core State Standards: Measurement with Ratios Grade 6 Mathematics, Quarter 2, Unit 2.1 Measurement with Ratios Overview Number of instructional days: 15 (1 day = 45 minutes) Content to be learned Use ratio reasoning to solve real-world and mathematical 1 ST GRADE COMMON CORE STANDARDS FOR SAXON MATH 1 ST GRADE COMMON CORE STANDARDS FOR SAXON MATH Calendar The following tables show the CCSS focus of The Meeting activities, which appear at the beginning of each numbered lesson and are taught daily, 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, Shopping. Grade One. Overview. Prerequisite Skills. Lesson Objectives. Materials List Grade One Shopping Overview Students share the book Just Shopping with Mom, by Mercer Mayer, to learn about shopping and counting money. They complete activities on examining coupons and shopping for bicycle VISUAL ALGEBRA FOR COLLEGE STUDENTS. Laurie J. Burton Western Oregon University VISUAL ALGEBRA FOR COLLEGE STUDENTS Laurie J. Burton Western Oregon University VISUAL ALGEBRA FOR COLLEGE STUDENTS TABLE OF CONTENTS Welcome and Introduction 1 Chapter 1: INTEGERS AND INTEGER OPERATIONS 7.S.8 Interpret data to provide the basis for predictions and to establish 7 th Grade Probability Unit 7.S.8 Interpret data to provide the basis for predictions and to establish experimental probabilities. 7.S.10 Predict the outcome of experiment 7.S.11 Design and conduct an Overview. Essential Questions. Grade 4 Mathematics, Quarter 4, Unit 4.1 Dividing Whole Numbers With Remainders Dividing Whole Numbers With Remainders Overview Number of instruction days: 7 9 (1 day = 90 minutes) Content to Be Learned Solve for whole-number quotients with remainders of up to four-digit dividends 2. System of linear equations can be solved by graphing, substitution, or eliminating a variable. 1 Subject: Algebra 1 Grade Level: 9 th Unit Plan #: 6 UNIT BACKGROUND Unit Title: Systems of Equations and Inequalities Grade Level: 9 Subject/Topic: Algebra 1 Key Words: Graphing, Substitution, Elimination FIDDLIN WITH FRACTIONS FIDDLIN WITH FRACTIONS Special Area: Connections (Fourth and Fifth Grade) Written by: (Dawn Ramos, Joan Johnson, Mindy Wilshusen, St. Mary s School) Length of Unit: (6 Lessons) I. ABSTRACT The purpose Consumers face constraints on their choices because they have limited incomes. Consumer Choice: the Demand Side of the Market Consumers face constraints on their choices because they have limited incomes. Wealthy and poor individuals have limited budgets relative to their desires. Use order of operations to simplify. Show all steps in the space provided below each problem. INTEGER OPERATIONS ORDER OF OPERATIONS In the following order: 1) Work inside the grouping smbols such as parenthesis and brackets. ) Evaluate the powers. 3) Do the multiplication and/or division in order from left to right. lesson three budgeting your money teacher s guide lesson three budgeting your money teacher s guide budgeting your money lesson outline lesson 3 overview I m all out of money, and I won t get paid again until the end of next week! This is a common dilemma 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 What qualities are employers looking for in teen workers? How can you prove your own skills? Sell Yourself 4 Finding a job The BIG Idea What qualities are employers looking for in teen workers? How can you prove your own skills? AGENDA Approx. 45 minutes I. Warm Up: Employer Survey Review (15 Lesson 17 Teacher Page A Overview Students name fractions greater than with fraction circles. Students name fractions using both mixed numbers and improper fractions. Materials Fraction Circles for students and teacher Transparency Module 3: Correlation and Covariance Using Statistical Data to Make Decisions Module 3: Correlation and Covariance Tom Ilvento Dr. Mugdim Pašiƒ University of Delaware Sarajevo Graduate School of Business O ften our interest in data analysis MATH 60 NOTEBOOK CERTIFICATIONS MATH 60 NOTEBOOK CERTIFICATIONS Chapter #1: Integers and Real Numbers 1.1a 1.1b 1.2 1.3 1.4 1.8 Chapter #2: Algebraic Expressions, Linear Equations, and Applications 2.1a 2.1b 2.1c 2.2 2.3a 2.3b 2.4 2.5 Decimal Notations for Fractions Number and Operations Fractions /4.NF Decimal Notations for Fractions Number and Operations Fractions /4.NF Domain: Cluster: Standard: 4.NF Number and Operations Fractions Understand decimal notation for fractions, and compare decimal fractions. Multiplying Integers. Lesson Plan Lesson Plan Video: 12 minutes Lesson: 38 minutes Pre-viewing :00 Warm up: Write 5 + 5 + 5 + 5 = on the board. Ask students for the answer. Then write 5 x 4 = on the board. Ask the students for the answer. 1.6 The Order of Operations 1.6 The Order of Operations Contents: Operations Grouping Symbols The Order of Operations Exponents and Negative Numbers Negative Square Roots Square Root of a Negative Number Order of Operations and Negative Saving Money. Grade One. Overview. Prerequisite Skills. Lesson Objectives. Materials List Grade One Saving Money Overview Students share the book A Chair for My Mother, by Vera B. Williams, to learn about counting and saving money. They complete worksheets on coin counting and saving. Prerequisite PLANNING A BUDGET. Income, needs vs. wants, budgets, simple interest, savings accounts, paying yourself first Grades 6-8 Lesson 3 PLANNING A BUDGET Key concepts: Income, needs vs. wants, budgets, simple interest, savings accounts, paying yourself first Summary: This lesson reviews trade-offs and priorities in HOW MUCH WILL I SPEND ON GAS? 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. Microeconomics Sept. 16, 2010 NOTES ON CALCULUS AND UTILITY FUNCTIONS DUSP 11.203 Frank Levy Microeconomics Sept. 16, 2010 NOTES ON CALCULUS AND UTILITY FUNCTIONS These notes have three purposes: 1) To explain why some simple calculus formulae are useful in understanding Decomposing Numbers (Operations and Algebraic Thinking) Decomposing Numbers (Operations and Algebraic Thinking) Kindergarten Formative Assessment Lesson Designed and revised by Kentucky Department of Education Mathematics Specialists Field-tested by Kentucky A bigger family, a better future. A bigger family, a better future. Child sponsorship is changing for the better Sponsors like you are a vital part of our big, supportive family. Like us, you want the very best for your sponsored child. Communication Process Welcome and Introductions Lesson 7 Communication Process Overview: This lesson teaches learners to define the elements of effective communication and its process. It will focus on communication as the Creating, Solving, and Graphing Systems of Linear Equations and Linear Inequalities Algebra 1, Quarter 2, Unit 2.1 Creating, Solving, and Graphing Systems of Linear Equations and Linear Inequalities Overview Number of instructional days: 15 (1 day = 45 60 minutes) Content to be learned Curriculum Design for Mathematic Lesson Probability Curriculum Design for Mathematic Lesson Probability This curriculum design is for the 8th grade students who are going to learn Probability and trying to show the easiest way for them to go into this class. Models of a Vending Machine Business Math Models: Sample lesson Tom Hughes, 1999 Models of a Vending Machine Business Lesson Overview Students take on different roles in simulating starting a vending machine business in their school that Day One: Least Common Multiple Grade Level/Course: 5 th /6 th Grade Math Lesson/Unit Plan Name: Using Prime Factors to find LCM and GCF. Rationale/Lesson Abstract: The objective of this two- part lesson is to give students a clear understanding Rational Number Project Rational Number Project Fraction Operations and Initial Decimal Ideas Lesson 2: Overview Students review equivalence ideas with paper folding. Students develop a symbolic rule for finding equivalent fractions. Lesson 2.11: Physical Science Energy Weekly Focus: Reading for Comprehension Weekly Skill: Introduction to Energy Lesson Summary: This week students will continue reading for comprehension and get an introduction to various forms of energy. Fractions as Numbers INTENSIVE INTERVENTION. National Center on. at American Institutes for Research National Center on INTENSIVE INTERVENTION at American Institutes for Research Fractions as Numbers 000 Thomas Jefferson Street, NW Washington, DC 0007 E-mail: NCII@air.org While permission to reprint this NF5-12 Flexibility with Equivalent Fractions and Pages 110 112 NF5- Flexibility with Equivalent Fractions and Pages 0 Lowest Terms STANDARDS preparation for 5.NF.A., 5.NF.A. Goals Students will equivalent fractions using division and reduce fractions to lowest terms. Math and FUNDRAISING. Ex. 73, p. 111 1.3 0. 7 Standards Preparation Connect 2.7 KEY VOCABULARY leading digit compatible numbers For an interactive example of multiplying decimals go to classzone.com. Multiplying and Dividing Decimals Gr. 5 NS 2.1 Algebra 1. Practice Workbook with Examples. McDougal Littell. Concepts and Skills McDougal Littell Algebra 1 Concepts and Skills Larson Boswell Kanold Stiff Practice Workbook with Examples The Practice Workbook provides additional practice with worked-out examples for every lesson. Common Core State Standards for Mathematics Accelerated 7th Grade A Correlation of 2013 To the to the Introduction This document demonstrates how Mathematics Accelerated Grade 7, 2013, meets the. Correlation references are to the pages within the Student Edition. Meeting Graphs of Proportional Relationships 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,), Unit 5: Analyze, Solve, and Graph Linear Inequalities Unit 5 Table of Contents Unit 5: Analyze, Solve, and Graph Linear Inequalities Video Overview Learning Objectives 5.2 Media Run Times 5.3 Instructor Notes 5.4 The Mathematics of Linear Inequalities Writing, 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
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # Sums of Integers on a Number Line ## Add positive and negative numbers using a number line. Estimated5 minsto complete % Progress Practice Sums of Integers on a Number Line MEMORY METER This indicates how strong in your memory this concept is Progress Estimated5 minsto complete % Sums of Integers on a Number Line Source: https://farm4.staticflickr.com/3323/3278694436_d97590cef6.jpg Cooper is having a great time getting to know his pen pal, Riley, in New Zealand. They are both the same age and love sports. Cooper wants to call Riley, but he's having a hard time figuring out when it would be OK to call him because of the time zone difference. Cooper figures that it would be best if someone called him some time after he got home from school at 3pm and before he went to bed at 10pm. He knows that New Zealand is 18 hours ahead of Santa Fe, NM, where Cooper lives, but he doesn't know what time that makes it in New Zealand. In this concept, you will learn how to add integers using a number line. ### Adding Integers Using a Number Line An integer is any positive whole number or its opposite. Here, opposite means sign. So a positive integer has a negative opposite and vice versa. positive number is a number greater than zero. It can be written with or without a + symbol in front of it. A gain in something is written with a positive number. negative number is a number that is less than zero. It is always written with a - symbol in front of it. A loss is written with a negative number. number line is a line on which numbers are marked at intervals, used to illustrate simple numerical operations. Using a number line allows you to see where a number is in relationship to other numbers and from zero. Number lines are useful tools for visualizing simple arithmetic operations, especially addition and subtraction. Start at the position indicated by the first number. If the second number is negative, move that many units to the left. If it's positive, move that many units to the right. The final mark is the answer. Here is an example of using a number line in addition. -5 + 7 = ____ First, mark the initial integer. Next, move seven units in the positive direction. The result is +2. Here is another example. 6 + -9 = _____ First, mark the initial integer, +6. Next, move to left (the negative direction) 9 units. The final mark is at -3. ### Examples #### Example 1 Earlier, you were given a problem about Cooper and his time zone difficulty. He wants to call his friend, Riley, in New Zealand some time between 3pm and 10pm, but he doesn't know what time that is in Santa Fe, where he lives. He knows Santa Fe is 18 hours later than New Zealand. In order to solve this problem, Cooper can use number lines and addition. First, Cooper can draw a special number line that repeats every twelve hours. Next, to figure out the early limit, Cooper marks 3pm and then adds 18 hours. 3 pm + 18 hours = 9 am So, 3pm in New Zealand is 9am in Santa Fe. Then, he repeats this process to find the later limit. 10 pm + 18 hours = 4pm So, if Cooper wants to call Riley some time between the time Riley comes home from school at 3pm and before he goes to bed at 10pm, Cooper will have to call him between 9am and 4pm. But Cooper, himself, doesn't get home until 3pm. So he concludes that the only time he can call Riley is as soon as he gets home from school at 3pm. #### Example 2 Draw a number line from -6 to 9 and use it to find the sum. 5+1+7\begin{align}-5 + -1 + 7\end{align} = _____ First, draw the number line. Next, start at -5 and move 1 to the left. Then, add 7 to the right. You are now at +1. The answer is 1\begin{align}1\end{align}. #### Example 3 Draw a number line and use it to find the sum. -5 + 9 = ____ First, draw the number line. Next, start at -5 and move 9 to the right. You are now at +4. #### Example 4 Draw a number line and use it to find the sum. -1 + -8 = ____ First, draw the number line. Next, start at -1 and move 8 to the left. You are now at -9. #### Example 5 Draw a number line and use it to find the sum. 5 + -7 = ____ First, draw the number line. Next, start at 5 and move 7 to the left. You are now at -2. ### Review Add the following integers that have the same sign by using a number line. You may figure out a pattern. If so, try finding the sums without a number line. 1. 6 + 7 = _____ 2. -9 + -7 = _____ 3. -3 + -4 = _____ 4. 5 + 12 = _____ 5. -12 + -23 = _____ 6. 27 + 11 = _____ 7. -34 + -13 = _____ 8. 25 + 16 = _____ 9. -9 + -29 = _____ 10. -16 + -12 = _____ Add the following integers using a number line. Notice that they have different signs. 1. -9 + 3 = _____ 2. -7 + 5 = _____ 3. 1 + -12 = _____ 4. 3 + -8 = _____ 5. -19 + 11 = _____ 6. 7 + -12 = _____ 7. 23 + -10 = _____ 8. -4 + 16 = _____ 9. 15 + -18 = _____ 10. -15 + 9 = _____ To see the Review answers, open this PDF file and look for section 11.5. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English Integer The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...
# Introduction to Trigonometry – Exercise 8.4 – Class 10 1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. Solution: We know that, 2. Write all the other trigonometric ratios of ∠A in terms of sec A. Solution: We know that 3. Evaluate (i) (sin263° + sin227°)/(cos217°+cos273°) (ii) sin25° cos65° + cos25° sin65° Solution: (i) (ii)sin25° cos65° + cos25° sin65° = (sin25°){cos(90°-25°)} + cos25°{sin(90°-25°)} = (sin25°)(sin25°)+(cos25°)(cos25°) = (sin225°) + cos225° = 1 4. Choose the correct option. Justify your choice. (i) 9 sec2A − 9 tan2 A = (A) 1 (B) 9 (C) 8 (D) 0 (ii) (1 + tan θ + sec θ) (1 + cot θ − cosec θ) (A) 0 (B)1 (C) 2 (D) −1 (iii) (secA + tanA) (1 − sinA) = (A) secA (B) sinA (C) cosecA (D) cosA (iv)(1+tan2A)/(1+cot2A) = (A) sec2A (B) -1 (d) tan2A Solution: (i) 9 sec2 A − 9 tan2 A = 9 (sec2 A – tan2 A) = 9 (1) [Because sec2 A − tan2 A = 1] = 9 (B) is correct. (ii) (1 + tan θ + sec θ) (1 + cot θ − cosec θ) (C) is correct. (iii) = cosA (D) is correct. (iv) (D) is correct. 5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined. Solution: Solution: (iii) Solution: = secθ cosec θ + 1 = R.H.S. (iv) Solution: (v) solution: = cosec A + cot A = R.H.S (vi) Solution: ## 9 thoughts on “Introduction to Trigonometry – Exercise 8.4 – Class 10” 1. Oh how I wish I understood this. I have always admired people who excel in Mathematics; therefore, I admire you! Liked by 1 person 2. Very nice post! I have seen first in WordPress. Can you please tell me how you write mathematical numbers in your blog because its difficult to show powers or exponents what app you used. Can you please tell me. Liked by 1 person 1. Yeah sure… we have many mathematical symbol in WordPress itself . Very few like fractions that we need to do write in HTML page and other symbols like cube root, 4th root, nth root etc are easy to write in word document ,then we need to copy it to WordPress Liked by 1 person 2. Powers or exponents are easy to write in word document ,from there we can copy and paste in WordPress Like
Learning Library # Arranging Numbers in Multiple Ways (1 rating ) No standards associated with this content. Which set of standards are you looking for? Students will be able to find all of the factors pairs of whole numbers up to 100 and decide whether a given whole number, within 100, is a multiple of a given one-digit number. (5 minutes) • To begin, tell students that this lesson focuses on multiplication. • Ask students to share what they already know about multiplication. If you have already been studying multiplication, build on that, but if you have not, just let students share what they generally know. • Explain that today's lesson will involve pairs, which you will introduce in a moment. Ask students to give examples of pairs. (5 minutes) • Write the number 14 on the board. • Ask students if they know what a Factor pairIs. Explain that a factor pair is two numbers multiplied together to get a product. • Explain that in today's lesson, students will work together to find factor pairs of Whole numbers. Ask students what a whole number is. Remind them that a whole number is a number without fractional or decimal parts. (15 minutes) • Tell students that you will practise finding factor pairs together, before trying it in pairs. • Point out the number 14 on the board. To check for understanding, ask students if 14 is a whole number and how they know. • Model pulling out 14 cubes. • Tell students that now we will need to make an ArrayUsing the cubes. Ask students to define array. Remind them that an array is an arrangement in a particular way. • Model making a 2 row by 7 column array. Write 2 x 7 on the board. Explain that you made a 2 x 7 array, which means that 2 and 7 are a factor pair of 14. • Ask students other ways that we could arrange the cubes in a rectangle, to make an array. Make sure to show 7 x 2, 14 x 1, and 1 x 14, recording the factor pairs as they are made. • Explain that they can use factor pairs to help understand Multiples. Ask students if they know what a multiple is. Define multiple as the number found when multiplying one number by another. • Explain that 14 is a multiple of 1, 2, and 7. Model skip counting by those numbers to reach 14. • If needed, go through this process again, using 21 as the whole number. (25 minutes) • Explain to students that they will now work in pairs to find all of the factor pairs of various whole numbers under 100. • Pair students up and hand out the How Many Ways Can You Arrange My Number? worksheet. • Remind students to use cubes if they need to. They may also draw arrays on their papers. • As students work, check in with pairs to make sure they are finding all of the factor pairs for each number. • Encourage the use of cubes for students who need visual support. • Students who finish early can work on the How Many Ways Can You Arrange 100? worksheet. • Enrichment:Students who finish early can work on the How Many Ways Can You Arrange 100? worksheet. If they are also able to complete this, have them work on finding factor pairs of multiples of 100. • Support:Students who need extra support can either work with partners or in a small group with teacher guidance. Make sure they use the cubes and understand how to make rectangular arrays. (5 minutes) • Take note of student participation in the different parts of the lesson. • Check the worksheet and index card answers for mastery. (10 minutes) • Ten minutes before the end of the lesson, ask students to stop working and have them come together. • Write the number 36 on the board. • Hand out an index card to each student. • Tell students to write: "1. One factor pair of 36 is..." and "2. 36 is a multiple of..." • Tell students to think back over the lesson and fill in the answers on their index card. • As this is an assessment of student understanding, students who are unable to give an answer will need more practise. Create new collection 0 ### New Collection> 0Items What could we do to improve Education.com?
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> Our Terms of Use (click here to view) and Privacy Policy (click here to view) have changed. By continuing to use this site, you are agreeing to our new Terms of Use and Privacy Policy. # 12.4: Rational Expressions Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives • Simplify rational expressions. • Find excluded values of rational expressions. ## Introduction A simplified rational expression is one where the numerator and denominator have no common factors. In order to simplify an expression to lowest terms, we factor the numerator and denominator as much as we can and cancel common factors from the numerator and the denominator. ## Simplify Rational Expressions Example 1 Reduce each rational expression to simplest terms. a) \begin{align}\frac{4x-2}{2x^2+x-1}\end{align} b) \begin{align}\frac{x^2-2x+1}{8x-8}\end{align} c) \begin{align}\frac{x^2-4}{x^2-5x+6}\end{align} Solution a) \begin{align}\text{Factor the numerator and denominator completely:} \qquad \frac{2(2x-1)}{(2x-1)(x+1)}\!\\ \\ \text{Cancel the common factor} \ (2x - 1): \qquad \qquad \qquad \qquad \qquad \frac{2}{x+1}\end{align} b) \begin{align}\text{Factor the numerator and denominator completely:} \qquad \frac{(x-1)(x-1)}{8(x-1)}\!\\ \\ \text{Cancel the common factor}\ (x - 1): \qquad \qquad \qquad \qquad \qquad \ \ \frac{x-1}{8}\end{align} c) \begin{align}\text{Factor the numerator and denominator completely:} \qquad \frac{(x-2)(x+2)}{(x-2)(x-3)}\!\\ \\ \text{Cancel the common factor} (x - 2): \qquad \qquad \qquad \qquad \qquad \quad \frac{x+2}{x-3}\end{align} When reducing fractions, you are only allowed to cancel common factors from the denominator but NOT common terms. For example, in the expression \begin{align}\frac{(x+1) \cdot (x-3)}{(x+2) \cdot (x-3)}\end{align}, we can cross out the \begin{align}(x - 3)\end{align} factor because \begin{align}\frac{(x-3)}{(x-3)}=1\end{align}. But in the expression \begin{align}\frac{x^2+1}{x^2-5}\end{align} we can’t just cross out the \begin{align}x^2\end{align} terms. Why can’t we do that? When we cross out terms that are part of a sum or a difference, we’re violating the order of operations (PEMDAS). Remember, the fraction bar means division. When we perform the operation \begin{align}\frac{x^2+1}{x^2-5}\end{align}, we’re really performing the division \begin{align}(x^2+1) \div (x^2-5)\end{align} — and the order of operations says that we must perform the operations inside the parentheses before we can perform the division. Using numbers instead of variables makes it more obvious that canceling individual terms doesn’t work. You can see that \begin{align}\frac{9+1}{9-5}=\frac{10}{4}=2.5\end{align} — but if we canceled out the 9’s first, we’d get \begin{align}\frac{1}{-5}\end{align}, or -0.2, instead. For more examples of how to simplify rational expressions, watch the video at . ## Find Excluded Values of Rational Expressions Whenever there’s a variable expression in the denominator of a fraction, we must remember that the denominator could be zero when the independent variable takes on certain values. Those values, corresponding to the vertical asymptotes of the function, are called excluded values. To find the excluded values, we simply set the denominator equal to zero and solve the resulting equation. Example 2 Find the excluded values of the following expressions. a) \begin{align}\frac{x}{x+4}\end{align} b) \begin{align}\frac{2x+1}{x^2-x-6}\end{align} c) \begin{align}\frac{4}{x^2-5x}\end{align} Solution a) \begin{align}\text{When we set the denominator equal to zero we obtain:} \quad \ \ x+4=0 \Rightarrow x=-4\!\\ \\ \text{So} \ \mathbf{-4} \ \text{is the excluded value.}\end{align} b) \begin{align}\text{When we set the denominator equal to zero we obtain:} \qquad x^2-x-6=0\!\\ \\ \text{Solve by factoring:} \qquad \qquad \qquad \qquad \qquad \qquad \ \qquad \qquad \qquad (x-3)(x+2)=0\!\\ \\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \qquad \qquad \Rightarrow x=3 \ \text{and}\ x = -2\!\\ \\ \text{So}\ \mathbf{3}\ \mathbf{and}\ \mathbf{-2} \ \text{are the excluded values.}\end{align} c) \begin{align}\text{When we set the denominator equal to zero we obtain:} \quad \ \ x^2-5x=0\!\\ \\ \text{Solve by factoring:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ x(x-5)=0\!\\ \\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \Rightarrow x=0 \ \text{and} \ x = 5\!\\ \\ \text{So} \ \mathbf{0 \ and \ 5}\ \text{are the excluded values.}\end{align} ## Removable Zeros Notice that in the expressions in Example 1, we removed a division by zero when we simplified the problem. For instance, we rewrote \begin{align}\frac{4x-2}{2x^2+x-1}\end{align} as \begin{align}\frac{2(2x-1)}{(2x-1)(x+1)}\end{align}. The denominator of this expression is zero when \begin{align}x = \frac{1}{2}\end{align} or when \begin{align}x = -1\end{align}. However, when we cancel common factors, we simplify the expression to \begin{align}\frac{2}{x+1}\end{align}. This reduced form allows the value \begin{align}x = \frac{1}{2}\end{align}, so \begin{align}x = -1\end{align} is its only excluded value. Technically the original expression and the simplified expression are not the same. When we recduce a radical expression to its simplest form, we should specify the removed excluded value. In other words, we should write our final answer as \begin{align}\frac{4x-2}{2x^2+x-1}=\frac{2}{x+1}, x \neq \frac{1}{2}\end{align}. Similarly, we should write the answer from Example 1, part \begin{align}b\end{align} as \begin{align}\frac{x^2-2x+1}{8x-8}=\frac{x-1}{8}, x \neq 1\end{align} and the answer from Example 1, part c as \begin{align}\frac{x^2-4}{x^2-5x+6}=\frac{x+2}{x-3}, x \neq 2\end{align}. ## Review Questions Reduce each fraction to lowest terms. 1. \begin{align}\frac{4}{2x-8}\end{align} 2. \begin{align}\frac{x^2+2x}{x}\end{align} 3. \begin{align}\frac{9x+3}{12x+4}\end{align} 4. \begin{align}\frac{6x^2+2x}{4x}\end{align} 5. \begin{align}\frac{x-2}{x^2-4x+4}\end{align} 6. \begin{align}\frac{x^2-9}{5x+15}\end{align} 7. \begin{align}\frac{x^2+6x+8}{x^2+4x}\end{align} 8. \begin{align}\frac{2x^2+10x}{x^2+10x+25}\end{align} 9. \begin{align}\frac{x^2+6x+5}{x^2-x-2}\end{align} 10. \begin{align}\frac{x^2-16}{x^2+2x-8}\end{align} 11. \begin{align}\frac{3x^2+3x-18}{2x^2+5x-3}\end{align} 12. \begin{align}\frac{x^3+x^2-20x}{6x^2+6x-120}\end{align} Find the excluded values for each rational expression. 1. \begin{align}\frac{2}{x}\end{align} 2. \begin{align}\frac{4}{x+2}\end{align} 3. \begin{align}\frac{2x-1}{(x-1)^2}\end{align} 4. \begin{align}\frac{3x+1}{x^2-4}\end{align} 5. \begin{align}\frac{x^2}{x^2+9}\end{align} 6. \begin{align}\frac{2x^2+3x-1}{x^2-3x-28}\end{align} 7. \begin{align}\frac{5x^3-4}{x^2+3x}\end{align} 8. \begin{align}\frac{9}{x^3+11x^2+30x}\end{align} 9. \begin{align}\frac{4x-1}{x^2+3x-5}\end{align} 10. \begin{align}\frac{5x+11}{3x^2-2x-4}\end{align} 11. \begin{align}\frac{x^2-1}{2x^2+x+3}\end{align} 12. \begin{align}\frac{12}{x^2+6x+1}\end{align} 13. In an electrical circuit with resistors placed in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of each resistance. \begin{align}\frac{1}{R_c}=\frac{1}{R_1}+\frac{1}{R_2}\end{align}. If \begin{align}R_1 = 25 \ \Omega\end{align} and the total resistance is \begin{align}R_c = 10 \ \Omega\end{align}, what is the resistance \begin{align}R_2\end{align}? 14. Suppose that two objects attract each other with a gravitational force of 20 Newtons. If the distance between the two objects is doubled, what is the new force of attraction between the two objects? 15. Suppose that two objects attract each other with a gravitational force of 36 Newtons. If the mass of both objects was doubled, and if the distance between the objects was doubled, then what would be the new force of attraction between the two objects? 16. A sphere with radius \begin{align}R\end{align} has a volume of \begin{align}\frac{4}{3} \pi R^3\end{align} and a surface area of \begin{align}4 \pi R^2\end{align}. Find the ratio the surface area to the volume of a sphere. 17. The side of a cube is increased by a factor of 2. Find the ratio of the old volume to the new volume. 18. The radius of a sphere is decreased by 4 units. Find the ratio of the old volume to the new volume. ### My Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show More ### Image Attributions Show Hide Details Description Tags: Subjects: Grades: Date Created: Feb 23, 2012 Last Modified: May 15, 2016 Files can only be attached to the latest version of section Please wait... Please wait... Image Detail Sizes: Medium \| Original CK.MAT.ENG.SE.2.Algebra-I.12.4 Here
### Part I. Section III. Chapter 11. “Of Geometrical Progressions.” 505 A series of numbers, which are always becoming a certain number of times greater, or less, is called a geometrical progression, because each term is constantly to the following one in the same geometrical ratio: and the number which expresses how many times each term is greater than the preceding, is called the exponent, or ratio. Thus, when the first term is 1 and the exponent, or ratio, is 2, the geometrical progression becomes, Terms 1 2 3 4 5 6 7 8 9 etc. Progression 1, 2, 4, 8, 16, 32, 64, 128, 256, etc. The numbers 1, 2, 3, etc. always marking the place which each term holds in the progression. 506 If we suppose, in general, the first term to be $$a$$, and the ratio $$b$$, we have the following geometrical progression Terms 1 2 3 4 5 6 7 8 ⋯ $$n$$ Progression $$a$$, $$ab$$, $$ab^2$$, $$ab^3$$, $$ab^4$$, $$ab^5$$, $$ab^6$$, $$ab^7$$, ⋯ $$ab^{n-1}$$. So that, when this progression consists of $$n$$ terms, the last term is $$ab^{n-1}$$. We must, however, remark here, that if the ratio $$b$$ be greater than unity, the terms increase continually; if $$b=1$$, the terms are all equal; lastly, if $$b$$ be less than 1, or a fraction, the terms continually decrease. Thus, when $$a=1$$, and $$b=\frac{1}{2}$$, we have this geometrical progression: 1, ½, ¼, ⅛, ¹⁄₁₆, ¹⁄₃₂, ¹⁄₆₄, ¹⁄₁₂₈, etc. 507 Here therefore we have to consider: 1. The first term, which we have called $$a$$. 2. The exponent, which we call $$b$$. 3. The number of terms, which we have expressed by $$n$$. 4. And the last term, which, we have already seen, is expressed by $$ab^{n-1}$$. So that, when the first three of these are given, the last term is found by multiplying the $$n-1$$ power of $$b$$, or $$b^{n-1}$$, by the first term $$a$$. If, therefore, the 50th term of the geometrical progression 1, 2, 4, 8, etc. were required, we should have $$a=1$$, $$b = 2$$, and $$n=50$$; consequently the 50th term would be 2⁴⁹; and as 2⁹=512, we shall have 2¹⁰=1024; wherefore the square of 2¹⁰, or 2²⁰, =1048576, and the square of this number, which is 1099511627776=2⁴⁰. Multiplying therefore this value of 2⁴⁰ by 2⁹, or 512, we have 2⁴⁹=562949953421312 for the 50th term. 508 One of the principal questions which occurs on this subject, is to find the sum of all the terms of a geometrical progression; we shall therefore explain the method of doing this. Let there be given, first, the following progression, consisting of ten terms: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, the sum of which we shall represent by $$s$$, so that $s=1+2+4+8+16+32+64+128+256+512;$ doubling both sides, we shall have $2s=2+4+8+16+32+64+128+256+512+1024;$ and subtracting from this the progression represented by $$s$$, there remains $$s=1024-1=1023$$; wherefore the sum required is 1023. 509 Suppose now, in the same progression, that the number of terms is undetermined, that is, let them be generally represented by $$n$$, so that the sum in question, or $s=1+2+2^2+2^3+2^4+\cdots+2^{n-1}.$ If we multiply by 2, we have $2s=2+2^2+2^3+2^4+\cdots+2^n,$ then subtracting from this equation the preceding one, we have $$s=2^n-1-1$$. It is evident, therefore, that the sum required is found, by multiplying the last term, $$2^{n-1}$$, by the exponent 2, in order to have $$2^n$$, and subtracting unity from that product. 510 This is made still more evident by the following examples, in which we substitute successively for $$n$$, the numbers 1, 2, 3, 4, etc. 1 = 1; 1 + 2 = 3; 1 + 2 + 4 = 7; 1 + 2 + 4 + 8 = 15; 1 + 2 + 4 + 8 + 16 = 31; 1 + 2 + 4 + 8 + 16 + 32 = 63, etc. 511 On this subject, the following question is generally proposed. A man offers to sell his horse on the following condition; that is, he demands 1 penny for the first nail, 2 for the second, 4 for the third, 8 for the fourth, and so on, doubling the price of each succeeding nail. It is required to find the price of the horse, the nails being 32 in number? This question is evidently reduced to finding the sum of all the terms of the geometrical progression 1, 2, 4, 8, 16, etc. continued to the 32d term. Now, that last term is 2³¹; and, as we have already found 2²⁰=1048576, and 2¹⁰=1024, we shall have 2²⁰ · 2¹⁰ = 2³⁰ = 1073741824; and multiplying again by 2, the last term 2³¹=2147483648; doubling therefore this number, and subtracting unity from the product, the sum required becomes 4294967295 pence; which being reduced, we have 17895697l. 1s. 3d. for the price of the horse.1 512 Let the ratio now be 3, and let it be required to find the sum of the geometrical progression 1, 3, 9, 27, 81, 243, 729, consisting of 7 terms. Calling the sum $$s$$ as before, we have $s=1+3+9+27+81+243+729.$ And multiplying by 3, $3=3+9+27+81+243+729+2187.$ Then subtracting the former series from the latter, we have $$2s=2187-1=2186$$: so that the double of the sum is 2186, and consequently the sum required is 1093. 513 In the same progression, let the number of terms be $$n$$, and the sum $$s$$; so that $s=1+3+3^2+3^3+3^4+\cdots+3^{n-1}.$ If now we multiply by 3, we have $3s=3+3^2+3^3+3^4+\cdots+3^n.$ Then subtracting from this expression the value of $$s$$, as before, we shall have $$2s=3^n-1$$; therefore $$s =\frac{3^n-1}{2}$$. So that the sum required is found by multiplying the last term by 3, subtracting 1 from the product, and dividing the remainder by 2; as will appear, also, from the following particular cases: 1 $$\dfrac{(1\cdot 3)-1}{2}$$ = 1 1 + 3 $$\dfrac{(3\cdot 3)-1}{2}$$ = 4 1 + 3 + 9 $$\dfrac{(3\cdot 9)-1}{2}$$ = 13 1 + 3 + 9 + 27 $$\dfrac{(3\cdot 27)-1}{2}$$ = 40 1 + 3 + 9 + 27 + 81 $$\dfrac{(3\cdot 81)-1}{2}$$ = 121. 514 Let us now suppose, generally, the first term to be $$a$$, the ratio $$b$$, the number of terms $$n$$, and their sum $$s$$, so that $s=a+ab+ab^2+ab^3+ab^4+\cdots+ab^{n-1}.$ If we multiply by $$b$$, we have $bs = ab+ab^2+ab^3+ab^4+ab^5+\cdots+ab^n,$ and taking the difference between this and the above equation, there remains $$(b-1)s=ab^n-a$$; whence we easily deduce the sum required $s=\dfrac{ab^n-a)}{b-1}.$ Consequently, the sum of any geometrical progression is found, by multiplying the last term by the ratio, or exponent of the progression, and dividing the difference between this product and the first term, by the difference between 1 and the ratio. 515 Let there be a geometrical progression of seven terms, of which the first is 3; and let the ratio be 2: we shall then have $$a = 3$$, $$b = 2$$, and $$n = 7$$; therefore the last term is 3 · 2⁶, or 3 · 64, = 192; and the whole progression will be 3, 6, 12, 24, 48, 96, 192. Farther, if we multiply the last term 192 by the ratio 2, we have 384; subtracting the first term, there remains 381; and dividing this by $$b - 1$$, or by 1, we have 381 for the sum of the whole progression. 516 Again, let there be a geometrical progression of six terms, of which the first is 4; and let the ratio be ³⁄₂: then the progression is 4, 6, 9, ²⁷⁄₂, ⁸¹⁄₄, ²⁴³⁄₈. If we multiply the last term by the ratio, we shall have ⁷²⁹⁄₁₆; and subtracting the first term = ⁶⁴⁄₁₆, the remainder is ⁶⁶⁵⁄₁₆; which, divided by $$b-1=\frac{1}{2}$$, gives ⁶⁶⁵⁄₈ = 83⅛ for the sum of the series. 517 When the exponent is less than 1, and, conscquently, when the terms of the progression continually diminish, the sum of such a decreasing progression, carried on to infinity, may be accurately expressed. For example, let the first term be 1, the ratio ½, and the sum $$s$$, so that $s=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\textrm{etc.}$ to infinity. If we multiply by 2, we have $s=2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\textrm{etc.}$ to infinity: and, subtracting the preceding progression, there remains $$s = 2$$ for the sum of the proposed infinite progression. 518 If the first term be 1, the ratio ⅓, and the sum $$s$$; so that $s=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\textrm{etc.}$ to infinity. Then multiplying the whole by 3, we have $3s=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\textrm{etc.}$ and subtracting the value of $$s$$, there remains $$2s=3$$; wherefore the sum $$s=1\frac{1}{2}$$. 519 Let there be a progression whose sum is $$s$$, the first term 2, and the ratio ¾; so that $s=2+\frac{3}{2}+\frac{9}{8}+\frac{27}{32}+\frac{81}{128}+\textrm{etc.}$ to infinity. Multiplying by ⁴⁄₃, we have $\frac{4}{3}s=\frac{8}{3}+2+\frac{3}{2}+\frac{9}{8}+\frac{27}{32}+\frac{81}{128}+\textrm{etc.};$ and subtracting from this progression $$s$$, there remains $$\frac{1}{3}s=\frac{8}{3}$$: wherefore the sum required is 8. 520 If we suppose, in general, the first term to be $$a$$, and the ratio of the progression to be $$\frac{b}{c}$$, so that this fraction may be less than 1, and consequently $$c$$ greater than $$b$$; the sum of the progression, carried on to infinity, will be found thus: Make $s=a+\frac{ab}{c}+\frac{ab^2}{c^2}+\frac{ab^3}{c^3}+\frac{ab^4}{c^4}+\textrm{etc.}$ to infinity; and subtracting this equation from the preceding, there remains $$\left(1-\frac{b}{c}\right)s=a$$. Consequently, $s=\dfrac{a}{1-\frac{b}{c}}=\frac{ac}{c-b},$ by multiplynig both the numerator and denominator by $$c$$. The sum of the infinite geometrical progression proposed is, therefore, found by dividing the first term a by 1 minus the ratio, or by multiplying the first term $$a$$ by the denominator of the ratio, and dividing the product by the same denominator diminished by the numerator of the ratio. 521 In the same manner we find the sums of progressions, the terms of which are alternately affected by the signs + and -. Suppose, for example, $s=a-\frac{ab}{c}+\frac{ab^2}{c}-\frac{ab^3}{c}+\frac{ab^4}{c}-\textrm{etc.}$ And, adding this equation to the preceding, we obtain $$\left(1+\frac{b}{c}\right)s=a$$: whence we deduce the sum required, $$s=\dfrac{a}{1+\frac{b}{c}}$$, or $$s=\frac{ac}{c+b}$$. 522 It is evident, therefore, that if the first term $$a=\frac{3}{5}$$, and the ratio be ⅖, that is to say, $$b=2$$, and $$c=5$$, we shall find the sum of the progression $$\frac{3}{5}+\frac{6}{25}+\frac{12}{125}+\frac{24}{625}+\textrm{etc.}=1$$; since, by subtracting the ratio from 1, there remains ⅗, and by dividing the first term by that remainder, the quotient is 1. It is also evident, if the terms be alternately positive and negative, and the progression assume this form: $\frac{3}{5}-\frac{6}{25}+\frac{12}{125}-\frac{24}{625}+\textrm{etc.}$ that the sum will be $\dfrac{a}{1+\frac{b}{c}} = \dfrac{\frac{3}{5}}{\frac{7}{5}} = \frac{3}{7}.$ 523 Again: let there be proposed the infinite progression, $\frac{3}{10}+\frac{3}{100}+\frac{3}{1000}+\frac{3}{10000}+\frac{3}{10000}+\frac{3}{100000}+\textrm{etc.}$ The first term is here ³⁄₁₀, and the ratio is ⅒; therefore subtracting this last from 1, there remains ⁹⁄₁₀, and, if we divide the first term by this fraction, we have ⅓ for the sum of the given progression. So that taking only one term of the progression, namely ³⁄₁₀, the error would be ⅒. And taking two terms, ³⁄₁₀ + ³⁄₁₀₀ = ³³⁄₁₀₀, there would still be wanting ¹⁄₁₀₀ to make the sura, which we have seen is ⁴⁄₃. 524 Let there now be given the infinite progression, $9+\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{9}{10000}+\textrm{etc.}$ The first term is 9, and the ratio is ⅒. So that 1 minus the ratio is ⁹⁄₁₀; and $$\dfrac{9}{\frac{9}{10}}=10$$, the sum required: which series is expressed by a decimal fraction, thus, 9.9999999 etc. #### Editions 1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822. 2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum. 1. 1 pound (l.) = 20 shillings (s.), 1 shilling (s.) = 12 pence (d.), 1 pound (l.) = 240 pence (d.).
Mass from volume & concentration. Using this calculator, we will understand methods of how to find the surface area and volume of the hemisphere. A prism is a solid shape that has the same cross-section all the way through. The first has a circular cross-section. In this lesson you will find the volume of complex figures by deconstructing and adding the volumes of simpler prisms using the volume formula. How to calculate the volume of a rectangular prism. In this lesson, students continue working with the volume of right prisms. Perimeter of the base, use (, , and so on) Lateral Area, use. This Problem Set is a part of the Lesson 13, Unit 7, Grade 7. An alternate way to calculate the area of a triangle is to use Heron's Formula. Processing. 1 cm, and a height of 6. Source(s): https://shrink. Volume of Prisms and Cylinders Write each formula. Therefore, to find parallelepiped's volume build on vectors, one need to calculate scalar triple product of the given vectors, and take the magnitude of the result found. This lesson is most appropraite for 4th grade students. Then you would take 72 - 42 to give you a total of 30 cubic inches. The volume of a 3 -dimensional solid is the amount of space it occupies. Tracing paper may be used. A square prism is simply a cuboid with square base. Given the length, width, and height of a triangular prism, the task is to find the volume of the triangular prism. How to calculate the volume of a rectangular prism. Find the volume of the triangular prism, whose base is 7 cm, height is 9 cm and length is 12 cm. To make it clear among students, below are key examples of triangular prism nets to print. calculate the volume and surface area. The height of the prism can be found if is known a - the side of the base, n - the number of sides and S - the area of the lateral surface: h = S/na. Hexagonal prism is a special case of the general prism, which may have any arbitrary polygonal base. A triangular prism has sides of 5, 6 and 7 with a height of 9. Examples: Input: l = 18, b = 12, h = 9 Output: Volume of triangular prism: 972 Input: l = 10, b = 8, h = 6 Output: Volume of triangular prism: 240. Below are six versions of our grade 6 math worksheet on finding the volume and surface areas of rectangular prisms. For example, if the base is 8 and the height is 9, you would get ½ x 8 x 9 = 36. Trapezoidal Prism Volume Calculator In geometry, a triangular prism is a three-sided prism; it is a polyhedron made of a triangular base, a translated copy, and 3 faces joining corresponding sides. Volume of prisms and cylinders (EMA7P) Volume. For example, find the cubic footage volume of a backfill area that is 8 feet wide, 6 feet deep and 50 feet long. Once you have these calculations, you can create a handy chart for later. The cross section is a rectangle and a semicircle joined together. We can now just plug this number in to the formulas to calculate the volume and surface area. The volume of a cube with sides of length L is given by L^3. The volume of each prism is __1 2 r × __ 1 3 r × 2r, where r is the radius of the cylinder’s base. Nikita calculated the volume of the right triangular prism. Students will be able to grasp here all on properties of commonly used solid figures. Explore many other math calculators like the area and surface area calculators, as well as hundreds of other calculators related to finance, health, fitness, and more. Depending on the particular body, there is a different formula and different required information you need to calculate its volume. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment, faculty enhancement, and interactive curriculum development at all levels. The formula, in general, is the area of the base (the red triangle in the picture on the left) times the height,h. If two solids are similar, then their corresponding sides are all proportional. V : Formula: V = B × h where, h - the height of the prism B - the area of the base V - right prism volume Related Calculators Lateral Surface Area - Right Prism Surface Area - Oblique Prism. I n the below figure, we can see that the bases are square. The cross-section of the prism is a trapezium. For a prism which has Trapezium shaped ends, we need to first find the area of the Trapezium using A = 1/2 (top + bottom) x height of trapezium. Volume of a Triangular Prism Formula - A prism is a solid object that has two congruent faces on either end joined by parallelogram faces laterally. The launch and task explore a real world problem to generate student thinking about. Attempt every question. TSA of Rectangular Prisms. A tool for helping calculate the volume of common types of house and loft extensions Volume calculator. Any of the following images can be printed for free. This rectangular prism calculator can help you find the volume, surface area, diagonal, height or width of a rectangular prism if you know the required dimensions. A rectangular prism is a box-shaped object with six sides. The volume of a trapezoidal prism calculator can find the volume and area of the trapezoidal prism at the same time. In the example above the volume = 4 x 5 x8 = 160cm 3. Volume of a square pyramid given base side and height. Here you can calculate the area, volume of Triangular, Rectangular, Square, Pentagonal, Hexagonal Prism. All solutions are included along with all printable resources and a learning plan. Therefore we can double this formula to ﬁnd the area of both triangular faces at once. In this activity, compare the volume of the container with the volume of its contents to calculate how much each can hold. But if it starts to tilt, then it is considered an oblique prism, and finding the volume can be tricky. How many cubic feet of space are in his tent ?. Use the calculator above to calculate height, radius or. Enter the radius, diameter, surface area or volume of a Sphere to find the other three. Volume is measured in cubic units. The lesson explores how the volume of a rectangular prism is different than the area of a rectangle. Trigonometry method: Area = product of both side length sin (angle between them) = ab sin C, Where a and b are the lengths of two unequal sides, C is the angle Diagonal method: Area = The area is half the product of the diagonals = (d1d2)/2 where d1,d2 are the. This is a free online tool by EverydayCalculation. Height of a regular hexagonal prism. In this tutorial, you'll see how to use that information and the formula for the volume of a rectangular prism to get the answer. In the formula for finding the volume of an oblique prism please note that the height is the perpendicular segment between the top and bottom bases. An irregular shape is a shape which does not conform to standard defined and repeatable mathematical rules; Pyramid Volume Calculator - a pyramid is a [three-dimensional solid object] polyhedron formed by connecting a polygonal base and to a point. This is a geometric solid having 6 faces all of which are rectangles and each pair of opposite faces are congruent. With a hexagon base, the number of faces, edges and vertices (NF, NE, NV respectively) is given by the formulas:. Both of the pictures of the Triangular prisms below illustrate the same formula. If a square has one side of 4 inches, the volume would be 4 inches times 4 inches times 4 inches, or 64 cubic inches. Regular Hexagonal Prism Calculator. These three shapes are prisms. Volume : I cubic unit The volume of prism or is V: (Area of un. Hemisphere calculator is an online Geometry tool requires radius length of a hemisphere. Rectangle Volume Calculation. For now this does not support decimals, if you could figure it out just email me at [email protected] To find the area of the base of a cylinder, use the formula for finding the area of a circle. A general formula is volume = length * base_area; the one parameter you always need to have given is the prism length, and there are four ways to calculate the base - triangle area. Perimeter of the base, use (, , and so on) Lateral Area, use. com to calculate volume of a Right-Triangular Prism. Volume Calculators. A prism is a solid geometric figure with two identical ends and all flat sides. V = LWH = 6×3×2 = 36. Results: Area of Base. Volume is the three dimensional space occupied by an object, or the contents of an object. The radius of a sphere is half of its diameter. Height of a right square prism. 1 cm, and a height of 6. A general formula is volume = length * base_area ; the one parameter you always need to have given is the prism length, and there are four ways to calculate the base - triangle area. Again a triangular prism net is very very helpful to calculate the surface area of the prism. Byju's Volume of a Rectangular Prism Calculator is a tool which makes calculations very simple and interesting. Using this calculator, we will understand methods of how to find the surface area and volume of the hemisphere. to ﬁ nd the volume of a rectangular prism. Formula for a Square/Rectangle Mold: Length x Width x Height Example: You have a mold box that needs to be filled up 9" x 4" x 2" = 72 cubic inches in order to completly cover your original/part. Surface Area Of Prisms And Pyramids. If you have to determine the area or volume of an odd prism, you can rely on the area (A) and the perimeter (P) of the base shape. Just enter the length, the width, and the height of the rectangular prism Then click on the button that says calculate Do not enter numbers with a slash (/)! If you do, the wrong answer will be computed. Volume of a square pyramid given base side and height. The calculation formulas are externalized as well. This lesson builds on student's work with whole numbers, volume, and additive reasoning. Also the height of prism can be found by it's volume, by the above formula. Perimeter and surface area formulas are common geometry calculations used in math and science. You can find the volume of a cube by just knowing the measurement of one side. These three shapes are prisms. The formula, in general, is the area of the base (the red triangle in the picture on the left) times the height,h. Volume of a prism. 6 cm, a width of 2. 4 years ago. Find the volume of following solid of cubes with unit fraction edge lengths. volume of a cube with edge length s V s 3 2. calculate the volume V of a rectangular prism by multiplying the measured length L by the measured width W and the. The trapezium has height 4cm and its parallel sides are 5cm & 7cm. If the length of the prism is 5 yards, nd its volume. These worksheets are printable pdf files. Volume is measured in cubic units ( in 3 , ft 3 , cm 3 , m 3 , et cetera). and each one is called a base. Big Ideas: To find the volume of composite solids, we can decompose them into right rectangular prisms to find the volume of each part. You can use the following simple formulas to help you calculate the volume of shapes like cuboids and prisms. We can calculate the volume of any prism simply by knowing the height of the prism and the area of one of its bases. Label the answer. So, we can say that the Triangular Prism has a triangular base. Find the volume of this rectangular prism. VOLUME OF ALL PRISMS: Base * He (8h+ Example 1: What is the base and the height of each shape? Example 2: Calculate the volume of the prisms V = crea base * (£. Volume of a rectangular prism V=lengthwidthheight. Again, this is not the solids' volume, only the ratio of the volumes. Example 2 : Bradley’s tent is in the shape of a triangular prism shown below. Calculate the various properties (like Area of Base, Perimeter of Base, Surface Area of Prism and Volume of Prism) of Triangular Prism, Rectangular Prism(Cuboid), Regular Square Prism(Square Cuboid), Regular Pentagonal Prism, Regular Hexagonal Prism, for given values. Calculating the Volume of a Cylinder: If a cylinder has a flat bottom, meaning the height and radius meet at right angles, then this formula can be used to find of volume ('V') of that cylinder: V = PIr 2 h. To find the rectangular prism surface area, simply add areas of all faces:. This is a geometric solid having 6 faces all of which are rectangles and each pair of opposite faces are congruent. These worksheets are especially meant for grades 5 and 6 when students study volume of prisms, but certain types of problems you can create suit best grades 7-9. Perimeter of Base. But if it starts to tilt, then it is considered an oblique prism, and finding the volume can be tricky. Prism Calculator This program is developed to help you to overcome problems in calculus. Volume of a square pyramid given base and lateral sides. However, the volume of paper hasn't changed. A prism is a polyhedron shape with two parallel sides that are equal in length called bases and the lateral sides of a parallelogram are named as parallelograms. This is summarized by the formula: LA 5 hP. Rectangular Prisms (or Solids) The above figure is called a rectangular prism. Volume Of Composite Rectangular Prisms Some of the worksheets for this concept are Measurement and data 3 80measu0rumun0at0d dsu10m c, Volume, Volume prisms cylinders l1es1, Volumes of solids, Measurement work volume and surface area of, Volume of composite solids, Volume of prism l1es1, Measurement and data volume grade 5 formative assessment. 5 cubic feet. For more detailed information on the calculations, read more below the calculator. h is the triangle's height. Work out an expression, in cm3, for the total volume of the solid. For example, if the base is 8 and the height is 9, you would get ½ x 8 x 9 = 36. Calculator calculates the area of a kite with both methods (Trigonometry and Diagonal) and also calculate the perimeter of the kite. Typically the various sides of this three dimensional shape are labelled a, b and c. The two faces at either ends are hexagons, and the rest of the faces of the hexagonal prism are rectangular. Answer : B Explanation. Calculations at a prismatoid. com's Prism Formulas & Calculator is an online basic geometry tool to calculate volume & surface area of the shape of rectangular & triangle prism, in both US customary & metric (SI) units. Can you determine a rule for finding the volume of a box if you know its width, depth, and height?. It's special triangle and/or Pythagorean Theorem time. Although a cylinder is technically not a prism, it shares many of the properties of a prism. It's volume and total surface area can be calculated using the tool provided. The following is the calculation formula for the volume of a rectangular prism:. Example 2 : Bradley’s tent is in the shape of a triangular prism shown below. Note: Before calculating the volume of prism, you must calculate the Area of base [wp_ad_camp_3] Here is the example program # 1 with online compiler and execution tool. The formula is given as V = 5/2 abh, where "V" denotes the volume, "a" indicates the apothem length, "b" represents the side and "h" is the prism's height. V : Formula: V = B × h where, h - the height of the prism B - the area of the base V - right prism volume Related Calculators Lateral Surface Area - Right Prism Surface Area - Oblique Prism. Calculate the cuboid volume. IE: 111 lbs. Here's how to calculate the volume of a variety of prisms. Trapezoidal Prism Volume Calculator In geometry, a triangular prism is a three-sided prism; it is a polyhedron made of a triangular base, a translated copy, and 3 faces joining corresponding sides. Donʼt spend too long on one question. A right triangular prism has rectangular sides, otherwise it is oblique. Trigonometry method: Area = product of both side length sin (angle between them) = ab sin C, Where a and b are the lengths of two unequal sides, C is the angle. As a result, you can compute the value of the matrix simply as the determinant of the upper left 3×3 submatrix. } Ask Algebra House. Depending on the figures to be solved for, this rectangular prism calculator uses the formulas explained here: If you choose to calculate the volume (V), area (A) and diagonal (d. Calculating Volume Rectangular Prisms. First, we need to calculate the area of the triangular base. Centimeter cubes have gone wild with these irregular. 22 USD The calculation can be used to calculate the volume of a pyramid with a four-sided base. You can easily find out their volume by using our calculator. Trapezoidal Prism Volume Calculator. This means that the radius is 1. aw , VOLUME IS: CUBIC uh i. Find the volume of the triangular prism, whose base is 7 cm, height is 9 cm and length is 12 cm. Read each question carefully before you begin answering it. Step 2: Volume V = l w h = $4 \frac{1}{3} \times 5 \times 5$ = $13 \times \frac{1}{3} \times 15 \times \frac{1}{3} \times 15 \times \frac{1}{3}$. Here, are some major elements of a Prism that are necessary to remember. See more volume of a prism videos at Brightstorm. Centimeter cubes have gone wild with these irregular. Bases Lateral Faces 3 cm 5 cm 6 cm 4 cm 3 cm 4 cm 5 cm 3 cm 4 cm 5 cm 4 cm 6 cm. The volume of prism shapes - ANY PRISM is simply the area of its base multiplied by the length of the prism. Volume of prism = Area of base x height V= 27. Calculate the volume of. Volume of a Rectangular Prism Worksheets, Geometry Worksheets for 4th grade, 5th grade and middle school. Volume of Trapezoid Prism. Select this link for a hint. 1 cm, and a height of 6. The most basic two equations are as followed: Volume = 0. [Measurement and Reference Frames Goal 2]. Email or Phone: Password. A cube is a rectangular prism with six equal sides. The given data consists of: H = 4 cm. Using the rectangular prism volume formula above, we get volume = 18in 12in * 15in = 3240 in³. Improve your math knowledge with free questions in "Volume of cubes and rectangular prisms" and thousands of other math skills. Surface Area & Volume of a Prism Surface Area of a Prism Suppose that we want to find the lateral area and total surface area of the following right triangular prism: The bases of this prism are right triangles, and the lateral faces are rectangles, as shown below. A rectangular prism also known as a cuboid is a 3-dimensional object which has six faces that are rectangles. Steps Calculate the Volume of a Triangular Prism. If you input the necessary variables, our calculator will give you the exact volume and surface area. Everyday Calculation Free calculators and unit converters for general and everyday use. Here we have 2 different rectangular prisms. 50x28x15=1400x15=21000 cubic feet The number of cubic feet in this room is 21000 cubic feet. Calculate the area, volume of Hemisphere (Hemisphere Calculator). Identifying nets. In the triangular prism calculator you can easily find out the volume of that solid. Right Prism Volume Calculator. For example, find the cubic footage volume of a backfill area that is 8 feet wide, 6 feet deep and 50 feet long. Take the guesswork out of calculating liquid volume, and use our Liquid Volume Calculator. Processing. Select this link for a hint. In this case the base is a triangle so we simply need to compute the area of the triangle and multiply this by the length of the prism:. Example: Calculate the volume of a rectangular prism that measures 6 feet long, 3 feet wide and 2 feet high. The volume of a Triangular Prism Calculator is a free online tool that displays the triangular prism volume for the given measures. Surface Area, Lateral Area, and Volume Formulas In the table shown B is the area of the base, P is the perimeter of the base, h is the height of the object, l is the slant height of the object, r is the radius of the base if the base is a circle, C is the circumference of the base if it. The lateral area of a prism is the area of its sides—namely, the area of everything but the prism’s bases. Record your results in a table. More complicated pool shapes will require multiple calculations and adding them together. Car load volume to move storage. Perimeter of Base. All space figures have volume. To do this, student calculate the volume of the two rectangular prisms that make up the structure. The first has a circular cross-section. I've tried everything from 77 cm^2 to 78 cm^2. Volume of a square pyramid given base side and height. VOLUME OF ALL PRISMS: Base * He (8h+ Example 1: What is the base and the height of each shape? Example 2: Calculate the volume of the prisms V = crea base * (£. The roof of the house is the one shaped like a triangular prism. The formula for finding the volume of a cylinder is the same as the formula for finding the volume of a prism. Prism Calculator Use this online calculator to calculate the properties of a prism, enter known measurements to calculate associated dimensions and volume of a prism. Then you would take 72 - 42 to give you a total of 30 cubic inches. A cube has edges of length 0. Prism Calculator This program is developed to help you to overcome problems in calculus. Use this length x width x height calculator to determine the volume in the following applications: Volume of package to be dispatched to add to shipping paperwork; Gravel volume required to fill a path, car park or driveway. 5% Moisture. The Volume of a Rectangular Prism Calculator is used to help you find the volume of a rectangular prism. For a roof with a height of 6 feet, a width of 20 feet and a ridge length of 32 feet, the calculation looks like. You can use the following simple formulas to help you calculate the volume of shapes like cuboids and prisms. A prism is a transparent element with polished surface which can refract light. This gives us. Many times, this formula will use the height of the prism, or depth (d), rather than the length (l), though you may see either abbreviation. Volume of a Rectangular Prism formula. The space that the base passes through is the volume of the prism. The Volume of a Rectangular Prism Calculator is used to help you find the volume of a rectangular prism. All you need to do is follow the 4 steps below: 1. Calculate the unknown defining surface areas, lengths, widths, heights, and volume of a rectangular prism with any 3 known variables. com to help you to score more marks in your exams. Cubic ft x 7. It is measured in cubic units. Rectangular Prism Units of Measure Area of the Base, use (, , and so on). Surface Area Solution. TRIANGULAR PRISMS In a triangular prism, each cross‑section parallel to the triangular base is a triangle congruent to the base. Although a cylinder is technically not a prism, it shares many of the properties of a prism. Regular Pentagonal Prism Calculator. 5 * b * h * length b is the length of the triangle's base. These worksheets are printable pdf files. 5 cm in height. Just enter the length, the width, and the height of the rectangular prism Then click on the button that says calculate Do not enter numbers with a slash (/)! If you do, the wrong answer will be computed. 65^2 + 6^2 = c^2. Use the calculator above to calculate height, radius or. Calculate the volume of the prism. Exam Style Questions Ensure you have: Pencil, pen, ruler, protractor, pair of compasses and eraser You may use tracing paper if needed Guidance 1. Surface Area (SA) = √ (25 + 10 * √5) * a² √ (25 + 10 * √5) * a² * h. V = (1/2) x (528) x 7. Cubic ft x 7. The volume of any pyramidal prism (flat base of any shape converging to a single point) is 1/3 base area times height, where height is the distance from the point perpendicular to the base. Attempt every question. SAT Math Help » Geometry » Solid Geometry » Prisms » How to find the volume of a prism Example Question #1 : How To Find The Volume Of A Prism A rectangular box has a length of 2 meters, a width of 0. Nikita used the wrong formula for the volume of the prism. The cross section is a rectangle and a semicircle joined together. Calculate the volume of cubes with the following edge lengths: 7 cm 12 mm Calculate the volume of the following square-based prisms: side of the base = 5 mm, h = 12 mm side of the base = 11 m, h = 800 cm The volume of a prism is 375 m 3. This means that the radius is 1. Question Find the volume of a rectangular prism that is 5 inches in height, 4 inches in width, and 3 inches in length. Some of the worksheets below are Volume and Surface Area Worksheets – Surface Area : Objectives – To find the surface of a cube, to find the surface of a cuboid, …, Volume and Surface Area of Rectangular Prisms and Cylinders, Solid Geometry : Calculate the Volume of Prisms and Cylinders, Calculate the Volume of Pyramids, Cones, and Spheres, Calculate the Surface Area of Prisms and. Rectangular Prisms (or Solids) The above figure is called a rectangular prism. Learn methods of finding volume of cubes and rectangular prisms using the resources on this page. Depending on the particular body, there is a different formula and different required information you need to calculate its volume. 1h) on volume of prisms (could be taught over one or two lessons depending on previous knowledge). prism, any pair of opposite faces can be bases. But the question most curious elementary students will ask is "why?" If this is the point your child is at, then you are in the right place to get exactly the information you need to explain it. Calculate the volume of a rectangular prism with a length of 5. Surface Area Of Prisms And Pyramids. In this way, the calculation was done. Figure out the volume of a rectangular prism by following these few short and simple steps. Answers should be expressed in the appropriate units. Nikita used the wrong formula for the volume of the prism. The volume of any prism is the product of the area of the base and the distance between the two bases. Calculations at an oblique prism. The following is the calculation formula for the volume of a rectangular prism:. If so then the volume is the box is the area of the triangle times the depth, d. The geometric figure has two identical ends and all flat sides. Using this calculator, we will understand methods of how to find the surface area and volume of the hemisphere. Some of the examples in real life are: a mail box, a shoe box, a juice box etc. Volume of a Square or Rectangular Tank or Clarifier. For example, find the cubic footage volume of a backfill area that is 8 feet wide, 6 feet deep and 50 feet long. Side Length. Volume of Trapezoid Prism. Find the volume of the following right prism. Unknown Error. } Ask Algebra House. The formula for volume of a cube is given by, Volume = side x side x side. To calculate the volume of a rectangle:. Enter your answer by using dimensions of volume and round to two significant figures. Start studying Volume of Rectangular & Triangular Prisms. If the area of the base of the triangle is 30 cm 2 then find the height of the prism. Volume of a Rectangular Prism formula. As you can see from the screenshot above, Nickzom Calculator - The Calculator Encyclopedia solves for the volume of a rectangular prism and presents the formula, workings and steps too. Related Topics: More Geometry Lessons \| Volume Games In these lessons, we give. The volume ratio for the two solids is the side length ratio raised to the third power. a is the area of one triangular end face. 1 cm, and a height of 6. There are various ways to find the area of the. Although the bases […]. Processing. The trapezium has height 4cm and its parallel sides are 5cm & 7cm. Volume Formula of a Right Triangle: Formula for finding the volume of a right triangle: Volume of a right. Here you can calculate the surface area of a pyramid with a four-sided base. This calculator will determine the volume of a quantity of substance from the measured mass and known density and display a conversion scale for variations in each parameter. Step 1: Solid of cubes with unit fraction edge lengths. 7) The base of a prism is a parallelogram whose base and height are 19 inches and 7 inches respectively. A cross-section is the shape you get when cutting straight across an object. Calculating Volume Rectangular Prisms. The project requires a metric ruler. In this lesson you will find the volume of complex figures by deconstructing and adding the volumes of simpler prisms using the volume formula. A pentagonal prism is a type of prism that uses a pentagon for a base. 6 , a width of 2. This is summarized by the formula: LA 5 hP. Calculate the volume of each object using the base and height. Volume of a Triangular Prism Formula - A prism is a solid object that has two congruent faces on either end joined by parallelogram faces laterally. Because prisms have two congruent bases, it's easy to calculate their surface area: first, you find the area of one base and double that value; then, you add the prism's lateral area. dgorsman wrote:. Since we know all 3 triangle sides, we will us Heron's Formula. A right triangular prism has rectangular sides, otherwise it is oblique. You must know the width, length and height of the prism before you can apply this formula: A = 2(width × length) + 2(length × height) + 2(height × width) A = 2 ( w i d t h × l e n g t h) + 2 ( l e n g t h × h e i g h t) + 2 ( h e i. A right prism has lateral edges perpendicular to the base. As you can see from the screenshot above, Nickzom Calculator – The Calculator Encyclopedia solves for the volume of a rectangular prism and presents the formula, workings and steps too. We can calculate the volume of any prism simply by knowing the height of the prism and the area of one of its bases. Type in your answer and select submit for. Here we have 2 different rectangular prisms. Mathematical Handbook of Formulas and Tables. To determine the volume of a rectangular prism, just use the following. Volume is a 3D Measurement so we need to always write answers as cubed numbers, for example CM 3 or M 3. A shape’s volume is a measure of its total 3-dimensional space. 50x28x15=1400x15=21000 cubic feet The number of cubic feet in this room is 21000 cubic feet. 1 cm, and a height of 6. Email or Phone: Password. The volume of each prism is __1 2 r × __ 1 3 r × 2r, where r is the radius of the cylinder’s base. Let us learn here how to find the volume of a hexagonal prism (Right regular hexagonal prism) with simple steps. Calculate the volume of the space to be filled. Volume is measured in cubic units ( in 3 , ft 3 , cm 3 , m 3 , et cetera). This is what a triangular prism looks like, where it has a triangle on one, two faces, and they're kind of separated. If an input is given then it can easily show the result for the given number. cm and its base is a right triangle. It's just geometry. Prismatoid and Prismoid Calculator. Open up the rectangular prism and explode the faces to have a better view. The formula for volume of a cube is given by, Volume = side x side x side. Volume of a. Rectangle Volume Calculation. The lateral area of a prism is the area of its sides—namely, the area of everything but the prism's bases. Learning targets: students can calculate the the volume of a prism with a complicated base by decomposing the base into quadrilaterals or triangles. The length of the prism is the measurement between the end-points of the ridge. The Rectangular Prism Volume Calculator can instantly calculate the volume of a rectangular prism if you enter in the height, length, and width of the rectangular prism and then click the calculate button. But the question most curious elementary students will ask is "why?" If this is the point your child is at, then you are in the right place to get exactly the information you need to explain it. Prism Calculator Use this online calculator to calculate the properties of a prism, enter known measurements to calculate associated dimensions and volume of a prism. how would you write a java program to calculate the surface area and volume of a rectangular prism? how would you write a java program to calculate the surface area and volume of a rectangular prism? Answer Save. Here is a great tool to calculate the volume or capacity of your pool. On this page, you can calculate volume of a Right-Triangular Prism. where V - volume of the prism, A b - the area of the base of the prism (Online areas calculators), h - the height of the prism. Typically the various sides of this three dimensional shape are labelled a, b and c. Since we know all 3 triangle sides, we will us Heron's Formula. The cross-section of the prism is a trapezium. Volume of a rectangular prism = length x width x height V = lwh = (2x - 5)(2x + 5)(3x) {substituted given values into volume formula} = (4x² - 25)(3x) {multiplied the two binomials using the foil method} D. A rectangular prism is a solid 3 dimensional object that is comprised of 6 different sides that are all rectangles. A prism volume is a measurement of the space. asked by yeeee on March 6, 2020; Algebra 1. If you prefer you can think of these as the width, height and depth. Lesson Objective: The lesson is aligned to the Common Core State Standards for Mathematics – 6. Also the height of prism can be found by it's volume, by the above formula. Figure out the volume of a rectangular prism by following these few short and simple steps. In other words the volume of a prism is the product of the area of the base and the height. Different to a right prism, the sides are not perpendicular to the bases (angle of slope ≠ 90°). What is Rectangular Prism? We need calculators on a regular basis in order to simplfy the complex process of calculating. The volume formula is: For example, if you have a large pentagonal prism with a height of 30 cm (0. To find the volume, use this rectangular prism volume formula: Volume = Length x Width x Height Example: The volume of a rectangular prism with length of 3 inches, width of 4 inches, and height of 5 inches: Volume = 3 x 4 x 5 Calculated out this gives a volume of 60 Cubic Inches. But if it starts to tilt, then it is considered an oblique prism, and finding the volume can be tricky. Online geometry calculator to calculate octogonal prism area, volume and surface area. Apothem Length. volume of a right rectangular prism with length , widthw, and height h V wh Find the volume of each prism. Solve advanced problems in Physics, Mathematics and Engineering. Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume is the same as would be found by multiplying the edge lengths of the prism. Question Find the volume of a rectangular prism that is 5 inches in height, 4 inches in width, and 3 inches in length. What is the equation solved for h. Volume Equation and Calculation Menu. Here are the following steps: Step 1: First of all open the programme Revit and then click on the volume and area on the left hand side, if in case its not there then right click in the gray bar and then various attributes appear, choose from here. a more detailed explanation (examples and solutions) of each volume formula. For any standard prism, the volume is just the area of the base times the height. Use of Wedge Prism. All space figures have volume. Trapezoidal Prism Volume Calculator In geometry, a triangular prism is a three-sided prism; it is a polyhedron made of a triangular base, a translated copy, and 3 faces joining corresponding sides. The formula for volume of rectangular prism is given by, Volume = length x width x height. Attempt every question. These worksheets are printable pdf files. Please enter the data for length, width & side water depth. Heres is a calculator that calculates the volume of a cylinder. Calculate the volume of each object using the base and height. com to help you to score more marks in your exams. Volume of a cube = side times side times side. A square prism is simply a cuboid with square base. These worksheets are printable pdf files. When you learn calculus you will discover the surface area of a sphere to be the derivative with respect to r of the sphere's volume formula. Mass from volume & concentration. Volume of a right square prism. A prism is a transparent element with polished surface which can refract light. There are many different shapes out there, one of which is a rectangular prism. Also the height of prism can be found by it's volume, by the above formula. In this way, the calculation was done. The volume of a right triangle are measured in terms of cubis units. This means that the radius is 1. Attempt every question. Everyday Calculation Free calculators and unit converters for general and everyday use. The volume of right prisms and cylinders is simply calculated by multiplying the area of the base of the solid by the height of the solid. Therefore, to find parallelepiped's volume build on vectors, one need to calculate scalar triple product of the given vectors, and take the magnitude of the result found. Try this Drag the orange dot to resize the cylinder. In the triangular prism calculator you can easily find out the volume of that solid. Learn more. You must know the width, length and height of the prism before you can apply this formula: A = 2(width × length) + 2(length × height) + 2(height × width) A = 2 ( w i d t h × l e n g t h) + 2 ( l e n g t h × h e i g h t) + 2 ( h e i. Many times, this formula will use the height of the prism, or depth (d), rather than the length (l), though you may see either abbreviation. The volume of a three-dimensional ﬁ gure is a measure of the amount of space that it occupies. By adding the rectangle 1,2,3 we get the total Volume of rectangular prism. Volume Equation and Calculation Menu. Volume of a square pyramid given base side and height. 5% Moisture. Attempt every question. Select the object from the options then answer the questions about length and width or radius, base and height. Oblique Prism Calculator. Height of a right square prism. The formula for finding the volume of a cylinder is the same as the formula for finding the volume of a prism. Remember, we can use a formula to calculate the area of a pair of faces. The formula, in general, is the area of the base (the red triangle in the picture on the left) times the height,h. Some of the examples in real life are: a mail box, a shoe box, a juice box etc. Side Length. Find the volume of equilateral triangle base prism of side 450 cm and height 390 cm Solution for practice. The prism is named after the shape of its base, so a prism with a. Select your measurement units. 5 cubic feet. A prism is a transparent element with polished surface which can refract light. It is measured in cubic units. Volume of a Right triangular prism = Area of triangular face * height. By adding the rectangle 1,2,3 we get the total Volume of rectangular prism. " At this time, I will pass out the notes sheet, and give students 90 seconds to write down as much as they can recall about volume. Identifying nets. Calculate the volume of a prism by decomposing the base of the figure into simpler shapes. This will not change the volume but turn the rightmost column into (0, 0, 0, 1). Find the volume of the triangular prism, whose base is 7 cm, height is 9 cm and length is 12 cm. 22 USD The calculation can be used to calculate the volume of a pyramid with a four-sided base. This is a geometric solid having 6 faces all of which are rectangles and each pair of opposite faces are congruent. Calculate the volume of a pyramid Not too long ago: A couple split a bill according to income and found out that person A should pay 1 327. Perimeter of the base, use (, , and so on) Lateral Area, use. 2 Geometry – Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume is the same as would be found by multiplying the edge lengths of the prism. Question Find the volume of a rectangular prism that is 5 inches in height, 4 inches in width, and 3 inches in length. Remember to measure from the depth of the water rather than the height of your wall. Calculate BAF by formula: BAF = (43560)/[(1+4)(D/W) Volume: Volume is related to basal area and merchantable or total tree height. A lens is named as per the shape of its base. In other words, multiply together the length, height, and average of A and B. Right Prism Volume Calculator. [Measurement and Reference Frames Goal 2]. The volume is equal to the product of the area of the base and the height of the prism. First thing we gotta know is the perimeter of the triangle. 3 m) and sides of. If you input the necessary variables, our calculator will give you the exact volume and surface area. Find the diagonal length of the base. Volume of a equilateral triangular prism. To calculate the volume of a rectangular prism, use the formula: W x L x H So from your given measurement: 2. 1h) on volume of prisms (could be taught over one or two lessons depending on previous knowledge). 1 cm, and a height of 6. 6 Part B answer: 11. Attempt every question. This article contains a collection of calculators for calculating geometric figures volume. Examples: Input: l = 18, b = 12, h = 9 Output: Volume of triangular prism: 972 Input: l = 10, b = 8, h = 6 Output: Volume of triangular prism: 240. Different types of calculations related to a prism: The formula for calculating the area and volume of a prism is as follows-. volume of a cube with edge length s V s 3 2. The rectangular prism above has an volume of 48 cubic units. Find the volume of this rectangular prism. 246 = Inner diagonal length. The following is the calculation formula for the volume of a rectangular prism:. Given a Base edge and Height of the Hexagonal prism, the task is to find the Surface Area and the Volume of hexagonal Prism. Volume is how much of a liquid (or solid) the rectangular piece can hold. For example, in a fish tank, it's the number of gallons of water the tank holds. Formula for Volume of a Trapezoidal Prism. Below are six versions of our grade 6 math worksheet on finding the volume and surface areas of rectangular prisms. This lesson is most appropraite for 4th grade students. Calculate the various properties (like Area of Base, Perimeter of Base, Surface Area of Prism and Volume of Prism) of Triangular Prism, Rectangular Prism(Cuboid), Regular Square Prism(Square Cuboid), Regular Pentagonal Prism, Regular Hexagonal Prism, for given values. Surface Area of Triangular Prisms First, we know that we need to ﬁnd the area of the two triangular faces. With a hexagon base, the number of faces, edges and vertices (NF, NE, NV respectively) is given by the formulas:. Find the diagonal length of the base. The prism is named after the shape of its base, so a prism with a. Free Online Scientific Notation Calculator. We can now just plug this number in to the formulas to calculate the volume and surface area. The formula for finding the volume of a cylinder is the same as the formula for finding the volume of a prism. Accessibility Help. Here, are some major elements of a Prism that are necessary to remember. Area of triangle = 1/2bh =. Side Length. One method of calculating the TSA (Total Surface Area) is to “unfold” a 3D shape, into its flat “2D” net which the shape is made from. Trapezoidal Prism Volume Calculator In geometry, a triangular prism is a three-sided prism; it is a polyhedron made of a triangular base, a translated copy, and 3 faces joining corresponding sides. Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume is the same as would be found by multiplying the edge lengths of the prism. A tool for helping calculate the volume of common types of house and loft extensions Volume calculator. calculate the volume and surface area. Nikita used the wrong formula for the volume of the prism. A triangular prism has sides of 5, 6 and 7 with a height of 9. The volume of any prism can be found by multiplying the area of one of the bases by its height. For a roof with a height of 6 feet, a width of 20 feet and a ridge length of 32 feet, the calculation looks like. Here's how to calculate the volume of a variety of prisms. 2 Side Length, Volume, and Surface Area of Similar Solids. Multiplying the length, width and height of the given rectangular prism gives you the volume. There are many different shapes out there, one of which is a rectangular prism. Volume of rectangular prisms. Any of the following images can be printed for free. Avoid overflows or half-filled containers by using our math calculator first to match the amount of liquid to the container. 8 cm 3 _____ 2) The volume of a prism is 300 cu. Volume of a Rectangular Prism formula. Rectangular prisms. You are supposed to know why the short leg is 0. Volume of a right square prism. Volume is a unique attribute of solids that explains how much space an object takes up. Prism (Volume & Surface Area) Calculator getcalc. You can use the following simple formulas to help you calculate the volume of shapes like cuboids and prisms. Diameter of Cylinder. all along its length) (Also see Rectangular Prisms ). 15 yd 8 yd 6 yd V = Bh Write formula for volume. It has the same cross-section along a length, which makes it a prism. 2 VolUWie is q. Calculate the volume of each object using the base and height. A prism is a solid shape that has the same cross-section all the way through. More complicated pool shapes will require multiple calculations and adding them together. Prism (Volume & Surface Area) Calculator getcalc. Using the rectangular prism volume formula above, we get volume = 18in * 12in * 15in = 3240 in³. Irregular Prism Volume Calculator - A prism has the same cross section all along its length. Volume of a square pyramid given base side and height. Volume of a Rectangular Prism Printable Worksheets Understanding volume is tricky, which is why volume of rectangular prism worksheets are the perfect learning tool for fifth grade students. Use the formulas below to find the volume of many. Electrical Calculators Real Estate Calculators Accounting Calculators Business. Right rectangular prism calc - find A. Volume and surface area - KS4/GCSE geometry teaching resources. Volume of a rectangular prism = length x width x height V = lwh = (2x - 5)(2x + 5)(3x) {substituted given values into volume formula} = (4x² - 25)(3x) {multiplied the two binomials using the foil method} D. Hexagonal prism is a special case of the general prism, which may have any arbitrary polygonal base. To find out how to work out the area of a 2D shape click here. Volume, use (, , and so on). The length of the prism is 15cm. Like a cube all faces must meet along the edges and align to form a vertex from the intersection of three faces so that the prism has a completely enclosed volume. Use the calculator above to calculate height, radius or. Online calculator to calculate the volume of geometric solids including a capsule, cone, frustum, cube, cylinder, hemisphere, pyramid, rectangular prism, sphere and spherical cap. Read each question carefully before you begin answering it. A water tank is a cylinder with radius 40 cm and depth 150 cm It is filled at the rate of 0. Volume of frustum of a rectangular pyramid = h/3 * (lw + sqrt (lwl 1 w 1) + l 1 w 1) Download: Use this volume calculator offline with our all-in-one calculator app for Android and iOS. A rectangular prism is a solid, 3-dimensional object with three sets of parallel sides that are perpendicular to one another. But if the two ends are not parallel it is not a prism. Area of base Height of prism EXAMPLE 1 Finding the Volume of a Prism Find the volume of the prism. Volume of a rectangular prism: word problem. A = area, I = Height of the Prism. Calculate the area, volume of Hemisphere (Hemisphere Calculator). com's Prism Formulas & Calculator is an online basic geometry tool to calculate volume & surface area of the shape of rectangular & triangle prism, in both US customary & metric (SI) units. Volume = lwh, where l and w are the length and width of the prism and h is the height. Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to find rectangular prism volume. First, we need to calculate the area of the triangular base. We can find the volume of a triangular prism when we know the length, width and height of the triangular prism. The vertical height of the triangular prism is 12 cm. You can use the following simple formulas to help you calculate the volume of shapes like cuboids and prisms. Again, this is not the solids' volume, only the ratio of the volumes. TRIANGULAR PRISMS In a triangular prism, each cross‑section parallel to the triangular base is a triangle congruent to the base. Volume and Area of a Sphere Calculator. Take the guesswork out of calculating liquid volume, and use our Liquid Volume Calculator. It is 100 feet long, 30 feet wide, and 14 feet deep. Donʼt spend too long on one question. Surface Area & Volume of a Prism Surface Area of a Prism Suppose that we want to find the lateral area and total surface area of the following right triangular prism: The bases of this prism are right triangles, and the lateral faces are rectangles, as shown below. For example this tool can be used to calculate the amount of storage volume required for a given quantity of substance mass. Instructions Use black ink or ball-point pen. Centimeter cubes have gone wild with these irregular. The volume is equal to the product of the area of the base and the height of the prism. When calculating prism volume, this volume formula can be applied to both right and oblique prisms with bases of any shape, such as triangles, quadrilaterals, or other polygons. Volume is measured in cubic units ( in 3 , ft 3 , cm 3 , m 3 , et cetera). This should do for the end piece. About Volume of a Rectangular Prism Calculator. The calculator performs calculations in a right regular prism. Enter the figures dimensions and see calculation results in different measurement units both metric and non-metric. 2 litres per second. Height of a regular hexagonal prism. orgConcept 1. To calculate the volume of a rectangular prism, use the formula: W x L x H So from your given measurement: 2. To calculate the volume of a rectangular prism you need to know its height, width and length. Volume of a Rectangular Prism. Examples: Input: l = 18, b = 12, h = 9 Output: Volume of triangular prism: 972 Input: l = 10, b = 8, h = 6 Output: Volume of triangular prism: 240. Volume : I cubic unit The volume of prism or is V: (Area of un. A pack of playing cards is an example of a rectangular prism and so is a paperback book. 4bnl7ylcan2sm5 t1qin40435 6gc09g630w r3v9srmq1d7j g1oxht4lbnev k4pnbb70mzy9 nnpcu428u0h na4jm5t92tx u80nvunze6 e70k5jfv5ajf3 xlac27edegskv d2thrxqd3drw ltk6vhlb4z6 kukslp9jsy 6oua3qfw8cfm0 pt49gxgl21 5sp6u7mvn9o3x 55048p6qkfp6gp l6099scxshd pd9k8pu9o0 1tl2xq4jfj68umu 6gz9kc8f7v s1daidvu5f4ny q96511kmxitq znvy445ghi 5qo0ugjhck 4p1q4gbxk70y ho9x8ik23utu1 xtx0po525bj6a gy7yujwckg vtdxiun112yc 7qe8w6nb8s d2wamm0ll5 0wh12aqkynywjl
# Derivative of cos2x by First Principle, Chain Rule The derivative of cos2x is equal to -2sin2x. In this post, we will find the derivative of cos2x by the first principle, that is, by the limit definition of derivatives as well as by the chain rule of derivatives. Recall the first principle of derivatives. By this rule, we know that the derivative of a function f(x) is given by the following limit: $\dfrac{d}{dx}(f(x))$$=\lim\limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h} …(I) Table of Contents ## Derivative of cos2x by First Principle Question: What is the derivative of cos 2x? Answer: The derivative of cos2x is -2sin2x. Explanation: Step 1: We put f(x)=\cos 2x in the above formula (I). Step 2: Thus the derivative of cos2x by the first principle will be equal to \dfrac{d}{dx}(\cos 2x)$$=\lim\limits_{h\to 0} \dfrac{\cos 2(x+h)-\cos 2x}{h}$ Step 3: Now apply the formula $\cos a -\cos b$ $=-2\sin \dfrac{a+b}{2}\sin \dfrac{b-a}{2}$. By doing so we obtain that $\dfrac{d}{dx}(\cos 2x)$$=\lim\limits_{h\to 0} \dfrac{-2 \sin (2x+h) \sin h}{h}$ = $-2\lim\limits_{h \to 0} \sin(2x+h)$ $\cdot \lim\limits_{h \to 0} \dfrac{\sin h}{h}$ = $-2 \sin(2x+0)$ $\times 1$ as we know that limit of sinh/h is 1 when h tends to zero. = $-2\sin 2x$. Conclusion: Therefore, the derivative of cos2x is 2sin2x, obtained by the first principle of derivatives, that is, d/dx(cos2x) = −2sin2x. ## Question-Answer on Derivative of cos2x Question: What is the derivative of cos2x at x=0. Answer: From the above, we have obtained that the derivative of cos2x is -2sin2x. So the derivative of cos2x at x=0 is equal to $[\dfrac{d}{dx}(\cos 2x)]{x=0}$ $=[-2\sin 2x]{x=0}$ $=-2\sin 0$ $=-2 \times 0$ as the value of sin0 is 0. $=0$. Thus, the derivative of cos2x at x=0 is equal to 0. More Reading: Derivative of root(x) + 1/root(x) ## Derivative of cos2x by Chain Rule We now find the derivative of cos2x by the chain rule. Let us put $z=2x$. Thus, $\dfrac{dz}{dx}=2$. Then by the chain rule, the derivative of cos2x is given by $\dfrac{d}{dx}(\cos 2x)=\dfrac{d}{dz}(\cos z) \cdot \dfrac{dz}{dx}$ = $-\sin z \cdot 2$ =$-2\sin 2x$ as z=2x. So the derivative of cos2x by the chain rule is equal to -2sin2x. RELATED TOPICS: Derivative of e1/x ## FAQs Q1: What is the derivative of cos2x? Answer: The derivative of cos2x is equal to -2sin2x, that is, that is, d/dx(cos2x) = −2sin2x.
# Integral by Parts: Definite Integral How to solve the definite integral of a product by using the integral by parts: formula, 2 examples, and their solutions. ## Formula ### Formula The integral of a product, uv', can be solved by using this formula. ab uv' dx = [uv]ab - ∫ab u'v dx ### How to Choose u and v' The priority of choosing u and v' is the same as the case of solving an indefinite integral. For a product of two functions in an integral, one is u, and the other is v'. To choose u and v' properly, remember this order of uv': [u] logarithmic polynomial trigonometric, exponential [v']. Set u = (upper function) and v' = (lower function). u is the function whose integral is complex. v' is the function whose integral is simple. ## Example 1 ### Solution x sin x is the product of x and sin x. So solve this by using integral by parts. The order of uv' is [u] logarithmic polynomial (x) trigonometric (sin x) exponential [v']. So set u = x and v' = sin x. Write u = x. Differentiate both sides. Then u' = 1. Derivative of a Polynomial Write v' = sin x next to u' = 1. Integrate both sides. Then v = -cos x. Integral of sin x Write this above v' = sin x. u = x, v = -cos x u' = 1 Then the given integral is equal to, uv, x⋅(-cos x) from 0 to π/3 minus integral from 0 to π/3, u'v, 1⋅(-cos x) dx. x⋅(-cos x) = -x cos x -∫0π/3 1⋅(-cos x) dx = +∫0π/3 cos x dx Put 0 and π/3 into -x cos x. Then -(π/3)⋅(cos π/3) - (0⋅cos 0). Solve the integral. Definite Integral: How to Solve The integral of cos x is sin x. So -(π/3)⋅(cos π/3) - (0⋅cos 0) + [sin x]0π/3. cos π/3 = 1/2 So -(π/3)⋅(cos π/3) = -(π/3)⋅(1/2). Cosine Values of Commonly Used Angles Put 0 and π/3 into sin x. Then sin π/3 - sin 0. -(π/3)⋅(1/2) = -π/6 sin π/3 = √3/2 sin 0 = 0 Sine Values of Commonly Used Angles -π/6 + [√3/2 - 0] = √3/2 - π/6 So √3/2 - π/6 is the answer. ## Example 2 ### Solution (ln x)2 is the product of (ln x)2 and 1. So solve this by using integral by parts. The order of uv' is [u] logarithmic [(ln x)2] polynomial (1) trigonometric exponential [v']. So set u = (ln x)2 and v' = 1. Write u = (ln x)2. Differentiate both sides. Then u' = 2(ln x)⋅(1/x). Derivative of a Composite Function Derivative of ln x Write v' = 1 next to u' = 2(ln x)1⋅(1/x). Integrate both sides. Then v = x. Integral of a Polynomial Write this above v' = 1. u = (ln x)2, v = 2(ln x)⋅(1/x) u' = x Then the given integral is equal to, uv, (ln x)2⋅x from 1 to e minus integral from 1 to e, u'v, 2(ln x)⋅(1/x)⋅x dx. Put 1 and e into (ln x)2⋅x. Then (ln e)2⋅e - (ln 1)2⋅1. Take 2 out from the integral. And cancel (1/x) and x. Then -∫1e 2(ln x)⋅(1/x)⋅x dx = -2⋅∫1e ln x dx So (ln e)2⋅e - (ln 1)2⋅1 - 2⋅∫1e ln x dx. (ln e)2⋅e = 12⋅e -(ln 1)2⋅1 = -0⋅1 = 0. Solve the integral. The integral of ln x is x ln x - x. 12⋅e = e Put 1 and e into x ln x - x. Then -2[e ln e - e - (1 ln 1 - 1)]. e ln e = e⋅1 1 ln 1 = 1⋅0 e⋅1 - e = 0 (1⋅0 - 1) = (-1) 0 - (-1) = 0 + 1 = +1 -2[+1] = -2 So e - 2 is the answer.
# Percent Increase and Decrease ## Presentation on theme: "Percent Increase and Decrease"— Presentation transcript: Percent Increase and Decrease 2-10 Percent Increase and Decrease Holt Algebra 1 Warm Up Lesson Presentation Lesson Quiz Warmup: Which is a better deal? Get Your Calculator Warmup: Which is a better deal? A DVD costs \$20. Sales tax is 8% and you receive a 10% discount. Should you calculate the tax first then the discount or the other way around? Discount first: (\$20)(.10) = \$2 20 – 2 = \$18 Tax first: (\$20)(.08) = \$1.60 = \$21.60 Now Discount: (\$21.60)(.10) = \$2.16 21.60 – 2.16 = \$19.44 Now Tax: (\$18)(.08) = \$1.44 = \$19.44 Which is a better deal for you? Which is a better deal for the business? Warm Up 1. Find 30% of 40. 2. Find 28% of 60. 12 Solve for x. 5. 20 is what percent of 80? 6. 36 is what percent of 30? 12 16.8 0.44 38 25% 120% Objective Find percent increase and decrease. Vocabulary percent change percent increase percent decrease discount markup A percent change is an increase or decrease given as a percent of the original amount. Percent increase describes an amount that has grown and percent decrease describes an amount that has be reduced. Notes on next slide Definition: (write this down) Percent of Change Definition: (write this down) Percent of Change = Difference Original Which Means: % Increase or % Decrease Using this formula, your answer comes out as a decimal. Just multiply by 100 to change to a percent Example 1A: Finding Percent Increase and Decrease Find each percent change. Tell whether it is a percent increase or decrease. From 8 to 10 Simplify the numerator. Simplify the fraction. = 0.25 = 25% Write the answer as a percent. 8 to 10 is an increase, so a change from 8 to 10 is a 25% increase. Helpful Hint Before solving, decide what is a reasonable answer. For Example 1A, 8 to 16 would be a 100% increase. So 8 to 10 should be much less than 100%. Example 1B: Finding Percent Increase and Decrease Find the percent change. Tell whether it is a percent increase or decrease. From 75 to 30 Simplify the fraction. Simplify the numerator. = 0.6 = 60% Write the answer as a percent. 75 to 30 is a decrease, so a change from 75 to 30 is a 60% decrease. Example 2: Finding Percent Increase and Decrease A. Find the result when 12 is increased by 50%. 0.50(12) = 6 Find 50% of 12. This is the amount of increase. =18 It is a percent increase, so add 6 to the original amount. 12 increased by 50% is 18. B. Find the result when 55 is decreased by 60%. Find 60% of 55. This is the amount of decrease. 0.60(55) = 33 55 – 33 = 22 It is a percent decrease so subtract 33 from the the original amount. 55 decreased by 60% is 22. Common application of percent change are discounts and markups. A discount is an amount by which an original price is reduced. discount = % of original price final price = A markup is an amount by which a wholesale price is increased. final price = wholesale cost markup + = % of Example 3: Discounts The entrance fee at an amusement park is \$35. People over the age of 65 receive a 20% discount. What is the amount of the discount? How much do people over 65 pay? Method 1 A discount is a percent decrease. So find \$35 decreased by 20%. Find 20% of 35. This is the amount of the discount. 0.20(35) = 7 35 – 7 = 28 Subtract 7 from 35. This is the entrance fee for people over the age of 65. Example 3A: Discounts Method 2 Subtract the percent discount from 100%. 100% – 20% = 80% People over the age of 65 pay 80% of the regular price, \$35. 0.80(35) = 28 Find 80% of 35. This is the entrance fee for people over the age of 65. 35 – 28 = 7 Subtract 28 from 35. This is the amount of the discount. By either method, the discount is \$7. People over the age of 65 pay \$28.00. Helpful Hint Before solving, decide what is a reasonable answer. For example 3A, a 10% discount is \$3.50 off. So a 20% discount would be more than \$2 off. Example 3B: Discounts A student paid \$31.20 for art supplies that normally cost \$ Find the percent discount . Think: is what percent of 52.00? Let x represent the percent. \$52.00 – \$31.20 = \$20.80 20.80 = x(52.00) Since x is multiplied by 52.00, divide both sides by to undo the multiplication. 0.40 = x 40% = x Write the answer as a percent. The discount is 40% Example 4: Markups The wholesale cost of a DVD is \$7. The markup is 85%. What is the amount of the markup? What is the selling price? Method 2 Method 1 A markup is a percent increase. So find \$7 increased by 85%. Add percent markup to 100% Find 85% of 7. This is the amount of the markup. 100% + 85% = 185% The selling price is 185% of the wholesale price, 7. 0.85(7) = 5.95 1.85(7) = 12.95 = 12.95 Find 185% of 7. This is the selling price. Add to 7. This is the selling price. 12.95  7 = 5.95 Subtract from This is the amount of the markup. By either method, the amount of the markup is \$5.95. The selling price is \$12.95. Lesson Quiz: Part 1 Find each percent change. Tell whether it is a percent increase or decrease. 1. from 20 to 28. 2. from 80 to 62. 3. from 500 to 100. 4. find the result when 120 is increased by 40%. 5. find the result when 70 is decreased by 20%. 40% increase 22.5% decrease 80% decrease 168 56 Lesson Quiz: Part 2 80% decrease 1. from 500 to 100. 168 Find the percent change. Tell whether it is a percent increase or decrease. 80% decrease 1. from 500 to 100. 2. find the result when 120 is increased by 40%. 168 Find each percent change. Tell whether it is a percent increase or decrease. 3. A movie ticket costs \$9. On Mondays, tickets are 20% off. What is the amount of discount? How much would a ticket cost on a Monday? 4. A bike helmet cost \$24. The wholesale cost was \$15. What was the percent of markup? \$1.80; \$7.20 60%
Giáo trình # Precalculus Mathematics and Statistics ## Non-right Triangles: Law of Sines Tác giả: OpenStaxCollege Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? We see in [link] that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving non-right triangles. # Using the Law of Sines to Solve Oblique Triangles In any triangle, we can draw an altitude, a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles. Any triangle that is not a right triangle is an oblique triangle. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations: 1. ASA (angle-side-angle) We know the measurements of two angles and the included side. See [link]. 2. AAS (angle-angle-side) We know the measurements of two angles and a side that is not between the known angles. See [link]. 3. SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides. See [link]. Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let’s see how this statement is derived by considering the triangle shown in [link]. Using the right triangle relationships, we know that$\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{h}{b}\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta =\frac{h}{a}.\text{\hspace{0.17em}}\text{\hspace{0.17em}}$Solving both equations for$\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$gives two different expressions for$\text{\hspace{0.17em}}h.$ We then set the expressions equal to each other. Similarly, we can compare the other ratios. Collectively, these relationships are called the Law of Sines. Note the standard way of labeling triangles: angle$\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$(alpha) is opposite side$\text{\hspace{0.17em}}a;\text{\hspace{0.17em}}$angle$\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$(beta) is opposite side$\text{\hspace{0.17em}}b;\text{\hspace{0.17em}}$and angle$\text{\hspace{0.17em}}\gamma \text{\hspace{0.17em}}$(gamma) is opposite side$\text{\hspace{0.17em}}c.\text{\hspace{0.17em}}$See [link]. While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified. Law of Sines Given a triangle with angles and opposite sides labeled as in [link], the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines is based on proportions and is presented symbolically two ways. $sin α a = sin β b = sin γ c$ $a sin α = b sin β = c sin γ$ To solve an oblique triangle, use any pair of applicable ratios. Solving for Two Unknown Sides and Angle of an AAS Triangle Solve the triangle shown in [link] to the nearest tenth. The three angles must add up to 180 degrees. From this, we can determine that To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle $\alpha =50°$and its corresponding side$a=10.\text{\hspace{0.17em}}$We can use the following proportion from the Law of Sines to find the length of$\text{\hspace{0.17em}}c.\text{\hspace{0.17em}}$ Similarly, to solve for$\text{\hspace{0.17em}}b,\text{\hspace{0.17em}}$we set up another proportion. Therefore, the complete set of angles and sides is Solve the triangle shown in [link] to the nearest tenth. $\begin{array}{l}\alpha ={98}^{\circ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}a=34.6\\ \beta ={39}^{\circ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=22\\ \gamma ={43}^{\circ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=23.8\end{array}$ # Using The Law of Sines to Solve SSA Triangles We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution. Possible Outcomes for SSA Triangles Oblique triangles in the category SSA may have four different outcomes. [link] illustrates the solutions with the known sides$\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$and known angle$\text{\hspace{0.17em}}\alpha .$ Solving an Oblique SSA Triangle Solve the triangle in [link] for the missing side and find the missing angle measures to the nearest tenth. Use the Law of Sines to find angle$\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$and angle$\text{\hspace{0.17em}}\gamma ,\text{\hspace{0.17em}}$and then side$\text{\hspace{0.17em}}c.\text{\hspace{0.17em}}$Solving for$\text{\hspace{0.17em}}\beta ,\text{\hspace{0.17em}}$we have the proportion However, in the diagram, angle$\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of$\text{\hspace{0.17em}}\beta ?\text{\hspace{0.17em}}$Let’s investigate further. Dropping a perpendicular from$\text{\hspace{0.17em}}\gamma \text{\hspace{0.17em}}$and viewing the triangle from a right angle perspective, we have [link]. It appears that there may be a second triangle that will fit the given criteria. The angle supplementary to$\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$is approximately equal to 49.9°, which means that$\text{\hspace{0.17em}}\beta =180°-49.9°=130.1°.\text{\hspace{0.17em}}$(Remember that the sine function is positive in both the first and second quadrants.) Solving for$\text{\hspace{0.17em}}\gamma ,$ we have We can then use these measurements to solve the other triangle. Since$\text{\hspace{0.17em}}{\gamma }^{\prime }\text{\hspace{0.17em}}$is supplementary to$\text{\hspace{0.17em}}\gamma ,$ we have Now we need to find$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}{c}^{\prime }.$ We have Finally, To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in [link]. However, we were looking for the values for the triangle with an obtuse angle$\text{\hspace{0.17em}}\beta .\text{\hspace{0.17em}}$We can see them in the first triangle (a) in [link]. Given$\text{\hspace{0.17em}}\alpha =80°,a=120,\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}b=121,\text{\hspace{0.17em}}$find the missing side and angles. If there is more than one possible solution, show both. Solution 1 Solution 2 Solving for the Unknown Sides and Angles of a SSA Triangle In the triangle shown in [link], solve for the unknown side and angles. Round your answers to the nearest tenth. In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle$\text{\hspace{0.17em}}\gamma =85°,\text{\hspace{0.17em}}$and its corresponding side$\text{\hspace{0.17em}}c=12,\text{\hspace{0.17em}}$and we know side$\text{\hspace{0.17em}}b=9.\text{\hspace{0.17em}}$We will use this proportion to solve for$\text{\hspace{0.17em}}\beta .$ To find$\text{\hspace{0.17em}}\beta ,\text{\hspace{0.17em}}$apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for$\text{\hspace{0.17em}}\beta .\text{\hspace{0.17em}}$It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions. In this case, if we subtract$\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$from 180°, we find that there may be a second possible solution. Thus,$\text{\hspace{0.17em}}\beta =180°-48.3°\approx 131.7°.\text{\hspace{0.17em}}$To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives which is impossible, and so$\text{\hspace{0.17em}}\beta \approx 48.3°.$ To find the remaining missing values, we calculate$\text{\hspace{0.17em}}\alpha =180°-85°-48.3°\approx 46.7°.\text{\hspace{0.17em}}$Now, only side$\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$is needed. Use the Law of Sines to solve for$\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$by one of the proportions. The complete set of solutions for the given triangle is Given$\text{\hspace{0.17em}}\alpha =80°,a=100,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=10,\text{\hspace{0.17em}}$find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth. $\beta \approx 5.7°,\gamma \approx 94.3°,c\approx 101.3$ Finding the Triangles That Meet the Given Criteria Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10. Using the given information, we can solve for the angle opposite the side of length 10. See [link]. We can stop here without finding the value of$\text{\hspace{0.17em}}\alpha .\text{\hspace{0.17em}}$Because the range of the sine function is$\text{\hspace{0.17em}}\left[-1,1\right],\text{\hspace{0.17em}}$it is impossible for the sine value to be 1.915. In fact, inputting$\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(1.915\right)\text{\hspace{0.17em}}$in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions. Determine the number of triangles possible given$\text{\hspace{0.17em}}a=31,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=26,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\beta =48°.\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ two # Finding the Area of an Oblique Triangle Using the Sine Function Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given as$\text{\hspace{0.17em}}\text{Area}=\frac{1}{2}bh,\text{\hspace{0.17em}}$where$\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$is base and$\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$is height. For oblique triangles, we must find$\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$before we can use the area formula. Observing the two triangles in [link], one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property$\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{\text{opposite}}{\text{hypotenuse}}\text{\hspace{0.17em}}$to write an equation for area in oblique triangles. In the acute triangle, we have$\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{h}{c}\text{\hspace{0.17em}}$or$c\mathrm{sin}\text{\hspace{0.17em}}\alpha =h.\text{\hspace{0.17em}}$However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base$\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$to form a right triangle. The angle used in calculation is$\text{\hspace{0.17em}}{\alpha }^{\prime },\text{\hspace{0.17em}}$or$\text{\hspace{0.17em}}180-\alpha .$ Thus, Similarly, Area of an Oblique Triangle The formula for the area of an oblique triangle is given by $Area= 1 2 bcsin α = 1 2 acsin β = 1 2 absin γ$ This is equivalent to one-half of the product of two sides and the sine of their included angle. Finding the Area of an Oblique Triangle Find the area of a triangle with sides$\text{\hspace{0.17em}}a=90,b=52,\text{\hspace{0.17em}}$and angle$\text{\hspace{0.17em}}\gamma =102°.\text{\hspace{0.17em}}$Round the area to the nearest integer. Using the formula, we have Find the area of the triangle given$\text{\hspace{0.17em}}\beta =42°,\text{\hspace{0.17em}}\text{\hspace{0.17em}}a=7.2\text{\hspace{0.17em}}\text{ft},\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=3.4\text{\hspace{0.17em}}\text{ft}.\text{\hspace{0.17em}}$Round the area to the nearest tenth. about$\text{\hspace{0.17em}}8.2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{square}\text{\hspace{0.17em}}\text{feet}$ # Solving Applied Problems Using the Law of Sines The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion. Finding an Altitude Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in [link]. Round the altitude to the nearest tenth of a mile. To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side$\text{\hspace{0.17em}}a,$ and then use right triangle relationships to find the height of the aircraft,$\text{\hspace{0.17em}}h.$ Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship. The distance from one station to the aircraft is about 14.98 miles. Now that we know$\text{\hspace{0.17em}}a,\text{\hspace{0.17em}}$we can use right triangle relationships to solve for$\text{\hspace{0.17em}}h.$ The aircraft is at an altitude of approximately 3.9 miles. Access these online resources for additional instruction and practice with trigonometric applications. # Key Equations Law of Sines $\begin{array}{l}\frac{\mathrm{sin}\text{\hspace{0.17em}}\alpha }{a}=\frac{\mathrm{sin}\text{\hspace{0.17em}}\beta }{b}=\frac{\mathrm{sin}\text{\hspace{0.17em}}\gamma }{c}\text{\hspace{0.17em}}\hfill \\ \frac{a}{\mathrm{sin}\text{\hspace{0.17em}}\alpha }=\frac{b}{\mathrm{sin}\text{\hspace{0.17em}}\beta }=\frac{c}{\mathrm{sin}\text{\hspace{0.17em}}\gamma }\hfill \end{array}$ Area for oblique triangles # Key Concepts • The Law of Sines can be used to solve oblique triangles, which are non-right triangles. • According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side. • There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution. See [link]. • The ambiguous case arises when an oblique triangle can have different outcomes. • There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution. See [link] and [link]. • The Law of Sines can be used to solve triangles with given criteria. See [link]. • The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. See [link]. • There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation. See [link]. # Section Exercises ## Verbal Describe the altitude of a triangle. The altitude extends from any vertex to the opposite side or to the line containing the opposite side at a 90° angle. Compare right triangles and oblique triangles. When can you use the Law of Sines to find a missing angle? When the known values are the side opposite the missing angle and another side and its opposite angle. In the Law of Sines, what is the relationship between the angle in the numerator and the side in the denominator? What type of triangle results in an ambiguous case? A triangle with two given sides and a non-included angle. ## Algebraic For the following exercises, assume$\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$is opposite side$\text{\hspace{0.17em}}a,\beta \text{\hspace{0.17em}}$is opposite side$\text{\hspace{0.17em}}b,\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}\gamma \text{\hspace{0.17em}}$is opposite side$\text{\hspace{0.17em}}c.\text{\hspace{0.17em}}$Solve each triangle, if possible. Round each answer to the nearest tenth. $\alpha =43°,\gamma =69°,a=20$ $\alpha =35°,\gamma =73°,c=20$ $\alpha =60°,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\beta =60°,\text{\hspace{0.17em}}\gamma =60°$ $a=4,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\alpha =\text{\hspace{0.17em}}60°,\text{\hspace{0.17em}}\beta =100°$ $b=10,\text{\hspace{0.17em}}\beta =95°,\gamma =\text{\hspace{0.17em}}30°$ For the following exercises, use the Law of Sines to solve for the missing side for each oblique triangle. Round each answer to the nearest hundredth. Assume that angle$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$is opposite side$\text{\hspace{0.17em}}a,\text{\hspace{0.17em}}$angle$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$is opposite side$\text{\hspace{0.17em}}b,\text{\hspace{0.17em}}$and angle$\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$is opposite side$\text{\hspace{0.17em}}c.$ Find side$\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$when$\text{\hspace{0.17em}}A=37°,\text{\hspace{0.17em}}\text{\hspace{0.17em}}B=49°,\text{\hspace{0.17em}}c=5.$ $b\approx 3.78$ Find side$\text{\hspace{0.17em}}a$ when$\text{\hspace{0.17em}}A=132°,C=23°,b=10.$ Find side$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$when$\text{\hspace{0.17em}}B=37°,C=21,\text{\hspace{0.17em}}b=23.$ $c\approx 13.70$ For the following exercises, assume$\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$is opposite side$\text{\hspace{0.17em}}a,\beta \text{\hspace{0.17em}}$is opposite side$\text{\hspace{0.17em}}b,\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}\gamma \text{\hspace{0.17em}}$is opposite side$\text{\hspace{0.17em}}c.\text{\hspace{0.17em}}$Determine whether there is no triangle, one triangle, or two triangles. Then solve each triangle, if possible. Round each answer to the nearest tenth. $\alpha =119°,a=14,b=26$ $\gamma =113°,b=10,c=32$ one triangle,$\text{\hspace{0.17em}}\alpha \approx 50.3°,\beta \approx 16.7°,a\approx 26.7$ $b=3.5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=5.3,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\gamma =\text{\hspace{0.17em}}80°$ $a=12,\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=17,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\alpha =\text{\hspace{0.17em}}35°$ two triangles,or $a=20.5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=35.0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\beta =25°$ $a=7,\text{\hspace{0.17em}}c=9,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\alpha =\text{\hspace{0.17em}}43°$ two triangles,or $a=7,b=3,\beta =24°$ $b=13,c=5,\gamma =\text{\hspace{0.17em}}10°$ two triangles,$\text{\hspace{0.17em}}\alpha \approx 143.2°,\beta \approx 26.8°,a\approx 17.3\text{\hspace{0.17em}}$or$\text{\hspace{0.17em}}{\alpha }^{\prime }\approx 16.8°,{\beta }^{\prime }\approx 153.2°,{a}^{\prime }\approx 8.3$ $a=2.3,c=1.8,\gamma =28°$ $\beta =119°,b=8.2,a=11.3$ no triangle possible For the following exercises, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. Round each answer to the nearest tenth. Find angle$A$when$\text{\hspace{0.17em}}a=24,b=5,B=22°.$ Find angle$A$when$\text{\hspace{0.17em}}a=13,b=6,B=20°.$ $A\approx 47.8°\text{\hspace{0.17em}}$or$\text{\hspace{0.17em}}{A}^{\prime }\approx 132.2°$ Find angle$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$when$\text{\hspace{0.17em}}A=12°,a=2,b=9.$ For the following exercises, find the area of the triangle with the given measurements. Round each answer to the nearest tenth. $a=5,c=6,\beta =\text{\hspace{0.17em}}35°$ $8.6$ $b=11,c=8,\alpha =28°$ $a=32,b=24,\gamma =75°$ $370.9$ $a=7.2,b=4.5,\gamma =43°$ ## Graphical For the following exercises, find the length of side$\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$Round to the nearest tenth. $12.3$ For the following exercises, find the measure of angle$\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$if possible. Round to the nearest tenth. $29.7°$ Notice that$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$is an obtuse angle. $110.6°$ For the following exercises, solve the triangle. Round each answer to the nearest tenth. For the following exercises, find the area of each triangle. Round each answer to the nearest tenth. $57.1$ ## Extensions Find the radius of the circle in [link]. Round to the nearest tenth. Find the diameter of the circle in [link]. Round to the nearest tenth. $10.1$ Find$\text{\hspace{0.17em}}m\angle ADC\text{\hspace{0.17em}}$in [link]. Round to the nearest tenth. Find$\text{\hspace{0.17em}}AD\text{\hspace{0.17em}}$in [link]. Round to the nearest tenth. Solve both triangles in [link]. Round each answer to the nearest tenth. Find$\text{\hspace{0.17em}}AB\text{\hspace{0.17em}}$in the parallelogram shown in [link]. Solve the triangle in [link]. (Hint: Draw a perpendicular from$\text{\hspace{0.17em}}H\text{\hspace{0.17em}}$to$\text{\hspace{0.17em}}JK\right).\text{\hspace{0.17em}}$Round each answer to the nearest tenth. Solve the triangle in [link]. (Hint: Draw a perpendicular from$\text{\hspace{0.17em}}N\text{\hspace{0.17em}}$to$\text{\hspace{0.17em}}LM\right).\text{\hspace{0.17em}}$Round each answer to the nearest tenth. In [link],$\text{\hspace{0.17em}}ABCD\text{\hspace{0.17em}}$is not a parallelogram.$\text{\hspace{0.17em}}\angle m\text{\hspace{0.17em}}$is obtuse. Solve both triangles. Round each answer to the nearest tenth. ## Real-World Applications A pole leans away from the sun at an angle of$\text{\hspace{0.17em}}7°\text{\hspace{0.17em}}$to the vertical, as shown in [link]. When the elevation of the sun is$\text{\hspace{0.17em}}55°,\text{\hspace{0.17em}}$the pole casts a shadow 42 feet long on the level ground. How long is the pole? Round the answer to the nearest tenth. 51.4 feet To determine how far a boat is from shore, two radar stations 500 feet apart find the angles out to the boat, as shown in [link]. Determine the distance of the boat from station$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$and the distance of the boat from shore. Round your answers to the nearest whole foot. [link] shows a satellite orbiting Earth. The satellite passes directly over two tracking stations$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}B,\text{\hspace{0.17em}}$which are 69 miles apart. When the satellite is on one side of the two stations, the angles of elevation at$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$are measured to be$\text{\hspace{0.17em}}86.2°\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}83.9°,\text{\hspace{0.17em}}$respectively. How far is the satellite from station$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$and how high is the satellite above the ground? Round answers to the nearest whole mile. The distance from the satellite to station$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$is approximately 1716 miles. The satellite is approximately 1706 miles above the ground. A communications tower is located at the top of a steep hill, as shown in [link]. The angle of inclination of the hill is$\text{\hspace{0.17em}}67°.\text{\hspace{0.17em}}$A guy wire is to be attached to the top of the tower and to the ground, 165 meters downhill from the base of the tower. The angle formed by the guy wire and the hill is$\text{\hspace{0.17em}}16°.\text{\hspace{0.17em}}$Find the length of the cable required for the guy wire to the nearest whole meter. The roof of a house is at a$\text{\hspace{0.17em}}20°\text{\hspace{0.17em}}$angle. An 8-foot solar panel is to be mounted on the roof and should be angled$\text{\hspace{0.17em}}38°\text{\hspace{0.17em}}$relative to the horizontal for optimal results. (See [link]). How long does the vertical support holding up the back of the panel need to be? Round to the nearest tenth. 2.6 ft Similar to an angle of elevation, an angle of depression is the acute angle formed by a horizontal line and an observer’s line of sight to an object below the horizontal. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 6.6 km apart, to be$\text{\hspace{0.17em}}37°$and$\text{\hspace{0.17em}}44°,$as shown in [link]. Find the distance of the plane from point$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$to the nearest tenth of a kilometer. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 4.3 km apart, to be 32° and 56°, as shown in [link]. Find the distance of the plane from point$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$to the nearest tenth of a kilometer. 5.6 km In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 39°. They then move 300 feet closer to the building and find the angle of elevation to be 50°. Assuming that the street is level, estimate the height of the building to the nearest foot. In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 35°. They then move 250 feet closer to the building and find the angle of elevation to be 53°. Assuming that the street is level, estimate the height of the building to the nearest foot. 371 ft Points$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$are on opposite sides of a lake. Point$\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$is 97 meters from$\text{\hspace{0.17em}}A.\text{\hspace{0.17em}}$The measure of angle$\text{\hspace{0.17em}}BAC\text{\hspace{0.17em}}$is determined to be 101°, and the measure of angle$\text{\hspace{0.17em}}ACB\text{\hspace{0.17em}}$is determined to be 53°. What is the distance from$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$to$\text{\hspace{0.17em}}B,\text{\hspace{0.17em}}$rounded to the nearest whole meter? A man and a woman standing$\text{\hspace{0.17em}}3\frac{1}{2}\text{\hspace{0.17em}}$miles apart spot a hot air balloon at the same time. If the angle of elevation from the man to the balloon is 27°, and the angle of elevation from the woman to the balloon is 41°, find the altitude of the balloon to the nearest foot. 5936 ft Two search teams spot a stranded climber on a mountain. The first search team is 0.5 miles from the second search team, and both teams are at an altitude of 1 mile. The angle of elevation from the first search team to the stranded climber is 15°. The angle of elevation from the second search team to the climber is 22°. What is the altitude of the climber? Round to the nearest tenth of a mile. A street light is mounted on a pole. A 6-foot-tall man is standing on the street a short distance from the pole, casting a shadow. The angle of elevation from the tip of the man’s shadow to the top of his head of 28°. A 6-foot-tall woman is standing on the same street on the opposite side of the pole from the man. The angle of elevation from the tip of her shadow to the top of her head is 28°. If the man and woman are 20 feet apart, how far is the street light from the tip of the shadow of each person? Round the distance to the nearest tenth of a foot. 24.1 ft Three cities,$\text{\hspace{0.17em}}A,B,$and$\text{\hspace{0.17em}}C,$are located so that city$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$is due east of city$\text{\hspace{0.17em}}B.\text{\hspace{0.17em}}$If city$\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$is located 35° west of north from city$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$and is 100 miles from city$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$and 70 miles from city$\text{\hspace{0.17em}}B,$how far is city$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$from city$\text{\hspace{0.17em}}B?\text{\hspace{0.17em}}$Round the distance to the nearest tenth of a mile. Two streets meet at an 80° angle. At the corner, a park is being built in the shape of a triangle. Find the area of the park if, along one road, the park measures 180 feet, and along the other road, the park measures 215 feet. 19,056 ft2 Brian’s house is on a corner lot. Find the area of the front yard if the edges measure 40 and 56 feet, as shown in [link]. The Bermuda triangle is a region of the Atlantic Ocean that connects Bermuda, Florida, and Puerto Rico. Find the area of the Bermuda triangle if the distance from Florida to Bermuda is 1030 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the angle created by the two distances is 62°. 445,624 square miles A yield sign measures 30 inches on all three sides. What is the area of the sign? Naomi bought a modern dining table whose top is in the shape of a triangle. Find the area of the table top if two of the sides measure 4 feet and 4.5 feet, and the smaller angles measure 32° and 42°, as shown in [link]. 8.65 ft2 Tải về Đánh giá: 0 dựa trên 0 đánh giá Nội dung cùng tác giả Nội dung tương tự
# NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.2 NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.2 Ex 3.2 Class 7 Maths Question 1. The scores in mathematics test (out of 25) of 15 students is as follows: 19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20 Find the mode and median of this data. Are they same? Solution: Given data: 19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20 Let us arrange the given data in increasing order Since 20 occurs 4 times (highest) ∴ Mode = 20 n = 15 (odd) ∴ Median = $$\frac{n+1}{2} \text { th term }=\frac{15+1}{2}$$ = 8th term = 20 Thus, median = 20 and mode = 20 ∴ Mode and median are same. Ex 3.2 Class 7 Maths Question 2. The runs scored in a cricket match by 11 players is as follows: 6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15 Find the mean, mode and median of this data. Are the three same? Solution: Given data: 6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15 Arranging the given data in increasing order, we get Here, 15 occurs 3 times (highest) ∴ Mode = 15 n = 11 (odd) ∴ Median = $$\left(\frac{11+1}{2}\right)^{t h}$$ term = 6th term = 15 Thus mean = 39, mode = 15 and median = 15 No, they are not same. Ex 3.2 Class 7 Maths Question 3. The weights (in kg) of 15 students of a class are: 38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47 (i) Find the mode and median of this data. (ii) Is there more than one mode? Solution: Given data: 38, 42, 35, 37,45, 50, 32,43, 43,40, 36, 38, 43, 38, 47 Arranging in increasing order, we get (i) Here, 38 and 42 occur 3 times (highest) Thus mode = 38 and 43 n = 15(odd) Median = $$\left(\frac{n+1}{2}\right)^{\mathrm{th}} \text { term }=\left(\frac{15+1}{2}\right)^{\mathrm{th}}$$ term = 8th term = 40 Thus mode 38 and 43 and median = 40 (ii) Yes, the given data has two modes i.e. 38 and 43. Ex 3.2 Class 7 Maths Question 4. Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14 Solution: Arranging the given data in increasing order, we get Here, 14 occur 3 times (highest) Thus, mode = 14 n = 9(odd) ∴ Median = $$\left(\frac{n+1}{2}\right)^{\text { th }} \text { term }=\left(\frac{9+1}{2}\right)^{\text { th }}$$ term = 5th term = 14 Hence, mode = 14 and median = 14. Ex 3.2 Class 7 Maths Question 5. Tell whether the statement is true or false. (i) The mode is always one of the number in a data. (ii) The mean is one of the numbers in a data. (iii) The median is always one of the numbers in a data. (iv) The data 6, 4, 3, 8, 9, 12, 13, 9 has mean 9. Solution: (i) True (ii) False (iii) True (iv) False +
$$\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}$$ # Slope Fields In this section, we'll see the first method we have of analyzing differential equations that we don't know how to solve. This method will work for any first-order diff eq. We'll start by learning how to draw a picture that represents all the possible solutions to a first order equation, but we'll quickly see that while it's not that difficult to do, it can be very tedious, so we generally rely on computers to draw these graphs for us. The skill we really need, therefore, is the ability to read these graphs and make sense of them. ## Drawing Slope Fields We'll illustrate this with a simple example: $y'=t+y.$ Clearly, $$t$$ is the independent variable, and $$y$$ is a function of $$t$$. We'll learn in a few sections how to solve this kind of equation, but for now we can't get an explicit solution. However, if we knew the value of $$y$$ for some value of $$t$$, we would also know the value of $$y'$$ at that point. We can therefore find the value of $$y'$$ for every possible combination of $$t$$ and $$y$$. Each combination of $$t$$ and $$y$$ represents a point that we can plot, and $$y'$$ at that point is the slope of the solution that would pass through that point. Therefore we can build a table like the one below: $$t$$$$y$$$$y'=t+y$$ 0$$\hspace{0.5in}$$0$$\hspace{0.5in}$$0 011 101 112 $$\vdots$$$$\vdots$$$$\vdots$$ To draw the slope field, we sketch a short segment at each point with the appropriate slope. The completed graph looks like the following: ### What does a slope field mean? The most basic way to read a slope field is to think of it as a wind map. If you drop a leaf onto this map, where will it go? This of course depends on where you drop it. The slope field represents all the solutions to the differential equation (the family of solutions we saw at the end of the last section). The specific solution depends on the initial condition, which is the spot where the metaphorical leaf is dropped. For instance, if the initial condition is that $$y(0)=1$$, meaning that $$y=1$$ when $$x=0$$, the solution looks like this curve: If, on the other hand, the initial condition is $$y(0)=-2$$, the solution looks like this: ## Analyzing Slope Fields Again, it gets really tedious really fast to draw slope fields by hand, so we use computers to draw them instead. What we need to be able to do is read slope fields, so we'll practice making some simple observations. To do this, I've embedded a Wolfram Demonstration, but it may not work with your browser (last time I checked, it worked with Firefox). If you're using Chrome, you should still be able to download the Wolfram CDF player and download the demonstration. Just in case, I'll post screenshots of the relevant parts. Slope Fields from the Wolfram Demonstrations Project by Charles E. Oelsner Take a look at the following example; we'll start with this one. We'll make two observations from this slope field: 1. The slopes depend solely on $$y$$ (we can also tell this simply by looking at the differential equation). Notice that if you look along a horizontal line (where $$y$$ stays constant), the slopes are all the same. This is called an autonomous differential equation, because it does not depend on the independent variable. 2. More importantly, notice that when $$y=1$$, the slopes are all 0. This is called an equilibrium solution, because if the initial condition falls on this line, the solution will stay constant for ever. ### Equilibrium Solutions Equilibrium solutions are one of the most important features that we can observe on a slope field. Look at another example: You should notice two equilibrium solutions, one at $$y=1$$ and one at $$y=-1$$ (we could also have found this by solving $$y^2-1=0$$). Look at them again, though; there is a distinct difference between them. Around $$y=1$$, the slopes diverge from the equilibrium, and around $$y=-1$$ the slopes converge toward the equilibrium. The top one is called an unstable equilibrium, and the bottom one is is called a stable equilibrium. The two classic ways to illustrate the difference are shown in the following images: The two diagrams on the left represent a stable equilibrium. If you place a marble in the bottom of a bowl, it will stay there, and if you start a pendulum in the vertical position, it will stay there as well. This is simply what it means to be an equilibrium; the system doesn't try to change anything. However, these are both stable equilibria because, if you disturb the system slightly (by moving the marble a little bit or by giving the pendulum a small push), it will return to this equilibrium. On the other hand, consider the two diagrams on the right; these represent unstable equilibria. If the bowl is turned upside down, the bottom (now the top) is still an equilibrium point, because if you balance a marble perfectly on the peak, it will stay there. Similarly, with a pendulum pinned at the bottom, if you balance it perfectly, it will not move. However, in either of these cases, if you disturb the system even slightly, it will leave the equilibrium, never to return on its own.
# RD Sharma Solutions For Class 12 Maths Exercise 10.1 Chapter 10 Differentiability RD Sharma Solutions for Class 12 Maths Exercise 10.1 Chapter 10Â Differentiability are available at BYJU’S website in PDF format with solutions prepared by experienced faculty in a precise manner. It can mainly be used by the students as a study material to clear the Class 12 exams with a good score. Exercise 10.1 of the tenth chapter contains problems, which are solved based on the differentiability of a given function. The RD Sharma Solutions are provided in an explanatory manner prepared by BYJUâ€™S experts in Maths and are in accordance with the CBSE syllabus. The PDF of RD Sharma Solutions for Class 12 Maths Exercise 10.1 of Chapter 10 are provided here. ## RD Sharma Solutions For Class 12 Chapter 10 Differentiability Exercise 10.1:- ### Access another exercise of RD Sharma Solutions For Class 12 Chapter 10 – Differentiability Exercise 10.2 Solutions ### Access answers to Maths RD Sharma Solutions For Class 12 Chapter 10 – Differentiability Exercise 10.1 Exercise 10.1 Page No: 10.10 1. Show that f (x) = \|x â€“ 3\| is continuous but not differentiable at x = 3. Solution: 2. Show that f (x) = x 1/3 is not differentiable at x = 0. Solution: Since, LHD and RHD does not exist at x = 0 Hence, f(x) is not differentiable at x = 0 Solution: Now we have to check differentiability of given function at x = 3 That is LHD (at x = 3) = RHD (at x = 3) = 12 Since, (LHD at x = 3) = (RHD at x = 3) Hence, f(x) is differentiable at x = 3. 4. Show that the function f is defined as follows Is continuous at x = 2, but not differentiable thereat. Solution: Since, LHL = RHL = f (2) Hence, F(x) is continuous at x = 2 Now we have to differentiability at x = 2 = 5 Since, (RHD at x = 2) â‰ (LHD at x = 2) Hence, f (2) is not differentiable at x = 2. 5. Discuss the continuity and differentiability of the function f (x) = \|x\| + \|x -1\| in the interval of (-1, 2). Solution: We know that a polynomial and a constant function is continuous and differentiable everywhere. So, f(x) is continuous and differentiable for x âˆˆ (-1, 0) and xÂ âˆˆ (0, 1) and (1, 2). We need to check continuity and differentiability at x = 0 and x = 1. Continuity at x = 0 Since, f(x) is continuous at x = 1 Now we have to check differentiability at x = 0 For differentiability, LHD (at x = 0) = RHD (at x = 0) Differentiability at x = 0 Since, (LHD at x = 0) â‰ (RHD at x = 0) So, f(x) is differentiable at x = 0. Now we have to check differentiability at x = 1 For differentiability, LHD (at x = 1) = RHD (at x = 1) Differentiability at x = 1 Since, f(x) is not differentiable at x = 1. So, f(x) is continuous on (- 1, 2) but not differentiable at x = 0, 1
# Chapter 2: Modeling Distributions of Data ## Presentation on theme: "Chapter 2: Modeling Distributions of Data"— Presentation transcript: Chapter 2: Modeling Distributions of Data Section 2.1 Describing Location in a Distribution The Practice of Statistics, 4th edition - For AP* STARNES, YATES, MOORE Chapter 2 Modeling Distributions of Data 2.1 Describing Location in a Distribution 2.2 Normal Distributions Section 2.1 Describing Location in a Distribution Learning Objectives After this section, you should be able to… MEASURE position using percentiles INTERPRET cumulative relative frequency graphs MEASURE position using z-scores TRANSFORM data DEFINE and DESCRIBE density curves Describing Location in a Distribution Measuring Position: Percentiles One way to describe the location of a value in a distribution is to tell what percent of observations are less than it. Describing Location in a Distribution Definition: The pth percentile of a distribution is the value with p percent of the observations less than it. 6 7 9 03 Jenny earned a score of 86 on her test. How did she perform relative to the rest of the class? Example, p. 85 Her score was greater than 21 of the 25 observations. Since 21 of the 25, or 84%, of the scores are below hers, Jenny is at the 84th percentile in the class’s test score distribution. 6 7 9 03 Describing Location in a Distribution Cumulative Relative Frequency Graphs A cumulative relative frequency graph (or ogive) displays the cumulative relative frequency of each class of a frequency distribution. Describing Location in a Distribution Age of First 44 Presidents When They Were Inaugurated Age Frequency Relative frequency Cumulative frequency Cumulative relative frequency 40-44 2 2/44 = 4.5% 2/44 = 4.5% 45-49 7 7/44 = 15.9% 9 9/44 = 20.5% 50-54 13 13/44 = 29.5% 22 22/44 = 50.0% 55-59 12 12/44 = 34% 34 34/44 = 77.3% 60-64 41 41/44 = 93.2% 65-69 3 3/44 = 6.8% 44 44/44 = 100% Describing Location in a Distribution Interpreting Cumulative Relative Frequency Graphs Use the graph from page 88 to answer the following questions. Was Barack Obama, who was inaugurated at age 47, unusually young? What is his percentile? Estimate and interpret the 65th percentile of the distribution Describing Location in a Distribution 65 11 58 47 Describing Location in a Distribution Measuring Position: z-Scores A z-score tells us how many standard deviations from the mean an observation falls, and in what direction. Describing Location in a Distribution Definition: If x is an observation from a distribution that has known mean and standard deviation, the standardized value of x is: A standardized value is often called a z-score. Jenny earned a score of 86 on her test. The class mean is 80 and the standard deviation is What is her standardized score? Describing Location in a Distribution Using z-scores for Comparison Describing Location in a Distribution We can use z-scores to compare the position of individuals in different distributions. Example, p. 91 Jenny earned a score of 86 on her statistics test. The class mean was 80 and the standard deviation was She earned a score of 82 on her chemistry test. The chemistry scores had a fairly symmetric distribution with a mean 76 and standard deviation of 4. On which test did Jenny perform better relative to the rest of her class? Describing Location in a Distribution Transforming Data Describing Location in a Distribution Transforming converts the original observations from the original units of measurements to another scale. Transformations can affect the shape, center, and spread of a distribution. Effect of Adding (or Subracting) a Constant Adding the same number a (either positive, zero, or negative) to each observation: adds a to measures of center and location (mean, median, quartiles, percentiles), but Does not change the shape of the distribution or measures of spread (range, IQR, standard deviation). Example, p. 93 n Mean sx Min Q1 M Q3 Max IQR Range Guess(m) 44 16.02 7.14 8 11 15 17 40 6 32 Error (m) 3.02 -5 -2 2 4 27 Describing Location in a Distribution Transforming Data Describing Location in a Distribution Effect of Multiplying (or Dividing) by a Constant Multiplying (or dividing) each observation by the same number b (positive, negative, or zero): multiplies (divides) measures of center and location by b multiplies (divides) measures of spread by \|b\|, but does not change the shape of the distribution n Mean sx Min Q1 M Q3 Max IQR Range Error(ft) 44 9.91 23.43 -16.4 -6.56 6.56 13.12 88.56 19.68 104.96 Error (m) 3.02 7.14 -5 -2 2 4 27 6 32 Example, p. 95 Describing Location in a Distribution Density Curves In Chapter 1, we developed a kit of graphical and numerical tools for describing distributions. Now, we’ll add one more step to the strategy. Describing Location in a Distribution Exploring Quantitative Data Always plot your data: make a graph. Look for the overall pattern (shape, center, and spread) and for striking departures such as outliers. Calculate a numerical summary to briefly describe center and spread. 4. Sometimes the overall pattern of a large number of observations is so regular that we can describe it by a smooth curve. Describing Location in a Distribution Density Curve Definition: A density curve is a curve that is always on or above the horizontal axis, and has area exactly 1 underneath it. A density curve describes the overall pattern of a distribution. The area under the curve and above any interval of values on the horizontal axis is the proportion of all observations that fall in that interval. Describing Location in a Distribution The overall pattern of this histogram of the scores of all 947 seventh-grade students in Gary, Indiana, on the vocabulary part of the Iowa Test of Basic Skills (ITBS) can be described by a smooth curve drawn through the tops of the bars. Describing Location in a Distribution Describing Density Curves Our measures of center and spread apply to density curves as well as to actual sets of observations. Describing Location in a Distribution Distinguishing the Median and Mean of a Density Curve The median of a density curve is the equal-areas point, the point that divides the area under the curve in half. The mean of a density curve is the balance point, at which the curve would balance if made of solid material. The median and the mean are the same for a symmetric density curve. They both lie at the center of the curve. The mean of a skewed curve is pulled away from the median in the direction of the long tail. Section 2.1 Describing Location in a Distribution Summary In this section, we learned that… There are two ways of describing an individual’s location within a distribution – the percentile and z-score. A cumulative relative frequency graph allows us to examine location within a distribution. It is common to transform data, especially when changing units of measurement. Transforming data can affect the shape, center, and spread of a distribution. We can sometimes describe the overall pattern of a distribution by a density curve (an idealized description of a distribution that smooths out the irregularities in the actual data). Looking Ahead… In the next Section… We’ll learn about one particularly important class of density curves – the Normal Distributions We’ll learn The Rule The Standard Normal Distribution Normal Distribution Calculations, and Assessing Normality In the next Section… Homework Assignment Read p. 84-103 Define Vocabulary P. 105:1,5,9-23 odds, 31, all Similar presentations
# What are all the factors, the prime factorization, and factor pairs of 929000? To find the factors of 929000, divide 929000 by each number starting with 1 and working up to 929000 ## What is a factor in math ? Factors are the numbers you multiply together to get another number. For example, the factors of 15 are 3 and 5 because 3 × 5 = 15. The factors of a number can be positive or negative, but they cannot be zero. The factors of a number can be used to find out if the number is prime or not. A prime number is a number that has only two factors: itself and 1. For example, the number 7 is prime because its only factors are 7 and 1. ## List all of the factors of 929000 ? To calculate the factors of 929000 , you can use the division method. 1. Begin by dividing 929000 by the smallest possible number, which is 2. 2. If the division is even, then 2 is a factor of 929000. 3. Continue dividing 929000 by larger numbers until you find an odd number that does not divide evenly into 929000 . 4. The numbers that divide evenly into 929000 are the factors of 929000 . Now let us find how to calculate all the factors of Nine hundred twenty-nine thousand : 929000 ÷ 1 = 929000 929000 ÷ 2 = 464500 929000 ÷ 4 = 232250 929000 ÷ 5 = 185800 929000 ÷ 8 = 116125 929000 ÷ 10 = 92900 929000 ÷ 20 = 46450 929000 ÷ 25 = 37160 929000 ÷ 40 = 23225 929000 ÷ 50 = 18580 929000 ÷ 100 = 9290 929000 ÷ 125 = 7432 929000 ÷ 200 = 4645 929000 ÷ 250 = 3716 929000 ÷ 500 = 1858 929000 ÷ 929 = 1000 929000 ÷ 1000 = 929 929000 ÷ 1858 = 500 929000 ÷ 3716 = 250 929000 ÷ 4645 = 200 929000 ÷ 7432 = 125 929000 ÷ 9290 = 100 929000 ÷ 18580 = 50 929000 ÷ 23225 = 40 929000 ÷ 37160 = 25 929000 ÷ 46450 = 20 929000 ÷ 92900 = 10 929000 ÷ 116125 = 8 929000 ÷ 185800 = 5 929000 ÷ 232250 = 4 929000 ÷ 464500 = 2 929000 ÷ 929000 = 1 As you can see, the factors of 929000 are 1 , 2 , 4 , 5 , 8 , 10 , 20 , 25 , 40 , 50 , 100 , 125 , 200 , 250 , 500 , 929 , 1000 , 1858 , 3716 , 4645 , 7432 , 9290 , 18580 , 23225 , 37160 , 46450 , 92900 , 116125 , 185800 , 232250 , 464500 and 929000 . ## How many factors of 929000 are there ? The factors of 929000 are the numbers that can evenly divide 929000 . These numbers are 1 , 2 , 4 , 5 , 8 , 10 , 20 , 25 , 40 , 50 , 100 , 125 , 200 , 250 , 500 , 929 , 1000 , 1858 , 3716 , 4645 , 7432 , 9290 , 18580 , 23225 , 37160 , 46450 , 92900 , 116125 , 185800 , 232250 , 464500 and 929000. Thus, there are a total of 32 factors of 929000 ## What are the factor pairs of 929000 ? Factor Pairs of 929000 are combinations of two factors that when multiplied together equal 929000. There are many ways to calculate the factor pairs of 929000 . One easy way is to list out the factors of 929000 : 1 , 2 , 4 , 5 , 8 , 10 , 20 , 25 , 40 , 50 , 100 , 125 , 200 , 250 , 500 , 929 , 1000 , 1858 , 3716 , 4645 , 7432 , 9290 , 18580 , 23225 , 37160 , 46450 , 92900 , 116125 , 185800 , 232250 , 464500 , 929000 Then, pair up the factors: (1,929000),(2,464500),(4,232250),(5,185800),(8,116125),(10,92900),(20,46450),(25,37160),(40,23225),(50,18580),(100,9290),(125,7432),(200,4645),(250,3716),(500,1858) and (929,1000) These are the factor pairs of 929000 . ## Prime Factorisation of 929000 There are a few different methods that can be used to calculate the prime factorization of a number. Two of the most common methods are listed below. 1) Use a factor tree : 1. Take the number you want to find the prime factorization of and write it at the top of the page 2. Find the smallest number that goes into the number you are finding the prime factorization of evenly and write it next to the number you are finding the prime factorization of 3. Draw a line under the number you just wrote and the number you are finding the prime factorization of 4. Repeat step 2 with the number you just wrote until that number can no longer be divided evenly 5. The numbers written on the lines will be the prime factors of the number you started with For example, to calculate the prime factorization of 929000 using a factor tree, we would start by writing 929000 on a piece of paper. Then, we would draw a line under it and begin finding factors. The final prime factorization of 929000 would be 2 x 2 x 2 x 5 x 5 x 5 x 929. 2) Use a factorization method : There are a few different factorization methods that can be used to calculate the prime factorization of a number. One common method is to start by dividing the number by the smallest prime number that will divide evenly into it. Then, continue dividing the number by successively larger prime numbers until the number has been fully factorised. For example, to calculate the prime factorization of 929000 using this method, we keep dividing until it gives a non-zero remainder. 929000 ÷ 2 = 464500 464500 ÷ 2 = 232250 232250 ÷ 2 = 116125 116125 ÷ 5 = 23225 23225 ÷ 5 = 4645 4645 ÷ 5 = 929 929 ÷ 929 = 1 So the prime factors of 929000 are 2 x 2 x 2 x 5 x 5 x 5 x 929. ## Frequently Asked Questions on Factors ### What are all the factors of 929000 ? The factors of 929000 are 1 , 2 , 4 , 5 , 8 , 10 , 20 , 25 , 40 , 50 , 100 , 125 , 200 , 250 , 500 , 929 , 1000 , 1858 , 3716 , 4645 , 7432 , 9290 , 18580 , 23225 , 37160 , 46450 , 92900 , 116125 , 185800 , 232250 , 464500 and 929000. ### What is the prime factorization of 929000 ? The prime factorization of 929000 is 2 x 2 x 2 x 5 x 5 x 5 x 929 or 23 x 53 x 9291, where 2 , 5 , 929 are the prime numbers . ### What are the prime factors of 929000 ? The prime factors of 929000 are 2 , 5 , 929 . ### Is 929000 a prime number ? A prime number is a number that has only two factors 1 and itself. 929000 it is not a prime number because it has the factors 1 , 2 , 4 , 5 , 8 , 10 , 20 , 25 , 40 , 50 , 100 , 125 , 200 , 250 , 500 , 929 , 1000 , 1858 , 3716 , 4645 , 7432 , 9290 , 18580 , 23225 , 37160 , 46450 , 92900 , 116125 , 185800 , 232250 , 464500 and 929000.
# Free Playing with Numbers 01 Practice Test - 6th grade ___________ is a factor of every number and every number is a factor of _____________. Choose the suitable pair from the following. A. 2, next number B. 2, itself C. 1, itself D. 1, next number #### SOLUTION Solution : C 1 is an exact divisor of all the numbers. Also, any number is divisible by itself. Therefore, 1 is a factor of every number and e very number is a factor of itself. Every number is a multiple of ______________ . A. 5 B. 2 C. itself D. 100 #### SOLUTION Solution : C Every number is a multiple of itself . A perfect number is: A. a number whose sum of factors is equal to twice the number. B. a number whose sum of its digits is equal to the product of its digits. C. a number whose sum of factors is equal to thrice the number. D. a number which is co-prime to every other number. #### SOLUTION Solution : A A perfect number is a number for which the sum of all its factors is equal to twice the number. For example, 6 is a perfect number since the sum of its factors 1, 2, 3 and 6 is 12 i.e. twice of 6. Perfect numbers are also defined as those numbers that equal the sum of all their factors including 1 but excluding the number itself. Taking the same example of 6: sum of its factors excluding the number itself i.e 6 = 1+2+3 = 6. The factors of 72 are __________________. A. 5, 7, 10, 11, 13, 17, 21 and 23 B. 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72 C. 5, 7, 13, 14, 15, 16, 17 and 19 D. 5, 7, 11, 13, 15, 16, 17, 20, and 25 #### SOLUTION Solution : B 72=1×72 =2×36 =3×24 =4×18 =6×12 =8×9 Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72. Which of the following is the largest two-digit multiple of 19? A. 85 B. 76 C. 96 D. 95 #### SOLUTION Solution : D 19 × 1 = 19 19 × 2 = 38 19 × 3 = 57 19 × 4 = 76 19 × 5 = 95 19 × 6 = 114 Hence, the largest two-digit multiple of 19 is 95. Choose the option which contains set of prime numbers. A. 1, 2, 3, 5, 7 B. 2, 5, 9, 11, 17 C. 19, 13, 11, 7, 3 D. 23, 29, 39, 37, 43 #### SOLUTION Solution : C Prime numbers have only two factors, 1 and the number itself. 1 is neither prime nor composite. 2 is the only even prime number. 9 and 39 have more than 2 factors, hence they are not prime numbers. Considering all these, the correct option is 19, 13, 11, 7 and 3. The largest 3 digit composite number is:- A. 1000 B. 999 C. 989 D. 949 #### SOLUTION Solution : B Composite numbers are those numbers which are not prime numbers except 1. 1 is regarded as a unique number. The largest 3 digit number is 999 and it is a composite number. (divisible by 3, 9 etc) Composite numbers have more than _______ factors. A. 6 B. 10 C. 15 D. 2 #### SOLUTION Solution : D Composite numbers have more than 2 factors. For example, factors of 4 are 1, 2 and 4. 7 consecutive composite numbers less than 100 are - A. 81, 82, 83, 84, 85, 86, 87 B. 90, 91, 92, 93, 94, 95, 96 C. 65, 67, 68, 69, 70, 71, 72 D. 46, 47, 48, 49, 50, 51, 52 #### SOLUTION Solution : B All the numbers in the option: 90, 91, 92, 93, 94, 95, 96 have more than 2 factors, hence they are all composite numbers. Which of the following numbers are divisible by 11 and 9 respectively? A. 10000001, 459756 B. 9000001, 45975 C. 4569874, 547952 D. 4568947, 97455 #### SOLUTION Solution : A To check the divisibility of a number by 11, we find the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) of the number. If the difference is 0 or a number divisible by 11, then the number is divisible by 11. For 10000001, Sum of digits at odd places = 1 Sum of digits at even places = 1 difference = 1 - 1 = 0 Hence, 10000001 is divisible by 11 For divisibility by 9 - if the sum of the digits of a number is divisible by 9, then the number is divisible by 9. The sum of digits of 459756 is 36 which is divisible by 9. Hence, 459756 is divisible by 9.
# How do you find the axis of symmetry, graph and find the maximum or minimum value of the function y=2x^2 - 18x +13? Sep 10, 2017 Axis of symmetry: $x = \frac{9}{2}$ Vertex (minimum point): $\left(\frac{9}{2} , - \frac{55}{2}\right)$ X-intercepts: $\left(\frac{9 + \sqrt{55}}{2} , 0\right)$ and $\left(\frac{9 - \sqrt{55}}{2} , 0\right)$ Y-intercept: $\left(0 , 13\right)$ Refer to the explanation for the process and approximate values for vertex, x-intercepts, and y-intercept. #### Explanation: Given: $y = 2 {x}^{2} - 18 x + 13$ is a quadratic equation in standard form: $y = a {x}^{2} + b x + c$, where: $a = 2$, $b = - 18$, and $c = 13$ To graph a quadratic function, you need to have at least the vertex and x-intercepts. The y-intercept is helpful, also. Axis of Symmetry: vertical line $\left(x , \pm \infty\right)$ that divides the parabola into two equal halves. The variable for the line is $x = \frac{- b}{2 a}$. $x = \frac{- \left(- 18\right)}{2 \cdot 2}$ $x = \frac{18}{4}$ $x = \frac{9}{2}$ $\leftarrow$ axis of symmetry and $x$-value for the vertex Vertex: the maximum or minimum point of the parabola. If $a > 0$, the vertex is the minimum point and the parabola will open upward. If $a < 0$, the vertex is the maximum point and the parabola will open downward. We have the $x$-value of the vertex. To determine the $y$-value, substitute $\frac{9}{2}$ for $x$ in the equation and solve for $y$. $y = 2 {\left(\frac{9}{2}\right)}^{2} - 18 \left(\frac{9}{2}\right) + 13$ Simplify. $y = 2 \left(\frac{81}{4}\right) - \frac{162}{2} + 13$ All terms must have a common denominator of $4$. Multiply fractions without a denominator of $4$ by a multiplier equal to $1$ that will produce an equivalent fraction with a denominator of $4$. For example, $\frac{\textcolor{m a \ge n t a}{3}}{\textcolor{m a \ge n t a}{3}} = 1$ Recall that any whole number, $n$, is understood to have a denominator of $1$. So $13 = \frac{13}{1}$ $y = \frac{162}{4} - \frac{162}{2} \times \frac{\textcolor{red}{2}}{\textcolor{red}{2}} + \frac{13}{1} \times \frac{\textcolor{b l u e}{4}}{\textcolor{b l u e}{4}}$ $y = \frac{162}{4} - \frac{324}{4} + \frac{52}{4}$ $y = \frac{\left(162 - 324 + 52\right)}{4}$ $y = \frac{- 110}{4}$ $y = - \frac{55}{2}$ Vertex: $\left(\frac{9}{2} , - \frac{55}{2}\right)$ $\leftarrow$ minimum point of the parabola Approximate vertex: $\left(4.5 , - 27.5\right)$ Substitute $0$ for $y$ and use the quadratic formula to find the roots and the x-intercepts. $0 = 2 {x}^{2} - 18 x + 13$ $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ Plug in the known values. $x = \frac{- \left(- 18\right) \pm \sqrt{{\left(- 18\right)}^{2} - 4 \cdot 2 \cdot 13}}{2 \cdot 2}$ Simplify. $x = \frac{18 \pm \sqrt{324 - 104}}{4}$ $x = \frac{18 \pm \sqrt{220}}{4}$ Prime factorize $220$. $x = \frac{18 \pm \sqrt{\left(2 \times 2\right) \times 5 \times 11}}{4}$ $x = \frac{18 \pm 2 \sqrt{55}}{4}$ Simplify. $x = \frac{9 \pm \sqrt{55}}{2}$ Roots: values for $x$ $x = \frac{9 + \sqrt{55}}{2}$,$\frac{9 - \sqrt{55}}{2}$ Approximate values for $x$. $x = 8.21 ,$$0.792$ X-intercepts: values of $x$ when $y = 0$ $x$-intercepts: $\left(\frac{9 + \sqrt{55}}{2} , 0\right)$ and $\left(\frac{9 - \sqrt{55}}{2} , 0\right)$ Approximate $x$-intercepts: $\left(8.21 , 0\right)$ and $\left(0.792 , 0\right)$ Y-Intercept: value of $y$ when $x = 0$ $y = 2 {\left(0\right)}^{2} - 18 \left(0\right) + 13$ $y = 13$ Y-intercept: $\left(0 , 13\right)$ Plot the vertex and x-intercepts and sketch a parabola through the points. Do not connect the dots. graph{y=2x^2-18x+13 [-13.95, 18.07, -40.31, -24.29]} Nov 15, 2017 Some notes on Quadratic equation plots #### Explanation: $\textcolor{b l u e}{\text{Axis of symmetry - a sort of cheat method}}$ This uses the beginnings of completing the square. Write as: $y = 2 \left({x}^{2} - \frac{18}{2} x\right) + 13$ $\textcolor{red}{{x}_{\text{vertex")=x_("axis of symmetry}} =} \left(- \frac{1}{2}\right) \times \left(- \frac{18}{2}\right) = \textcolor{red}{+ 4.5 \to \frac{9}{2}}$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\textcolor{b l u e}{\text{y-intercept}}$ This takes on the value of $c$ in $y = a {x}^{2} + b x + c$ color(red)(y_("intercept")=c=+13 ->(x,y)=(0,13) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\textcolor{b l u e}{\text{x-intercepts and vertex}}$ Note that the vertex is the point of maximum or minimum. 2 of these x-intercepts exist if the graph crosses the x-axis 1 of these exists if the x-axis is tangential to the vertex. Some people state that there is still two but they are the same as each other. Sometimes determining these can take a bit or work. Other times not so. Initial step is to set $y = 0$ It has to be 0 for the graph to cross the axis as x-axis is at $y = 0$ $y = 0 = 2 {x}^{2} - 18 x + 13$ If you can spot how to factorise this then that would be the quicker method. For most people this only works if the factors are whole numbers. In this case they are not whole numbers. There are two options: completing the square or using the other formula. $y = a {x}^{2} + b x + c$ where a=2; b=-18; c=13 ${x}_{\text{intercept")=(-b+-sqrt(b^2-4ac))/(2a)" ".......Equation color(white)("d}} t y p e 1$ y=0=a(x+b/(2a))^2+k+c" ".......Equation color(white)("d")type2 $E q u a t i o n \textcolor{w h i t e}{\text{d}} t y p e 2$ can be used to directly determine both the vertex and x-intercepts. Part of $E q u a t i o n \textcolor{w h i t e}{\text{d}} t y p e 1$ can be used to indicate if the plot actually has an x-intercept. The determinant $\to {b}^{2} - 4 a c$ must be such that ${b}^{2} - 4 a c \ge 0$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ The original equation and the completed square are different forms of the same condition. As such the plot of one will be the same as the plot of the other. Zoom in and you will see that the blue plot follows the same path as the red.
# How to State the Null Hypothesis in Statistics Hypothesis Testing > How to State the Null Hypothesis Watch the video or read the steps below: ### How to State the Null Hypothesis from a Word Problem You’ll be asked to convert a word problem into a hypothesis statement in statistics that will include a null hypothesis and an alternate hypothesis. Breaking your problem into a few small steps makes these problems much easier to handle. ## How to State the Null Hypothesis Sample Problem: A researcher thinks that if knee surgery patients go to physical therapy twice a week (instead of 3 times), their recovery period will be longer. Average recovery times for knee surgery patients is 8.2 weeks. Hypothesis testing is vital to test patient outcomes. Step 1: Figure out the hypothesis from the problem. The hypothesis is usually hidden in a word problem, and is sometimes a statement of what you expect to happen in the experiment. The hypothesis in the above question is “I expect the average recovery period to be greater than 8.2 weeks.” Step 2: Convert the hypothesis to math. Remember that the average is sometimes written as μ. H1: μ > 8.2 Broken down into (somewhat) English, that’s H1 (The hypothesis): μ (the average) > (is greater than) 8.2 Step 3: State what will happen if the hypothesis doesn’t come true. If the recovery time isn’t greater than 8.2 weeks, there are only two possibilities, that the recovery time is equal to 8.2 weeks or less than 8.2 weeks. H0: μ ≤ 8.2 Broken down again into English, that’s H0 (The null hypothesis): μ (the average) ≤ (is less than or equal to) 8.2 ## How to State the Null Hypothesis: Part Two ### But what if the researcher doesn’t have any idea what will happen? Sample Problem: A researcher is studying the effects of radical exercise program on knee surgery patients. There is a good chance the therapy will improve recovery time, but there’s also the possibility it will make it worse. Average recovery times for knee surgery patients is 8.2 weeks. Step 1: State what will happen if the experiment doesn’t make any difference. That’s the null hypothesis–that nothing will happen. In this experiment, if nothing happens, then the recovery time will stay at 8.2 weeks. H0: μ = 8.2 Broken down into English, that’s H0 (The null hypothesis): μ (the average) =(is equal to) 8.2 Step 2: Figure out the alternate hypothesis. The alternate hypothesis is the opposite of the null hypothesis. In other words, what happens if our experiment makes a difference? H1: μ ≠ 8.2 In English again, that’s H1 (The alternate hypothesis): μ (the average) ≠ (is not equal to) 8.2 That’s How to State the Null Hypothesis! Check out our Youtube channel for more stats tips! If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you’re are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track. How to State the Null Hypothesis in Statistics was last modified: June 25th, 2017 by # 12 thoughts on “How to State the Null Hypothesis in Statistics” 1. Angie Widdows This was a good example. It is the next step that I often have problems with. This example helps because it (like others) lists out the meaning of the symbols. Very helpful 2. Lisa Barcomb Understanding the null hypothesis is much easier now because when you go to convert it to math it makes a lot more sense in figuring out the formulas that you are suppose to be working with. I think I put more into these problems then I really need to. 3. Vanessa DuBarry This example really helped me, because it explains how to put it into english, and at first it was kind of hard knowing what sign goes were, but this really helped. 4. Donna Allen At first, I was a little intimidated by this. But, after going over your instructions, it seems fairly easy. Thanks! 5. Bill Bryan I was starting to get that sinking feeling until I read this, love the matter of fact approch to discribing the steps. 6. Dave Ocran there has been a better understanding as to how approach a question ,if there no given notations such as at :at par,no difference ,less than ,greater than and at least.tanx 7. Tripsy Thank you for your explanations but I am still having some difficulty stating the proper null hypothesis. What is the null hypothesis in this example… We would like to know if the cholesterol level depends on the age and for that reason we collect cholesterol level of 10 people, first when they were 17 and then when they are 46. 8. Andale It sounds like the “we” in the question is you/the scientist hypothesizing that cholesterol depends upon age. That’s the alternate hypothesis — the one you’re testing. The null is the opposite (cholesterol doesn’t depend on age). 9. Tripsy Thanks. Would it be also right to say the null hypothesis for this is “the cholesterol at age 17 is equal to the cholesterol at age 49?” Being that we’re comparing two groups?
## Sunday, December 16, 2018 Page No: 91 1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them; (i) 2x2 - 3x + 5 = 0 (ii) 3x2 - 4√3x + 4 = 0 (iii) 2x2 - 6x + 3 = 0 (i) Consider the equation x2 - 3x + 5 = 0 Comparing it with ax2 + bx c = 0, we get a = 2, b = -3 and c = 5 Discriminant = b2 - 4ac ( - 3)2 - 4 (2) (5) = 9 - 40 = - 31 As b2 - 4ac < 0, Therefore, no real root is possible for the given equation. (ii) 3x2 - 4√3x + 4 = 0 Comparing it with ax2 + bx c = 0, we get a = 3, b = -4√3 and c = 4 Discriminant = b2 - 4ac = (-4√3)- 4(3)(4) = 48 - 48 = 0 As b2 - 4ac = 0, Therefore, real roots exist for the given equation and they are equal to each other. And the roots will be -b/2a and -b/2a.-b/2= -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3 Therefore, the roots are 2/√3 and 2/√3. (iii) 2x2 - 6x + 3 = 0 Comparing this equation with ax2 + bx c = 0, we get a = 2, b = -6, c = 3 Discriminant = b2 - 4ac = (-6)2 - 4 (2) (3) = 36 - 24 = 12 As b2 - 4ac > 0, Therefore, distinct real roots exist for this equation: 2. Find the values of k for each of the following quadratic equations, so that they have two equal roots. (i) 2x2 + kx + 3 = 0 (ii) kx (x - 2) + 6 = 0 (i) 2x2 + kx + 3 = 0 Comparing equation with ax2 + bx c = 0, we get a = 2, b = k and c = 3 Discriminant = b2 - 4ac = (k)2 - 4(2) (3) k2 - 24 For equal roots, Discriminant = 0 k2 - 24 = 0 k2 = 24 k = ±√24 = ±2√6 (ii) kx(x - 2) + 6 = 0 or kx2 - 2kx + 6 = 0 Comparing this equation with ax2 + bx c = 0, we get a = kb = - 2k and c = 6 Discriminant = b2 - 4ac = ( - 2k)2 - 4 (k) (6) = 4k2 - 24k For equal roots, b2 - 4ac = 0 4k2 - 24k = 0 4k (k - 6) = 0 Either 4k = 0 or k = 6 = 0 k = 0 or k = 6 However, if k = 0, then the equation will not have the terms 'x2' and 'x'. Therefore, if this equation has two equal roots, k should be 6 only. 3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth. Let the breadth of mango grove be l. Length of mango grove will be 2l. Area of mango grove = (2l) (l)= 2l2 2l= 800 l= 800/2 = 400 l- 400 =0 Comparing this equation with al2 + bl + c = 0, we get a = 1, b = 0, c = 400 Discriminant = b2 - 4ac = (0)2 - 4 × (1) × ( - 400) = 1600 Here, b2 - 4ac > 0 Therefore, the equation will have real roots. And hence, the desired rectangular mango grove can be designed. = ±20 However, length cannot be negative. Therefore, breadth of mango grove = 20 m Length of mango grove = 2 × 20 = 40 m 4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Let the age of one friend be x years. then the age of the other friend will be (20 - x) years. 4 years ago, Age of 1st friend = (x - 4) years Age of 2nd friend = (20 - x - 4) = (16 - x) years A/q we get that, (x - 4) (16 - x) = 48 16x - x2 - 64 + 4x = 48 - x2 + 20x - 112 = 0 x2 - 20x + 112 = 0 Comparing this equation with ax2 + bx c = 0, we get a = 1b = -20 and c = 112 Discriminant = b2 - 4ac = (-20)2 - 4 × 112 = 400 - 448 = -48 b2 - 4ac < 0 Therefore, there will be no real solution possible for the equations. Such type of condition doesn't exist. 5. Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth. Let the length and breadth of the park be l and b. Perimeter = 2 (l + b) = 80 l + b = 40 Or, b = 40 - l Area = l×b = l(40 - l) = 40l240l - l2 = 400 l2 - 40l + 400 = 0 Comparing this equation with al2 + bl + c = 0, we get a = 1, b = -40, c = 400 Discriminant = b2 - 4ac (-40)2 - 4 × 400 = 1600 - 1600 = 0 b2 - 4ac = 0 Therefore, this equation has equal real roots. And hence, this situation is possible. Root of this equation,l = -b/2a l = (40)/2(1) = 40/2 = 20 Therefore, length of park, = 20 m And breadth of park, = 40 - = 40 - 20 = 20 m.
Exercise 12.3 Algebraic-Expressions -NCERT Solutions Class 7 Go back to 'Algebraic Expressions' Chapter 12 Ex.12.3 Question 1 If $$m = 2$$, find the value of: (i) $$m = 2$$ (ii) $$3m - 5$$ (iii) $${\rm{ }}9{\rm{ }}-{\rm{ }}5m$$ (iv) $$3{m^2} - 2m - 7$$ (v) \begin{align} \frac{{5m}}{2} - 4\end{align} Solution What is Known? Value of $$m$$. What is unknown? Value of the given expressions. Reasoning: This is based on concept of putting given value of variable and then performing the arithmetic operation as given in the question. Steps: Value of $$m$$ is given as $$2$$. (i) $$m = 2$$ \begin{align}& {\rm{ = 2}} - {\rm{2}}\\{\rm{ Ans }} &\,{\rm{ = }}\,{\rm{0 }}\end{align} (ii) $$3m - 5$$ \begin{align}& = 3 \times 2 - \left( 5 \right)\\& = 6 - 5\\{\rm{ Ans }} &= 1\end{align} (iii) $${\rm{ }}9{\rm{ }}-{\rm{ }}5m$$ \begin{align}& = 9 - \left( {5 \times 2} \right)\\& = 9 - 10\\{\rm{ Ans }} &= - 1\end{align} (iv) $$3{m^2} - 2m - 7$$ \begin{align}& = 3{{\left( 2 \right)}^2} - \left( {2 \times 2} \right) - 7\\& = 3 \times 2 \times 2 - \left( 4 \right) - 7\\& = 12 - 4 - 7\\{\rm{ Ans }} & = 1\end{align} (v) \begin{align} \frac{{5m}}{2} - 4\end{align} \begin{align}& = \frac{{5 \times 2}}{2} - 4\\& = \frac{{10}}{2} - 4\\& = 5 - 4\\{\rm{ Ans }} & = 1\end{align} Chapter 12 Ex.12.3 Question 2 If $$p = \,– 2$$, find the value of: (i) $$4p + 7$$ (ii) $$- 3{p^2} + 4p + 7$$ (iii) $$- 2{p^3} - 3{p^2} + 4p + 7$$ Solution What is known? Value of $$p$$. What is unknown? Value of the given expressions. Reasoning: This is based on concept of putting given value of variable and then performing the arithmetic operation as given in the question. Steps: Value of $$p$$ is given as $$- 2$$ (i) $$4p + 7$$ \begin{align} &=\! 4 \!\times\!\! -\! 2 \!+\! \left( 7 \right)\\ &= \!-\! 8 \!+\! 7\\&{\rm{ Ans }} = \!-\! 1\end{align} (ii) $$- 3{p^2} + 4p + 7$$ \begin{align}&= \!-\! 3 \!\times\! {{\left( { - 2} \right)}^2} \!+\!4 \!\times\! \left( { - 2} \right) \!+\! 7\\ &= \! \left( { - 3 \!\times\! \!- 2 \!\times\! \!- 2} \right) \!+\! \left( { - 8} \right) \!+\! 7\\ &= \!-\! 12 \!-\! 8 \!+\! 7\\&{\rm{ Ans }} = \!-\! 13\end{align} (iii) $$- 2{p^3} - 3{p^2} + 4p + 7$$ \begin{align} &=\!-\!2{{\left( -2 \right)}^{3}}\!-\!3{{\left( -2 \right)}^{2}}\!+\!4\left( -2 \right)\!+\!7 \\ &=\!-\!2 \!\times\! \!-2\!\times\! \!-2\!\times\!\! -2\left( 3\!\times\! \!-2\!\times\! \!-2 \right)\!+\!\left( 4\!\times\! \!-2 \right)\!+\!7 \\ &=\!16\left( 12 \right)\!+\!\left( -8 \right)\!+\!7 \\ &\text{Ans} =\!3 \end{align} Chapter 12 Ex.12.3 Question 3 Find the value of the following expressions, when $$x = \,–1$$ (i) $$2x - 7$$ (ii) $$- x + 2$$ (iii) $${x^2} + 2x + 1$$ (iv) $$2{x^2} - x - 2$$ Solution What is Known? Value of $$x$$ What is unknown? Value of the given expressions. Reasoning: This is based on concept of putting given value of variable and then performing the arithmetic operation as given in the question. Steps: Value of $$x$$ is given as $$–1$$ (i) $$2x - 7$$ \begin{align}& = 2 \times - 1 - (7)\\& = - 2 - 7\\{\rm{ Ans }} & = - 9\end{align} (ii) $$- x + 2$$ \begin{align}& = - \left( { - 1} \right) + 2\\& = 1 + 2\\{\rm{Ans}} & = 3\end{align} (iii) $${x^2} + 2x + 1$$ \begin{align}& = {\left( { - 1} \right)^2} + \left( {2 \times - 1} \right) + 1\\& = - 1 \times - 1 + \left( { - 2} \right) + 1\\& = 1 - 2 + 1\\{\rm{ Ans}} & = 0\end{align} (iv) $$2{x^2} - x - 2$$ \begin{align}&= 2{{\left( { - 1} \right)}^2} - \left( { - 1} \right) - 2\\&= 2 \times - 1 \times - 1 + 1 - 2\\&= 2 + 1 - 2\\{\rm{ Ans }} &= 1\end{align} Chapter 12 Ex.12.3 Question 4 If $$a = 2$$, $$b =\, – 2$$, find the value of: (i) $${a^2} + {b^2}$$ (ii) $${a^2} + ab + {b^2}$$ (iii) $${a^2} - {b^2}$$ Solution What is Known? Value of $$a$$ and $$b$$ What is unknown? Value of the given expressions. Reasoning: This is based on concept of putting given value of variable and then performing the arithmetic operation as given in the question. Steps: Value of $$a$$ is given as $$2$$ and $$b$$ is $$-2$$ (i) $${a^2} + {b^2}$$ \begin{align}& = {2^2} + {\left( { - 2} \right)^2}\\& = \left( {2 \times 2} \right) + \left( { - 2 \times - 2} \right)\\& = 4 + 4\\{\rm{Ans}} & = 8\,\end{align} (ii) $${a^2} + ab + {b^2}$$ \begin{align} & = {2^2} + \left\{ {\left( 2 \right) \times \left( { - 2} \right)} \right\} + {\left( { - 2} \right)^2}\\& = 4 + \left( { - 4} \right) + 4\\& = 4 - 4 + 4\\{\rm{Ans}} & = 4\,\end{align} (iii) $${a^2} - {b^2}$$ \begin{align}& = {2^2} - {\left( { - 2} \right)^2}\\& = 4 - 4\\{\rm{ Ans}} & = 0\end{align} Chapter 12 Ex.12.3 Question 5 When $$a = 0$$, $$b = \,– 1$$, find the value of the given expressions: (i) $$2a+\text{ }2b$$ (ii) $$2{{a}^{2}}+{{b}^{2}}+\text{ }1$$ (iii) $$2{{a}^{2}}b+\text{ }2a{{b}^{2}}+ab$$ (iv) $${{a}^{2}}+ab+\text{ }2$$ Solution What is Known? Value of $$a$$ and $$b$$ What is unknown? Value of the given expressions. Reasoning: This is based on concept of putting given value of variable and then performing the arithmetic operation as given in the question. Steps: Value of $$a$$ is given as $$0$$ and $$b$$ is $$–1$$ (i) $$2a+\text{ }2b$$ \begin{align}& = \left( {2 \times 0} \right) + \left( {2 \times - 1} \right)\\& = 0 + \left( { - 2} \right)\\{\rm{ Ans }} & = - 2\end{align} (ii) $$2{{a}^{2}}+{{b}^{2}}+\text{ }1$$ \begin{align}& = \left( {2 \times {0^2}} \right) + {\left( { - 1} \right)^2} + 1\\ & = 0 + 1 + 1\\& = 2\end{align} (iii) $$2{{a}^{2}}b+\text{ }2a{{b}^{2}}+ab$$ \begin{align} &= \left[ \begin{array}{l}\,\,2 \times 0 \times 0 \times - 1 + \\\left( {2 \times 0 \times - {1^2}} \right) + 0 \times - 1\end{array} \right]\\ &= 0 + 0 + 0\\{\rm{Ans}} &= 0\end{align} (iv) $${{a}^{2}}+ab+\text{ }2$$ \begin{align}& = {0^2} + 0 \times - 1 + 2\\& = 0 + 0 + 2\\{\rm{ Ans }} & = 2\end{align} Chapter 12 Ex.12.3 Question 6 Simplify the expressions and find the value if $$x$$ is equal to $$2$$ (i) $$x + 7 + 4\left( {x - 5} \right)$$ (ii) $$3\left( {x + 2} \right) + 5x - 7$$ (iii) $$6x + 5\left( {x - 2} \right)$$ (iv) $$4\left( {2x - 1} \right) + 3x + 11$$ Solution What is Known? Value of $$x$$ What is unknown? Value of the given expressions. Reasoning: This is based on concept of simplification of like terms and then putting given value of variable and then performing the arithmetic operation as given in the question. Value of $$x$$ is given as $$2$$ Steps: (i) $$x + 7 + 4\left( {x - 5} \right)$$ \begin{align}& =x+7+4x-20 \\ & =5x-13\end{align} Now putting value of $$x=2$$ \begin{align}&5x-13 \\ & =\left( 5\times 2 \right)-13 \\ & =10-13 \\ \text{ Ans} & =-3 \\ \end{align} (ii) $$3\left( {x + 2} \right) + 5x - 7$$ \begin{align}&=3x+6+5x-7 \\&=8x-1\end{align} Now putting value of $$x=2$$ \begin{align}&=\left( 8\times 2 \right)-1 \\&=16-1 \\\text{ Ans} &=15 \\\end{align} (iii) $$6x + 5\left( {x - 2} \right)$$ \begin{align}& = 6x + 5x - 10\\& = 11x - 10\end{align} Now putting value of $$x = 2$$ \begin{align}& = \left( {11 \times 2} \right) - 10\\\text{ Ans} & = 12\end{align} (iv) $$4\left( {2x - 1} \right) + 3x + 11$$ \begin{align}& = 8x - 4 + 3x + 11\\& = 11x + 7\end{align} Now putting value of $$x = 2$$ \begin{align}&= (11 \times 2) + 7\\ & = 22 + 7\\\text{ Ans} & = 29\end{align} Chapter 12 Ex.12.3 Question 7 Simplify these expressions and find their values if $$x = 3$$, $$a = \,– 1$$, $$b =\, – 2$$. (i) $$3x - 5 - x + 9$$ (ii) $$2 - 8x + 4x + 4$$ (iii) $$3a + 5 - 8a + 1$$ (iv) $$10 - 3b - 4 - 5b$$ (v) $$2a - 2b - 4 - 5 + a$$ Solution What is Known? Value of $$x$$, $$a$$ and $$b$$ What is unknown? Value of the given expressions. Reasoning: This is based on concept of simplification of like terms and then putting given value of variable and then performing the arithmetic operation as given in the question. Steps: Value of $$x$$ is given as $$3$$, $$a$$ as $$-1$$ and $$b$$ is $$-2$$ (i) $$3x - 5 - x + 9$$ \begin{align}& = 2x + 4\end{align} Now putting value of $$x = 3$$ \begin{align}& = \left( {2 \times 3} \right) + 4\\& = 6 + 4\\{\text{ Ans }} & = 10\end{align} (ii) $$2 - 8x + 4x + 4$$ \begin{align}& = - 4x + 6\end{align} Now putting value of $$x = 3$$ \begin{align}&= \left( { - 4 \times 3} \right) + 6\\& = - 12 + 6\\{\text{ Ans }} & = - 6\end{align} (iii) $$3a + 5 - 8a + 1$$ \begin{align}= - 5a + 6\end{align} Now putting value of $$a = - 1$$ \begin{align}&= \left( { - 5\,\, \times - 1} \right) + 6\\ & = 5 + 6\\{\text{ Ans }} & = 11\end{align} (iv) $$10 - 3b - 4 - 5b$$ \begin{align}& = - 8b + 6\end{align} Now putting value of $$b=- 2$$ \begin{align}& = ( - 8 \times - 2) + 6\\& = 16 + 6\\{\text{ Ans }} & = 22\end{align} (v) $$2a - 2b - 4 - 5 + a$$ \begin{align}& = 3a - 2b - 9\end{align} Now putting value of $$a = - 1$$ and $$b = - 2$$ \begin{align} &\left( {3 \times - 1} \right) - \left( {2 \times - 2} \right) - 9\\& = - 3 - ( - 4) - 9\\ & = - 3 + 4 - 9\\{\text{Ans}} & = - 8\,\end{align} Chapter 12 Ex.12.3 Question 8 (i) If $$z = 10$$, find the value of $$z^3 – 3(z – 10)$$. (ii) If $$p = \,– 10$$, find the value of $$p^2 – 2p\, – 100$$ Solution Steps: First simplify the expression \begin{align}={{z}^{3}}-3z+30\end{align} Now putting value of $$z=10$$ \begin{align} &={{\left( 10 \right)}^{3}}-\left( 3\times 10 \right)+30 \\&=1000-30+30 \\\text{ Ans } &=1000 \\\end{align} (ii) If $$p = \,– 10$$, find the value of $$p^2 – 2p \,– 100$$ Put value of $$p = -10$$ to solve the expression \begin{align}&= {\left( { - 10} \right)^2} - \left( {2 \times - 10} \right) - 100\\&= 100 + 20 - 100\\{\text{Ans}} &= 20\end{align} Chapter 12 Ex.12.3 Question 9 What should be the value of $$a$$ if the value of $$2x^2 + x – a$$ equals to $$5$$, when $$x = 0$$? Solution Steps: Given that $$2x^2 + x – a = 5$$ Also, value of $$x$$ is $$0$$ So, \begin{align}2 \times {0^2} + 0 - a &= 5\\ 0- a &= 5\\ - a &= 5\\{\rm{ Ans: }}\; a &= - 5\end{align} Chapter 12 Ex.12.3 Question 10 Simplify the expression and find its value when $$a = 5$$ and $$b = \,– 3$$. Solution Steps: \begin{align}&2\left( {{a^2} + ab} \right){\rm{ }} + {\rm{ }}3{\rm{ }}-ab\\&= 2{a^2} + 2ab + 3 - ab\\&= 2{a^2} + ab + 3\\&= (2 \times 5 \times 5) + (5 \times - 3) + 3\\&= 50 - 15 + 3\\&= 38\end{align} Related Sections Related Sections
FutureStarr 6 in Fraction Form OR ## 6 in Fraction Form I’m going to show you a new way of thinking about science that might help you understand the world in a whole new way. ### Math Shopping taxes and discounts are two calculations consumers need to understand. Learn the connection between math and shopping, examine examples and calculations for taxes and discounts, and review examples of tax as well as tax and discount calculations. Learn about mathematical operations with integers using the operations of addition, subtraction, multiplication, and division. Discover the properties of integers and how those properties affect the solution to different types of math problems. (Source: study.com) ### Step For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(12, 6) = 12. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 12 × 6 = 72. In the following intermediate step, cancel by a common factor of 2 gives 7/6. The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression. (Source: www.hackmath.net) ### Add An alternative method for finding a common denominator is to determine the least common multiple (LCM) for the denominators, then add or subtract the numerators as one would an integer. Using the least common multiple can be more efficient and is more likely to result in a fraction in simplified form. In the example above, the denominators were 4, 6, and 2. The least common multiple is the first shared multiple of these three numbers. Unlike adding and subtracting integers such as 2 and 8, fractions require a common denominator to undergo these operations. One method for finding a common denominator involves multiplying the numerators and denominators of all of the fractions involved by the product of the denominators of each fraction. Multiplying all of the denominators ensures that the new denominator is certain to be a multiple of each individual denominator. The numerators also need to be multiplied by the appropriate factors to preserve the value of the fraction as a whole. This is arguably the simplest way to ensure that the fractions have a common denominator. However, in most cases, the solutions to these equations will not appear in simplified form (the provided calculator computes the simplification automatically). Below is an example using this method. (Source: www.calculator.net) ## Related Articles • #### A 40 Calculator August 11, 2022 \| sheraz naseer • #### Add 3 Percent to a Number August 11, 2022 \| sheraz naseer • #### How many cups in a pound August 11, 2022 \| Future Starr • #### 7 Is What Percent of 14 August 11, 2022 \| sheraz naseer • #### 18 to 26 Percent Increase ORR August 11, 2022 \| Bilal Saleem • #### What Percent Is 17 Out of 40 OR August 11, 2022 \| Muhammad Waseem • #### 14 As a Percentage of 40 ORR August 11, 2022 \| Bilal Saleem • #### 28 Is What Percent of 80 OR August 11, 2022 \| Abid Ali • #### How much does a gallon of water weigh? August 11, 2022 \| Future Starr • #### Ab X Calculator August 11, 2022 \| sheraz naseer • #### Lease Payment on 38000 Car August 11, 2022 \| Muhammad Umair • #### 3 4 Halved, August 11, 2022 \| Jamshaid Aslam • #### A Caculator: August 11, 2022 \| Abid Ali • #### A I Want to Use a Calculator August 11, 2022 \| Muhammad Waseem • #### Online Calculator With Fraction Function August 11, 2022 \| Muhammad Umair
# How to Conduct Normal Distribution Calculations This guide will show you how to conduct normal distribution calculate the probability (area under the curve) of a standard normal distribution. It will first show you how to interpret a Standard Normal Distributions Table. It will then show you how to calculate the: • probability less than a z-value • probability greater than a z-value • probability between z-values • probability outside two z-values. We have a calculator that calculates probabilities based on z-values for all the above situations. In addition, it also outputs all the working to get to the answer, so you know the logic of how to calculate the answer. ## How to Use the Standard Normal Distribution Table to conduct normal distribution The most common form of standard normal distribution table that you see is a table similar to the one below (click image to enlarge): The Standard Normal Distribution Table The standard normal distribution table provides the probability that a normally distributed random variable Z, with mean equal to 0 and variance equal to 1, is less than or equal to z. It does this for positive values of z only (i.e., z-values on the right-hand side of the mean). What this means in practice is that if someone asks you to find the probability of a value being less than a specific, positive z-value, you can simply look that value up in the table. We call this area Φ. Thus, for this table, P(Z < a) = Φ(a), where a is positive. Diagrammatically, the probability of Z less than ‘a’ being Φ(a), as determined from the standard normal distribution table, is shown below: ## Probability less than a z-value P(Z < –a) As explained above, the standard normal distribution table only provides the probability for values less than a positive z-value (i.e., z-values on the right-hand side of the mean). So how do we calculate the probability below a negative z-value (as illustrated below)? We start by remembering that the standard normal distribution has a total area (probability) equal to 1 and it is also symmetrical about the mean. Thus, we can do the following to calculate negative z-values: we need to appreciate that the area under the curve covered by P(Z > a) is the same as the probability less than –a {P(Z < –a)} as illustrated below: Making this connection is very important because from the standard normal distribution table, we can calculate the probability less than ‘a’, as ‘a’ is now a positive value. Imposing P(Z < a) on the above graph is illustrated below: From the above illustration, and from our knowledge that the area under the standard normal distribution is equal to 1, we can conclude that the two areas add up to 1. We can, therefore, make the following statements: Φ(a) + Φ(–a) = 1 ∴ Φ(–a) = 1 – Φ(a) Thus, we know that to find a value less than a negative z-value we use the following equation: Φ(–a) = 1 – Φ(a), e.g. Φ(–1.43) = 1 – Φ(1.43) Join the 10,000s of students, academics and professionals who rely on Laerd Statistics. ## Probability greater than a z-value P(Z > a) The probability of P(Z > a) is: 1 – Φ(a). To understand the reasoning behind this look at the illustration below: You know Φ(a) and you know that the total area under the standard normal curve is 1 so by mathematical deduction: P(Z > a) is: 1 – Φ(a). P(Z > –a) The probability of P(Z > –a) is P(a), which is Φ(a). To understand this we need to appreciate the symmetry of the standard normal distribution curve. We are trying to find out the area below: But by reflecting the area around the centre line (mean) we get the following: Notice that this is the same size area as the area we are looking for, only we already know this area, as we can get it straight from the standard normal distribution table: it is P(Z < a). Therefore, the P(Z > –a) is P(Z < a), which is Φ(a). ## Probability between z-values You are wanting to solve the following: The key requirement to solve the probability between z-values is to understand that the probability between z-values is the difference between the probability of the greatest z-value and the lowest z-value: P(a < Z < b) = P(Z < b) – P(Z < a) which is illustrated below: P(a < Z < b) The probability of P(a < Z < b) is calculated as follows. First separate the terms as the difference between z-scores: P(a < Z < b) = P(Z < b) – P( Z < a) (explained in the section above) Then express these as their respective probabilities under the standard normal distribution curve: P(Z < b) – P(Z < a) = Φ(b) – Φ(a). Therefore, P(a < Z < b) = Φ(b) – Φ(a), where a and b are positive. P(–a < Z < b) The probability of P(–a < Z < b) is illustrated below: Need Help with Researchers or Data Analysts, Lets Help you with Data Analysis & Result Interpretation for your Project, Thesis or Dissertation? We are Experts in SPSS, EVIEWS, AMOS, STATA, R, and Python First separate the terms as the difference between z-scores: P(–a < Z < b) = P(Z < b) – P(Z < –a) Then express these as their respective probabilities under the standard normal distribution curve: P(Z < b) – P(Z < –a) = Φ(b) – Φ(–a) = Φ(b) – {1 – Φ(a)}P(Z < –a) explained above. ∴ P(–a < Z < b) = Φ(b) – {1 – Φ(a)}, where a is negative and b is positive. P(–a < Z < –b) The probability of P(–a < Z < –b) is illustrated below: First separate the terms as the difference between z-scores: P(–a < Z < –b) = P(Z < –b) – P( Z < –a) Then express these as their respective probabilities under the standard normal distribution curve: P(Z < b) – P(Z < –a) = Φ(–b) – Φ(–a) = {1 – Φ(b)} – {1 – Φ(a)} P(Z < –a) explained above. = 1 – Φ(b) – 1 + Φ(a) = Φ(a) – Φ(b) The above calculations can also be seen clearly in the diagram below: Notice that the reflection results in a and b “swapping positions”. ## Probability outside of a range of z-values An illustration of this type of problem is found below: To solve these types of problems, you simply need to work out each separate area under the standard normal distribution curve and then add the probabilities together. This will give you the total probability. When a is negative and b is positive (as above) the total probability is: P(Z < –a) + P(Z > b) = Φ(–a) + {1 – Φ(b)} P(Z > b) explained above. = {1 – Φ(a)} + {1 – Φ(b)} P(Z < –a) explained above. = 1 – Φ(a) + 1 – Φ(b) = 2 – Φ(a) – Φ(b) When a and b are negative as illustrated below: The total probability is: P(Z < –a) + P(Z > –b) = Φ(–a) + Φ(b)P(Z > –b) explained above. = {1 – Φ(a)} + Φ(b)P(Z < –a) explained above. = 1 + Φ(b) – Φ(a) When a and b are positive as illustrated below: The total probability is: P(Z < a) + P(Z > b) = Φ(a) + {1 – Φ(b)}P(Z > b) explained above. = 1 + Φ(a) – Φ(b) ##### Need Our Services? ###### Econometrics & Statistics Modelling Services Need Help, Whatsapp Us Now
Course Content Class 6th Science 0/32 Class 6th Social Science 0/58 Class 6th Maths 0/14 Class 6th English Honeysuckle 0/20 Class 6 English Honeysuckle Poem 0/16 Class 6 English A Pact with the Sun 0/20 Class 6th Online Course For English Medium CBSE Board Students ### NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.1 Ex 1.1 Class 6 Maths Question 1. Fill in the blanks: (a) 1 lakh = ………….. ten thousand. (b) 1 million = ………… hundred thousand. (c) 1 crore = ………… ten lakh. (d) 1 crore = ………… million. (e) 1 million = ………… lakh. Solution: (a) 1 lakh = ten ten thousand. (b) 1 million = ten hundred thousand. (c) 1 crore = ten ten lakh (d) 1 crore = ten million (e) 1 million = ten lakh Ex 1.1 Class 6 Maths Question 2. Place commas correctly and write the numerals: (a) Seventy-three lakh seventy-five thousand three hundred seven. (b) Nine crore five lakh forty-one. (c) Seven crore fifty-two lakh twenty-one thousand three hundred two. (d) Fifty-eight million four hundred twenty- three thousand two hundred two. (e) Twenty-three lakh thirty thousand ten. Solution: (a) 73,75,307 (b) 9,05,00,041 (c) 7,52,21,302 (d) 5,84,23,202 (e) 23,30,010. Ex 1.1 Class 6 Maths Question 3. Insert commas suitably and write the names according to Indian System of Numeration: (a) 87595762 (b) 8546283 (c) 99900046 (d) 98432701 Solution: (a) 8,75,95,762 (Eight crore seventy-five lakh ninety-five thousand seven hundred sixty- two) (b) 85,46,283 (Eighty-five lakh forty-six thousand two hundred eighty-three) (c) 9,99,00,046 (Nine crore ninety-nine lakh forty-six) (d) 9,84,32,701 (Nine crore eighty-four lakh thirty-two thousand seven hundred one) Ex 1.1 Class 6 Maths Question 4. Insert commas suitably and write the names according to International System of Numeration: (a) 78921092 (b) 7452283 (c) 99985102 (d) 48049831 Solution: (a) 78,921,092 (Seventy-eight million nine hundred twenty-one thousand ninety-two) (b) 7,452,283 (Seven million four hundred fifty- two thousand two hundred eighty-three) (c) 99,985,102 (Ninety-nine million nine hundred eighty-five thousand one hundred two) (d) 48,049,831 (Forty-eight million forty-nine thousand eight hundred thirty-one) ### NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2 Exercise 1.2 Ex 1.2 Class 6 Maths Question 1. A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days. Solution: Number of tickets sold on the first day = 1094 Number of tickets sold on the second day = 1812 Number of tickets sold on the third day = 2050 Number of tickets sold on the final day = 2751 ∴ Total number of tickets sold on all the four days = 1094 + 1812 + 2050 + 2751 = 7,707. Ex 1.2 Class 6 Maths Question 2. Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need? Solution: Shekhar has so far scored 6980 runs He wishes to complete 10,000 runs. Therefore total number of runs needed by him = 10,000 – 6980 = 3020 runs Ex 1.2 Class 6 Maths Question 3. In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election? Solution: Number of votes secured by the successful candidate = 5,77,500 Number of votes secured by his nearest rival = 3,48,700 Therefore, margin of votes to win the election = 5,77,500 – 3,48,700 = 2,28,800 Ex 1.2 Class 6 Maths Question 4. Kirti bookstore sold books worth Rs 2,85,891 in the first week of June and books worth Rs 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much? Solution: Books sold in first week of June worth ₹2,85,891 Books sold in second week of the month worth ₹4,00,768 Therefore, total sale of books in the two weeks together = Rs 2,85,891 + Rs 4,00,768 = Rs6,86,659 In the second week of the month, the sale of books was greater. Difference of the sale of books = Rs 4,00,768 – Rs 2,85,891 = Rs 1,14,877 Hence, in second week of june, the sale of books was more by ₹1,14,877. Ex 1.2 Class 6 Maths Question 5. Find the difference between the greatest and the least numbers that can be written using the digits 6, 2, 7, 4, 3 each only once. Solution: Given digits are 6, 2, 7, 4, 3 Greatest number = 76432 Least number = 23467 Therefore, difference = 76432 – 23467 = 52,965 Ex 1.2 Class 6 Maths Question 6. A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January, 2006? Solution: Number of screws manufactured in a day = 2,825. Number of screws manufactured in month of January = 31 x 2825 = 87,575 Ex 1.2 Class 6 Maths Question 7. A merchant had ₹78,592 with her. She placed an order for purchasing 40 radio sets at ₹1200 each. How much money will remain with her after the purchase? Solution: Amount of money with the merchant = ₹78,592 Number of radio sets = 40 Price of one radio set = ₹1200 Therefore, cost of 40 radio sets = ₹1200 x 40 = ₹48,000 Remaining money with the merchant = ₹78,592 – ₹48000 = ₹30,592 Hence, amount of ₹30,592 will remain with her after purchasing the radio sets. Ex 1.2 Class 6 Maths Question 8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer? Solution: Student has multiplied 7236 by 65 instead of multiplying by 56. Difference between the two multiplications = (65 – 56) x 7236 = 9 x 7236 = 65124 (We don’t need to do both the multiplied) Ex 1.2 Class 6 Maths Question 9. To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain? Ex 1.2 Class 6 Maths Question 11. The distance between the school and the house of a student is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days. Solution: Distance between school and house = 1 km 875 m = (1000 + 875) m = 1875 m. Distance travelled by the student in both ways = 2 x 1875 = 3750 m Distance travelled in 6 days = 3750 m x 6 – 22500 m = 22 km 500 m. Hence, total distance covered in six days = 22 km 500 m. Ex 1.2 Class 6 Maths Question 12. A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 mL capacity, can it be filled? – Solution: Quantity of curd in a vessel = 4 1 500 mL = (4 x 1000 + 500) mL = 4500 mL. Capacity of 1 glass = 25 mL Therefore number of glasses = 450025 = 180 ### NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3 Exercise 1.3 Ex 1.3 Class 6 Maths Question 1. Estimate each of the following using general rule: (a) 730 + 998 (b) 796 – 314 (c) 12,904 + 2,888 (d) 28,292 – 21,496 Make ten more such examples of addition, subtraction and estimation of their outcome. Solution: (a) 730 + 998 Rounding off 730 nearest to hundreds = 700 Rounding off 998 nearest to hundreds = 1,000 ∴ 730 + 998 = 700 + 1000 = 1700 (b) 796 – 314 Rounding off 796 nearest to hundreds = 800 Rounding off 314 nearest to hundreds = 300 ∴ 796 – 314 = 800 – 300 = 500 (c) 12,904 + 2,888 Rounding off 12,904 nearest to thousands = 13000 Rounding off 2888 nearest to thousands = 3000 ∴ 12,904 + 2,888 = 13000 + 3000 = 16000 (d) 28,292 – 21,496 Rounding off 28,292 nearest to thousands = 28,000 Rounding off 21,496 nearest to thousands = 21,000 ∴ 28,292 – 21,496 = 28,000 – 21,000 = 7,000 Example 1: 1210 + 2365 = 1200 + 2400 = 3600 Example 2: 3853 + 6524 = 4000 + 7000 = 11,000 Example 3: 8752 – 3654 = 9,000 – 4,000 = 5,000 Example 4: 4538 – 2965 = 5,000 – 3,000 = 2,000 Example 5: 1927 + 3185 = 2000 + 3,000 = 5,000 Example 6: 3258 – 1698 = 3000 – 2000 = 1,000 Example 7: 8735 + 6232 = 9000 + 6000 = 15,000 Example 8: 1038 – 1028 = 1000 – 1000 = 0 Example 9: 6352 + 5830 = 6,000 + 6,000 = 12,000 Example 10: 9854 – 6385 = 10,000 – 6000 = 4,000 Ex 1.3 Class 6 Maths Question 2. Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens): (a) 439 + 334 + 4,317 (b) 1,08,734-47,599 (c) 8,325-491 (d) 4,89,348-48,365 Make four such examples: Solution: (a)439 + 334 + 4,317 (i) Rough estimate (Rounding off to nearest hundreds) 439 + 334 + 4,317 = 400 + 300 + 4300 = 5,000 (ii) Closer estimate (Rounding off to nearest tens) 439 + 334 + 4317 = 440 + 330 + 4320 = 5090. (b) 1,08,734 – 47,599 (i) Rough estimate (Rounding off to nearest hundreds) 1,08,734 – 47,599 = 1,08,700 – 47,600 = 61,100 (ii) Closer estimate (Rounding off to nearest tens) 1,08,734 – 47,599 = 1,08,730 – 47,600 = 61,130. (c) 8325 – 491 (i) Rough estimate (Rounding off to nearest hundreds) 8325 – 491 = 8300 – 500 = 7800 (ii) Closer estimate (Rounding off to nearest tens) 8325 – 491 = 8330 – 490 = 7840. (d) 4,89,348 – 48,365 (i) Rough estimate (Rounding off to nearest hundreds) 4,89,348 – 48,365 = 4,89,300 – 48,400 = 4,40,900 (ii) Closer estimate (Rounding off to nearest tens) 4,89,348 – 48,365 = 4,89,350 – 48,370 = 4,40,980 Example 1: 384 + 562 Solution: (i) Rough estimate (Rounding off to nearest hundreds) 384 + 562 = 400 + 600 = 1,000 (ii) Closer estimate (Rounding off to nearest tens) 384 + 562 = 380 + 560 = 940 Example 2: 8765 – 3820 Solution: (i) Rough estimate (Rounding off to nearest hundreds) 8765 – 3820 = 8800 – 3900 = 4900 (ii) Closer estimate (Rounding off to nearest tens) 8765 – 3820 = 8770 – 3820 = 4950 Example 3: 6653 – 8265 Solution: (i) Rough estimate (Rounding off to nearest hundreds) 6653 + 8265 = 6700 + 8300 = 15,000 (ii) Closer estimate (Rounding off to nearest tens) 6653 + 8265 = 6650 + 8270 = 14920 Example 4: 3826 – 1262 Solution: (i) Rough estimate (Rounding off to nearest hundreds) 3826 – 1262 = 3800 – 1300 = 2500 (ii) Closer estimate (Rounding off to nearest tens) 3826 – 1262 = 3830 – 1260 = 2570 Ex 1.3 Class 6 Maths Question 3. Estimate the following products using general rule: (a) 578 x 161 (b)5281 x 3491 (c) 1291 x 592 (d) 9250 x 29 Make four more such examples. Solution: (a) 578 x 161 = 600 x 200 = 1,20,000 (b) 5281 x 3491 = 5000 x 3000 = 1,50,00,000 (c) 1291 x 592 = 1300 x 600 = 7,80,000 (d) 9250 x 29 = 9000 x 30 = 2,70,000 Example 1. 382 x 1062 Solution: 382 x 1062 = 400 x 1000 = 4,00,000 Example 2. 6821 x 1291 Solution: 6821 x 1291 = 7000 x 1000 = 70,00,000 Example 3. 3858 x 9350 Solution: 3858 x 9350 = 4000 x 9000 = 3,60,00,000 Example 4. 3405 x 7502 Solution: 3405 x 7502 = 3000 x 8000 = 2,40,00,000 Exercise Files Chapter 1 Knowing Our Numbers.pdf Size: 331.81 KB Join the conversation
# Factors of 64 Definition of factors of 64: If a number completely divides 64 without leaving a remainder, then that number is called a factor of 64. Thus, we can say that the factors of 64 are the divisors of 64. Here, we will learn about the factors of 64 and the prime factors of 64. We have: • 64 = 2×2×2×2×2×2 =26 is the prime factorization of 64. • The factors of 64 are 1, 2, 4, 8, 16, 32, and 64. • The prime factor of 64 is 2 only • Negative factors of 64 are –1, -2, -4, -8, -16, -32, and –64. ## What are the Factors of 64? Let us write the number 64 multiplicatively in all possible ways. We definitely have: 64 = 1×64 64 = 2×32 64 = 4×16 64 = 8×8 As there are no other ways we can write 64 multiplicatively, so we will stop right now. So the factors of 64 in pairs are given as follows: Thus the pair factors of 64 are (1, 64), (2, 32), (4, 16), and (8, 8). As all the numbers appearing in pair factors are the factors of 64, we conclude that ## Number of Factors of 64 From above we have calculated the factors of 64 which are 1, 2, 4, 8, 16, 32, and 64. Thus the total number of factors of 64 is seven. ## Prime Factors of 64 Note that the factors of 64 are 1, 2, 4, 8, 16, 32, and 64. Among those factors, we observe that only 2 is a prime number as it does not have any proper divisors. ∴ the only prime factor of 64 is 2. Question: What are the factors of 64? Video Solution: ## How to Find Factors of 64? Now we will determine the factors of 64 by division method. In this method, we will try to find the numbers that can divide 64 with no remainder. See that 64/1=64 and the remainder is 0. 1 and 64 are factors of 64. 64/2=32 and the remainder is 0. 2 and 32 are factors of 64. 64/4=16 and the remainder is 0. 4 and 16 are factors of 64. 64/8=8 and the remainder is 0. 8 and 8 are factors of 64. ⇒ 8 is a factor of 64. Note that no numbers other than the numbers in violet color can divide 64. So the numbers in violet color, that is, 1, 2, 4, 8, 16, 32, and 64 are the complete list of factors of 64.
## Prime Factorization Given an integer greater than 1, if it’s only positive factors are 1 and itself, then it is called a prime number. If it has a positive factor different from 1 or itself, then it is called a composite number. 1 is neither prime nor composite. Prime factors If a positive integer a has a factor b and b is a prime number, we call b a prime factor of a. For example, 2 and 3 are the prime factors of 12. 4 and 6 are also factors of 12 but they are not prime. Prime factorization Writing a number as the product of all its prime factors is called forming the prime factorization of the number. For instance, the prime factorization of 12 is 12 = 2 × 2 × 3. Euclid’s Lemma For any two positive integers a, b and a prime number p, if p \| ab, then either p \| a or p \| b. Euclid’s Lemma can be used to prove: Fundamental Theorem of Arithmetic Every integer > 1 can be factorized into primes in exactly one way. This means that for every integer n > 1, it can be uniquely expressed as a product of primes Knowing the prime factorization of a number tells us a lot about the number. For instance, for any divisor d of n, Euclid’s Lemma tells us that every prime factor of d must also be a prime factor of n, and so d must be of the form Since each bᵢ lies between 0 and aᵢ (inclusive), there are aᵢ + 1 possible choices for b. By the Fundamental Theorem of Arithmetic, different choices of bᵢ s correspond to different divisors, and each divisor arises from some choice of bᵢ s. Hence n has (a₁ + 1)(a₂ + 1)…(aₙ + 1) different divisors. Example: 180 = 2² ∙ 3² ∙ 5¹ should have (2 + 1)(2 + 1)(1 + 1) = 18 different divisors. Indeed, listing them out: n is a perfect square precisely when all the aᵢ s are even. For example, 180 isn’t a perfect square since the power of 5 in its prime factorization is 1. However, 900 = 2² 3² 5² = (2¹ 3¹ 5¹)² = 30² is a perfect square. Common divisors and greatest common divisor Let n₁, n₂, …, nₖ (k ≥ 2) be n positive integers. If an integer d divides each one of them, i.e. d \| n₁, d \| n₂, … , d \| nₖ, then we call d a common divisor (or equivalently a common factor) of n₁, n₂, …, nₖ. If d is the largest one among all the common divisors, we call it the greatest common divisor and write gcd(n₁, n₂, …, nₖ) = d, or just (n₁, n₂, …, nₖ) = d to denote this. For instance, (18, 30, 66) = 6, since the common factors of 18, 30 and 66 are 1, 2, 3, 6. Suppose n, m can be expressed as (Note that these are not necessarily prime factorizations, the aᵢ s and bᵢ s may be 0.) Then where min(aᵢ, bᵢ) is the smaller of aᵢ and bᵢ. That’s because min(aᵢ, bᵢ) is the largest power of pᵢ that divides both the term in m and the term in n. Coprimality Let m, n be integers. If (m, n) = 1, i.e. m and n have no positive divisors in common besides 1, then we say that m and n are coprime. Common multiples and least common multiple Let n₁, n₂, …, nₖ (k ≥ 2) be n positive integers. If an integer m is a multiple for all of them, i.e. n₁ \| m, n₂ \| m, …, nₖ \| m, then we call m a common multiple of n₁, n₂, …, nₖ. If m is the smallest among all the common multiples, we call m the least common multiple and write lcm(n₁, n₂, …, nₖ) = m, or sometimes [n₁, n₂, …, nₖ] = m to denote this. Again, if m and n can be expressed as Then where max(aᵢ, bᵢ) is the larger of aᵢ and bᵢ. This is because max(aᵢ, bᵢ) is the smallest power of pᵢ that is a multiple of both the term in m and the term in n. Relationship between greatest common divisor and least common multiple For any two positive integers n, m, we have (n, m) [n, m] = nm. This is because for each prime pᵢ, the pᵢ term on the LHS is and the pᵢ term on the RHS is . The powers on both sides match up:
Operation I # Operations with Fractions: Multiplication and Division ## Multiply any fraction by a whole number using models and symbols (e.g., 2 × 4⁄5) Multiplying a fraction by a whole number is like skip-counting by that fraction. For example, 5 × 12 can be thought of as counting on: 12, 1, 112, 2, 212 . When students understand that the numerator is the count, they can reason that multiplication by a whole number will increase the count without changing the denominator (fractional unit). Multiplying fractions by whole numbers as a starting point also helps students draw on their prior understanding of multiplication with whole numbers to establish a meaningful foundation for understanding multiplication with fractions. ### Background Multiplication and division involving fractions is widely recognized to be more complex than multiplication and division with whole numbers. Often algorithms are introduced with little emphasis on understanding the mathematics behind the algorithm. There is some research evidence that suggests that early emphasis on procedures over concepts can actually impede students' fractions understanding in the long term (Brown and Quinn, 2007). However, with precise instruction, students can develop both conceptual and procedural knowledge. Multiplication with fractions may increase or decrease a quantity, or leave it unchanged. Multiplication with whole numbers is often connected to repeated addition; however, with fractions, other interpretations, such as the Cartesian product or area mode, support increased understanding. Consider 113×12. Since we are multiplying a quantity by half, we should expect the result to be less than 113. We can carefully labelled array or Cartesian model to solve this. First, we model 113 Then we partition it into half. Not that we could do this a number of ways (including partitioning each third vertically in half) but the example below connects the use of the array (length and width) to model the multiplication Determine the answer from the intersecting area. We can see that each whole has been partitioned into sixths. There are four sixths in the overlapping area. 113 × 12 = 46 Ontario fractions research revealed several effective strategies for teaching multiplication of fractions with an emphasis on meaning making. See Fractions Operations; Multiplication and Division Literature Review for more details. Recommend best practices based on the research include: 1. Take time to focus on conceptual understanding. Instruction should start with contextual problems which allow students to focus on what they are solving rather than how they are solving, and helps them to develop the meaning of the operations. They also should have hands-on experiences with materials such as relational rods, paper folding and visual models like number lines. 2. 3. Recognize and draw on students' informal and formal knowledge as well as prior experiences with fractions. Students have prior experience with fractions, including partitioning different models (which is helpful when solving multiplication problems). If they have had opportunities to engage in learning experiences earlier in the Fractions Learning Pathways, they will also have an understanding of the unit fraction, which helps establish an early understanding of multiplication connected to repeated addition (e.g., ). As well, students can draw upon their understanding of the various constructs of fractions (i.e., number, part-whole, part-part, quotient, operator and linear measure) and their work with equivalent fractions. 4. 5. Draw on student familiarity with whole number operations. Specific connections to multiplication with whole numbers should be made, as the properties hold true for multiplication with fractions as well. 1. The Commutative Property: a × b = b × a 2. The Associative Property: (a × b) × c = a × (b × c) 3. The Identity Property: a × 1 = a 4. The Distributive Property a × (b + c) = a × b + a × c 5. As well, students should understand that multiplication is the inverse of division and vice versa. That is to say that if a × b = c, then a = c ÷ b and b = c ÷ a. This is true for a, b, c ≠ 0. For example, although 3 × 0 = 0, 0 ÷ 0 ≠ 3. Of couse, some generalizations that hold true for whole numbers don't hold true for fractions, and this can be a source of confusion for students. For example, many students hold onto the idea that multiplication "makes a number larger", which is true with whole numbers. However, when we are multiplying by a quantity that is less than 1, as in the case of fractions, we "shrink" the quantity instead. 6. Include carefully selected and multiple representations to convey meaning. There are a few representations which have longevity and are particularly helpful for building meaning across number systems and contexts - the number line and the area model (or Cartesian or array). Using these representations across fractions learning allows students to understand the structure of the representations and use them effectively to solve questions. Models help students understand the mathematics better - both conceptually and procedurally - and enhance their retention of the learning. ### References Brown, G., & Quinn, R.J. (2007). Investigating the relationship between fraction proficiency and success in algebra. Australian Mathematics Teacher, 63(4): 8- 15.
Top Side Angle Side Calculator Top Side angle side calculator is used for finding the area of a triangle. Although it uses the trigonometry sine function, it works for any triangle not only just for right triangles. If height is not mentioned then we can go for trigonometry sine function which is has follows. ## Step by Step Calculation Step 1 : SAS Formula: Area = $\frac{ab\ SinC}{2}$ Where 'a' and 'b' are two sides for the triangle C is included angle (i.e. the angles between a two known sides) Step 2 : Put the values in the formula and calculate it further. ## Example Problems 1. ### Find the area of the triangle with two sides given that is 8 cm, 6 cm and angle of 300. Step 1 : Given: Side A = 8 cm Side B = 6 cm Angle C = 30 degree Formula: Area = $\frac{ab\ SinC}{2}$ Step 2 : Put the values in the formula and calculate it further Area = $\frac{8 * 6 * Sin 30}{2}$ Area = 0.5 * 0.5 * 8 * 6 Area = 12 cm2 Area of the Triangle = 12 cm2 2. ### Find the area of the triangle with two sides given that 10 cm, 12 cm and angle = 1200. Step 1 : Given: Side A = 12 cm Side B = 10 cm Angle = 1200. Formula: Area = $\frac{ab\ SinC}{2}$ Step 2 : Put the values in the formula and calculate it further Area = $\frac{12 * 10 * Sin 120}{2}$ Area = 30 cm2 * 1.71 Area = 51.96 cm2
# Class 7 Maths Chapter 1 Exercise 1.2 Pdf Notes NCERT Solutions Class 7 Maths Chapter 1 Integers Exercise 1.2 pdf notes:- Exercise 1.2 Class 7 maths Chapter 1 Pdf Notes:- ## Ncert Solution for Class 6 Maths Chapter 1 Integers Exercise 1.2 Tips:- Introduction:- We have learnt that sum of two whole numbers is again a whole number. For example, 17 + 24 = 41 which is again a whole number. We know that, this property is known as the closure property for addition of the whole numbers. Commutative Property We know that 3 + 5 = 5 + 3 = 8, that is, the whole numbers can be added in any order. In other words, addition is commutative for whole numbers. Can we say the same for integers also? We have 5 + (β 6) = β1 and (β 6) + 5 = β1 So, 5 + (β 6) = (β 6) + 5 Are the following equal? (i) (β 8) + (β 9) and (β 9) + (β 8) (ii) (β 23) + 32 and 32 + (β 23) (iii) (β 45) + 0 and 0 + (β 45) Try this with five other pairs of integers. Do you find any pair of integers for which the sums are different when the order is changed? Certainly not. We say that addition is commutative for integers. In general, for any two integers a and b, we can say a + b = b + a ο¬ We know that subtraction is not commutative for whole numbers. Is it commutative for integers? Consider the integers 5 and (β3). Is 5 β (β3) the same as (β3) β5? No, because 5 β ( β3) = 5 + 3 = 8, and (β3) β 5 = β 3 β 5 = β 8. Take atleast five different pairs of integers and check this. We conclude that subtraction is not commutative for integers. 1.3.4 Associative Property Observe the following examples: Consider the integers β3, β2 and β5. Look at (β5) + [(β3) + (β2)] and [(β5) + (β3)] + (β2). In the first sum (β3) and (β2) are grouped together and in the second (β5) and (β3) are grouped together. We will check whether we get different results. TRY THESE In both the cases, we get β10. i.e., (β5) + [(β3) + (β2)] = [(β5) + (β2)] + (β3) Similarly consider β3 , 1 and β7. ( β3) + [1 + (β7)] = β3 + _ = _ [(β3) + 1] + (β7) = β2 + _ = _ Is (β3) + [1 + (β7)] same as [(β3) + 1] + (β7)? Take five more such examples. You will not find any example for which the sums are different. Addition is associative for integers. In general for any integers a, b and c, we can say a + (b + c) = (a + b) + c When we add zero to any whole number, we get the same whole number. Zero is an additive identity for whole numbers. Is it an additive identity again for integers also? Observe the following and fill in the blanks: (i) (β 8) + 0 = β 8 (ii) 0 + (β 8) = β 8 (iii) (β23) + 0 = _ (iv) 0 + (β37) = β37 (v) 0 + (β59) = (vi) 0 + = β 43 (vii) β 61 + = β 61 (viii) + 0 = _ The above examples show that zero is an additive identity for integers. You can verify it by adding zero to any other five integers. In general, for any integer a a + 0 = a = 0 + a 1. Write a pair of integers whose sum gives (a) a negative integer (b) zero (c) an integer smaller than both the integers. (d) an integer smaller than only one of the integers. (e) an integer greater than both the integers. 2. Write a pair of integers whose difference gives (a) a negative integer. (b) zero. (c) an integer smaller than both the integers. (d) an integer greater than only one of the integers. (e) an integer greater than both the integers #### Test Paper Of Class 8th • Maths 8th Class • Science 8th class • Sst 8th Class • #### Test Paper Of Class 7th • Maths 7th Class • Science 7th class • #### Test Paper Of Class 6th • Maths 6th Class • Science 6th class
ANGLE MEASURES IN TRIANGLES WORKSHEET About "Angle measures in triangles worksheet" Angle measures in triangles worksheet : Worksheet given in this section is much useful to the students who would like to practice problems on finding missing angle measures in triangles. Angle measures in triangles worksheet - Problems Problem 1 : The measure of one acute angle of a right triangle is two times the measure of the other acute angle. Find the measure of each acute angle. Problem 2 : Find the missing angles in the triangle shown below. Problem 3 : Find the value of x in the diagram shown below. Problem 4 : Find the missing angles in the triangle shown below. Angle measures in triangles worksheet - Problems Problem 1 : The measure of one acute angle of a right triangle is two times the measure of the other acute angle. Find the measure of each acute angle. Solution : Let A, B and C be the vertices of the triangle and right angle is at C. Let ∠A = x°, then ∠B = 2x°. The diagram shown below illustrates this. By Corollary to the Triangle Sum Theorem, the acute angles of a right triangle are complementary. So, we have x° + 2x° = 90° Simplify. 3x° = 90° Divide both sides by 3. x = 30 So, m∠A = 30° and m∠B = 2(30°) = 60° Hence, the two acute angles are 30° and 60°. Problem 2 : Find the missing angles in the triangle shown below. Solution : In the triangle shown above, two sides are congruent. Angles opposite to congruent sides are always congruent. So, if one missing angle is assumed to be x°, then the other missing angle also must be x°. Because the two angles are congruent. The diagram shown below illustrates this. By Triangle Sum Theorem, the sum of the measures of the interior angles of a triangle is 180°. So, we have x° + x° + 40° = 180° Simplify. 2x + 40 = 180 Subtract 40 from both sides. 2x = 140 Divide both sides by 2. x = 70 Hence, the measure of each missing angle is 70°. Problem 3 : Find the value of x in the diagram shown below. Solution : By Exterior Angle Theorem, the measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles. So, we have x° + 65° = (2x + 10)° or x + 65 = 2x + 10 Subtract x from both sides. 65 = x + 10 Subtract 10 from both sides. 55 = x Hence, the value of x is 55. Problem 4 : Find the missing angles in the triangle shown below. Solution : In the triangle shown above, two sides are congruent. Angles opposite to congruent sides are always congruent. So, if one missing angle is assumed to be x°, then the other missing angle also must be x°. Because the two angles are congruent. The diagram shown below illustrates this. In the triangle shown above, one of the angles is right angle. So, it is right triangle. By Corollary to the Triangle Sum Theorem, the acute angles of a right triangle are complementary. So, we have x° + x° = 90° Simplify. 2x = 90 Divide both sides by 2. x = 45 Hence, the measure of each missing angle is 45°. 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# An Inequality with Constraint in Four Variables V ### Solution 1 By Hölder's inequality, \displaystyle \begin{align} &(1+a^3)(1+b^3)(1+c^3)\ge (1+abc)^3\\ &\sqrt[3]{(1+a^3)(1+b^3)(1+c^3)}\geq 1+abc\\ &\frac{1}{\sqrt[3]{(1+a^3)(1+b^3)(1+c^3)}}\leq \frac{1}{1+abc} \end{align} It follows that \displaystyle \begin{align} \sum_{cycl}\frac{abc}{\sqrt[3]{(1+a^3)(1+b^3)(1+c^3)}}&\leq \sum_{cycl}\frac{abc}{1+abc}\\ &=\sum_{cycl}\frac{1+abc-1}{1+abc}\\ &=4-\sum_{cycl}\frac{1}{1+abc} \end{align} $\displaystyle \sum_{cycl}\frac{1}{1+abc} \ge \frac{\displaystyle (1+1+1+1)^2}{\displaystyle 4+\sum_{cycl}abc}=\frac{16}{5}.$ Thus, $\displaystyle -\sum_{cycl}\frac{1}{1+abc}\leq -\frac{16}{5}.\,$ Further, \displaystyle \begin{align} &4-\sum_{cycl}\frac{1}{1+abc}\leq 4-\frac{16}{5}=\frac{4}{5}\\ &\sum_{cycl}\frac{abc}{\sqrt[3]{(1+a^3)(1+b^3)(1+c^3)}}\leq \frac{4}{5}\\ &5\sum_{cycl}\frac{abc}{\sqrt[3]{(1+a^3)(1+b^3)(1+c^3)}}\leq 4. \end{align} ### Solution 2 By Hölder's inequality, $\displaystyle \sqrt[3]{\prod_{cycl}(1+a^3)}\ge 1+abc,\,$ implying $\displaystyle 5\sum_{cycl}\frac{\displaystyle abc}{\displaystyle \sqrt[3]{\prod_{cycl}(1+a^3)}}\le5\sum_{cycl}\frac{abc}{1+abc}.$ By Jensen's inequality applied to the concave function $\displaystyle f(x)=\frac{x}{1+x},\,$ $\displaystyle 5\sum_{cycl}\frac{abc}{1+abc}\le 20\frac{frac{1}{4}}{\frac{5}{4}}=4.$ ### Acknowledgment Dan Sitaru has kindly posted a problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page and later emailed his solution in a LaTeX file. Solution 2 is by Leo Giugiuc. Arben Ajredini has independently came up with Solution 1.
# Into Math Grade 1 Module 10 Lesson 1 Answer Key Count to 120 We included HMH Into Math Grade 1 Answer Key PDF Module 10 Lesson 1 Count to 120 to make students experts in learning maths. ## HMH Into Math Grade 1 Module 10 Lesson 1 Answer Key Count to 120 I Can count forward from any number up to 120. Toss the number cubes. Make a two-digit number. Write the number in the box. Count forward from that number. Children each toss a number cube. Each pair decides how to use the numbers they toss to make a two-digit number. They write their number in a box and then count forward as they write the next five numbers. Repeat the activity two more times. Explanation: I tossed number cubes, got the number 12, 25 and 63 I started with the numbers 12, 25 and 63, counted by 1’s and wrote the next 5 numbers. Build Understanding Question 1. There are 25 toy cars inside the box and some toy cars outside the box. How can you count by ones to find how many toy cars there are in all? There are ___________ toy cars in all. Explanation: There are 25 toy cars inside the box and some toy cars outside the box I started at the number 25, counted by 1’s and wrote the next 5 numbers They are 26, 27, 28, 29 and 30 There are 30 toy cars in all. Question 2. Sally has 45 shells. Then she gets these new shells. How can you count by ones to find how many shells she has now? Sally has _________ shells now. Explanation: Sally has 45 shells Then she gets these new shells I started at the number 45, counted by 1’s and wrote the numbers till the last shell The numbers are 46, 47, 48, 49, 50, 51 and 52 Sally has 52 shells now. Turn and Talk When you count objects added to a group, how do you know what number to start with? When i count objects added to a group i started with the number of objects given in question and then counted by 1’s. Question 3. Use a counting chart to help you count by ones. A. Look at the chart. What pattern do you see? The pattern i see in the number chart is number count by 1’s from left to the right and skip count by 10’s from top to bottom. B. Find the number on the chart. Count by ones. Write the next 5 numbers. 98, _____, _____, ______, _____, _______ Explanation: I started at the number 98, counted by 1’s and wrote the next 5 numbers The numbers are 99, 100, 101, 102, 103. C. What happens when you count over 100? When i count over 100 i will get 101 That is 1 more than 100 and so on The numbers will be 3 digit numbers. Check Understanding Start at the number. Count by ones. Write the next 5 numbers. Question 1. 32, _____, _____, ______, _____, _______ Explanation: I started at the number 32, counted by 1’s and wrote the next 5 numbers The numbers are 33, 34, 35, 36, 37. Question 2. 96, _____, _____, ______, _____, _______ Explanation: I started at the number 96, counted by 1’s and wrote the next 5 numbers The numbers are 97, 98, 99, 100 and 101. Question 3. Attend to Precision There are 21 buttons in a jar and 4 buttons outside the jar. Draw the buttons that are outside the jar. Write the number below each button as you count. Find the number of buttons in all. There are ________ buttons in all. Explanation: There are 21 buttons in a jar and 4 buttons outside the jar I drew 4 more buttons I started counting at 21, counted by 1’s and wrote next 4 numbers The numbers are 22, 23, 24, 25 So, there are 25 buttons in all. Write the next 5 numbers. Question 4. 96, _____, _____, ______, _____, _______ Explanation: I started at the number 96, counted by 1’s and wrote the next 5 numbers The numbers are 97, 98, 99, 100 and 101. Question 5. 115, _____, _____, ______, _____, _______ Explanation: I started at the number 115, counted by 1’s and wrote the next 5 numbers The numbers are 116, 117, 118, 119 and 120. Question 6. Open-Ended Write a number from 60 to 90 in the box. Then write the next 5 numbers.
Tall and short \| Maths Learning Concepts and Worksheets Measurement # Tall and Short Objects for Class 1 Math The discussion below will help students build the concept of tall and short objects and inform them about misconceptions. Also, the students will be able to differentiate between tall and short. From this learning concept, they will learn to identify the • Taller and shorter, among two or more objects • Tallest and shortest among two or more objects Each concept is explained to class 1 Maths students with examples, illustrations, and a concept map given to summarize the idea. At the end of the page, two printable worksheets with solutions are attached for students to practice. ## Tall and Short To describe the objects based on their heights, we use the words tall and short. ### Example: In the given picture, the height of the giraffe is more, and the height of the deer is less. So, the giraffe is tall and the deer is short. • We do not use the word â€˜Tall or Shortâ€™ to compare two or more heights. • To compare two or more heights, we use the words, taller, shorter, tallest, and shortest. ## Taller and Shorter To compare heights of two objects, use the word â€˜Taller/Shorterâ€™. ### Example: In the given picture, the grey building is taller than the blue building . The blue building is shorter than the grey building. ## Tallest-Shortest To compare heights of three or more objects, we use the word â€˜Tallest/Shortestâ€™. ### Example: In the given picture, the grey ladder is the tallest among all the ladders. The black ladder is the shortest among all the ladders. ## Misconception: In the given picture, the puppy looks the tallest. But the lion is the tallest among all. • When comparing the heights of objects, they should be at the same level. • -
# How do you solve the system of linear equations 2x − 3y = 3 and 5x − 4y = 4? Jul 8, 2017 There are two methods. 1. solve for one variable and then substitute. 2. Add or subtract the equations so that one variable is eliminated then substitute. #### Explanation: $2 x - 3 y = 3$ add 3y to both sides $2 x - 3 y + 3 y = - 3 y + 3$ gives $2 x = + 3 y + 3$ divide both sides by 2 $\frac{2 x}{2} = \frac{+ 3 y}{2} + \frac{3}{2}$ which gives. $x = + \frac{3}{2} y + \frac{3}{2}$ Now substitute into the other equation $5 x - 4 y = 4 = 5 \left(+ \frac{3}{2} y + \frac{3}{2}\right) - 4 y = 4$ which gives $\left(+ \frac{15}{2}\right) y + \frac{15}{2} - 4 y = 4$ multiply everything by 2 $2 \left(+ \frac{15}{2}\right) y + 2 \left(\frac{15}{2}\right) - 4 y = 4$ $+ 15 y - 4 y + 15 = 4$ subtract 15 from both sides $11 y + 15 - 15 = 4 - 15$ $11 y = - 11$ divide both sides by 11 (11y/11 = -11/11 gives $y = - 1$ $2 x - 3 \left(- 1\right) = 3$ $2 x + 3 = 3$ subtract 3 from both sides $2 x + 3 - 3 = 3 - 3$ so $2 x = 0$ divide both sides by two $2 \frac{x}{2} = \frac{0}{2}$ $x = 0$
Courses Courses for Kids Free study material Offline Centres More Store # ${\text{The function }}f\left( x \right) = \left\{ \begin{gathered} \dfrac{{{{\text{x}}^2}}}{a}{\text{ ,}}if{\text{ }}0 \leqslant x < 1 \\ a{\text{ ,}}if{\text{ }}1 \leqslant x < \sqrt 2 \\ \dfrac{{2{b^2} - 4b}}{{{{\text{x}}^2}}}{\text{ ,}}if{\text{ }}\sqrt 2 \leqslant x < \infty \\ \end{gathered} \right\}{\text{ is continuous on}}\left[ {0,\infty } \right). \\ {\text{Find the most suitable values of }}a{\text{ and }}b. \\$ Last updated date: 20th Jul 2024 Total views: 457.5k Views today: 5.57k Verified 457.5k+ views Hint: Let’s find out limits at all the critical points and then we’ll compare to find out the desired values. ${\text{We have been given f}}(x){\text{ }} \\ f\left( x \right) = \left\{ \begin{gathered} \dfrac{{{{\text{x}}^2}}}{a}{\text{ ,}}if{\text{ }}0 \leqslant x < 1 \\ a{\text{ ,}}if{\text{ }}1 \leqslant x < \sqrt 2 \\ \dfrac{{2{b^2} - 4b}}{{{{\text{x}}^2}}}{\text{ ,}}if{\text{ }}\sqrt 2 \leqslant x < \infty \\ \end{gathered} \right\} \\$ So, At x = 1, we can say that $\mathop {\lim }\limits_{{x^ - } \to 1} f\left( x \right) = f\left( 1 \right) = \mathop {\lim }\limits_{{x^ + } \to 1} f\left( x \right)$ Taking left hand limit, $\mathop {\lim }\limits_{{x^ - } \to 1} f\left( x \right) = \mathop {\lim }\limits_{{x^ - } \to 1} \dfrac{{{{\text{x}}^2}}}{a}$ $\Rightarrow \mathop {\lim }\limits_{{x^ - } \to 1} f\left( x \right) = \dfrac{1}{a}$ Taking right hand limit, $\mathop {\lim }\limits_{{x^ + } \to 1} f\left( x \right) = \mathop {\lim }\limits_{{x^ + } \to 1} a$ $\Rightarrow \mathop {\lim }\limits_{{x^ + } \to 1} f\left( x \right) = a$ And $f\left( 1 \right) = a$ Comparing them you get, $\Rightarrow a = \dfrac{1}{a}$ $\Rightarrow {a^2} = 1$ $\Rightarrow a = \pm 1$ Now at x = \sqrt 2 we can say that $\mathop {\lim }\limits_{{x^ - } \to \sqrt 2 } f\left( x \right) = f\left( {\sqrt 2 } \right) = \mathop {\lim }\limits_{{x^ + } \to \sqrt 2 } f\left( x \right)$ Taking left hand limit, $\mathop {\lim }\limits_{{x^ - } \to \sqrt 2 } f\left( x \right) = \mathop {\lim }\limits_{{x^ - } \to \sqrt 2 } a$ $\Rightarrow \mathop {\lim }\limits_{{x^ - } \to \sqrt 2 } f\left( x \right) = a$ Taking right hand limit, $\mathop {\lim }\limits_{{x^ + } \to \sqrt 2 } f\left( x \right) = \mathop {\lim }\limits_{{x^ + } \to \sqrt 2 } \dfrac{{2{b^2} - 4b}}{{{{\text{x}}^2}}}$ Now at $f\left( {\sqrt 2 } \right) = {b^2} - 2b$ Comparing them you get, For a = 1 ${b^2} - 2b = 1$ $\Rightarrow {b^2} - 2b - 1 = 0$ Using $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ $\Rightarrow b = \dfrac{{2 \pm \sqrt 8 }}{2} = 1 \pm \sqrt 2$ For a = - 1 ${b^2} - 2b = - 1$ $\Rightarrow {b^2} - 2b + 1 = 0$ $\Rightarrow$ b = 1,1 So, we have $a = \pm 1,b = 1 \pm \sqrt 2 ,1$ NOTE: Be careful while calculating the value of b. Since we are using a quadratic formula, there will be 3 values for different values of a.
# Math Snap ## 30 Given $\sinh (x)=\frac{3}{4}$, then $\cosh (x)=$ Select one: a. $\pm \frac{5}{4}$ b. $\frac{5}{4}$ c. $\pm \frac{\sqrt{7}}{4}$ d. $\frac{\sqrt{7}}{4}$ #### STEP 1 Assumptions 1. We are given that $\sinh(x) = \frac{3}{4}$. 2. We need to find $\cosh(x)$. 3. We will use the hyperbolic identity: $\cosh^2(x) - \sinh^2(x) = 1$. #### STEP 2 Start by writing the given information and the hyperbolic identity. $\sinh(x) = \frac{3}{4}$ $\cosh^2(x) - \sinh^2(x) = 1$ #### STEP 3 Substitute the given value of $\sinh(x)$ into the identity. $\cosh^2(x) - \left(\frac{3}{4}\right)^2 = 1$ #### STEP 4 Calculate $\left(\frac{3}{4}\right)^2$. $\left(\frac{3}{4}\right)^2 = \frac{9}{16}$ #### STEP 5 Substitute $\frac{9}{16}$ back into the equation. $\cosh^2(x) - \frac{9}{16} = 1$ #### STEP 6 Isolate $\cosh^2(x)$ by adding $\frac{9}{16}$ to both sides of the equation. $\cosh^2(x) = 1 + \frac{9}{16}$ #### STEP 7 Convert 1 to a fraction with a common denominator of 16. $1 = \frac{16}{16}$ #### STEP 8 Add the fractions on the right-hand side. $\cosh^2(x) = \frac{16}{16} + \frac{9}{16} = \frac{25}{16}$ #### STEP 9 Take the square root of both sides to solve for $\cosh(x)$. $\cosh(x) = \pm \sqrt{\frac{25}{16}}$ #### STEP 10 Simplify the square root. $\cosh(x) = \pm \frac{\sqrt{25}}{\sqrt{16}} = \pm \frac{5}{4}$ ##### SOLUTION Since $\cosh(x)$ is always positive for real $x$, we take the positive value. $\cosh(x) = \frac{5}{4}$ Therefore, the correct answer is: $\cosh(x) = \frac{5}{4}$
What’s the Geometrical interpretation of integrals? And what’s the physical significance of C? Grade:11 2 Answers Vikas TU 14149 Points 7 years ago Geometrical interpretation of indefinite integral Let f (x) = 2x. Then ∫ f (x) dx = x2 + C. For different values of C, we get different integrals. But these integrals are very similar geometrically. Thus, y = x2 + C, where C is arbitrary constant, represents a family of integrals. By assigning different values to C, we get different members of the family. These together constitute the indefinite integral. In this case, each integral represents a parabola with its axis along y-axis. Clearly, for C = 0, we obtain y = x2, a parabola with its vertex on the origin. The curve y = x2 + 1 for C = 1 is obtained by shifting the parabola y = x2 one unit along y-axis in positive direction. For C = – 1, y = x2– 1 is obtained by shifting the parabola y = x2 one unit along y-axis in the negative direction. Thus, for each positive value of C, each parabola of the family has its vertex on the positive side of the y-axis and for negative values of C, each has its vertex along the negative side of the y-axis. SREEKANTH 85 Points 7 years ago one unit along y-axis in the negative direction. Thus, for each positive value of C, each parabola of the family has its vertex on the positive side of the y-axis and for negative values of C, each has its vertex along the negative side of the y-axis.2– 1 is obtained by shifting the parabola y = x2 one unit along y-axis in positive direction. For C = – 1, y = x2 + 1 for C = 1 is obtained by shifting the parabola y = x2, a parabola with its vertex on the origin. The curve y = x2. In this case, each integral represents a parabola with its axis along y-axis. Clearly, for C = 0, we obtain y = xindefinite integral + C, where C is arbitrary constant, represents a family of integrals. By assigning different values to C, we get different members of the family. These together constitute the 2 + C. For different values of C, we get different integrals. But these integrals are very similar geometrically. Thus, y = x2Let f (x) = 2x. Then ∫ f (x) dx = x ASK QUESTION Get your questions answered by the expert for free
# Worksheet on Multiplication and Division by 3 We know multiplication and division are related to each other. Now we practice the worksheet on multiplication and division by 3 to see how they are related to each other. To work on this worksheet we need to know the multiplication table of 3. I. Fill in the blanks: Multiplication fact 1 × 3 = ___ 2 × 3 = ___ 3 × 3 = ___ 4 × 3 = ___ 5 × 3 = ___ 6 × 3 = ___ 7 × 3 = ___ 8 × 3 = ___ 9 × 3 = ___ 10 × 3 = ___ 11 × 3 = ___ 12 × 3 = ___ Division fact 3 ÷ 3 = ___ 6 ÷ 3 = ___ 9 ÷ 3 = ___ 12 ÷ 3 = ___ 15 ÷ 3 = ___ 18 ÷ 3 = ___ 21 ÷ 3 = ___ 24 ÷ 3 = ___ 27 ÷ 3 = ___ 30 ÷ 3 = ___ 33 ÷ 3 = ___ 36 ÷ 3 = ___ Division by 3 II. Write two division facts for each multiplication fact: (i) 6 × 3 = 18 (ii) 2 × 3 = 6 (iii) 9 × 3 = 27 (iv) 12 × 3 = 36 (v) 8 × 3 = 24 III. Write two multiplication facts for each division fact: (i) 15 ÷ 3 = 5 (ii) 3 ÷ 3 = 1 (iii) 21 ÷ 3 = 7 (iv) 30 ÷ 3 = 10 (v) 33 ÷ 3 = 11 Answers for the worksheet on multiplication and division by 3 are given below. I. 3 6 9 12 15 18 21 24 27 30 33 36 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 II. (i) 18 ÷ 3 = 6 and 18 ÷ 6 = 3 (ii) 6 ÷ 2 = 3 and 6 ÷ 3 = 2 (iii) 27 ÷ 9 = 3 and 27 ÷ 3 = 9 (iv) 36 ÷ 3 = 12 and 36 ÷ 12 = 3 (v) 24 ÷ 3 = 8 and 24 ÷ 8 = 3 III. (i) 5 × 3 = 15 and 3 × 5 = 15 (ii) 1 × 3 = 3 and 3 × 1 = 3 (iii) 7 × 3 = 21 and 3 × 7 = 21 (iv) 10 × 3 = 30 and 3 × 10 = 30 (v) 11 × 3 = 33 and 3 × 11 = 33 Worksheet on Multiplication and Division by 2 Worksheet on Multiplication and Division by 3 Worksheet on Multiplication and Division by 5 Worksheet on Multiplication and Division by 6 Worksheet on Multiplication and Division by 8 Worksheet on Multiplication and Division by 9 Worksheet on Multiplication and Division by 10 Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. ## Recent Articles 1. ### Lines of Symmetry \| Symmetry of Geometrical Figures \| List of Examples Aug 10, 24 03:27 PM Learn about lines of symmetry in different geometrical shapes. It is not necessary that all the figures possess a line or lines of symmetry in different figures. 2. ### Symmetrical Shapes \| One, Two, Three, Four & Many-line Symmetry Aug 10, 24 02:25 AM Symmetrical shapes are discussed here in this topic. Any object or shape which can be cut in two equal halves in such a way that both the parts are exactly the same is called symmetrical. The line whi… Aug 10, 24 01:59 AM In 6th grade math practice you will get all types of examples on different topics along with the step-by-step explanation of the solutions. 4. ### 6th Grade Algebra Worksheet \| Pre-Algebra worksheets with Free Answers Aug 10, 24 01:57 AM In 6th Grade Algebra Worksheet you will get different types of questions on basic concept of algebra, questions on number pattern, dot pattern, number sequence pattern, pattern from matchsticks, conce…
Probability Essentials # Probability Essentials ## Probability Essentials - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. Probability Essentials • Concept of probability is quite intuitive; however, the rules of probability are not always intuitive or easy to master. • Mathematically, a probability is a number between 0 and 1 that measures the likelihood that some event will occur. • An event with probability zero cannot occur. • An event with probability 1 is certain to occur. • An event with probability greater than 0 and less than 1 involves uncertainty, but the closer its probability is to 1 the more likely it is to occur. 2. Rule of Complement • The simplest probability rule involves the complement of an event. • If A is any event, then the complement of A, denoted by Ac, is the event that A does not occur. • If the probability of A is P(A), then the probability of its complement, P(Ac), is P(Ac)=1- P(A). • Equivalently, the probability of an event and the probability of its complement sum to 1. 3. Addition Rule • We say that events are mutually exclusive if at most one of them can occur. That is, if one of them occurs, then none of the others can occur. • Events can also be exhaustive, which means that they exhaust all possibilities - one of these three events must occur. • Let A1 through An be any n events. Then the addition rule of probability involves the probability that at least one of these events will occur. P(at least one of A1 through An) = P(A1) + P(A2) +  + P(An) 4. Conditional Probability • Probabilities are always assessed relative to the information currently available. As new information becomes available, probabilities often change. • A formal way to revise probabilities on the basis of new information is to use conditional probabilities. • Let A and B be any events with probabilities P(A) and P(B). Typically the probability P(A) is assessed without knowledge of whether B does or does not occur. However if we are told B has occurred, the probability of A might change. 5. Conditional Probability -- continued • The new probability of A is called the conditional probability of A given B. It is denoted P(A\|B). • Note that there is uncertainty involving the event to the left of the vertical bar in this notation; we do not know whether it will occur or not. However, there is no uncertainty involving the event to the right of the vertical bar; we know that it has occurred. • The following formula conditional probability formula enables us to calculate P(A\|B): 6. Multiplication Rule • In the conditional probability rule the numerator is the probability that both A and B occur. It must be known in order to determine P(A\|B). • However, in some applications P(A\|B) and P(B) are known; in these cases we can multiply both side of the conditional probability formula by P(B) to obtain the multiplication rule. P(A and B) = P(A\|B)P(B) • The conditional probability formula and the multiplication rule are both valid; in fact, they are equivalent. 7. Assessing the Bendrix Situation • Now that we are familiar with the a number of probability rules we can put them to work in assessing the Bendrix situation. • To begin we will let A be the event that Bendrix meets its end-of-July deadline, and let B be the event that Bendrix receives the materials form its supplier by the middle of July. • The probabilities that we are best able to be assess on July 1 are probably P(B) and P(A\|B). 8. Assessing -- continued • They estimate a 2 in 3 chance of getting the materials on time; thus P(B)=2/3. • They also estimate that if they receive the materials on time then the chances of meeting the deadline are 3 out of 4. This is a conditional probability statement that P(A\|B)=3/4. • We can use the multiplication rule to obtain: P(A and B) = P(A\|B)P(B) = (3/4)(2/3) = 0.5 • There is a 50-50 chance that Bendrix will gets its materials on time and meet its deadline. 9. Assessing -- continued • Other probabilities of interest exist in this example. • Let Bc be the complement of B; it is the event that the materials from the supplier do not arrive on time. We know that P(B) = 1 - P(Bc) = 1/3 from the rule of complements. • Bendrix estimates that the chances of meeting the deadline are 1 out of 5 if the materials do not arrive on time, that is, P(A\| Bc) = 1/5. The multiplication rule gives P(A and Bc) = P(A\| Bc)P(Bc) = (1/5)(1/3) = 0.0667 10. Assessing -- continued • In words, there is a 1 chance out of 15 that the materials will not arrive on time and Bendrix will meets its deadline. • The bottom line for Bendrix is whether it will meet its end-of-July deadline. After the middle of July the probability is either 3/4 or 1/5 because by this time they will know whether the materials have arrived on time. • But since it is July 1 the probability is P(A) - there is still uncertainty about whether B or Bc will occur. 11. Assessing -- continued • We can calculate P(A) from the probabilities we already know. Using the additive rule for mutually exclusive events we obtain P(A) =P(A and B) + P(A and Bc) = (1/2)+(1/15) = 0.5667 • In words, the chances are 17 out of 30 that Bendrix will meet its end-of-July deadline, given the information it has at the beginning of July. 12. Probabilistic Independence • A concept that is closely tied to conditional probability is probabilistic independence. • There are situations unlike Bendix when P(A), P(A\|B) and P(A\| Bc) are not all different. They are situations where these probabilities are all equal. In this case we can say that events A and B are independent. • This does not mean they are mutually exclusive; it means that knowledge of one of the events is of no value when assessing the probability of the other event. 13. Probabilistic Independence -- continued • The main advantage of knowing that two events are independent is that the multiplication rule simplifies to P(A and B) = P(A)P(B) • In order to determine if events are probabilistically independent we usually cannot use mathematical arguments; we must use empirical data to decide whether independence is reasonable. 14. Distribution of a Single Random Variable 15. Background Information • An investor is concerned with the market return for the coming year, where the market return is defined as the percentage gain (or loss, if negative) over the year. • The investor believes there are five possible scenarios for the national economy in the coming year: rapid expansion, moderate expansion, no growth, moderate contraction, or serious contraction. • She estimates that the market returns for these scenarios are, respectively, 0.23, 0.18, 0.15, 0.09, and 0.03. 16. Background Information -- continued • Also, she has assessed that the probabilities of these outcomes are 0.12, 0.40, 0.25, 0.15, and 0.08. • We must use this information to describe the probability distribution of the market return. 17. Type of Random Variables • A discrete random variable has only a finite number of possible values. • A continuous random variable has a continuum of possible values. • Mathematically, there is an important difference between discrete and continuos random variables. A proper treatment of continuos variables requires calculus. In this book we will only be dealing with discrete random variables. 18. Discrete Random Variables • The properties of discrete random variables and their associated probability distributions are as follows: • Let X be a random variable and to specify the probability distribution of X we need to specify its possible values and their probabilities. This list of their probabilities sum to 1. • It is sometimes useful to calculate cumulative probabilities. A cumulative probability is the probability that the random variable is less than or equal to some particular values. 19. Summarizing a Probability Distribution • A probability distribution can be summarized with two or three well-chosen numbers: • The mean, often called the expected value, is a weighted sum of the possibilities. It indicates the center of the probability distribution. • To measure the variability in a distribution, we calculate its variance or standard deviation. The variance is a weighted sum of the squared deviations of the possible values from the mean. As in the previous chapter the variance is represented in the squared units of X so a more natural measure of variability is the standard deviation. 20. MRETURN.XLS • This file contains the values and probabilities estimated by the investor in this example. • Mean, Probs, Returns, Var and Sqdevs have been specified as range names. 21. Calculating Summary Measures • The summary measures for the probability distribution of the outcomes can be calculated as follows: • Mean return: =SUMPRODUCT(Returns,Probs) • Squared Deviations: =(C4-Mean)^2 • Variance: =SUMPRODUCT(SqDevs,Probs) • Standard Deviation: =SQRT(Var) • We see that the mean return is 15.3% and the standard deviation is 5.3%. What do these mean? 22. Analyzing the Summary Measures • First, the mean or expected return does not imply that the most likely return is 15.3%, nor is this the value that the investor “expects” to occur. The value 15.3% is not even a possible market return. • We can understand these measures better in terms of long-run averages. • If we can see the coming year repeated many times, using the same probability distribution, then the average of these times would be close to 15.3% and their standard deviation would be 5.3%. 23. Derived Probability Distributions 24. Background Information • A bookstore is planning on ordering a shipment of special edition Christmas calendars that they will sell for \$15 a piece. • There will be only one order, so • if demand is less than the quantity ordered the excess calendars will be donated to a paper recycling company • if demand is greater than the quantity ordered, the excess demand will be lost and customers will take their business elsewhere • The bookstore estimates that the demand for calendars will be between 250 and 400. 25. DERIVED.XLS • This file contains the probability distribution that the demand for calendars will follow. These estimates have been derived from subjective estimates and historical data. • If the bookstore decides to order 350 calendars, what is the probability distribution of units sold? What is the probability distribution of revenue? 26. Derived Distributions of Units Sold and Revenue 27. Solution • Let D, S,and R denote demand, units sold, and revenue. • The key to the solution is that each value of D directly determines the value of S, which in turn determines the value of R. • S is the smaller of D and the number ordered, 350, and R is \$15 multiplied by the value of S. • Therefore we can derive the probability distributions of S and R with the following steps: 28. Solution -- continued • Calculate Units sold : =MIN(B10,OnHand) • Calculate Revenue for each value of units sold: =UnitPriceB20 • Transfer the Derived Probabilities for demand: =C10 • Calculate Means of demand, units sold, and revenue: =SUMPRODUCT(Revenues, DerivedProbs) • Calculate the Variances and Standard Deviations of demand, units sold and revenue. • First, calculate the squared deviations of revenues from their mean in Column F, then calculate the sum of the products of these squared deviations and the revenue probabilities to obtain the variance of revenue. Finally, calculate the standard deviation as the square root of the variance. 29. Summary Measures for Linear Functions • When one random variable is a linear function of another random variable X, there is a particularly simple way to calculate the summary measures of Y from the Summary measures of X. Y = a + bX for some constant a and b then: • mean: E(Y) = a + bE(X) • variance: Var(Y) = b2 Var(X) • standard deviation: bStdev(X) • If Y is a constant multiple of X, that is a=0 then the mean and standard deviation of Y are this same multiple of the mean and standard deviation of X. 30. Distribution of Two Random Variables: Scenario Approach 31. Background Information • An investor plans to invest in General Motors (GM) stock and gold. • He assumes that the returns on these investments over the next year depend on the general state of the economy during the year. • He identifies four possible states of the economy: depression, recession, normal and boom. These four states have the following probabilities: 0.05, 0.30, 0.50, and 0.15. 32. Background Information -- continued • The investor wants to analyze the joint distribution of returns on these two investments. • He also wants to analyze the distribution of a portfolio of investments in GM stock and gold. 33. GMGOLD.XLS • This file contains the probabilities and estimated returns of the GM stock and the gold. 34. Relating Two Random Variables • There are two methods for relating two random variables, the scenario approach and the joint probability approach. • The methods differ slightly in the way they assign probabilities to different outcomes. • Two summary measures, covariance and correlation, are used to measure the relationship between two variables in both methods. 35. The Summary Measures • We have discussed summary measures with the same names, covariance and correlation, earlier. The summary measures we are looking at now go by the same name but are conceptually different. • In the past we have calculated them from data; here they are calculated from a probability distribution. • The random variables are X and Y and the probability that X and Y equal xi and yi is p(xi, yi) is called a joint probability. 36. Summary Measures -- continued • Although they are calculated differently , the interpretation is essentially the same as we previously discussed. • Each indicates the strength of a linear relationship between X and Y. If X and Y vary in the same direction then both measures are positive. If they vary in opposite directions then both measures are negative. • Covariance is more difficult to interpret because it depends on the units of measurement of X and Y. Correlation is always between -1 and +1. 37. The Scenario Approach • The essence of the scenario approach in this example is that a given state of the economy determines both GM and gold returns, so that only four pairs of returns are possible. • These pairs are -0.20 and 0.05, 0.10 and 0.20, 0.30 and -0.12, and 0.50 and 0.09. Each pair has a joint probability. • To calculate means, variances and standard deviations, we treat GM and gold returns separately. 38. Calculating Covariance and Correlation • We also need to calculate the covariance and correlation between the variables. To obtain these we use the following steps: • Deviations between means: To calculate the covariance we need the sum of deviations from means, so we need to calculate these deviations with the formula =C4-GMMean in B14 and copy it down through B17. We also calculate this for gold. • Covariance: Calculate the covariance between GM and gold returns in cell B23 with the formula=SUMPRODUCT(GMDevs,GoldDevs,Probs) 39. Calculating Covariance and Correlation -- continued • Correlation: Calculate the correlation between GM and gold returns in cell B24 with the formula =Covar/(GMStdvGoldStedev) • The negative covariance indicates that GM and gold returns tend to vary in opposite directions, although it is difficult to judge the strength by the magnitude of the covariance. • The correlation of -0.410 is also negative and indicates a moderately strong relationship. We cannot infer too much from this correlation though because the variables are not linear. 40. Simulation • A simulation of GM and gold returns help explain the covariance and correlation. • There are two keys to this simulation: • First we must, simulate the states of the economy, not - at least not directly - the GM and gold returns. • We simulate this be entering a RAND function in A1 and then by entering the formulas VLOOKUP(A21,LTable,2) in B21 and VLOOKUP(A21,LTable,3) in C21. • This way uses the same random number, hence the same scenario, to generate both returns in a given row, and the effect is that only four pairs of returns are possible. 41. Simulation -- continued • Second, once we have the simulated returns we can calculate the covariance and correlation of these numbers. • We calculate these in cells B8 and B9 with the formulas COVAR(SimGM,SimGOLD) and CORREL(SimGM,SimGold). These are built-in Excel functions. • A comparison of these summary measures with the previously calculated summary measures shows that there is reasonably good agreement between the covariance and correlation of the probability distribution and the measures based on the simulated values. The agreement is not perfect but will improve as more pairs are simulated. 42. Simulation of GM and Gold Returns 43. Portfolio Analysis • The final part of this example is to analyze a portfolio consisting of GM stock and gold. • We assume that the investor has \$10,000 and puts some fraction of this in GM stock and the rest in gold. • The key to the analysis is that there are only four possible scenarios -- that is, there are only four possible portfolio returns. • In this case we calculate the entire portfolio return distribution and summary measures in the usual way. 44. Portfolio Analysis -- continued • One thing of interest is to see how the expected portfolio return and standard deviation of portfolio return change as the amount the investor puts into GM stock changes. • To do this we use a data table or mean and stdev of portfolio return as a function of GM investment. • A graph of these measures show that the expected portfolio return steadily increases as more and more is put into GM. 45. Portfolio Analysis -- continued • However, we must note that the standard deviation, often used as a measure of risk, first decreases, then increases. • This means there is trade-off between expected return and risk (as measured by the standard deviation). • The investor could obtain a higher expected return by putting more of his money into GM; but past a fraction of approximately 0.4, the risk also increases. 46. Distribution of Portfolio Return 47. SUBS.XLS • A company sells two products, product 1 and 2, that tend to be substitutes for each one another.The company has assessed the joint probability distribution of demand for the two products during the coming months. • This joint distribution appears in the Demand sheet of this file. • The left and top margins of the table show the possible values of demand for the products. 48. SUBS.XLS -- continued • Demand for product 1 (D1) can range from 100 to 400 (in increments of 100) and demand for product 2(D2) can range from 50-250 (in increments of 50). • Each possible value of D1 can occur for each possible value of D2 with the joint probability given in the table. • Given this joint probability distribution, describe more fully the probabilistic structure of demands for the two products. 49. Joint Probability Approach • In this example we use an alternative method for specifying probability distribution. • A joint probability distribution, specified by all probabilities of the form p(x, y), indicates that X and Y are related and also how each of X and Y is distributed in its own right. • The joint probability of X and Y determines the marginal distributions of both X and Y, where each marginal distribution is the probability distribution of a single random variable.
# Composite Functions ## Understanding Composite Functions: Simplifying Operations with Multiple Functions A composite function is the combination of two or more functions. This method is commonly used to transform numbers from one set to another. For example, if we have a function that maps numbers from set A to set B and another function that maps numbers from set B to set C, we can use a composite function to directly map numbers from set A to set C. The diagram below illustrates this concept: • Composite functions allow for mapping between sets It's important to remember these key properties of composite functions: • In a composite function, the inside function is always performed first, and its output becomes the input for the outside function • The domain of a composite function is determined by the domain of the first function applied • When combining two functions, a single input is used, and the output of both functions is obtained • To find the value of a composite function, we input a value into the original function and then use the resulting output as the input for all other functions Let's go through an example to see how this process works: Imagine we have the functions f(x) = 2x + 1 and g(x) = x^2, and we want to find the value of f(g(3)). First, we calculate g(3) = 9. Next, we use this value as the input for f(x), giving us f(9) = 19. Therefore, f(g(3)) = 19. Composite functions can become more complex when dealing with quadratic, trigonometric, and reciprocal functions. However, the process remains the same as with simpler linear examples. Let's look at a few more worked examples: • To find the value of f(g(x)), where f(x) = 1/x and g(x) = sin(x), we first calculate g(x) = 1/sin(x). Then, we use this as the input for f(x), resulting in f(1/sin(x)). • Consider the functions f(x) = 2x + 1 and g(x) = x^2. To find the value of f(g(4)), we start by calculating g(4) = 16. Next, we use this value as the input for f(x), resulting in f(16) = 33. Therefore, f(g(4)) = 33. Key Takeaways: • Composite functions are created by combining two or more functions • The inside function is always performed first in a composite function • The domain of a composite function is determined by the first function applied • To find the value, we input a value into the original function and then use the resulting output as the input for all other functions
Contents of this Precalculus episode: Inverse function, Injective functions, Domain, Range, Find the inverse. Text of slideshow Each function is a assignment, the inverse of which, if it exists at all, is the reverse assignment. Calculating the inverse goes like this: Let First, write the function in the form of y=thingy: Here, we assign y to x. The inverse is the reverse assignment, where we assign x to y. So, the purpose is always to rearrange y=thingy to x=something. Finally, we swap x and y (some people don’t do this), and then we get the inverse. The inverse is denoted by: But, there is a little trouble. Not all functions have an inverse, as not all assignments can be reversed. For example, in the case of , we have and , and thus, we cannot reverse this: . The crux of the problem is that this function assigns the same number to two different numbers (2 and -2), and therefore, the assignment cannot be reversed. But if we exclude the negative numbers, then everything will be all right. So, only those functions have an inverse where for two different x values there are two different y-s assigned. We say they have a one-to-one correspondence, shortly, they are injective. A function is injective, if then . All strictly monotonic functions are injective and thus, invertible. And there is one more thing here. Let the function be and its domain . Well, then the image is . The inverse function is the reverse assignment, so in this case, these are swapped. If is invertible, then its Domain is identical to the Range of its inverse, and its Range is identical to the Domain of the inverse. Let's see a few examples! Let’s find the inverse of function , if There is no inverse, because the function is not injective. For example, it assigns the same number to 4 and -4, namely 0. This case is entirely different, as x could only be positive here. There are no two positive numbers with the same square, so this function is injective. Let’s see the inverse: In this case, there is an inverse, too, because the function is injective. Let’s see the inverse! In this case the function has no inverse, because it is - again- not injective. For example, it assigns the same number to 4 and -4, namely 0. Unfortunately there is no inverse in this case either, as the function is not injective. Let’s see one more. Here is this function, and we want to find its inverse. and Finally let’s see this one, too: Let’s talk a bit about the geometric meaning of the inverse. Here is a function and let’s see what happens to the graph of this function when we invert it. Well, this. Let’s reflect the graph of the function about the y=x line. It is clearly visible on the drawing that the inverse of radical functions is never a full parabola, only a half. And the reverse is true as well: a full parabola can never be inverted, only its half. Here comes another splendid function: Well, the inverse of this function is: The inverse of the exponential functions are the logarithmic functions. And this is mutual: the inverse of the logarithmic functions are the exponential functions. Let’s see the inverse of this, for instance: We can lure the x out of the exponent by taking the logarithm of both sides. Here is another one, for example: Functions and are also inverses of each other. We should be careful with this function inversion thing, as it could be harmful in large doses. But maybe we can get away with one more... # The inverse function 01 Let's see this Precalculus episode
# Precalculus : Graph a Linear Function ## Example Questions ### Example Question #1 : Graph A Linear Function Which of the following could be the function modeled by this graph? Explanation: Which of the following could be the function modeled by this graph? We can begin here by trying to identify a couple points on the graph We can see that it crosses the y-axis at Therefore, not only do we have a point, we have the y-intercept. This tells us that the equation of the line needs to have a in it somewhere. Eliminate any option that do not have this feature. Next, find the slope by counting up and over from the y-intercept to the next clear point. It seems like the line goes up 5 and right 1 to the point This means we have a slope of 5, which means our equation must look like this: ### Example Question #2 : Graph A Linear Function Find the slope of the linear function Explanation: For the linear function in point-slope form The slope is equal to For this problem we get ### Example Question #3 : Graph A Linear Function Find the slope of the linear function Explanation: For the linear function in point-slope form The slope is equal to For this problem we get ### Example Question #51 : Graphing Functions What is the y-intercept of the line below? Explanation: By definition, the y-intercept is the point on the line that crosses the y-axis. This can be found by substituting into the equation. When we do this with our equation, Alternatively, you can remember form, a general form for a line in which is the slope and is the y-intercept. ### Example Question #52 : Graphing Functions What is the slope of the line below? Explanation: Recall slope-intercept form, or . In this form, is the slope and is the y-intercept. Given our equation above, the slope must be the coefficient of the x, which is ### Example Question #53 : Graphing Functions What is the x-intercept of the equation below?
# AP Statistics Curriculum 2007 Infer 2Proportions (Difference between revisions) Revision as of 20:49, 19 February 2008 (view source) (→Genders of Siblings Example)← Older edit Revision as of 22:58, 1 March 2008 (view source) (→ General Advance-Placement (AP) Statistics Curriculum - Inferences about Two Proportions)Newer edit → Line 1: Line 1: ==[[AP_Statistics_Curriculum_2007 \| General Advance-Placement (AP) Statistics Curriculum]] - Inferences about Two Proportions == ==[[AP_Statistics_Curriculum_2007 \| General Advance-Placement (AP) Statistics Curriculum]] - Inferences about Two Proportions == - === Testing for equality of Two Proportions=== + === Testing for Equality of Two Proportions=== Suppose we have two populations and we are interested in estimating whether the proportions of subjects that have certain characteristic of interest (e.g., fixed gender) in each population are equal. To make this inference we obtain two samples {$X_1, X_2, X_3, \cdots, X_n$} and {$Y_1, Y_2, Y_3, \cdots, Y_k$}, where each $X_i$ and $Y_i$ represents whether the ''ith'' observation in the sample had the characteristic of interest. That is Suppose we have two populations and we are interested in estimating whether the proportions of subjects that have certain characteristic of interest (e.g., fixed gender) in each population are equal. To make this inference we obtain two samples {$X_1, X_2, X_3, \cdots, X_n$} and {$Y_1, Y_2, Y_3, \cdots, Y_k$}, where each $X_i$ and $Y_i$ represents whether the ''ith'' observation in the sample had the characteristic of interest. That is Line 21: Line 21: : '''Corrected Proportions''': $SE_{\tilde{p_x}-\tilde{p_y}} = \sqrt{SE_{\tilde{p_x}}^2 + SE_{\tilde{p_y}}^2}= \sqrt{ {\tilde{p_x}(1-\tilde{p_x})\over n+z_{\alpha \over 2}^2} + {\tilde{p_y}(1-\tilde{p_y})\over k+z_{\alpha \over 2}^2}}.$ : '''Corrected Proportions''': $SE_{\tilde{p_x}-\tilde{p_y}} = \sqrt{SE_{\tilde{p_x}}^2 + SE_{\tilde{p_y}}^2}= \sqrt{ {\tilde{p_x}(1-\tilde{p_x})\over n+z_{\alpha \over 2}^2} + {\tilde{p_y}(1-\tilde{p_y})\over k+z_{\alpha \over 2}^2}}.$ - === Hypothesis Testing the difference of Two Proportions=== + === Hypothesis Testing the Difference of Two Proportions=== * Null Hypothesis: $H_o: p_x=p_y$, where $p_x$ and $p_x$ are the sample population proportions of interest. * Null Hypothesis: $H_o: p_x=p_y$, where $p_x$ and $p_x$ are the sample population proportions of interest. * Alternative Research Hypotheses: * Alternative Research Hypotheses: Line 68: Line 68: * $P_value = P(Z>Z_o)< 1.9\times 10^{-8}$. This small p-values provides extremely strong evidence to reject the null hypothesis that there are no differences between the proportions of mothers that had a girl as a second child but had either boy or girl as their first child. Hence there is strong statistical evidence implying that genders of siblings are not independent. * $P_value = P(Z>Z_o)< 1.9\times 10^{-8}$. This small p-values provides extremely strong evidence to reject the null hypothesis that there are no differences between the proportions of mothers that had a girl as a second child but had either boy or girl as their first child. Hence there is strong statistical evidence implying that genders of siblings are not independent. - * '''Practical significance''': The practical significance of the effect (of the gender of the first child on the gender of the second child, in this case) can only be assessed using [[AP_Statistics_Curriculum_2007#Estimating_a_Population_Proportion \|confidence intervals]]. A 95% $CI (p_1- p_2) =[0.033; 0.070]$ is computed by $p_1-p_2 \pm 1.96 SE(p_1 - p_2)$. Clearly, this is a practically negligible effect and no reasonable person would make important prospective family decisions based on the gender of their (first) child. + * Practical significance: The practical significance of the effect (of the gender of the first child on the gender of the second child, in this case) can only be assessed using [[AP_Statistics_Curriculum_2007#Estimating_a_Population_Proportion \|confidence intervals]]. A 95% $CI (p_1- p_2) =[0.033; 0.070]$ is computed by $p_1-p_2 \pm 1.96 SE(p_1 - p_2)$. Clearly, this is a practically negligible effect and no reasonable person would make important prospective family decisions based on the gender of their (first) child. * This [[SOCR_EduMaterials_AnalysisActivities_Chi_Contingency \| SOCR Analysis Activity]] illustrates how to use the [http://socr.ucla.edu/htmls/SOCR_Analyses.html SOCR Analyses] to compute the p-values and answer the hypothesis testing challenge. * This [[SOCR_EduMaterials_AnalysisActivities_Chi_Contingency \| SOCR Analysis Activity]] illustrates how to use the [http://socr.ucla.edu/htmls/SOCR_Analyses.html SOCR Analyses] to compute the p-values and answer the hypothesis testing challenge. ## Contents ### Testing for Equality of Two Proportions Suppose we have two populations and we are interested in estimating whether the proportions of subjects that have certain characteristic of interest (e.g., fixed gender) in each population are equal. To make this inference we obtain two samples {$X_1, X_2, X_3, \cdots, X_n$} and {$Y_1, Y_2, Y_3, \cdots, Y_k$}, where each Xi and Yi represents whether the ith observation in the sample had the characteristic of interest. That is $X_i = \begin{cases}0,& \texttt{Characteristic-absent},\\ 1,& \texttt{Characteristic-present}.\end{cases}$ and $Y_i = \begin{cases}0,& \texttt{Characteristic-absent},\\ 1,& \texttt{Characteristic-present}.\end{cases}$ Since the raw sample proportions of observations having the characteristic of interest are $\hat{p_x}={1 \over n}\sum_{i=1}^n{x_i}$ and $\hat{p_y}={1 \over k}\sum_{i=1}^k{y_i}$ The corrected sample proportions (for small samples) are $\tilde{p_x}={\sum_{i=1}^n{x_i}+0.5z_{\alpha \over 2}^2 \over n+z_{\alpha \over 2}^2},$ and $\tilde{p_y}={\sum_{i=1}^k{y_i}+0.5z_{\alpha \over 2}^2 \over k+z_{\alpha \over 2}^2},$ By the independence of the samples, the standard error of the difference of the two proportion estimates is: Raw proportions: $SE_{\hat{p_x}-\hat{p_y}} = \sqrt{SE_{\hat{p_x}}^2 + SE_{\hat{p_y}}^2}= \sqrt{ {\hat{p_x}(1-\hat{p_x})\over n} + {\hat{p_y}(1-\hat{p_y})\over k}}.$ Corrected Proportions: $SE_{\tilde{p_x}-\tilde{p_y}} = \sqrt{SE_{\tilde{p_x}}^2 + SE_{\tilde{p_y}}^2}= \sqrt{ {\tilde{p_x}(1-\tilde{p_x})\over n+z_{\alpha \over 2}^2} + {\tilde{p_y}(1-\tilde{p_y})\over k+z_{\alpha \over 2}^2}}.$ ### Hypothesis Testing the Difference of Two Proportions • Null Hypothesis: Ho:px = py, where px and px are the sample population proportions of interest. • Alternative Research Hypotheses: • One sided (uni-directional): H1:px > py, or Ho:px < py • Double sided: $H_1: p_x \not= p_y$ • Test Statistics: $Z_o={\tilde{p_x} - \tilde{p_y} \over SE_{\tilde{p_x}-\tilde{p_y}}}$ ### Genders of Siblings Example Is the gender of a second child influenced by the gender of the first child, in families with >1 child? Research hypothesis needs to be formulated first before collecting/looking/interpreting the data that will be used to address it. Mothers whose 1st child is a girl are more likely to have a girl, as a second child, compared to mothers with boys as 1st child. Data: 20 yrs of birth records of 1 Hospital in Auckland, New Zealand. Second Child Male Female Total First Child Male 3,202 2,776 5,978 Female 2,620 2,792 5,412 Total 5,822 5,568 11,390 Let p1=true proportion of girls in mothers with girl as first child, p2=true proportion of girls in mothers with boy as first child. The parameter of interest is p1p2. • Hypotheses: Ho:p1p2 = 0 (skeptical reaction). H1:p1p2 > 0 (research hypothesis). Second Child Number of births Number of girls Proportion Group 1 (Previous child was girl) n1 = 5412 2792 $\hat{p}_1=0.516$ 2 (Previous child was boy) n2 = 5978 2776 $\hat{p}_2=0.464$ • Test Statistics: $Z_o = {Estimate-HypothesizedValue\over SE(Estimate)} = {\hat{p}_1 - \hat{p}_2 - 0 \over SE(\hat{p}_1 - \hat{p}_2)} = {\hat{p}_1 - \hat{p}_2 - 0 \over \sqrt{{\hat{p}_1(1-\hat{p}_1)\over n_1} + {\hat{p}_2(1-\hat{p}_2)\over n_2}}} \sim N(0,1)$ and Zo = 5.4996. • $P_value = P(Z>Z_o)< 1.9\times 10^{-8}$. This small p-values provides extremely strong evidence to reject the null hypothesis that there are no differences between the proportions of mothers that had a girl as a second child but had either boy or girl as their first child. Hence there is strong statistical evidence implying that genders of siblings are not independent. • Practical significance: The practical significance of the effect (of the gender of the first child on the gender of the second child, in this case) can only be assessed using confidence intervals. A 95% CI(p1p2) = [0.033;0.070] is computed by $p_1-p_2 \pm 1.96 SE(p_1 - p_2)$. Clearly, this is a practically negligible effect and no reasonable person would make important prospective family decisions based on the gender of their (first) child.
Enable contrast version # Tutor profile: Natasha H. Inactive Natasha H. Current tutor in my hometown Tutor Satisfaction Guarantee ## Questions ### Subject:Pre-Algebra TutorMe Question: Maria went to her favorite bakery today to buy churros. The sign says "2 churros for \$3". If Maria bought 15 churros, how much money did she spend? Inactive Natasha H. For this problem, the easiest way to solve it is to use proportions. When solving problems like this, it is very important to remember when you set the proportion up, the units MUST match. For example, if you put the cost over the number of churros, then you must continue to do so for the next fraction. If the units do not match, then your answer will be incorrect, unless you get lucky. So lets begin to set this up. 2/\$3 = 15/\$x. We are trying to find the cost, so we will call that \$x. When doing proportions, you will cross multiply. So you will multiply 2(x) giving you 2x and 15(\$3) giving you 45. Now you have 2x=45. You want to isolate the variable to get your answer and to do that, you will divide 2 on both sides. When you divide 45 by 2, your answer is \$22.5. Normally for word problems, it's a good habit to write out a full sentence. Don't forget about that! Some teachers are very picky. ### Subject:Basic Math TutorMe Question: 2/3 + 1/12 = ? Inactive Natasha H. When adding and subtracting fractions, the first thing you want to look at is the denominator. In order to add and subtract fractions, the denominators must be the same number. In this example, the denominators are not the same numbers, so you will need to do some multiplying to convert 2/3 to get 12 on the bottom. So you know that 3x4 will give you 12, so you will multiple 2/3 top and bottom by 4. After multiplying you will get 8/12 + 1/12, which will give you 9/12. If the question asks to simplify, then don't forget to simplify. However, if the question does not specify further, then leaving the answer at 9/12 should be fine, unless other directed by your teacher. ### Subject:Algebra TutorMe Question: Find the distance between the points (-4 , -5) and (-1 , -1). Inactive Natasha H. First, understand what the question is asking. The question asks to find the distance, meaning they want you to find the slope. Slope is rise over run or in other words Y2-Y1 divided by X2-X1. Coordinate points are (x,y), so lets go ahead and label (-4,-5) as (x1,x2) and (-1,-1) as (x2,y2). After we do that we get -1-(-5) divided by -1-(-4), which will give us -1+5 over -1+4. We get that answer because negative times negative gives us positive. Then we proceed to adding the numbers and we get 4/3. ## Contact tutor Send a message explaining your needs and Natasha will reply soon. Contact Natasha
# Soal Matematika Kelas 6 Sd Bab 6 SiSTEM Koordinat - Www.Bimbelbrilian.Com Approved & Edited by ProProfs Editorial Team The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes. \| By Catherine Halcomb Catherine Halcomb Community Contributor Quizzes Created: 1379 \| Total Attempts: 5,981,962 Questions: 30 \| Attempts: 5,815 Settings . • 1. ### Perhatikan gambar peta desa di bawah ini untuk mengisi soal nomor 1 – 6! Keterangan : P = Letak rumah Andi Q = Letak rumah Alia R = Letak rumah Sinta S = Letak rumah Siska X = Letak rumah Budi Y = Letak rumah Angga Z = Letak rumah Rahma Letak rumah Andi adalah pada koordinat .... • A. E3 • B. B4 • C. M5 • D. P4 D. P4 Explanation The correct answer is P4 because the location of Andi's house is indicated by the letter P on the map. The number 4 represents the column number, indicating that Andi's house is located in column 4. Therefore, the correct coordinate for Andi's house is P4. Rate this question: • 2. ### Perhatikan gambar peta desa di bawah ini untuk mengisi soal nomor 1 – 6! Keterangan : P = Letak rumah Andi Q = Letak rumah Alia R = Letak rumah Sinta S = Letak rumah Siska X = Letak rumah Budi Y = Letak rumah Angga Z = Letak rumah Rahma Rumah Siska terleatak pada koordinat .... • A. N3 • B. N4 • C. J4 • D. B2 A. N3 Explanation The correct answer is N3 because it is the coordinate where the house of Siska is located on the village map. Rate this question: • 3. ### Perhatikan gambar peta desa di bawah ini untuk mengisi soal nomor 1 – 6! Keterangan : P = Letak rumah Andi Q = Letak rumah Alia R = Letak rumah Sinta S = Letak rumah Siska X = Letak rumah Budi Y = Letak rumah Angga Z = Letak rumah Rahma Rumah Rahmad terletak pada koordinat .... • A. Q4 • B. J2 • C. J1 • D. E3 C. J1 Explanation The correct answer is J1. Based on the given information, the location of Rahmad's house can be determined by the coordinates J1 on the map. Rate this question: • 4. ### Perhatikan gambar peta desa di bawah ini untuk mengisi soal nomor 1 – 6! Keterangan : P = Letak rumah Andi Q = Letak rumah Alia R = Letak rumah Sinta S = Letak rumah Siska X = Letak rumah Budi Y = Letak rumah Angga Z = Letak rumah Rahma Rumah yang terletak pada koordinat B2 adalah rumah .... • A. Budi • B. Angga • C. Sinta • D. Alia A. Budi Explanation The correct answer is Budi because the house located at coordinate B2 is represented by the letter X, which corresponds to Budi in the given key. Rate this question: • 5. ### Perhatikan gambar peta desa di bawah ini untuk mengisi soal nomor 1 – 6! Keterangan : P = Letak rumah Andi Q = Letak rumah Alia R = Letak rumah Sinta S = Letak rumah Siska X = Letak rumah Budi Y = Letak rumah Angga Z = Letak rumah Rahma Rumah yang terletak pada koordinat J4 adalah rumah .... • A. Siska • B. Budi • C. Alia • D. Rahma C. Alia Explanation The correct answer is Alia because the house located at coordinate J4 on the map is labeled as Q, which represents the location of Alia's house. Rate this question: • 6. ### Perhatikan gambar peta desa di bawah ini untuk mengisi soal nomor 1 – 6! Keterangan : P = Letak rumah Andi Q = Letak rumah Alia R = Letak rumah Sinta S = Letak rumah Siska X = Letak rumah Budi Y = Letak rumah Angga Z = Letak rumah Rahma Rumah yang terletak pada baris ke-3 adalah rumah .... • A. Rahma dan Alia • B. Budi dan Siska • C. Angga dan Alia • D. Angga dan Siska D. Angga dan Siska Explanation The correct answer is "Angga dan Siska". This can be determined by looking at the map and identifying the houses located on the third row. The houses labeled "Angga" and "Siska" are the only ones on the third row, so they are the correct answer. Rate this question: • 7. ### Garis sumbu X pada bidang koordinat merupakan garis .... • A. Vertikal • B. Tengah • C. Pinggir • D. Horizontal D. Horizontal Explanation The correct answer is horizontal because the X-axis on a coordinate plane represents the horizontal direction. It is the line that runs left to right and is perpendicular to the Y-axis. The X-axis is used to plot the values of the independent variable in a graph or coordinate system. Rate this question: • 8. ### Perhatikan bidang koordinat dibawah ini untuk mengisi soal nomor 8 – 15! Koordinat titik A adalah .... • A. (6,-2) • B. (6,3) • C. (3,6) • D. (-3,6) D. (-3,6) Explanation The correct answer is (-3,6) because it is the only option that matches the given coordinates. The x-coordinate is -3 and the y-coordinate is 6, which is consistent with the answer choice. The other options have different x or y coordinates that do not match the given coordinates. Rate this question: • 9. ### Perhatikan bidang koordinat dibawah ini Koordinat titik C adalah ..... • A. (2,5) • B. (5,2) • C. (-2,-5) • D. (-5,-2) B. (5,2) Explanation The coordinate point C is given as (5,2). This means that the x-coordinate of C is 5 and the y-coordinate is 2. Rate this question: • 10. ### Perhatikan bidang koordinat dibawah ini Kooordinat titik F adalah .... • A. (-4,-5) • B. (-4,-5) • C. (2,-6) • D. (-3,1) A. (-4,-5) Explanation The coordinate point F is given as (-4,-5). Rate this question: • 11. ### Perhatikan bidang koordinat dibawah ini Titik yang terletak pada koordinat (2,-6) adalah .... • A. Titik A • B. Titik B • C. Titik D • D. Titik E D. Titik E Explanation The point located at coordinates (2,-6) is represented by the point E on the coordinate plane. Rate this question: • 12. ### Perhatikan bidang koordinat dibawah ini untuk mengisi soal nomor 8 – 15! Titik yang terletak pada koordinat (-3,1) adalah .... • A. Titik F • B. Titik E • C. Titik G • D. Titik H C. Titik G Explanation The point (-3,1) is located in the third quadrant of the coordinate plane. In this quadrant, both the x-coordinate and the y-coordinate are negative. Among the given options, the only point that satisfies this condition is Titik G. Therefore, the correct answer is Titik G. Rate this question: • 13. ### Perhatikan bidang koordinat dibawah ini untuk mengisi soal nomor 8 – 15! Titik yang terletak pada garis X = 5 adalah .... • A. Titik A dan B • B. Titik C dan D • C. Titik E dan F • D. Titik G dan H B. Titik C dan D Explanation The correct answer is "Titik C dan D" because the points on the line X = 5 have their x-coordinate fixed at 5, while the y-coordinate can vary. Therefore, only the points with x-coordinate equal to 5, which are Titik C and D, are located on the line X = 5. Rate this question: • 14. ### Perhatikan bidang koordinat dibawah ini untuk mengisi soal nomor 8 – 15! Titik yang terletak pada garis Y = 3 adalah .... • A. Titik A dan H • B. B dan C • C. B dan H • D. F dan B C. B dan H Explanation The correct answer is B and H. This is because the given equation Y = 3 represents a horizontal line passing through the y-coordinate 3. The points A and H lie on this line as their y-coordinate is 3. Therefore, the points B and H are the correct answer options. Rate this question: • 15. ### Perhatikan bidang koordinat dibawah ini untuk mengisi soal nomor 8 – 15! Titik A dan titik G sama-sama menempati .... • A. Y = 4 • B. X = 3 • C. Y = -3 • D. X = -3 D. X = -3 Explanation The correct answer is X = -3 because the point A and point G both have the same x-coordinate, which is -3. This means that both points lie on the vertical line that passes through the point (-3, 0) on the coordinate plane. The y-coordinate of the points does not matter in this case because the question is specifically asking about the x-coordinate. Rate this question: • 16. ### Perhatikan gambar di bawah ini untuk mengisi soal nomor 16 – 18 ! Bangun yang dibentuk dari gambar di atas adalah ..... • A. Persegi panjang • B. Segitiga • C. Jajar genjang • D. Trapesium D. Trapesium Explanation The shape formed by the given figure is a trapezium. A trapezium is a quadrilateral with one pair of parallel sides. In the given figure, the top and bottom sides are parallel, while the other two sides are not. Therefore, the correct answer is trapezium. Rate this question: • 17. ### Perhatikan gambar di bawah ini untuk mengisi soal nomor 16 – 18 ! Titik-titik koordinat di bawah ini yang tidak sesuai gambar di atas adalah .... • A. A (-3,3) • B. (3,3) • C. C (-5,5) • D. D (-5.-5) C. C (-5,5) Explanation The given answer, C (-5,5), is incorrect because it does not match the coordinates shown in the image. In the image, there are only positive values for both the x and y coordinates, while C (-5,5) has a negative x coordinate. Therefore, C (-5,5) is the coordinate that does not match the image. Rate this question: • 18. ### Perhatikan gambar di bawah ini untuk mengisi soal nomor 16 – 18 ! Luas bangun di atas adalah .... • A. 64 satuan • B. 60 satuan • C. 62 satuan • D. 58 satuan A. 64 satuan Explanation The correct answer is 64 satuan because the shape in the image appears to be a square, and the area of a square is calculated by multiplying the length of one side by itself. Since the length of one side is not given in the question, we can assume that it is equal to the width of the shape. Therefore, the area of the shape is equal to the width multiplied by the width, which gives us 64 satuan. Rate this question: • 19. ### Jika Titik-titik A, B dan C dihubungkan dengan garis-garis. Maka akan membentuk sebuah segitiga yang mempunyai luas .... • A. 30 satuan • B. 16,5 satuan • C. 15 satuan • D. 33 satuan B. 16,5 satuan Explanation The correct answer is 16,5 satuan. The question states that if points A, B, and C are connected with lines, it will form a triangle. The area of a triangle is calculated by multiplying the base by the height and dividing it by 2. Since the question does not provide the base and height of the triangle, it is not possible to calculate the exact area. Therefore, the best explanation for the correct answer is that it is based on an assumption or previous knowledge about the specific triangle being referred to in the question. Rate this question: • 20. ### Perhatikangambar di bawah ini untuk mengisi soal nomor 19 – 21! Agar gambar titikt-titik di atas menjadi sebuah gambar layang-layang, maka koordinat titik D adalah berada pada titik .... • A. (4, -3) • B. (-3, 4) • C. (-2, 4) • D. (4, -2) D. (4, -2) Explanation To form a rhombus, the opposite sides must be parallel and equal in length. Looking at the given coordinates, the points (4, -3), (-3, 4), and (-2, 4) do not form a rhombus because their opposite sides are not parallel. However, the points (4, -2) and (-2, 4) do form a rhombus because their opposite sides are parallel and equal in length. Therefore, the correct answer is (4, -2). Rate this question: • 21. ### Perhatikangambar di bawah ini untuk mengisi soal nomor 19 – 21! Agar gambar titikt-titik di atas menjadi sebuah gambar jajar genjang, maka koordinat titik D adalah berada pada titik .... • A. (-2,-1) • B. (-1,-2) • C. (-4,-3) • D. (-3,-4) B. (-1,-2) Explanation The correct answer is (-1,-2) because in order for the dots to form a parallelogram, the opposite sides must be parallel and equal in length. By looking at the given coordinates, the opposite sides of the parallelogram formed by the dots are (-2,-1) to (-1,-2) and (-4,-3) to (-3,-4). These sides are parallel and equal in length, satisfying the condition for a parallelogram. Therefore, the correct coordinate for point D is (-1,-2). Rate this question: • 22. ### Bangun yang terbentuk dari titik P (2,1), Q (10,1), R (10,5) dan S (2,5) adalah bangun .... • A. Persegi • B. Persegi panjang • C. Layang-layang • D. Belah ketupat B. Persegi panjang Explanation The given points P (2,1), Q (10,1), R (10,5), and S (2,5) form a rectangle. A rectangle is a quadrilateral with all four angles equal to 90 degrees. The sides opposite to each other are equal in length, which is true for the given points. Therefore, the correct answer is "Persegi panjang" which translates to "rectangle" in English. Rate this question: • 23. ### Bangun yang terbentuk dari titik A (-4,1), B (-1,4) , C (4,4) dan D (1,1) adalah bangun .... • A. Segiempat • B. Layang-layang • C. Jajar genjang • D. Segitiga C. Jajar genjang Explanation The given points A, B, C, and D form a parallelogram because opposite sides are parallel and equal in length. A parallelogram is a type of quadrilateral, and one specific type of parallelogram is a parallelogram with equal sides, which is known as a rhombus. However, since the question does not specify that the sides are equal, the correct answer is a general term for a quadrilateral with opposite sides parallel, which is a jajar genjang in Indonesian. Rate this question: • 24. ### Bangun yang terbentuk dari titik M (0,3) , N (0,-3) dan O (7,0) adalah bangun .... • A. Segitiga sama sisi • B. Segitiga sembarang • C. Segitiga sama kaki • D. Segitiga siku-siku C. Segitiga sama kaki Explanation The given points M (0,3), N (0,-3), and O (7,0) form a triangle. To determine the type of triangle formed, we can calculate the lengths of the sides. The distance between M and N is 6 units (3-(-3)), the distance between N and O is 7 units (7-0), and the distance between O and M is also 7 units (7-0). Since all three sides have different lengths, the triangle formed is not equilateral. However, since two sides have the same length (7 units), the triangle is isosceles or "segitiga sama kaki" in Indonesian. Rate this question: • 25. ### Luas bangun di atas adalah .... • A. 40 satuan • B. 80 satuan • C. 100 satuan • D. 120 satuan B. 80 satuan Explanation The given answer, 80 satuan, is the correct answer because it is the only option that matches the given information. The question asks for the "luas bangun di atas" (area of the shape above), but the shape or any other information about it is not provided. Therefore, without any additional information, it is not possible to determine the correct answer. Rate this question: • 26. ### Keliling bangun di atas adalah .... • A. 30 satuan • B. 10 satuan • C. 36 satuan • D. 80 satuan C. 36 satuan Explanation The correct answer is 36 satuan because the question is asking for the perimeter of a shape, and 36 satuan is the only option that represents a measurement of perimeter. The other options do not make sense in the context of finding the perimeter of a shape. Rate this question: • 27. ### Titik-titik koordina yang tidak sesuai dengan gambar di atas adalah .... • A. A (-4,3) • B. B (-4,5) • C. C (5,6) • D. D (6,-3) C. C (5,6) Explanation The given answer, C (5,6), is incorrect because the coordinates (5,6) do not match the points in the picture above. Rate this question: • 28. ### Jika pada gambar bidang koordinat di atas di tambahi satu titik pada titik koordinat (9,-3) maka bangun yang terbentuk dari titik-titik di atas adalah bangun .... • A. Layang-layang • B. Segilima • C. Trapesium • D. Jajargenjang C. Trapesium Explanation Adding a point at coordinates (9,-3) to the given coordinate plane would form a trapezium. Rate this question: • 29. • A. (2,6) • B. (6,-2) • C. (-2,6) • D. (-2,-6) B. (6,-2) Explanation When a point is reflected across the x-axis, the y-coordinate remains the same, but the sign changes. In this case, the original point is (6,2), and when it is reflected across the x-axis, the y-coordinate becomes -2. Therefore, the reflected point will be (6,-2). Rate this question: • 30. • A. (3,7 ) • B. (-7,-3) • C. (-7,3) • D. (3,-7) C. (-7,3) Explanation When a point is reflected across the y-axis, the x-coordinate remains the same but the sign of the y-coordinate is changed. In this case, the original point (7,3) is reflected across the y-axis, resulting in the point (-7,3). Therefore, the correct answer is (-7,3). Rate this question: Quiz Review Timeline + Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness. • Current Version • Mar 22, 2023 Quiz Edited by ProProfs Editorial Team • Oct 06, 2018 Quiz Created by Catherine Halcomb
Related Search Web Results brainly.com/question/17686758 The prime factorization of fifty is 2 times 25. Step-by-step explanation: So, the prime factors of 50 are 2 × 5 × 5 or 2 × 52, where 2 and 5 are the prime numbers. researchmaniacs.com/Calculator/PrimeFactors/Prime-Factorization-of-50.html To find all the prime factors of 50, divide it by the lowest prime number possible. Then divide that result by the lowest prime number possible. Keep doing this until the result itself is a prime number. The prime factorization of 50 will be all the prime numbers you used to divide, in addition to the last result, which is a prime number. The prime factorization of 50 is 2 × 5 × 5. To use a factor tree to find the prime factorization of a number, x, we use the following... See full answer below. The prime factorization of 50 is 2x5x5 or 2 x 52. The prime factorization of 18 is 2x3x3 or 2 x 32. The prime factorization of 32 is 2x2x2x2x2 or 25. Prime factorization is expressing a positive integer as product of its prime factors. Consider, for example the number 100. The factors of 100 are: 1, 2, 4, 5, 10, 20, 50 and 100. brilliant.org/wiki/prime-factorization A prime factor tree provides a pictorial representation of the prime factorization for a positive integer. Starting with the given integer N N N at the top of the tree, two branches are drawn toward two positive factors of N. N. N. The process is repeated for the numbers at the end of each branch that is drawn until each "leaf" is a prime number. www.mathportal.org/calculators/numbers-calculators/prime-factorization... Finding prime factorization and factor tree. Example: Find prime factorization of 60. Step 1: Start with any number that divides 60, in this we will use 10. So, $\color{blue}{60 = 6 \cdot 10}$. www.mathwarehouse.com/arithmetic/numbers/prime-number/prime-factorization... The prime factorization calculator can: Calculate the prime factorization of the number you type (Numbers above 10 million may or may not time out. Calculating the prime factorization of large numbers is not easy, but the calculator can handle pretty darn big ones!) Determine whether or not a number is prime; Create sieve of Erasthones for the ... www.wikihow.com/Find-Prime-Factorization Here's how to find the GCF of 30 and 36, using prime factorization: Find the prime factorizations of the two numbers. The prime factorization of 30 is 2 x 3 x 5. The prime factorization of 36 is 2 x 2 x 3 x 3. Find a number that appears on both prime factorizations. Cross it out once on each list and write it on a new line. www.dummies.com/education/math/using-prime-factorizations Using prime factorization to find the LCM. One method for finding the least common multiple (LCM) of a set of numbers is to use the prime factorizations of those numbers. Here’s how: List the prime factors of each number. Suppose you want to find the LCM of 18 and 24. List the prime factors of each number: 18 = 2 × 3 × 3. 24 = 2 × 2 × 2 × 3 Related Search
# Math fact: The sum of any number of consecutive odd whole numbers, beginning with 1, is a perfect square e.g. 1+3=4, 1+3+5=9, 1+3+5+7=16. ## Presentation on theme: "Math fact: The sum of any number of consecutive odd whole numbers, beginning with 1, is a perfect square e.g. 1+3=4, 1+3+5=9, 1+3+5+7=16."— Presentation transcript: Math fact: The sum of any number of consecutive odd whole numbers, beginning with 1, is a perfect square e.g. 1+3=4, 1+3+5=9, =16 Example 1: Estimating Square Roots of Numbers The 55 is between two integers. Name the integers. Explain your answer. 55 36, 49, 64, 81 List perfect squares near 55. 49 < 55 < 64 Find the perfect squares nearest 55. 49 < 55 < 64 Find the square roots of the perfect squares. 7 < 55 < 8 55 is between 7 and 8 because 55 is between 49 and 64. Example 2: Approximating Square Roots to the Nearest Hundredth Approximate √135 to the nearest hundredth. Step 1 Find the value of the whole number. 121 < 135 < 144 Find the perfect squares nearest 135. √121 < √135 < 144 Find the square roots of the perfect squares. 11 < 135 < 12 The number will be between 11 and 12. The whole number part of the answer is 11. Example 2 Continued Approximate √135 to the nearest hundredth. Step 2 Find the value of the decimal. Find the difference between the given number, 135, and the lower perfect square. 135 – 121 = 14 Find the difference between the greater perfect square and the lower perfect square. 144 – 121 = 23 1423 Write the difference as a ratio. Divide to find the approximate decimal value. 14 ÷ 23 ≈ 0.609 Approximate √135 to the nearest hundredth. Example 2 Continued Approximate √135 to the nearest hundredth. Step 3 Find the approximate value. Combine the whole number and decimal. = ≈ 11.61 Round to the nearest hundredth. The approximate value of to the nearest hundredth is Example 3: Using a Calculator to Estimate the Value of a Square Root Use a calculator to find Round to the nearest tenth. 600 ≈ … Use a calculator. 600 ≈ 24.5 Round to the nearest tenth. 600 rounded to the nearest tenth is 24.5. Similar presentations
# Sal's Sandwich Investigation 555 Words3 Pages Sal's Sandwich Shop sells wraps and sandwiches as part of its lunch specials. The profit on every sandwich is \$2 and the profit on every wrap is \$3. Sal made a profit of \$1,470 from lunch specials last month. The equation 2x + 3y = 1,470 represents Sal's profits last month, where x is the number of sandwich lunch specials sold and y is the number of wrap lunch specials sold. Change the equation to slope-intercept form. Identify the slope and y-intercept of the equation. Be sure to show all your work. To change the or any equation to slope-intercept form or to identify the Y intercept you must follow some steps to make sure you get the right answer. For this I am going to just break down the equation and show how I get my answer. First, we are going to just write out the equation, simple enough.…show more content… (-2) – 2 To find the value of 3y all we must do is add 1,470 and the negative factor of 2x (–2x) + 1,470 = {3y} Let “/” represent a fraction. Now we are going to find the value of 3y/3 (–2x/3) + (1,470/3) = {3y/3} Then we are going to find the value of Y (–2x/3) + 490 = {y} For this word problem, after following the steps to find it, the Y-intercept is {490} and the slope is {-2/3x} Describe how you would graph this line using the slope-intercept method. Be sure to write using complete sentences. Well in this case I guess I would find and graph the Y-intercept. And then I would take the pointer and move three times to the right of the graph and then two points to the bottom Write the equation in function notation. Explain what the graph of the function represents. Be sure to use complete sentences. For this I am going to write the equation as so down below: F(x) = -2x/3 +
Go Math Grade 2 Answer Key Chapter 5 2-Digit Subtraction answer key is useful for students who are preparing for their examinations and can download this pdf for free of cost. In this chapter, each and every question was explained in detail which helps students to understand easily. Go Math Grade 2 Answer Key Chapter 5 2-Digit Subtraction explains different types of questions on 2 Digit Subtraction. In this chapter, we can see different topics on Break Apart Ones to Subtract, Break Apart Numbers to Subtract, Model Regrouping for Subtraction, Model and Record 2-Digit Subtraction, 2-Digit Subtraction, etc. Those topics were being set up by the mathematical professionals as indicated by the most recent release. Look down this page to get the answers to all the inquiries. Click on the links to look at the subjects shrouded in this chapter 2-Digit Subtraction. Chapter: 9- 2-Digit Subtraction Lesson 1: Algebra • Break Apart Ones to Subtract Lesson 2: Algebra • Break Apart Numbers to Subtract Lesson 3: Model Regrouping for Subtraction Lesson 4: Model and Record 2-Digit Subtraction Lesson 5: 2-Digit Subtraction Lesson 5.5 2-Digit Subtraction 2-Digit Subtraction Homework & practice 5.5 Lesson 6: Practice 2-Digit Subtraction Mid-Chapter Checkpoint 2-Digit Subtraction Mid-Chapter Checkpoint Lesson 7: Rewrite 2-Digit Subtraction Lesson 8: Add to Find Differences Lesson 9: Problem Solving • Subtraction Lesson 10: Algebra • Write Equations to Represent Subtraction Lesson 11: Solve Multistep problems Chapter 5 Review/Test 2-Digit Subtraction Chapter 5 Review Test There are hundreds of different kinds of dragonflies. If 52 dragonflies are in a garden and 10 fly away, how many dragonflies are left? How many are left if 10 more fly away? The total number of files left is 32 files. Explanation: As there are hundreds of different kinds of dragonflies and if 52 dragonflies are in a garden and 10 flies away, so the number of files left is 52 – 10= 42 files and again there are 10 more flies away. So there will be 42 – 10= 32 files left. ### 2-Digit Subtraction Show What You Know Subtraction Patterns Subtract 2. Complete each subtraction sentence. Question 1. Question 2. 6 – __ = ___ On subtracting 6 – 2 we will get the result as 4. Explanation: By subtracting 2 with 6 we will get the result as 6 – 2= 4. Question 3. 5 – __ = __ On subtracting 5 – 2 we will get the result as 3. Explanation: By subtracting 2 with 5 we will get the result as 5 – 2= 3. Question 4. 4 – ___ = __ On subtracting 4 – 2 we will get the result as 2. Explanation: By subtracting 2 with 4 we will get the result as 4 – 2= 2. Question 5. 3 – __ = __ On subtracting 3 – 2 we will get the result as 1. Explanation: By subtracting 2 with 3 we will get the result as 3 – 2= 1. Question 6. 2 – _ = __ On subtracting 2 – 2 we will get the result as 0. Explanation: By subtracting 2 with 2 we will get the result as 2 – 2= 0. Subtraction Facts Write the difference. Question 7. On subtracting 8 – 5 we will get the result as 3. Explanation: By subtracting 5 with 8 we will get the result as 8 – 5= 3. Question 8. On subtracting 14 – 6 we will get the result as 8. Explanation: By subtracting 6 with 14 we will get the result as 14 – 6= 8. Question 9. On subtracting 9 – 6 we will get the result as 3. Explanation: By subtracting 6 with 9 we will get the result as 9 – 6= 3. Question 10. On subtracting 16 – 7 we will get the result as 9. Explanation: By subtracting 7 with 16 we will get the result as 16 – 7= 9. Question 11. On subtracting 12 – 6 we will get the result as 6. Explanation: By subtracting 6 with 12 we will get the result as 12 – 6= 6. Question 12. On subtracting 10 – 8 we will get the result as 2. Explanation: By subtracting 8 with 10 we will get the result as 10 – 8= 2. Tens and Ones Write how many tens and ones are in each model. Question 13. 54 __ tens __ ones The number of tens is 5 and the number of ones is 4. Explanation: In the above image, we can see 10 blocks with 5 rows. Which are 10 × 5= 50 blocks, so there are 5 tens and we can see 4 blocks which are 4 ones. So the total number of blocks is 50 + 4= 54 blocks. Question 14. 45 __ tens __ ones The number of tens is 4 and the number of ones is 5. Explanation: In the above image, we can see 10 blocks with 4 rows. Which are 10 × 4= 40 blocks, so there are 4 tens and we can see 5 blocks which are 5 ones. So the total number of blocks is 40 + 5= 45 blocks. ### 2-Digit Subtraction Vocabulary Builder Visualize It Fill in the boxes of the graphic organizer. Understand Vocabulary Draw a line to complete the sentence. 1. A digit can be 0,1,2,3,4,5,6,7,8, or 9. 2. You can regroup to trade 10 ones for 1 ten. 3. 20 ones are the same as 2 tens. Explanation: 1. A digit is a single symbol that is used to make numerals, so the digits can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 are the ten digits which we will use every day. 2. Here, regrouping is defined as the process of making and then carrying out the operation like addition with two-digit numbers or larger than the two-digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. We can regroup to trade 10 ones for 1 ten. And we can regroup to trade 10 ones for 1 ten. 3. 20 ones are the same as 2 tens. As two tens mean 2 × 10= 20 and 20 × 1= 20, so 20 ones are the same as 2 tens. ### 2-Digit Subtraction Game: Subtraction Search Materials Search • 3 sets of number cards 4 – 9 • 18 Play with a partner. 1. Shuffle all the cards. Place them face down in one stack. 2. Take one card. Find a square with a subtraction problem with this number as the difference. Your partner checks your answer. 3. If you are correct, place a on that square. If there is no match, skip your turn. 4. Take turns. The first player to have on all the squares wins. ### 2-Digit Subtraction Vocabulary Game Going Places with GOMATH! Words Bingo For 3 to 6 players Materials • 1 set of word cards • 1 Bingo board for each player • game markers How to Play 1. The caller chooses a word card and reads the word. Then the caller puts the word card in a second pile. 2. Players put a marker on the word each time they find it on their Bingo boards. 3. Repeat steps 1 and 2 until a player marks 5 boxes in a line going down, across, or on a slant and calls “Bingo.” 4. Check the answers. Have the player who said “Bingo” read the words aloud while the caller checks the word cards in the second pile. The Write Way Reflect Choose one idea. Write about it in the space below. • Explain how drawing quick pictures helps you add 2-digit numbers. • Tell about all the different ways you can add 2-digit numbers. • Write about a time that you helped explain something to a classmate. What was your classmate having trouble with? How did you help him or her? ### Lesson 5.1 Algebra • Break Apart Ones to Subtract Essential Question How does breaking apart a number make subtracting easier? Listen and Draw Write two addends for each sum. The two addends for the first image are 3 and 4. The two addends for the second image are 4 and 5. The two addends for the third image are 2 and 3. The two addends for the fourth image are 3 and 3. The two addends for the fifth image are 2 and 2. The two addends for the sixth image are 4 and 4. Explanation: As addend can be defined as the numbers are added together to get the sum. Here the given sum is 7, so to get the sum 7 we will take the two addends as 3 + 4 so we can get the sum as 7. The given sum is 9, so to get the sum 9 we will take the two addends 4 + 5 so we can get the sum as 9. The given sum is 5, so to get the sum 5 we will take the two addends 2 + 3 so we can get the sum as 5. The given sum is 6, so to get the sum 6 we will take the two addends 3 + 3 so we can get the sum as 6. the given sum is 4, so to get the sum 4 we will take the two addends 2 + 2 so we can get the sum as 4. The given sum is 8, so to get the sum 8 we will take the two addends 4 + 4 so we can get the sum as 8. Math Talk MATHEMATICAL PRACTICES Describe how you chose addends for each sum. The addends are chosen by the given sum. As addend can be defined as the numbers are added to together to get the sum. So we have chosen the two addends by the given sum. Model and Draw Break apart ones. Subtract in two steps. So, 63 − 7 = ___. The subtraction of 63 – 7 is 56. Explanation: Here, we have started at 63 and subtracted 3, to subtract 3 on the number line jump makes a jump from 63 to 60 and the size of the jump is 3 and we will get 60. Then we have to subtract 4, so we will start from 60 on the number line jump makes a jump from 60 to 56 and the size of the jump is 4 and we will get 56. So the subtraction of 63 – 7 is 56. Share and Show MATH BOARD Break apart ones to subtract. Write the difference. Question 1. The subtraction of 55 – 8 is 47. Explanation: Here, we have started at 55 and subtracted 5, to subtract 5 on the number line jump makes a jump from 55 to 50 and the size of the jump is 5 and we will get 50. Then we have to subtract 3, so we will start from 50 on the number line jump makes a jump from 50 to 47 and the size of the jump is 3 and we will get 47. So the subtraction of 55 – 8 is 47. Question 2. The subtraction of 42 – 5 is 37. Explanation: Here, we have started at 42 and subtracted 2, to subtract 2 on the number line jump makes a jump from 42 to 40 and the size of the jump is 2 and we will get 40. Then we have to subtract 3, so we will start from 40 on the number line jump makes a jump from 40 to 37 and the size of the jump is 3 and we will get 37. So the subtraction of 42 – 5 is 37. Question 3. 41 – 9 = __ The subtraction of 41 – 9 is 32. Explanation: Here, we have started at 41 and subtracted 4, to subtract 4 on the number line jump makes a jump from 41 to 37 and the size of the jump is 4 and we will get 37. Then we have to subtract 5, so we will start from 37 on the number line jump makes a jump from 37 to 32 and the size of the jump is 5 and we will get 32. So the subtraction of 41 – 9 is 32. Question 4. 53 – 6 = __ The subtraction of 53 – 6 is 47. Explanation: Here, we have started at 53 and subtracted 3, to subtract 3 on the number line jump makes a jump from 53 to 50 and the size of the jump is 3 and we will get 50. Then we have to subtract 3, so we will start from 50 on the number line jump makes a jump from 50 to 47 and the size of the jump is 3 and we will get 47. So the subtraction of 53 – 6 is 47. Question 5. 44 – 7 = __ The subtraction of 44 – 7 is 37. Explanation: Here, we have started at 44 and subtracted 3, to subtract 3 on the number line jump makes a jump from 44 to 41 and the size of the jump is 3 and we will get 41. Then we have to subtract 4, so we will start from 41 on the number line jump makes a jump from 41 to 37 and the size of the jump is 4 and we will get 37. So the subtraction of 44 – 7 is 37. Question 6. 52 – 8 = __ The subtraction of 52 – 8 is 44. Explanation: Here, we have started at 52 and subtracted 4, to subtract 4 on the number line jump makes a jump from 52 to 48 and the size of the jump is 4 and we will get 48. Then we have to subtract 4, so we will start from 48 on the number line jump makes a jump from 48 to 44 and the size of the jump is 4 and we will get 44. So the subtraction of 52 – 8 is 44. Break apart ones to subtract. Write the difference. Question 7. The subtraction of 75 – 7 is 68. Explanation: Here, we have started at 75 and subtracted 3, to subtract 3 on the number line jump makes a jump from 75 to 72 and the size of the jump is 3 and we will get 72. Then we have to subtract 4, so we will start from 72 on the number line jump makes a jump from 72 to 68 and the size of the jump is 4 and we will get 68. So the subtraction of 75 – 7 is 68. Question 8. The subtraction of 86 – 8 is 78. Explanation: Here, we have started at 86 and subtracted 4, to subtract 4 on the number line jump makes a jump from 86 to 82 and the size of the jump is 4 and we will get 82. Then we have to subtract 4, so we will start from 82 on the number line jump makes a jump from 82 to 78 and the size of the jump is 4 and we will get 78. So the subtraction of 86 – 8 is 78. Question 9. 82 – 5 = __ The subtraction of 82 – 5 is 77. Explanation: Here, we have started at 82 and subtracted 3, to subtract 3 on the number line jump makes a jump from 82 to 79 and the size of the jump is 3 and we will get 79. Then we have to subtract 2, so we will start from 79 on the number line jump makes a jump from 79 to 77 and the size of the jump is 2 and we will get 77. So the subtraction of 82 – 5 is 77. Question 10. 83 – 7 = __ The subtraction of 83 – 7 is 76. Explanation: Here, we have started at 83 and subtracted 3, to subtract 3 on the number line jump makes a jump from 83 to 80 and the size of the jump is 3 and we will get 80. Then we have to subtract 4, so we will start from 80 on the number line jump makes a jump from 80 to 76 and the size of the jump is 4 and we will get 76. So the subtraction of 83 – 7 is 76. Question 11. 72 – 7 = __ The subtraction of 72 – 7 is 65. Explanation: Here, we have started at 72 and subtracted 3, to subtract 3 on the number line jump makes a jump from 72 to 69 and the size of the jump is 3 and we will get 69. Then we have to subtract 4, so we will start from 69 on the number line jump makes a jump from 69 to 65 and the size of the jump is 4 and we will get 65. So the subtraction of 72 – 7 is 65. Question 12. 76 – 9 = __ The subtraction of 76 – 9 is 67. Explanation: Here, we have started at 76 and subtracted 5, to subtract 5 on the number line jump makes a jump from 76 to 71 and the size of the jump is 5 and we will get 71. Then we have to subtract 4, so we will start from 71 on the number line jump makes a jump from 71 to 67 and the size of the jump is 4 and we will get 67. So the subtraction of 76 – 9 is 67. Question 13. 85 – 8 = __ The subtraction of 85 – 8 is 77. Explanation: Here, we have started at 85 and subtracted 4, to subtract 4 on the number line jump makes a jump from 85 to 81 and the size of the jump is 4 and we will get 81. Then we have to subtract 4, so we will start from 81 on the number line jump makes a jump from 81 to 77 and the size of the jump is 4 and we will get 77. So the subtraction of 85 – 8 is 77. Question 14. 71 – 6 = __ The subtraction of 71 – 6 is 65. Explanation: Here, we have started at 71 and subtracted 3, to subtract 3 on the number line jump makes a jump from 71 to 68 and the size of the jump is 3 and we will get 68. Then we have to subtract 3, so we will start from 68 on the number line jump makes a jump from 68 to 65 and the size of the jump is 3 and we will get 65. So the subtraction of 71 – 6 is 65. Question 15. THINK SMARTER Cheryl brought 27 bagels for the bake sale. Mike brought 24 bagels. They sold all but 9 of them. How many bagels did they sell? __ bagels The total number of bagels sold is 42 bagels. Explanation: Cheryl brought 27 bagels for the bake sale and Mike brought 24 bagels, so the total number of bagels did Mike and Cheryl bought is 27 + 24= 51 bagels. And they sold all but 9 bagels are left with them, so the number of bagels sold is 51 – 9= 42 bagels. So the total number of bagels sold is 42 bagels. Question 16. MATHEMATICAL PRACTICE Analyze Lexi has 8 fewer crayons than Ken. Ken has 45 crayons. How many crayons does Lexi have? __ crayons The number of crayons Lexi had is 37 crayons. Explanation: As Lexi has 8 fewer crayons than Ken and Ken has 45 crayons, so Lexi has 45 – 8= 37 crayons. Problem Solving • Applications Write or draw to explain. Question 17. Cheryl built a toy train with 27 train cars. Then she added 18 more train cars. How many train cars are on the toy train now? __ train cars The number of train cars is on the toy train is 9 train cars. Explanation: Cheryl built a toy train with 27 train cars and then she added 18 more train cars, so the number of train cars are on the toy train is 27 – 18= 9 train cars. Question 18. MATHEMATICAL PrACTICE Analyze Samuel had 46 marbles. He gave some marbles to a friend and has 9 marbles left. How many marbles did Samuel give to his friend? __ marbles The number of marbles did Samuel gave to his friend is 37 marbles. Explanation: Samuel had 46 marbles and he gave some marbles to a friend and has 9 marbles left, so the number of marbles did Samuel gave to his friend is 46 – 9= 37 marbles. So the number of marbles did Samuel gave to his friend is 37 marbles. Question 19. THINK SMARTER Matthew had 73 blocks. He gave 8 blocks to his sister. How many blocks does Matthew have now? Draw or write to show how to solve the problem. Matthew has __ blocks now. The total number of blocks Matthew had is 65 blocks. Explanation: Matthew had 73 blocks and he gave 8 blocks to his sister, so the number of blocks did Matthew had is 73 – 8= 65 blocks. So after giving 8 blocks to his sister Matthew had 65 blocks. By subtracting the number of blocks did Matthew had and the number of blocks did he gave to his sister we had solved the problem. TAKE HOME ACTIVITY • Ask your child to describe how to find 34 − 6. The subtraction of 6 from 34 is 28. And there are two tens and eight ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 6 from 34, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 6 from 34 is 34 – 6= 28 and there are two tens and eight ones. ### Algebra • Break Apart Ones to Subtract Homework & Practice 5.1 Break apart ones to subtract. Write the difference. Question 1. 36 – 7 = __ The subtraction of 36 – 7 is 29. Explanation: Here, we have started at 36 and subtracted 3, to subtract 3 on the number line jump makes a jump from 36 to 33 and the size of the jump is 3 and we will get 33. Then we have to subtract 4, so we will start from 33 on the number line jump makes a jump from 33 to 29 and the size of the jump is 4 and we will get 29. So the subtraction of 36 – 7 is 56. Question 2. 35 – 8 = __ The subtraction of 35 – 8 is 27. Explanation: Here, we have started at 35 and subtracted 4, to subtract 4 on the number line jump makes a jump from 35 to 31 and the size of the jump is 4 and we will get 31. Then we have to subtract 4, so we will start from 31 on the number line jump makes a jump from 31 to 27 and the size of the jump is 4 and we will get 27. So the subtraction of 35 – 8 is 27. Question 3. 37 – 9 = __ The subtraction of 37 – 9 is 28. Explanation: Here, we have started at 37 and subtracted 4, to subtract 4 on the number line jump makes a jump from 37 to 33 and the size of the jump is 4 and we will get 33. Then we have to subtract 5, so we will start from 33 on the number line jump makes a jump from 33 to 28 and the size of the jump is 5 and we will get 28. So the subtraction of 37 – 9 is 28. Question 4. 41 – 6 = __ The subtraction of 41 – 6 is 35. Explanation: Here, we have started at 41 and subtracted 3, to subtract 3 on the number line jump makes a jump from 41 to 38 and the size of the jump is 3 and we will get 38. Then we have to subtract 3, so we will start from 35 on the number line jump makes a jump from 38 to 35 and the size of the jump is 3 and we will get 35. So the subtraction of 41 – 6 is 35. Question 5. 44 – 5 = __ The subtraction of 44 – 5 is 39. Explanation: Here, we have started at 44 and subtracted 3, to subtract 3 on the number line jump makes a jump from 44 to 41 and the size of the jump is 3 and we will get 41. Then we have to subtract 2, so we will start from 41 on the number line jump makes a jump from 41 to 39 and the size of the jump is 2 and we will get 39. So the subtraction of 44 – 5 is 39. Question 6. 33 – 7 = __ The subtraction of 33 – 7 is 26. Explanation: Here, we have started at 33 and subtracted 3, to subtract 3 on the number line jump makes a jump from 33 to 30 and the size of the jump is 3 and we will get 30. Then we have to subtract 4, so we will start from 30 on the number line jump makes a jump from 30 to 26 and the size of the jump is 4 and we will get 26. So the subtraction of 33 – 7 is 26. Question 7. 32 – 4 = __ The subtraction of 32 – 4 is 28. Explanation: Here, we have started at 32 and subtracted 2, to subtract 2 on the number line jump makes a jump from 32 to 30 and the size of the jump is 2 and we will get 30. Then we have to subtract 2, so we will start from 30 on the number line jump makes a jump from 30 to 28 and the size of the jump is 2 and we will get 28. So the subtraction of 32 – 4 is 28. Question 8. 31 – 6 = __ The subtraction of 31 – 6 is 25. Explanation: Here, we have started at 31 and subtracted 3, to subtract 3 on the number line jump makes a jump from 31 to 28 and the size of the jump is 3 and we will get 28. Then we have to subtract 3, so we will start from 28 on the number line jump makes a jump from 28 to 25 and the size of the jump is 3 and we will get 25. So the subtraction of 31 – 6 is 25. Problem Solving Choose a way to solve. Write or draw to explain. Question 9. Beth had 44 marbles. She gave 9 marbles to her brother. How many marbles does Beth have now? __ marbles. The total number of marbles does Beth had is 35 marbles. Explanation: Beth had 44 marbles and she gave 9 marbles to her brother and the remaining marbles do Beth had are 44 – 9= 35 marbles. Question 10. WRITE Math Draw a number line and show how to find the difference for 24 – 6 using the break apart method in this lesson. Lesson Check Question 1. What is the difference? 58 – 9 = __ The difference of 58 – 9 is 49. Explanation: Here, we have started at 58 and subtracted 4, to subtract 4 on the number line jump makes a jump from 58 to 54 and the size of the jump is 4 and we will get 54. Then we have to subtract 5, so we will start from 54 on the number line jump makes a jump from 54 to 49 and the size of the jump is 5 and we will get 49. So the subtraction of 58 – 9 is 49. Spiral Review Question 2. What is the difference? 14 – 6 = __ The difference of 14 – 6 is 8. Explanation: Here, we have started at 14 and subtracted 3, to subtract 3 on the number line jump makes a jump from 14 to 11 and the size of the jump is 3 and we will get 11. Then we have to subtract 3, so we will start from 11 on the number line jump makes a jump from 11 to 3 and the size of the jump is 3 and we will get 8. So the subtraction of 14 – 6 is 8. Question 3. What is the sum? 3 + 6 + 2 = ___ The sum of the three numbers is 11. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the three numbers 3 + 6 + 2 is 11. Question 4. What is the sum? 64 + 7 = __ The sum of the two numbers is 71. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the two numbers 64 + 7 is 71. Question 5. What is the sum? 56 + 18 = __ The sum of the two numbers is 74. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the two numbers 56 + 18 is 74. ### Lesson 5.2 Algebra • Break Apart Numbers to Subtract Essential Question How does breaking apart a number make subtracting easier? Listen and Draw Draw jumps on the number line to show how to break apart the number to subtract. Math Talk MATHEMATICAL PRACTICES Describe a Method For one of the problems, describe what you did. Model and Draw Break apart the number you are subtracting into tens and ones. Subtract 10. Next, subtract 2 to get to 60. Then subtract 5 more. So, 72 – 12 = ___ Share and Show MATH BOARD Break apart the number you are subtracting. Write the difference. Question 1. The subtraction of 43 – 18 is 25. Explanation: Here, we will break apart the number that we are subtracting. So first we will break into tens and ones. So here we will break 18 as 10 and 8 and we will subtract 10 and we will get the result as 33. Then we will start from 33 and subtract 3 to get to 30 and then subtract 5 more. So we will get the result as 25, and we will place it on the number line. So the subtraction of 43 – 18 is 25. Question 2. The subtraction of 45 – 14 is 31. Explanation: Here, we will break apart the number that we are subtracting. So first we will break into tens and ones. So here we will break 14 as 10 and 4 and we will subtract 10 and we will get the result as 35. Then we will start from 35 and subtract 2 to get to 33 and then subtract 2 more. So we will get the result as 31, and we will place it on the number line. So the subtraction of 45 – 14 is 31. Question 3. 46 – 17 = __ The subtraction of 46 – 17 is 29. Explanation: Here, we will break apart the number that we are subtracting. So first we will break into tens and ones. So here we will break 17 as 10 and 7 and we will subtract 10 and we will get the result as 36. Then we will start from 33 and subtract 3 to get to 30 and then subtract 5 more. So we will get the result as 25, and we will place it on the number line. So the subtraction of 46 – 17 is 29. Question 4. 44 – 16 = __ The subtraction of 44 – 16 is 28. Explanation: Here, we will break apart the number that we are subtracting. So first we will break into tens and ones. So here we will break 18 as 10 and 6 and we will subtract 10 and we will get the result as 34. Then we will start from 34 and subtract 4 to get to 30 and then subtract 2 more. So we will get the result as 28, and we will place it on the number line. So the subtraction of 44 – 16 is 28. Break apart the number you are subtracting. Write the difference. Question 5. The subtraction of 57 – 15 is 42. Explanation: Here, we will break apart the number that we are subtracting. So first we will break into tens and ones. So here we will break 15 as 10 and 5 and we will subtract 10 and we will get the result as 47. Then we will start from 47 and subtract 3 to get to 44 and then subtract 2 more. So we will get the result as 42, and we will place it on the number line. So the subtraction of 57 – 15 is 42. Question 6. The subtraction of 63 – 17 is 46. Explanation: Here, we will break apart the number that we are subtracting. So first we will break into tens and ones. So here we will break 17 as 10 and 7 and we will subtract 10 and we will get the result as 53. Then we will start from 53 and subtract 3 to get to 50 and then subtract 4 more. So we will get the result as 46, and we will place it on the number line. So the subtraction of 63 – 17 is 46. Question 7. 68 – 19 = __ The subtraction of 68 – 19 is 49. Explanation: Here, we will break apart the number that we are subtracting. So first we will break into tens and ones. So here we will break 19 as 10 and 9 and we will subtract 10 and we will get the result as 58. Then we will start from 58 and subtract 8 to get to 50 and then subtract 1 more. So we will get the result as 49, and we will place it on the number line. So the subtraction of 68 – 19 is 49. Question 8. 61 – 18 = __ The subtraction of 61 – 18 is 43. Explanation: Here, we will break apart the number that we are subtracting. So first we will break into tens and ones. So here we will break 18 as 10 and 8 and we will subtract 10 and we will get the result as 51. Then we will start from 51 and subtract 1 to get to 50 and then subtract 7 more. So we will get the result as 43, and we will place it on the number line. So the subtraction of 61 – 18 is 43. Question 9. THINK SMARTER Jane has 53 toys in a box. She takes some toys out. Now there are 36 toys in the box. How many toys did Jane take out of the box? The number of toys did Jane take out of the box is 17 toys. Explanation: As Jane has 53 toys in a box and she takes some toys out, so Now there are 36 toys in the box. So to find how many toys did Jane take out of the box we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So The subtraction of 36 from 53 is 53 – 36= 17. So the number of toys did Jane take out of the box is 17 toys. Question 10. GO DEEPER Look at Tom’s steps to solve a problem. Solve this problem in the same way. 42 – 15 = ? The subtraction of 42 – 15 is 27. Explanation: As we can see in the above image that the subtraction is done by the break apart number process, so the given problem is 42 – 15, so here we will break the 15 into 10 and 5, and then we will break 5 apart into 3 and 2. So first we will subtract 10 then the result will be 42 – 10 is 32 and then we will 2 then the result will be 32 – 2= 30. Now we will subtract 3 then the result will be 30 – 3= 27. So the subtraction of 42 – 15 is 27. Problem Solving • Applications Question 11. 38 people are in the library. Then 33 more people go into the library. How many people are in the library now? __ people The total number of people in the library is 71 people. Explanation: As there are 38 people in the library and then 33 more people go into the library, so to find the number of people in the library we will perform addition. So we will add 38 + 33 is 71. The total number of people in the library is 71 people. Question 12. MATHEMATICAL PRACTICE Analyze Alex has 24 toys in a chest. He takes some toys out of the chest. Then there are 16 toys in the chest. How many toys did he take out of the chest? __ toys There are 8 toys did Alex take out of the chest. Explanation: As Alex has 24 toys in a chest and he takes some toys out of the chest. Then there are 16 toys in the chest, so to find the number of toys did Alex has to take out of the chest we will perform subtraction. So we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 16 from 24 is 24 – 16= 8 and there are 8 toys did Alex take out of the chest. Question 13. THINK SMARTER Gail has two piles of newspapers. There are 32 papers in the first pile. There are 19 papers in the second pile. How many more papers are in the first pile than in the second pile? __ more papers Write or draw to explain how you solved the problem. The number of papers more in the first pile than the second pile is 13 more papers. Explanation: As Gail has two piles of newspapers and there are 32 papers in the first pile and there are 19 papers in the second pile. So to know how many papers are in the first pile than the second pile we will perform subtraction which is 32 – 19= 13 more papers. TAKE HOME ACTIVITY • Ask your child to write a subtraction story that uses 2-digit numbers. We will perform subtraction by breaking the two-digit number into tens place digit and ones place digit. Then the other number will be placed below the number. If the minuend number is less than the subtrahend number then we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So by this, we will perform the subtraction. ### Algebra • Break Apart Numbers to Subtract Homework & practice 5.2 Break apart the number you are subtracting. Write the difference. Question 1. 81 – 14 = __ The subtraction of 81 – 14 is 67. Explanation: Here, we have started at 81 and subtracted 7, to subtract 7 on the number line jump makes a jump from 81 to 74 and the size of the jump is 7 and we will get 74. Then we have to subtract 7, so we will start from 74 on the number line jump makes a jump from 74 to 67 and the size of the jump is 7 and we will get 67. So the subtraction of 81 – 14 is 67. Question 2. 84 – 16 = __ The subtraction of 84 – 16 is 68. Explanation: Here, we have started at 84 and subtracted 8, to subtract 8 on the number line jump makes a jump from 84 to 76 and the size of the jump is 8 and we will get 76. Then we have to subtract 8, so we will start from 76 on the number line jump makes a jump from 76 to 68 and the size of the jump is 8 and we will get 68. So the subtraction of 84 – 16 is 68. Question 3. 77 – 14 = __ The subtraction of 77 – 14 is 63. Explanation: Here, we have started at 77 and subtracted 7, to subtract 7 on the number line jump makes a jump from 77 to 70 and the size of the jump is 7 and we will get 70. Then we have to subtract 7, so we will start from 70 on the number line jump makes a jump from 70 to 63 and the size of the jump is 7 and we will get 63. So the subtraction of 77 – 14 is 63. Question 4. 83 – 19 = __ The subtraction of 83 – 19 is 64. Explanation: Here, we have started at 83 and subtracted 10, to subtract 10 on the number line jump makes a jump from 83 to 73 and the size of the jump is 10 and we will get 73. Then we have to subtract 9, so we will start from 73 on the number line jump makes a jump from 73 to 64 and the size of the jump is 9 and we will get 64. So the subtraction of 83 – 19 is 64. Question 5. 81 – 17 = __ The subtraction of 81 – 17 is 64. Explanation: Here, we have started at 81 and subtracted 10, to subtract 10 on the number line jump makes a jump from 81 to 71 and the size of the jump is 10 and we will get 71. Then we have to subtract 7, so we will start from 71 on the number line jump makes a jump from 71 to 64 and the size of the jump is 7 and we will get 64. So the subtraction of 81 – 17 is 64. Question 6. 88 – 13 = __ The subtraction of 88 – 13 is 75. Explanation: Here, we have started at 88 and subtracted 3, to subtract 3 on the number line jump makes a jump from 83 to 80 and the size of the jump is 3 and we will get 80. Then we have to subtract 10, so we will start from 80 on the number line jump makes a jump from 80 to 75 and the size of the jump is 10 and we will get 75. So the subtraction of 88 – 13 is 75. Question 7. 84 – 19 = __ The subtraction of 84 – 19 is 65. Explanation: Here, we have started at 84 and subtracted 10, to subtract 10 on the number line jump makes a jump from 84 to 74 and the size of the jump is 10 and we will get 74. Then we have to subtract 9, so we will start from 74 on the number line jump makes a jump from 74 to 65 and the size of the jump is 9 and we will get 65. So the subtraction of 84 – 19 is 65. Question 8. 86 – 13 = __ The subtraction of 86 – 13 is 73. Explanation: Here, we have started at 86 and subtracted 7, to subtract 7 on the number line jump makes a jump from 86 to 79 and the size of the jump is 7 and we will get 79. Then we have to subtract 6, so we will start from 79 on the number line jump makes a jump from 79 to 73 and the size of the jump is 6 and we will get 73. So the subtraction of 86 – 13 is 73. Question 7. 84 – 19 = __ The subtraction of 84 – 19 is 65. Explanation: Here, we have started at 84 and subtracted 10, to subtract 10 on the number line jump makes a jump from 84 to 74 and the size of the jump is 10 and we will get 74. Then we have to subtract 9, so we will start from 74 on the number line jump makes a jump from 74 to 65 and the size of the jump is 9 and we will get 65. So the subtraction of 84 – 19 is 65. Question 8. 86 – 18 = __ The subtraction of 86 – 18 is 68. Explanation: Here, we have started at 84 and subtracted 9, to subtract 9 on the number line jump makes a jump from 86 to 77 and the size of the jump is 9 and we will get 77. Then we have to subtract 9, so we will start from 77 on the number line jump makes a jump from 77 to 68 and the size of the jump is 9 and we will get 68. So the subtraction of 86 – 18 is 68. Problem Solving Solve. Write or draw to explain. Question 9. Mr. Pearce bought 43 plants. He gave 14 plants to his sister. How many plants does Mr. Pearce have now? __ plants The number of plants does Mr. Pearce had is 29 plants. Explanation: Mr. Pearce bought 43 plants and he gave 14 plants to his sister. So the number of plants does Mr. Pearce has now is 43 – 14. Here, we have started at 43 and subtracted 7, to subtract 7 on the number line jump makes a jump from 43 to 36 and the size of the jump is 7 and we will get 36. Then we have to subtract 7, so we will start from 36 on the number line jump makes a jump from 7 to 29 and the size of the jump is 7 and we will get 29. So the subtraction of 43 – 14 is 29. So the number of plants does Mr. Pearce had is 29 plants. Question 10. WRITE Math Draw a number line and show how to find the difference for 36 – 17 using the break apart method in this lesson. Lesson Check Question 1. What is the difference? 63 – 19 = __ The subtraction of 63 – 19 is 44. Explanation: Here, we have started at 63 and subtracted 10, to subtract 10 on the number line jump makes a jump from 63 to 53 and the size of the jump is 10 and we will get 53. Then we have to subtract 9, so we will start from 53 on the number line jump makes a jump from 53 to 44 and the size of the jump is 9 and we will get 44. So the subtraction of 63 – 19 is 44. Spiral Review Question 2. What is the sum? The sum of the two numbers is 37. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the two numbers 14 + 23 is 37. Question 3. What is the sum? 8 + 7 = _ The sum of the two numbers is 15. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the two numbers 8 + 7 is 15. Question 4. Write a related subtraction fact for 6 + 8 = 14. ____ Question 5. John has 7 kites. Annie has 4 kites. How many kites do they have altogether? __ kites The number of kites does they together have is 7 + 4= 11 kites. Explanation: John has 7 kites and Annie has 4 kites, so the number of kites do they together have is 7 + 4= 11 kites. ### Lesson 5.3 Model Regrouping for Subtraction Essential Question When do you regroup in subtraction? We will regroup in subtraction when the minuend digit is less than the subtrahend digit. Then we will take carry forward to the lesser digit and then we will perform subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. Listen and Draw Use to model the problem. Draw quick pictures to show your model. Math Talk MATHEMATICAL PRACTICES Describe why you traded a tens block for 10 ones blocks. Model and Draw How do you subtract 26 from 53? The subtraction of 53 and 26 is 27. Explanation: To perform subtraction for 26 from 53 we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So The subtraction of 26 from 53 is 27. Step 1 Show 53. Are there enough ones to subtract 26? Step 2 If there are not enough ones, regroup 1 ten as 10 ones. Step 3 Subtract 6 ones from 13 ones. Step 4 Subtract the tens. Write the tens and ones. Write the difference. Share and Show MATH BOARD Draw to show the regrouping. Write the difference two ways. Write the tens and ones. Write the number. Question 1. Subtract 13 from 41. __ tens __ ones __ The subtraction of 13 from 41 is 28. And there are two tens and eight ones. Explanation: In the above image, we can see that there are four tens blocks and one one’s block. So here we need to subtract 13 from 41, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 13 from 41 is 41 – 13= 28 and there are two tens and eight ones. Question 2. Subtract 9 from 48. __ tens __ ones __ The subtraction of 9 from 48 is 39. And there are three tens and nine ones. Explanation: In the above image, we can see that there is four tens blocks and eight one’s block. So here we need to subtract 9 from 48, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 9 from 48 is 48 – 9= 39 and there are three tens and nine ones. Question 3. Subtract 28 from 52. __ tens __ ones __ The subtraction of 28 from 52 is 24. And there are two tens and four ones. Explanation: In the above image, we can see that there is five tens blocks and two one’s block. So here we need to subtract 28 from 48, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 28 from 52 is 52 – 28= 24 and there are two tens and four ones. Draw to show the regrouping. Write the difference two ways. Write the tens and ones. Write the number. Question 4. Subtract 8 from 23 __ tens __ ones __ The subtraction of 8 from 23 is 15. And there are one ten and five ones. Explanation: In the above image, we can see that there is one tens block and five ones block. So here we need to subtract 8 from 23, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 8 from 23 is 23 – 8= 15 and there are one ten and five ones. Question 5. Subtract 36 from 45. __ tens __ ones __ The subtraction of 36 from 45 is 9. And there are no tens and nine ones. Explanation: In the above image, we can see that there is five tens blocks and two one’s block. So here we need to subtract 36 from 45, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 36 from 45 is 45 – 36= 9 and there are no tens and nine ones. Question 6. Subtract 6 from 43. __ tens __ ones __ The subtraction of 6 from 43 is 37. And there are three tens and seven ones. Explanation: In the above image, we can see that there is four tens blocks and three one’s block. So here we need to subtract 6 from 43, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 6 from 43 is 43 – 6= 37 and there are three tens and seven ones. Question 7. Subtract 39 from 67 __ tens __ ones __ The subtraction of 39 from 67 is 28. And there are two tens and eight ones. Explanation: In the above image, we can see that there is six tens blocks and seven ones block. So here we need to subtract 28 from 48, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 39 from 67 is 67 – 39= 28 and there are two tens and eight ones. Question 8. Subtract 21 from 50. __ tens __ ones __ The subtraction of 21 from 50 is 29. And there are two tens and nine ones. Explanation: In the above image, we can see that there is five tens blocks and zero ones in ones block. So here we need to subtract 21 from 50, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 21 from 50 is 50 – 21= 29 and there are two tens and nine ones. Question 9. Subtract 29 from 56 __ tens __ ones __ The subtraction of 29 from 56 is 27. And there are two tens and seven ones. Explanation: In the above image, we can see that there are five tens blocks and two one’s blocks. So here we need to subtract 28 from 48, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 29 from 56 is 56 – 29= 27 and there are two tens and seven ones. Question 10. GO DEEPER Draw to find what number was subtracted from 53. Subtract __ from 53. 3 tens 4 ones 34 Problem Solving • Applications Write or draw to explain. Question 11. THINK SMARTER Billy has 18 fewer marbles than Sara. Sara has 34 marbles. How many marbles does Billy have? __ marbles The total number of marbles does Billy had is 16 marbles. Explanation: As Billy has 18 fewer marbles than Sara and Sara has 34 marbles, so here fewer marbles means we will do subtraction. So the number of marbles do Billy had is 34 – 18= 16 marbles. Here we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the total number of marbles does Billy had is 16 marbles. Question 12. THINK SMARTER There are 67 toy animals in the store. Then the clerk sells 19 toy animals. How many toy animals are in the store now? Draw to show how to find the answer. __ toy animals The number of toys in the store is 48 toys. Explanation: As there are 67 toy animals in the store and then the clerk sells 19 toy animals, so to find how many toys in the store is we will perform subtraction. Here we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the number of toys in the store is 67 – 19= 48 toys. Describe how you solved the problem. __________________________ __________________________ __________________________ The problem is solved by using regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. TAKE HOME ACTIVITY • Ask your child to write a subtraction story and then explain how to solve it. We will perform subtraction by breaking the two-digit number into tens place digit and ones place digit. Then the other number will be placed below the number. If the minuend number is less than the subtrahend number then we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So by this, we will perform the subtraction. ### Model Regrouping for Subtraction Homework & Practice 5.3 Draw to show the regrouping. Write the difference two ways. Write the tens and ones. Write the number. Question 1. Subtract 9 from 35 __ tens __ ones __ The subtraction of 9 from 35 is 26. And there are two tens and six ones. Explanation: In the above image, we can see that there are three tens blocks and five one’s block. So here we need to subtract 9 from 35, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 9 from 35 is 35 – 9= 26 and there are two tens and six ones. Question 2. Subtract 14 from 52. __ tens __ ones __ The subtraction of 9 from 35 is 26. And there are two tens and six ones. Explanation: In the above image, we can see that there are three tens blocks and five one’s block. So here we need to subtract 9 from 35, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 9 from 35 is 35 – 9= 26 and there are two tens and six ones. Problem Solving Choose a way to solve. Write or draw to explain. Question 3. Explanation: As Mr. Ortega made 51 cookies and he gave 14 cookies away. So to find the number of cookies does Ortega had we will do regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 51 – 14 is 37 cookies. The cookies that Question 4. WRITE Math Draw a quick picture for 37. Draw to show how you would subtract 19 from 37. Write to explain what you did. __________________________ __________________________ __________________________ The subtraction of 37 – 19 is 18. The subtraction is done by the regrouping of subtraction. Explanation: The subtraction of 37 – 19 is 18. The subtraction is done by the regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 37 – 19 is 18. Lesson Check Question 1. Subtract 9 from 36. What is the difference? The subtraction of 9 from 36 is 27. And there are two tens and seven ones. Explanation: In the above image, we can see that there are three tens blocks and six one’s blocks. So here we need to subtract 9 from 36, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 9 from 36 is 36 – 9= 27 and there are two tens and seven ones. Question 2. Subtract 28 from 45. What is the difference? The subtraction of 28 from 45 is 17. And there are one ten and seven ones. Explanation: In the above image, we can see that there are four tens blocks and five one’s blocks. So here we need to subtract 28 from 45, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 28 from 45 is 45 – 28= 17 and there are one ten and seven ones. Spiral Review Question 3. What is the difference? 51 – 8 = __ The subtraction of 8 from 51 is 43. And there are four tens and three ones. Explanation: In the above image, we can see that there are three tens blocks and six one’s blocks. So here we need to subtract 8 from 51, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 8 from 51 is 51 – 8= 43 and there are four tens and three ones. Question 4. What is the sum? 38 + 35 = __ The sum of the two numbers is 73. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the two numbers 38 + 35 is 73. Question 5. What is the sum? The sum of the three numbers is 90. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the three numbers 63 + 18 + 9 is 90. ### Lesson 5.4 Model and Record 2-Digit Subtraction Essential Question How do you record 2-digit subtraction? We will perform subtraction by breaking the two-digit number into tens place digit and ones place digit. Then the other number will be placed below the number. If the minuend number is less than the subtrahend number then we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So by this, we will perform the subtraction. Listen and Draw Use to model the problem. Draw quick pictures to show your model. Math Talk MATHEMATICAL PRACTICE Explain a Method Did you trade blocks in your model? Explain why or why not. Model and Draw Trace over the quick pictures in the steps. Subtract. Share and Show MATH BOARD Draw a quick picture to solve. Write the difference. Question 1. The subtraction of 15 from 47 is 32. And there are three ten and two ones. Explanation: In the above image, we can see that there are four tens and seven ones. So here we need to subtract 15 from 47, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 15 from 47 is 47 – 15= 32 and there are three ten and two ones. Question 2. The subtraction of 18 from 32 is 14. And there are one ten and four ones. Explanation: In the above image, we can see that there are three tens and five ones. So here we need to subtract 18 from 32, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 18 from 32 is 32 – 18= 14 and there are one ten and four ones. Draw a quick picture to solve. Write the difference. Question 3. The subtraction of 29 from 35 is 6. And there are zero tens and six ones. Explanation: In the above image, we can see that there are three tens and five ones. So here we need to subtract 29 from 35, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 29 from 35 is 35 – 29= 6 and there are zero tens and six ones. Question 4. The subtraction of 5 from 28 is 23. And there are two tens and three ones. Explanation: In the above image, we can see that there are two tens and eight ones. So here we need to subtract 5 from 28, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 5 from 28 is 28 – 5= 23 and there are two ten and three ones. Question 5. The subtraction of 26 from 53 is 27. And there are two tens and seven ones. Explanation: In the above image, we can see that there are five tens and three ones. So here we need to subtract 26 from 53, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 26 from 53 is 53 – 26= 27 and there are two tens and seven ones. Question 6. The subtraction of 13 from 32 is 19. And there are one ten and nine ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 13 from 32, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 13 from 32 is 32 – 13= 19 and there are one ten and nine ones. Question 7. GO DEEPER There are 16 robins in the trees. 24 more fly in. Then 28 robins fly away. How many robins are still in the trees? __ robbins The total number of robins are still on the trees is 12 robins. Explanation: As there are 16 robins on the tree and 24 more fly-ins. So the number of robins is 24 + 16= 40. Now 28 robins fly away, so the total number of robins are still on the trees is 40 – 28= 12 robins. Problem Solving • Applications Question 8. THINK SMARTER Claire’s puzzle has 85 pieces. She has used 46 pieces so far. How many puzzle pieces have not been used yet? __ puzzle pieces The number of puzzle pieces that have not been used is 39 puzzle pieces. Explanation: As Claire’s puzzle has 85 pieces and she has used 46 pieces so far. So the number of puzzle pieces that have not been used is 85 – 46= 39 puzzle pieces. Question 9. MATHEMATICAL PRACTICE Analyze There were some people at the park. 24 people went home. Then there were 19 people at the park. How many people were at the park before? __ people The total number of people at the park before is 43 people. Explanation: As there were some people at the park and 24 people went home. Then there were 19 people at the park, so the total number of people at the park before is 24 + 19= 43 people. Question 10. THINK SMARTER Mr. Sims has a box of 44 erasers. He gives 28 erasers to his students. How many erasers does Mr. Sims have now? Show how you solved the problem. __ erasers The number of erasers does Mr. Sims has now is 16 erasers. Explanation: As Mr. Sims has a box of 44 erasers and he gives 28 erasers to his students. So the number of erasers does Mr. Sims has now is 44 – 28= 16 erasers. Here we have solved by using regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 28 from 44 is 44 – 28= 16 erasers. The number of erasers does Mr. Sims has now is 16 erasers. TAKE HOME ACTIVITY • Write 73 − 28 on a sheet of paper. Ask your child if he or she would regroup to find the difference. The subtraction of 73 – 28 is 45. Explanation: The subtraction of 73 – 28 is 45. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 73 – 28 is 45. ### Model and Record 2-Digit Subtraction Homework & Practice 5.4 Draw a quick picture to solve. Write the difference. Question 1. The subtraction of 17 from 43 is 26. And there are two tens and six ones. Explanation: In the above image, we can see that there are four tens and three ones. So here we need to subtract 17 from 43, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 17 from 43 is 43 – 17= 26 and there are two tens and six ones. Question 2. The subtraction of 29 from 38 is 9. And there are zero tens and nine ones. Explanation: In the above image, we can see that there are three tens and eight ones. So here we need to subtract 29 from 38, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 29 from 38 is 38 – 29= 9 and there are zero tens and nine ones. Problem Solving Solve. Write or draw to explain. Question 3. Kendall has 63 stickers. Her sister has 57 stickers. How many more stickers does Kendall have than her sister? __ more stickers. The number of stickers does Kendall has than her sister is 6 more stickers. Explanation: As Kendall has 63 stickers and her sister has 57 stickers, so the number of stickers does Kendall has than her sister is 63 – 57= 6 more stickers. Here we will perform regrouping of subtraction. Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 57 from 63 is 63 – 57= 9 stickers. Question 4. WRITE Math Draw a quick picture to show the number 24. Then draw a quick picture to show 24 after you have regrouped 1 ten as 1-ones. Explain how both pictures show the same number, 24. __________________________ __________________________ __________________________ Lesson Check Question 1. What is the difference? The subtraction of 18 from 47 is 29. And there are two tens and nine ones. Explanation: In the above image, we can see that there are four tens and seven ones. So here we need to subtract 18 from 47, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 18 from 47 is 47 – 18= 29 and there are two tens and nine ones. Question 2. What is the difference? The subtraction of 29 from 33 is 4. And there are zero tens and four ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 29 from 33, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 29 from 33 is 33 – 29= 4 and there are zero tens and four ones. Spiral Review Question 3. What is the difference? 10 – 6 = __ The difference is 4. Explanation: The difference between 10 and 6 is 4. Question 4. What is the sum? 16 + 49 = __ The sum of the two numbers is 65. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the two numbers 16 + 49 is 65. Question 5. What is the sum? 28 + 8 = __ The sum of the two numbers is 36. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the two numbers 28 + 8 is 36. Question 6. What is the difference? 52 – 6 = __ The difference is 46. Explanation: The difference between 52 and 6 is 46. ### Lesson 5.5 2-Digit Subtraction Essential Question How do you record the steps when subtracting 2-digit numbers? We will perform subtraction by breaking the two-digit number into tens place digit and ones place digit. Then the other number will be placed below the number. If the minuend number is less than the subtrahend number then we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So by this, we will perform the subtraction. Listen and Draw Draw a quick picture to model each problem. Math Talk MATHEMATICAL PRACTICES Use Reasoning Explain how you know when to regroup. We will perform regrouping when the minuend number is less than the subtrahend number then we will perform regrouping of subtraction. And Regrouping in subtraction is a process of exchanging one ten into ten ones. Model and Draw Share and Show MATH BOARD Regroup if you need to. Write the difference. Question 1. The subtraction of 14 from 31 is 19. And there are one ten and nine ones. Explanation: In the above image, we can see that there are three tens and one in ones place. So here we need to subtract 14 from 31, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 14 from 31 is 31 – 14= 19 and there are one ten and nine ones. Question 2. The subtraction of 21 from 56 is 35. And there are three tens and five ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 21 from 56, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 21 from 56 is 56 – 21= 35 and there are three tens and five ones. Question 3. The subtraction of 35 from 72 is 37. And there are three tens and seven ones. Explanation: In the above image, we can see that there are seven tens and two ones. So here we need to subtract 35 from 72, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 35 from 72 is 72 – 35= 37 and there are three tens and seven ones. Regroup if you need to. Write the difference. Question 4. The subtraction of 14 from 23 is 9. And there are zero tens and nine ones. Explanation: In the above image, we can see that there are two tens and three one’s. So here we need to subtract 14 from 23, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 14 from 23 is 23 – 14= 9 and there are zero tens and nine ones. Question 5. The subtraction of 57 from 87 is 30. And there are three ten and zero ones. Explanation: In the above image, we can see that there are eight tens and seven one’s. So here we need to subtract 57 from 87, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 57 from 87 is 87 – 57= 30 and there are three ten and zero ones. Question 6. The subtraction of 18 from 34 is 16. And there are one ten and six ones. Explanation: In the above image, we can see that there are three tens and four one’s. So here we need to subtract 18 from 32, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 18 from 34 is 34 – 18= 16 and there are one ten and six ones. Question 7. The subtraction of 13 from 61 is 48. And there are four ten and eight ones. Explanation: In the above image, we can see that there are six tens and one ones. So here we need to subtract 13 from 61, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 13 from 61 is 61 – 13= 48 and there are four ten and eight ones. Question 8. The subtraction of 18 from 45 is 27. And there are two tens and seven ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 13 from 32, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 18 from 45 is 45 – 18= 27 and there are two ten and nine ones. Question 9. The subtraction of 36 from 52 is 16. And there are one ten and six ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 36 from 52, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 36 from 52 is 52 – 36= 16 and there are one ten and six ones. Question 10. The subtraction of 13 from 32 is 19. And there are one ten and nine ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 13 from 32, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 13 from 32 is 32 – 13= 19 and there are one ten and nine ones. Question 11. The subtraction of 43 from 75 is 32. And there are three tens and two ones. Explanation: In the above image, we can see that there are seven tens and five ones. So here we need to subtract 43 from 75, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 43 from 75 is 75 – 43= 32 and there are three tens and two ones. Question 12. The subtraction of 27 from 56 is 29. And there are two tens and nine ones. Explanation: In the above image, we can see that there are five tens and six ones. So here we need to subtract 27 from 56, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 27 from 56 is 56 – 27= 29 and there are two tens and nine ones. Question 13. The subtraction of 29 from 94 is 65. And there are six tens and five ones. Explanation: In the above image, we can see that there are nine tens and four one’s. So here we need to subtract 29 from 94, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 29 from 94 is 94 – 29= 65 and there are six tens and five ones. Question 14. The subtraction of 39 from 87 is 48. And there are four tens and eight ones. Explanation: In the above image, we can see that there are eight tens and seven one’s. So here we need to subtract 39 from 87, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 39 from 87 is 87 – 39= 48 and there are four tens and eight ones. Question 15. The subtraction of 46 from 83 is 37. And there are three tens and seven ones. Explanation: In the above image, we can see that there are eight tens and three one’s. So here we need to subtract 46 from 83, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 46 from 83 is 83 – 46= 37 and there are three tens and seven ones. Question 16. THINK SMARTER Spencer wrote 5 fewer stories than Katie. Spencer wrote 18 stories. How many stories did Katie write? __ stories The number of stories written by Katie is 13 stories. Explanation: As Spencer wrote 5 fewer stories than Katie and Spencer wrote 18 stories. As Spencer wrote 5 fewer stories than Katie so we will perform subtraction. So the number of stories written by Katie is 18 – 5= 13 stories. Problem Solving • Applications Question 17. MATHEMATICAL PRACTICE Explain a Method Circle the problems below that you could use mental math to solve. 54 – 10 = __ 63 – 27 = __ 93 – 20 = __ 39 – 2 = __ 41 – 18 = __ 82 – 26 = __ __________________________ __________________________ __________________________ Explanation: Question 18. THINK SMARTER There are 34 chickens in the barn. If 16 chickens go outside into the yard, how many chickens will still be in the barn? Circle the number from the box to make the sentence true. The number of chickens will be in the barn will be 18 chickens. Explanation: As there are 34 chickens in the barn and if 16 chickens go outside into the yard then the number of chickens will be in the barn will be 34 – 16= 18 chickens. Here we have performed regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the number of chickens will be in the barn will be 34 – 16= 18 chickens. TAKE HOME ACTIVITY • Ask your child to write a 2-digit subtraction problem with no regrouping needed. Have your child explain why he or she chose those numbers. The subtraction of 56 – 23 is 33. Explanation: Given that we need to perform the subtraction without regrouping, so we need to take the minuend number greater than the subtrahend. Then we need not perform the regrouping, so we will take the numbers to be subtracted be 56 – 23= 33. ### 2-Digit Subtraction Homework & practice 5.5 Regroup if you need to. Write the difference. Question 1. The subtraction of 28 from 47 is 19. And there are one ten and nine ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 28 from 47, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 28 from 47 is 47 – 28= 19 and there are one ten and nine ones. Question 2. The subtraction of 18 from 33 is 15. And there are one ten and five ones. Explanation: In the above image, we can see that there are three tens and three ones. So here we need to subtract 18 from 33, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 18 from 33 is 33 – 18= 15 and there are one ten and five ones. Question 3. The subtraction of 14 from 28 is 19. And there are one ten and four ones. Explanation: In the above image, we can see that there are two tens and eight ones. So here we need to subtract 14 from 28, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 14 from 28 is 28 – 14= 14 and there are one ten and four ones. Question 4. The subtraction of 19 from 66 is 47. And there are four ten and seven ones. Explanation: In the above image, we can see that there are six tens and six ones. So here we need to subtract 19 from 66, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 19 from 66 is 66 – 19= 47 and there are four ten and seven ones. Question 5. The subtraction of 26 from 77 is 51. And there are five tens and one in one’s place. Explanation: In the above image, we can see that there are seven tens and seven ones. So here we need to subtract 26 from 77, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 26 from 77 is 77 – 26= 51 and there are five tens and nine one in one’s place. Question 6. The subtraction of 34 from 58 is 24. And there are two tens and four ones. Explanation: In the above image, we can see that there are five tens and eight ones. So here we need to subtract 34 from 58, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 34 from 58 is 58 – 34= 24 and there are two tens and four ones. Question 7. The subtraction of 25 from 52 is 27. And there are two tens and seven ones. Explanation: In the above image, we can see that there are five tens and five ones. So here we need to subtract 25 from 52, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 25 from 52 is 52 – 25= 27 and there are two tens and seven ones. Question 8. The subtraction of 49 from 87 is 38. And there are three tens and eight ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 49 from 87, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 49 from 87 is 87 – 49= 38 and there are three tens and eight ones. Problem Solving Solve. Write or draw to explain. Question 9. Mrs. Paul bought 32 erasers. She gave 19 erasers to students. How many erasers does she still have? The number of erasers Mrs. Paul had is 13 erasers. Explanation: As Mrs. Paul bought 32 erasers and she gave 19 erasers to students. So to find how many erasers does she still has is we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the number of erasers Mrs. Paul had is 32 – 9= 13 erasers. Question 10. WRITE Math Write a few sentences about different ways to show subtraction for a problem like 32 – 15. __________________________ __________________________ Lesson Check Question 1. What is the difference? The subtraction of 39 from 48 is 9. And there are zero tens and nine ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 39 from 48, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 39 from 48 is 48 – 39= 9 and there are zero tens and nine ones. Question 2. What is the difference? The subtraction of 66 from 84 is 18. And there are one ten and eight ones. Explanation: In the above image, we can see that there are eight tens and four ones. So here we need to subtract 66 from 84, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 66 from 84 is 84 – 66= 18 and there are one ten and eight ones. Spiral Review Question 3. What is the difference? The subtraction of 19 from 32 is 13. And there are one ten and three ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 13 from 32, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 19 from 32 is 32 – 19= 13 and there are one ten and three ones. Question 4. Write an addition fact that will give the same sum as 8 + 7. 10 + __ The number to get the sum as 15 is we should add 5 to the number 10. Explanation: The sum of 8 + 7 is 15. So to get the 15 as a result of the given number, we will add 5 to the number 10. So we will get the result as 15. Question 5. 27 boys and 23 girls go on a field trip to the museum. How many children go to the museum? __ children The number of children who went to the museum is 50 children. Explanation: As there are 27 boys 23 girls on the field trip to the museum, so the number of children who went to the museum is 27 + 23= 50 children Question 6. There were 17 berries in the basket. Then 9 berries are eaten. How many berries are there now? __ berries The number of berries will be 8 berries. Explanation: As there are 17 berries in the basket and 9 berries are eaten, then the number of berries will be 17 – 9= 8 berries. Here we have performed regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the number of berries will be 17 – 9= 8 berries ### Lesson 5.6 Practice 2-Digit Subtraction Essential Question How do you record the steps when subtracting 2-digit numbers? We will perform subtraction by breaking the two-digit number into tens place digit and ones place digit. Then the other number will be placed below the number. If the minuend number is less than the subtrahend number then we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So by this, we will perform the subtraction. Listen and Draw Choose one way to solve the problem. Draw or write to show what you did. Math Talk MATHEMATICAL PRACTICES Describe a different way that you could have solved the problem. Model and Draw Carmen had 50 game cards. Then she gave 16 game cards to Theo. How many game cards does Carmen have now? The number of game cards will be 50 – 16= 34 game cards. Explanation: As the Carmen had 50 game cards and she gave 16 game cards to Theo, then the number of game cards will be 50 – 16= 34 game cards. Here we have performed regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the number of game cards will be 50 – 16= 34 game cards. Step 1 Look at the ones. There are not enough ones to subtract 6 from 0. So, regroup Step 2 Subtract the ones. 10 – 6 = 4 Step 3 Subtract the tens. 4 – 1 = 3 Share and Show MATH BOARD Write the difference. Question 1. The subtraction of 19 from 38 is 19. And there are one ten and nine ones. Explanation: In the above image, we can see that there are three tens and eight ones. So here we need to subtract 19 from 38, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 19 from 38 is 38 – 19= 19 and there are one ten and nine ones. Question 2. The subtraction of 32 from 65 is 33. And there are three tens and three ones. Explanation: In the above image, we can see that there are six tens and five ones. So here we need to subtract 32 from 65, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 32 from 65 is 65 – 32= 33 and there are three tens and three ones. Question 3. The subtraction of 12 from 50 is 38. And there are three tens and eight ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 12 from 50, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 12 from 50 is 50 – 12= 38 and there are three tens and eight ones. Question 4. The subtraction of 4 from 23 is 19. And there are one ten and nine ones. Explanation: In the above image, we can see that there are two tens and three ones. So here we need to subtract 4 from 23, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 4 from 23 is 23 – 4= 19 and there are one ten and nine ones. Question 5. The subtraction of 38 from 70 is 32. And there are three tens and two ones. Explanation: In the above image, we can see that there are seven tens and zero one. So here we need to subtract 38 from 70, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 38 from 70 is 70 – 38= 32 and there are three tens and two ones. Question 6. The subtraction of 17 from 52 is 35. And there are three tens and five ones. Explanation: In the above image, we can see that there are five tens and two one. So here we need to subtract 17 from 52, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 17 from 52 is 52 – 17= 35 and there are three tens and five ones. Write the difference. Question 7. The subtraction of 24 from 41 is 17. And there are one ten and seven ones. Explanation: In the above image, we can see that there are four tens and one in ones place. So here we need to subtract 24 from 41, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 24 from 41 is 41 – 24= 17 and there are one ten and nine ones. Question 8. The subtraction of 16 from 58 is 42. And there are four tens and two ones. Explanation: In the above image, we can see that there are five tens and eight ones. So here we need to subtract 16 from 58, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 16 from 58 is 58 – 16= 42 and there are four tens and two ones. Question 9. The subtraction of 13 from 60 is 47. And there are four tens and seven ones. Explanation: In the above image, we can see that there are six tens and zero ones. So here we need to subtract 13 from 60, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 13 from 60 is 60 – 13= 47 and there are four tens and seven ones. Question 10. The subtraction of 47 from 52 is 5. And there are zero tens and five ones. Explanation: In the above image, we can see that there are five tens and two ones. So here we need to subtract 47 from 52, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 47 from 52 is 52 – 47= 5 and there are zero tens and five ones. Question 11. The subtraction of 46 from 72 is 26. And there are two tens and six ones. Explanation: In the above image, we can see that there are seven tens and two ones So here we need to subtract 46 from 72, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 46 from 72 is 72 – 46= 26 and there are two tens and six ones. Question 12. The subtraction of 6 from 37 is 31. And there are three tens and one in ones place. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 6 from 37, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 6 from 37 is 37 – 6= 31 and there are three tens and one in ones place. Question 13. The subtraction of 46 from 74 is 28. And there are two tens and eight ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 46 from 74, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 46 from 74 is 74 – 46= 28 and there are two tens and eight ones. Question 14. The subtraction of 18 from 90 is 72. And there are seven tens and two ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 18 from 90, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 18 from 90 is 90 – 18= 72 and there are seven tens and two ones. Question 15. GO DEEPER Write the missing numbers in the subtraction problems. The regrouping for each problem is shown. The missing values is 75 and 28 and 83 and 58. Explanation: Here, to find the missing values we will find with the help of the carry forward numbers. As the carry forward are 6 and 15 so the minuend is 75 as the result is 47, so the other number will be 75 – 47= 28. As the carry forward are 7 and 13 so the minuend is 83 as the result is 25, so the other number will be 83 – 58= 25. Question 16. THINK SMARTER Adam takes 38 rocks out of a box. There are 23 rocks left in the box. How many rocks were in the box to start? __ rocks The number of rocks were in the box to start is 61 rocks. Explanation: Adam takes 38 rocks out of a box and there are 23 rocks left in the box. So to find the number of rocks were in the box we will add the rocks that adam had took and the number of rocks left in the box. So the number of rocks were in the box to start is 38 + 28= 61 rocks. TAKE HOME ACTIVITY • Ask your child to show you one way to find 80 − 34. The subtraction of 34 from 80 is 46. And there are one ten and nine ones. Explanation: In the above image, we can see that there are eight tens and zero ones. So here we need to subtract 34 from 80, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 34 from 80 is 80 – 34= 46 and there are four tens and six ones. ### Practice 2-Digit Subtraction Homework & Practice 5.6 Write the difference. Question 1. The subtraction of 18 from 50 is 32. And there are three tens and two ones. Explanation: In the above image, we can see that there are five tens and zero ones. So here we need to subtract 18 from 50, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 18 from 50 is 50 – 18= 32 and there are three tens and two ones. Question 2. The subtraction of 13 from 32 is 19. And there are one ten and nine ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 13 from 32, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 13 from 32 is 32 – 13= 19 and there are one ten and nine ones. Question 3. The subtraction of 13 from 32 is 19. And there are one ten and nine ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 13 from 32, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 13 from 32 is 32 – 13= 19 and there are one ten and nine ones. Question 4. The subtraction of 13 from 32 is 19. And there are one ten and nine ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 13 from 32, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 13 from 32 is 32 – 13= 19 and there are one ten and nine ones. Question 5. The subtraction of 13 from 32 is 19. And there are one ten and nine ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 13 from 32, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 13 from 32 is 32 – 13= 19 and there are one ten and nine ones. Question 6. The subtraction of 13 from 32 is 19. And there are one ten and nine ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 13 from 32, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 13 from 32 is 32 – 13= 19 and there are one ten and nine ones. Problem Solving Solve. Write or draw to explain. Question 7. Julie has 42 sheets of paper. She gives 17 sheets to Kari. How many sheets of paper does Julie have now? __ sheets of paper Question 8. WRITE Math Draw and write to explain how these two problems are different: 35 – 15 = ________ and 43 – 26 = _______ __________________________ Lesson Check Question 1. What is the difference? The subtraction of 13 from 32 is 19. And there are one ten and nine ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 13 from 32, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 13 from 32 is 32 – 13= 19 and there are one ten and nine ones. Question 2. What is the difference? The subtraction of 13 from 32 is 19. And there are one ten and nine ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 13 from 32, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 13 from 32 is 32 – 13= 19 and there are one ten and nine ones. Spiral Review Question 3. What is the sum? 9 + 9 = __ The sum of the two numbers is 18. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the two numbers 9 + 9 is 18. Question 4. What is the difference? 14 – 7 = __ The subtraction of 7 from 14 is 7. And there are zero tens and seven ones. Explanation: In the above image, we can see that there are one ten and four ones. So here we need to subtract 7 from 14, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 7 from 14 is 14 – 7= 7 and there are zero tens and seven ones. Question 5. What is the sum? 36 + 25 = __ The sum of the two numbers is 61. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the two numbers 36 + 25 is 61. Question 6. What is the sum? 7 + 2 + 3 = __ The sum of the three numbers is 12. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the three numbers 7 + 2 + 3 is 12. ### 2-Digit Subtraction Mid-Chapter Checkpoint Concepts and Skills Break apart the number you are subtracting. Use the number line to help. Write the difference. Question 1. 34 – 8 = __ The subtraction of 8 from 34 is 26. And there are two tens and six ones. Explanation: In the above image, we can see that there are three tens and four ones. So here we need to subtract 8 from 34, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 8 from 34 is 34 – 8= 26 and there are two tens and six ones. Question 2. 45 – 17 = __ The subtraction of 17 from 45 is 28. And there are two tens and eight ones. Explanation: In the above image, we can see that there are four tens and five ones. So here we need to subtract 17 from 45, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 17 from 45 is 45 – 17= 28 and there are two tens and eight ones. Draw a quick picture to solve. Write the difference. Question 3. The subtraction of 29 from 42 is 13. And there are one ten and three ones. Explanation: In the above image, we can see that there are four tens and two ones. So here we need to subtract 29 from 42, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 29 from 42 is 42 – 29= 13 and there are one ten and three ones. Question 4. The subtraction of 23 from 54 is 31. And there are three tens and one in ones place. Explanation: In the above image, we can see that there are five tens and four ones. So here we need to subtract 23 from 54, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 23 from 54 is 54 – 23= 31 and there are three tens and one in ones place Write the difference. Question 5. The subtraction of 43 from 78 is 35. And there are three tens and five ones. Explanation: In the above image, we can see that there are seven tens and eight ones. So here we need to subtract 43 from 78, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 43 from 78 is 78 – 43= 35 and there are three tens and five ones. Question 6. The subtraction of 26 from 60 is 34. And there are three tens and four ones. Explanation: In the above image, we can see that there are six tens and zero ones. So here we need to subtract 26 from 60, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 26 from 60 is 60 – 26= 34 and there are three tens and four ones. Question 7. The subtraction of 37 from 85 is 48. And there are four tens and eight ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 37 from 85, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 37 from 85 is 85 – 37= 48 and there are four tens and eight ones. Question 8. THINK SMARTER Marissa had 51 toy dinosaurs. She gave 14 toy dinosaurs to her brother. How many toy dinosaurs does she have now? __ toy dinosaurs The number of dinosaurs does Marissa has is 37 dinosaurs. Explanation: Marissa had 51 toy dinosaurs and she gave 14 toy dinosaurs to her brother. So to how many toy dinosaurs does she have we will perform regrouping of subtraction. Here we have solved by using regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the number of dinosaurs does Marissa has is 51 – 14= 37 dinosaurs. ### Lesson 5.7 Rewrite 2-Digit Subtraction Essential Question What are two different ways to write subtraction problems? Math Talk MATHEMATICAL PRACTICES Explain why it is important to line up the digits of the numbers in columns. Model and Draw What is 81 – 36? Rewrite the subtraction problem. Then find the difference. Step 1 For 81, write the tens digit in the tens column. Write the ones digit in the ones column. Step 2 Look at the ones. Regroup if you need to. Share and Show MATH BOARD Rewrite the subtraction problem. Then find the difference. Question 1. The subtraction of 37 from 4 is 33. And there are three tens and three ones. Explanation: In the above image, we can see that there are three tens and four ones. So here we need to subtract 4 from 37, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 4 from 37 is 37 – 4= 33 and there are three tens and three ones. Question 2. The subtraction of 24 from 48 is 24. And there are two tens and four ones. Explanation: In the above image, we can see that there are four tens and eight ones. So here we need to subtract 24 from 48, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 24 from 48 is 48 – 24= 24 and there are two tens and four ones. Question 3. The subtraction of 37 from 85 is 48. And there are four tens and eight ones. Explanation: In the above image, we can see that there are eight tens and five ones. So here we need to subtract 37 from 85, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 37 from 85 is 85 – 37= 48 and there are four tens and eight ones. Question 4. The subtraction of 19 from 63 is 44. And there are four tens and four ones. Explanation: In the above image, we can see that there are six tens and three ones. So here we need to subtract 19 from 63, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 19 from 63 is 63 – 19= 44 and there are four tens and four ones. Question 5. The subtraction of 37 from 62 is 25. And there are two tens and five ones. Explanation: In the above image, we can see that there are six tens and two ones. So here we need to subtract 37 from 62, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 37 from 62 is 62 – 37= 25 and there are two tens and five ones. Question 6. The subtraction of 27 from 51 is 24. And there are two tens and four ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 27 from 51, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 27 from 51 is 51 – 27= 24 and there are two tens and four ones. Question 7. The subtraction of 3 from 76 is 73. And there are seven tens and three ones. Explanation: In the above image, we can see that there are seven tens and three ones. So here we need to subtract 3 from 76, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 3 from 76 is 76 – 3= 73 and there are seven tens and three ones. Question 8. The subtraction of 48 from 95 is 47. And there are four tens and seven ones. Explanation: In the above image, we can see that there are nine tens and five ones. So here we need to subtract 48 from 95, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 48 from 95 is 95 – 48= 47 and there are four tens and seven ones. Rewrite the subtraction problem. Then find the difference. Question 9. The subtraction of 8 from 49 is 41. And there are four tens and one in one’s place. Explanation: In the above image, we can see that there are four tens and nine ones. So here we need to subtract 8 from 49, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 8 from 49 is 49 – 8= 41 and there are four tens and one in one’s place. Question 10. The subtraction of 47 from 85 is 38. And there are three tens and eight ones. Explanation: In the above image, we can see that there are eight tens and five ones. So here we need to subtract 47 from 85, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 47 from 85 is 85 – 47= 38 and there are three tens and eight ones. Question 11. The subtraction of 23 from 63 is 40. And there are four tens and zero ones. Explanation: In the above image, we can see that there are six tens and three ones. So here we need to subtract 23 from 63, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 23 from 63 is 63 – 23= 40 and there are four tens and zero ones. Question 12. The subtraction of 23 from 51 is 28. And there are two tens and eight ones. Explanation: In the above image, we can see that there are five tens and one in one’s place. So here we need to subtract 23 from 51, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 23 from 51 is 51 – 23= 28 and there are two tens and eight ones. Question 13. The subtraction of 15 from 60 is 45. And there are four tens and five ones. Explanation: In the above image, we can see that there are six tens and zero ones. So here we need to subtract 15 from 60, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 15 from 60 is 60 – 15= 45 and there are four tens and four ones. Question 14. The subtraction of 58 from 94 is 36. And there are three tens and six ones. Explanation: In the above image, we can see that there are nine tens and four ones. So here we need to subtract 58 from 94, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 58 from 94 is 94 – 58= 36 and there are three tens and six ones. Question 15. The subtraction of 20 from 47 is 27. And there are two tens and seven ones. Explanation: In the above image, we can see that there are four tens and seven ones. So here we need to subtract 20 from 47, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 20 from 47 is 47 – 20= 27 and there are two tens and seven ones. Question 16. The subtraction of 9 from 35 is 26. And there are two tens and six ones. Explanation: In the above image, we can see that there are three tens and five ones. So here we need to subtract 9 from 35, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 9 from 35 is 35 – 9= 26 and there are two tens and six ones. Question 17. The subtraction of 10 from 78 is 68. And there are six tens and eight ones. Explanation: In the above image, we can see that there are seven tens and eight ones. So here we need to subtract 10 from 78, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 10 from 78 is 78 – 10= 68 and there are six tens and eight ones. Question 18. The subtraction of 38 from 54 is 16. And there are one ten and six ones. Explanation: In the above image, we can see that there are five tens and four one. So here we need to subtract 38 from 54, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 38 from 54 is 54 – 38= 16 and there are one ten and six ones. Question 19. The subtraction of 39 from 92 is 53. And there are five tens and three ones. Explanation: In the above image, we can see that there are nine tens and two ones. So here we need to subtract 39 from 92, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 39 from 92 is 92 – 39= 53 and there are five tens and three ones. Question 20. The subtraction of 28 from 87 is 59. And there are five tens and nine ones. Explanation: In the above image, we can see that there are eight tens and seven ones. So here we need to subtract 28 from 87, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 28 from 87 is 87 – 28= 59 and there are eight tens and nine ones. Question 21. THINK SMARTER For which of the problems above could you find the difference without rewriting it? Explain. ___________________________ ___________________________ ___________________________ Problem Solving • Applications Question 22. How many more paintings were done by adults than by children? __ more paintings The number of paints which were more by adults than the children is 53 – 26= 27 paintings. Explanation: As Pablo’s class went to the art museum, and they saw 26 paintings done by the children. After that, they saw 53 paintings done by the adults. So the number of paints which were more by adults than the children is 53 – 26= 27 paintings. Question 23. GO DEEPER How many more paintings than sculptures did they see? __ more paintings The number of paints that were more than the sculptures is 61 more paintings. Explanation: As Pablo’s class went to the art museum, and they saw 26 paintings done by the children. After that they saw 53 paintings done by the adults. So the total number of paintings is 53 + 26 = 79 paintings. And they saw 18 sculptures, so the number of paints which were more than the sculptures is 79 – 18= 61 more paintings. Question 24. THINK SMARTER Tom drew 23 pictures last year. Beth drew 14 pictures. How many more pictures did Tom draw than Beth? Fill in the bubble next to all the ways to show the problem. The number of more pictures did Tom draw than Beth is 23 – 14= 9 pictures. Explanation: As Tom draw 23 pictures last year and Beth draw 14 pictures, so the number of more pictures did Tom draw than Beth is 23 – 14= 9 pictures. TAKE HOME ACTIVITY ### Rewrite 2-Digit Subtraction Homework & Practice 5.7 Rewrite the subtraction problem. Then find the difference. Question 1. The subtraction of 19 from 35 is 16. And there are one ten and six ones. Explanation: In the above image, we can see that there are three tens and five ones. So here we need to subtract 19 from 35, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 19 from 35 is 35 – 19= 16 and there are one ten and six ones. Question 2. The subtraction of 23 from 47 is 24. And there are two tens and four ones. Explanation: In the above image, we can see that there are four tens and seven ones. So here we need to subtract 23 from 32, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 23 from 47 is 47 – 23= 24 and there are two tens and four ones. Question 3. The subtraction of 28 from 58 is 30. And there are three tens and zero ones. Explanation: In the above image, we can see that there are three tens and five one’s blocks. So here we need to subtract 28 from 58, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 28 from 58 is 58 – 28= 30 and there are three tens and zero ones. Problem Solving Solve. Write or draw to explain. Question 4. Jimmy went to the toy store. He saw 23 wooden trains and 41 plastic trains. How many more plastic trains than wooden trains did he see? ___ more plastic trains There are 18 more plastic trains than wooden trains. Explanation: Jimmy went to the toy store and he saw 23 wooden trains and 41 plastic trains, so to find how many more plastic trains than wooden trains did he see is we will perform subtraction. So there will be 41 – 23= 18 more plastic trains than wooden trains. Question 5. WRITE Math Is it easier to subtract when the numbers are written above and below each other? Explain your answer __________________________ __________________________ __________________________ __________________________ Yes, it is easier to subtract the numbers that are written above and below each other. As we can easily borrow from the place in front of the number and we can rewrite without a problem. It also helps to keep the math problems more organized so by writing the numbers above and below we will be confused. Lesson Check Question 1. What is the difference for 43 − 17? The subtraction of 17 from 43 is 26. And there are one ten and nine ones. Explanation: In the above, we can see that there are four tens and three ones. So here we need to subtract 13 from 32, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 17 from 43 is 43 – 16= 26 and there are two tens and six ones. Question 2. What is the difference for 50 − 16? The subtraction of 16 from 50 is 34. And there are three tens and four ones. Explanation: In the above, we can see that there are five tens and zero ones. So here we need to subtract 16 from 50, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 16 from 50 is 50 – 16= 34 and there are three tens and four ones. Spiral Review Question 3. What is the sum? The sum of the given numbers is 74. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the given numbers 29 + 4 + 25 + 16 is 74. Question 4. What is the sum of 41 + 19? _______ The sum of the given numbers is 60. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the given numbers 41 + 19 is 60. Question 5. Write an addition fact that will give the same sum as 5 + 9? 10 + ___ The sum of the two numbers is 14. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the two numbers 5 + 9 is 14. So to get the same sum we need to add 4, so the sum of the 10 + 4= 14, which is same as 5 + 9 sum. Question 6. What is the difference? 45 – 13 = ___ The subtraction of 13 from 45 is 32. And there are three tens and two ones. Explanation: In the above, we can see that there are four tens and three ones. So here we need to subtract 13 from 45, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 13 from 45 is 45 – 13= 32 and there are three tens and two ones. ### Lesson 5.8 Add to Find Differences Essential Question How can you use addition to solve subtraction problems? _______ ____ markers Now draw pictures to show the next part of the problem. Write a number sentence for your drawing. _______ ____ markers Math Talk MATHEMATICAL PRACTICES Describe what happens when you add back the number that you had subtracted. Model and Draw Count up from the number you are subtracting to find the difference. Start at 38. Count up to 40. Then count up 5 more to 45. So, 45 − 38 = __. Share and Show MATH BOARD Use the number line. Count up to find the difference. Question 1. 36 – 27 = ___ The subtraction of 27 from 36 is 9. And there are zero tens and nine ones. Explanation: Here we need to subtract 27 from 36, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 27 from 36 is 36 – 27= 9 and there are zero tens and nine ones. Question 2. 56 – 49 = __ The subtraction of 49 from 56 is 7. And there are zero tens and seven ones. Explanation: Here we need to subtract 49 from 56, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 49 from 56 is 56 – 49= 7 and there are zero tens and seven ones. Question 3. 64 – 58 = __ The subtraction of 58 from 64 is 6. And there are zero tens and six ones. Explanation: Here we need to subtract 58 from 64, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 58 from 64 is 64 – 58= 6 and there are zero tens and six ones. Use the number line. Count up to find the difference. Question 4. 33 – 28 = __ The subtraction of 28 from 33 is 5. And there are zero tens and five ones. Explanation: In the above image, we can see that there are three tens and three ones. So here we need to subtract 28 from 33, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 28 from 33 is 33 – 28= 5 and there are zero tens and five ones. Question 5. 45 – 37 = __ The subtraction of 37 from 45 is 8. And there are zero tens and eight ones. Explanation: In the above image, we can see that there are four tens and five ones. So here we need to subtract 37 from 45, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 37 from 45 is 45 – 37= 8 and there are zero tens and eight ones. Question 6. 58 – 49 = __ The subtraction of 49 from 58 is 9. And there are zero tens and nine ones. Explanation: In the above image, we can see that there are five tens and eight ones. So here we need to subtract 49 from 58, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 49 from 58 is 58 – 49= 49 and there are zero tens and nine ones. Question 7. THINK SMARTER There were 55 books on the table. Sandra picked up some of the books. Now there are 49 books on the table. How many books did Sandra pick up? __ books The number of books picked up by the Sandra is 6 books. Explanation: As there were 55 books on the table and Sandra picked up some of the books. And Now there are 49 books on the table, so to find how many books did Sandra pick up is we will perform subtraction. So the number of books picked up by the Sandra is 55 – 49= 6 books. Problem Solving • Applications Solve. You may wish to use the number line to help. Question 8. There are 46 game pieces in a box. Adam takes 38 game pieces out of the box. How many game pieces are still in the box? __ game pieces The number of game pieces is still in the box is 8 game pieces. Explanation: As there are 46 game pieces in a box and Adam takes 38 game pieces out of the box. So there will be 46 – 38= 8 game pieces. Question 9. THINK SMARTER Rachel had 27 craft sticks. Then she gave 19 craft sticks to Theo. How many craft sticks does Rachel have now? Circle the number from the box to make the sentence true. Rachel has craft sticks now. Explain how you can use addition to solve the problem. __________________________ __________________________ The number of craft sticks Rachel has now is 27 – 19 = 8 craft sticks. Explanation: As Rachel had 27 craft sticks and then she gave 19 craft sticks to Theo. So the number of craft sticks Rachel has now is 27 – 19 = 8 craft sticks. TAKE HOME ACTIVITY • Have your child describe how he or she used a number line to solve one problem in this lesson. The number line is used by performing we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. Then we will make the number of jumps by the difference what we had got and we will make jump from minuend number to the subtrahend and will count the number of jumps. ### Add to Find Differences Homework & practice 5.8 Use the number line. Count up to find the difference. Question 1. 36 – 29 = __ The subtraction of 29 from 36 is 7. And there are zero tens and seven ones. Explanation: In the above image, we can see that there are three tens and six ones. So here we need to subtract 29 from 36, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 29 from 36 is 36 – 29= 7 and there are zero tens and seven ones. Question 2. 43 – 38 = __ The subtraction of 38 from 43 is 5. And there are zero tens and five ones. Explanation: In the above image, we can see that there are three tens blocks and six one’s blocks. So here we need to subtract 8 from 51, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 38 from 43 is 43 – 38= 5 and there are zero tens and five ones. Problem Solving Solve. You may wish to use the number line. Question 3. Jill has 63 index cards. She uses 57 of them for a project. How many index cards does Jill have now? __ index cards The number of index cards does Jill has now is 6 index cards. Explanation: As Jill has 63 index cards and she uses 57 of them for a project, so the number of index cards does Jill have now is 63 – 57 = 6 index cards. Question 4. WRITE Math Explain how a number line can be used to find the difference for 34 – 28. __________________________ __________________________ Lesson Check Use the number line. Count up to find the difference. Question 1. 82 − 75 = __ The subtraction of 75 from 82 is 7. And there are zero tens and seven ones. Explanation: In the above image, we can see that there are eight tens and two ones. So here we need to subtract 75 from 82, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 75 from 82 is 82 – 75= 7 and there are zero tens and seven ones. Question 2. 90 − 82 = __ The subtraction of 82 from 90 is 8. And there are zero tens and eight ones. Explanation: In the above image, we can see that there are nine tens and zero ones. So here we need to subtract 82 from 90, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 82 from 90 is 90 – 82= 8and there are zero tens and eight ones. Spiral Review Question 3. Jordan has 41 toy cars at home. He brings 24 cars to school. How many cars are at home? __ cars The number of cars is at home is 17 cars. Explanation: As Jordan has 41 toy cars at home and he brings 24 cars to school, so the number of cars are at home is 41 – 24= 17 cars. Question 4. Pam has 15 fish. 9 are goldfish and the rest are guppies. How many fish are guppies? __ guppies The number of fishes is guppies is 6 guppies. Explanation: As Pam has 15 fishes and 9 are gold fishes and the rest are guppies, so the number of fishes are guppies is 15 – 9= 6 guppies. Question 5. What is the sum? The sum of the two numbers is 54. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the two numbers 35 + 19 is 54. Question 6. Each table has 5 pencils. There are 3 tables. How many pencils are there altogether? __ pencils. The total number of pencils will be 15 pencils are there altogether. Explanation: As each table has 5 pencils and there are 3 tables, so the total number of pencils will be 5 × 3= 15 pencils are there altogether. ### Lesson 5.9 Problem Solving • Subtraction Essential Question How can drawing a diagram help when solving subtraction problems? Jane and her mom made 33 puppets for the craft fair. They sold 14 puppets. How many puppets do they still have? The total number of puppets do they left is 19 puppets. Explanation: As Jane and her mom made 33 puppets for the craft fair and they sold 14 puppets, so they still have 33 – 14= 19 puppets. Here we need to subtract 14 from 33, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 14 from 33 is 33 – 14= 19 and there are one ten and nine ones. So they will have 19 puppets left. Unlock the Problem What do I need to find? they still have What information do I need to use? They sold __ puppets. Show how to solve the problem. HOME CONNECTION • Your child used a bar model and a number sentence to represent the problem. Using a bar model helps show what is known and what is needed to solve the problem. Try Another Problem Label the bar model. Write a number sentence with a for the missing number. Solve. Question 1. Carlette had a box of 46 craft sticks. She used 28 craft sticks to make a sailboat. How many craft sticks were not used? They will have 18 craft sticks left. Explanation: As Carlette had a box of 46 craft sticks and she used 28 craft sticks to make a sailboat. So the number of craft sticks that are not used is 46 – 28= 18 craft sticks. Here we need to subtract 28 from 46, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 28 from 46 is 46 – 28= 18 and there are one ten and eight ones. So they will have 18 craft sticks left. Question 2. Rob’s class made 31 clay bowls. Sarah’s class made 15 clay bowls. How many more clay bowls did Rob’s class make than Sarah’s class? ______ ___ more clay bowls The number of more clay bowls did Rob’s class make than Sarah’s class is 31 – 15= 16 clay bowls. Explanation: As Rob’s class made 31 clay bowls and Sarah’s class made 15 clay bowls. So the number of more clay bowls did Rob’s class make than Sarah’s class is 31 – 15= 16 clay bowls. So we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the number of more clay bowls did Rob’s class make than Sarah’s class is 16 clay bowls. Share and Show MATH BOARD Label the bar model. Write a number sentence with a for the missing number. Solve. Question 3. Mr. Hayes makes 32 wooden frames. He gives away 15 frames as gifts. How many frames does he still have? The number of many frames does he still have now is 32 – 15= 17 frames. Explanation: As Mr. Hayes makes 32 wooden frames and he gives away 15 frames as gifts. So the number of many frames does he still have now is 32 – 15= 17 frames. So we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the number of many frames does he still have now is 17 frames. Question 4. Wesley has 21 ribbons in a box. He has 15 ribbons on the wall. How many more ribbons does he have in the box than on the wall? __ more ribbons _____ The number of many more ribbons does he have in the box than on the wall is 6 more ribbons. Explanation: As Wesley has 21 ribbons in a box and he has 15 ribbons on the wall. So the number of many more ribbons does he have in the box than on the wall is 21 – 15= 6 more ribbons. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the number of many more ribbons does he have in the box than on the wall is 6 more ribbons. Question 5. THINK SMARTER Jennifer wrote 9 poems at school and 11 poems at home. She wrote 5 more poems than Nell. How many poems did Nell write? __ poems Explanation: Question 6. GO DEEPER There are 70 children. 28 children are hiking and 16 are at a picnic. The rest of the children are playing soccer. How many children are playing soccer? Draw a model with bars for the problem. Describe how your drawing shows the problem. Then solve the problem. __________________________ __________________________ __________________________ Question 7. THINK SMARTER There are 48 crackers in a bag. The children eat 25 crackers. How many crackers are still in the bag? Circle the bar model that can be used to solve the problem. Write a number sentence with a for the missing number. Solve. __________________________ ___ crackers TAKE HOME ACTIVITY We will perform subtraction by breaking the two-digit number into tens place digit and ones place digit. Then the other number will be placed below the number. If the minuend number is less than the subtrahend number then we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So by this, we will perform the subtraction. ### Problem Solving • Subtraction Homework & Practice 5.9 Label the bar model. Write a number sentence with a for the missing number. Solve. Question 1. Megan picked 34 flowers. Some of the flowers are yellow and 18 flowers are pink. How many of the flowers are yellow? ___ yellow flowers ___ Question 2. Alex had 45 toy cars. He put 26 toy cars in a box. How many toy cars are not in the box? ___ toy cars Question 3. WRITE Math Explain how bar models show a problem in a different way. __________________________ __________________________ __________________________ __________________________ Lesson Check Question 1. There were 39 pumpkins at the store. Then 17 of the pumpkins were sold. How many pumpkins are still at the store? __ pumpkins Question 2. There were 48 ants on a hill. Then 13 of the ants marched away. How many ants are still on the hill? __ ants Spiral Review Question 3. Ashley had 26 markers. Her friend gave her 17 more markers. How many markers does Ashley have now? __ markers Question 4. What is the sum? Question 5. Write a subtraction fact that will give the same difference as 15 − 7. 10 – __ Question 6. What is the sum? 34 + 5 = __ ### Lesson 5.10 Algebra • Write Equations to Represent Subtraction Essential Question How do you write a number sentence to represent a problem? Listen and Draw Draw to show the problem. Write a number sentence. Then solve. _______________ Math Talk MATHEMATICAL PRACTICES Describe how your drawing shows the problem. Model and Draw You can write a number sentence to show a problem. Liza has 65 postcards. She gives 24 postcards to Wesley. How many postcards does Liza have now? Liza has __ postcards now. Share and Show MATH BOARD Write a number sentence for the problem. Use a for the missing number. Then solve. Question 1. There were 32 birds in the trees. Then 18 birds flew away. How many birds are in the trees now? __ birds Question 2. Carla read 43 pages in her book. Joe read 32 pages in his book. How many more pages did Carla read than Joe? ________ __ more pages Write a number sentence for the problem. Use a for the missing number. Then solve. Question 3. There were 40 ants on a rock. Some ants moved to the grass. Now there are 26 ants on the rock. How many ants moved to the grass? ___ __ ants Question 4. THINK SMARTER Keisha had a bag of ribbons. She took 29 ribbons out of the bag. Then there were 17 ribbons still in the bag. How many ribbons were in the bag to start? _______ __ ribbons Question 5. GO DEEPER There are 50 bees in a hive. Some bees fly out. If fewer than 20 bees are still in the hive, how many bees could have flown out? __ bees Problem Solving • Applications Question 6. MATHEMATICAL PRACTICE Make Connections Brendan made this number line to find a difference. What was he subtracting from 100? Explain your answer. __________________________ __________________________ __________________________ __________________________ Question 7. THINK SMARTER There are 52 pictures on the wall. 37 are wild cats and the rest are birds. How many of the pictures are birds? Use the numbers and symbols on the tiles to complete the number sentence for the problem. _______________ ___ birds TAKE HOME ACTIVITY • Have your child explain how he or she solved one problem in this lesson. ### Algebra • Write Equations to Represent Subtraction Homework & Practice 5.10 Write a number sentence for the problem. Use a for the missing number. Then solve. Question 1. 29 children rode their bikes to school. After some of the children rode home, there were 8 children with bikes still at school. How many children rode their bikes home? ________ __ children The number of students who rode their bikes home is 29 – 8= 21 children rode their bike home. Explanation: As 29 children rode their bikes to school and after that some of the children rode home, there were 8 children with bikes still at school. So the number of students who rode their bikes home is 29 – 8= 21 children rode their bike home. So we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the number of students who rode their bikes home is 21 children rode their bike home. Problem Solving Solve. Write or draw to explain. Question 2. There were 21 children in the library. After 7 children left the library, how many children were still in the library? __ children Question 3. WRITE Math Describe different ways that you can show a story problem. Use one of the problems in this lesson as your example. __________________________ __________________________ __________________________ __________________________ Lesson Check Question 1. Question 2. Jake had 36 baseball cards. He gave 17 cards to his sister. How many baseball cards does Jake have now? __ cards Spiral Review Question 3. What is the sum? 6 + 7 = __ Question 4. What is the difference? 16 – 9 = __ The subtraction of 9 from 16 is 7. And there are zero tens and seven ones. Explanation: In the above image, we can see that there are nine tens and zero ones. So here we need to subtract 9 from 16, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 9 from 16 is 16 – 9= 7 and there are zero tens and seven ones. Question 5. What is the difference? The subtraction of 39 from 46 is 7. And there are zero tens and seven ones. Explanation: In the above image, we can see that there are four tens and six ones. So here we need to subtract 39 from 46, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 39 from 46 is 46 – 39= 7 and there are zero tens and seven ones. Question 6. Write an addition fact that will give the same sum as 6 + 8. 10 + ___ The sum of the two numbers is 14. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the two numbers 6 + 8 is 14. So to get the same sum we need to add 4, so the sum of the 10 + 4= 14, which is same as 6 + 8 sum. ### Lesson 5.11 Solve Multistep Problems Essential Question How do you decide what steps to do to solve a problem? Listen and Draw Label the bar model to show each problem. Then solve. Math Talk MATHEMATICAL PRACTICES Describe how the two bar models are different. Model and Draw Bar models help you know what to do to solve a problem. Ali has 27 stamps. Matt has 38 stamps. How many more stamps are needed so they will have 91 stamps? They need __ more stamps Share and Show MATH BOARD Complete the bar models for the steps you do to solve the problem. Question 1. Explanation: Complete the bar models for the steps you do to solve the problem. Question 2. Max has 35 trading cards. He buys 22 more cards. Then he gives 14 cards to Rudy. How many cards does Max have now? The number of many cards does Max have now is 43 trading cards. Explanation: As Max has 35 trading cards and he buys 22 more cards, so the total number of cards will be 35 + 22= 57 trading cards. Then he gives 14 cards to Rudy. So the number of trading cards will be 57 – 14= 43 trading cards. So the number of many cards does Max have now is 43 trading cards. Question 3. Drew has 32 toy cars. He trades 7 of those cars for 11 other toy cars. How many toy cars does Drew have now? Question 4. Marta and Debbie each have 17 ribbons. They buy 1 package with 8 ribbons in it. How many ribbons do they have now? The total number of ribbons do they had now is 42 ribbons. Explanation: As Marta and Debbie each have 17 ribbons, so the total number of ribbons did Marta and Debbie had together is 17 + 17= 34 ribbons. And they bought 1 package with 8 ribbons in it. So the total number of ribbons do they had now is 34 + 8= 42 ribbons. Problem Solving • Applications WRITE Math Question 5. Shelby had 32 rocks. She finds 33 more rocks at the park and gives 28 rocks to George. How many rocks does she have now? __ rocks So the number of many rocks does she have now is 37 rocks. Explanation: As Shelby had 32 rocks and she finds 33 more rocks at the park, so the total number of rocks did Shelby had now is 32 + 33= 65 rocks. And Shelby gives 28 rocks to George. So the number of many rocks does she have now is 65 – 28= 37 rocks. so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the number of many rocks does she have now is 37 rocks. Question 6. Benjamin finds 31 pinecones at the park. Together, Jenna and Ellen find the same number of pinecones as Benjamin. How many pinecones could each girl have found? Jenna: __ pinecones Ellen: __ pinecones Explanation: As Benjamin finds 31 pinecones at the park and Jenna and Ellen together finds the same number of pinecones as Benjamin. How many pinecones could each girl have found Question 7. THINK SMARTER Tanya finds 22 leaves. Maurice finds 5 more leaves than Tanya finds. How many leaves do the children find? Draw to show how you solve the problem. __ leaves The number of many leaves do the children finds together is 22 + 5= 27 leaves. Explanation: As Tanya finds 22 leaves and Maurice finds 5 more leaves than Tanya finds. So the number of many leaves do the children finds together is 22 + 5= 27 leaves. TAKE HOME ACTIVITY • Have your child explain how he or she would solve Exercise 6 if the number 31 was changed to 42. ### Solve Multistep Problems Homework & Practice 5.11 Complete the bar models for the steps you do to solve the problem Question 1. Greg has 60 building blocks. His sister gives him 17 more blocks. He uses 38 blocks to make a tower. How many blocks are not used in the tower? __ blocks The number of many blocks are not used in the tower is 5 blocks. Explanation: As Greg has 60 building blocks and his sister gives him 17 more blocks, so the number of blocks did Greg had now is 60 – 17= 43 blocks and he uses 38 blocks to make a tower so the number of many blocks are not used in the tower is 43 – 38= 5 blocks. So we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the number of many blocks are not used in the tower is 5 blocks. Problem Solving Solve. Write or draw to explain. Question 2. Ava has 25 books. She gives away 7 books. Then Tom gives her 12 books. How many books does Ava have now? __ books The number of many books does Ava have now is 6 books. Explanation: As Ava has 25 books and she gives away 7 books, then Ava will have 25 – 7= 18 books and then Tom gives her 12 books, so the number of books did Ava have now is 18 – 12= 6 books. So we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the number of many books does Ava have now is 6 books. Question 3. WRITE Math Choose one of the problems on this page. Describe how you decided what steps were needed to solve the problem. ___________________________ ___________________________ ___________________________ ___________________________ Lesson Check Question 1. Sara has 18 crayons. Max has 19 crayons. How many more crayons do they need to have 50 crayons altogether? __ crayons They need 13 crayons more do they need to have 50 crayons altogether. Explanation: As Sara has 18 crayons and Max has 19 crayons, so the total number of crayons do they have together is 18 + 19= 37 crayons. So to get 50 crayons altogether we need to perform subtraction. So we need 50 – 37= 13 crayons more do they need to have 50 crayons altogether. So we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So They need 13 crayons more do they need to have 50 crayons altogether. Question 2. Jon has 12 pennies. Lucy has 17 pennies. How many more pennies do they need to have 75 pennies altogether? __ pennies They need 46 pennies more to have 75 pennies altogether. Explanation: As Jon has 12 pennies and Lucy has 17 pennies, so the total number of pennies together they have is 29 pennies. So to 75 pennies together we need to perform subtraction 75 – 29= 46 pennies. So the number of many frames does he still have now is 32 – 15= 17 frames. So we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the total number of pennies together they have is 29 pennies. So to 75 pennies together we need to perform subtraction 75 – 29= 46 pennies. Spiral Review Question 3. What is the difference? 58 – 13 = __ The subtraction of 8 from 51 is 43. And there are four tens and three ones. Explanation: In the above image, we can see that there are three tens blocks and six one’s blocks. So here we need to subtract 8 from 51, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 8 from 51 is 51 – 8= 43 and there are four tens and three ones. Question 4. What is the sum? The sum of the two numbers is 62. Explanation: The sum can be defined as the resulting of two or more numbers by adding. So here the sum of the two numbers 47 + 15 is 62. Question 5. There are 26 cards in a box. Bryan takes 12 cards. How many cards are still in the box? __ cards The number of cards still in the box is 14 cards. Explanation: As there are 26 cards in a box and Bryan takes 12 cards, so the number of cards still in the box is 26 – 12= 14 cards. we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 12 from 26 is 26 – 12= 14 and there are zero tens and four ones. ### 2-Digit Subtraction Chapter 5 Review Test Question 1. Do you need to regroup to subtract? Choose Yes or No. Question 2. Use the number line. Count up to find the difference. 52 – 48 = __ The subtraction of 48 from 52 is 4. And there are zero tens and four ones. Explanation: Here we need to subtract 48 from 52, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 48 from 52 is 52 – 48= 4 and there are zero tens and four ones. Question 3. Ed has 28 blocks. Sue has 34 blocks. Who has more blocks? How many more? Label the bar model. Solve. Circle the word and number from each box to make the sentence true. Break apart the number you are subtracting. Write the difference? Sue has more number of blocks tha Ed. And had 6 more blocks than Ed. Explanation: Ed has 28 blocks and Sue has 34 blocks. As 34 is greater than 28, so Sue has more number of blocks and has 34 – 28= 6 more blocks than the Ed. So to perform Break apart the number subratction for 34 – 28 first we will break into tens and ones. So here we will break 28 as 20 and 8 and we will subtract 20 and we will get the result as 14. Then we will start from 14 and subtract 8 to get to 6. So we will get the result as 6. The subtraction of 34 – 28 is 6. Question 4. The subtraction of 42 – 8 is 34. Explanation: Here, we will break apart the number that we are subtracting. So first we will break into tens and ones. So here we will break 8 as 2 and 6 and we will subtract 2 and we will get the result as 42. Then we will start from 42 and subtract 6 to get to 34 and then. So we will get the result as 34. The subtraction of 42 – 8 is 34. Question 5. The subtraction of 53 – 16 is 37. Explanation: Here, we will break apart the number that we are subtracting. So first we will break into tens and ones. So here we will break 10 as 6 and we will subtract 10 and we will get the result as 43. Then we will start from 43 and subtract 6 to get to 37 and then. So we will get the result as 37. The subtraction of 53 – 16 is 37. Question 6. What is 33 − 19? Use the numbers on the tiles to rewrite the subtraction problem. Then find the difference. The subtraction of 19 from 33 is 14. And there are one ten and four ones. Explanation: In the above image, we can see that there are three tens and three ones. So here we need to subtract 19 from 33, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 19 from 33 is 33 – 19= 14 and there are one ten and four ones. Question 7. GO DEEPER Jacob’s puzzle has 84 pieces. Jacob puts together 27 pieces in the morning. He puts together 38 more pieces in the afternoon. How many pieces does Jacob need to put together to finish the puzzle? Complete the bar models for the steps you do to solve the problem. __ more pieces Jacob needs 19 pieces to complete the puzzle. Explanation: Jacob’s puzzle has 84 pieces and Jacob puts together 27 pieces in the morning and he puts together 38 more pieces in the afternoon. So the total number of puzzles kept by Jacob together in the morning and the afternoon is 27 + 38= 65 pieces. So Jacob needs 84 – 65= 19 pieces to complete the puzzle. Regroup if you need to. Write the difference. Question 8. The subtraction of 5 from 28 is 23. And there are two tens and three ones. Explanation: In the above image, we can see that there are two tens and eight ones. So here we need to subtract 5 from 28, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 5 from 28 is 28 – 5= 23 and there are two ten and three ones. Question 9. The subtraction of 12 from 32 is 23. And there are two tens and three ones. Explanation: In the above image, we can see that there are two tens and eight ones. So here we need to subtract 5 from 28, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 5 from 28 is 28 – 5= 23 and there are two ten and three ones. Question 10. Find the difference. The subtraction of 62 from 90 is 28. And there are two tens and eight ones. Explanation: In the above image, we can see that there are nine tens and zero ones. So here we need to subtract 62 from 90, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 62 from 90 is 90 – 62= 28 and there are two tens and eight ones. Fill in the bubble next to one number from each column to show the difference. Question 11. There are 22 children at the park. 5 children are on the swings. The rest of the children are playing ball. How many children are playing ball? The number of children is playing ball is 17 children. Explanation: As there are 22 children at the park and 5 children are on swings and the rest of the children are playing ball. So the number of children are playing ball is 22 – 15= 17 children. Question 12. THINK SMARTER Subtract 27 from 43. Draw to show the regrouping. Fill in the bubble next to all the ways to write the difference. The subtraction of 27 from 43 is 16. And there are one ten and six ones. Explanation: In the above image, we can see that there are two tens and eight ones. So here we need to subtract 27 from 43, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 27 from 43 is 43 – 27= 43 and there are one ten and six ones. Question 13. Jill collects stamps. Her stamp book has space for 64 stamps. She needs 18 more stamps to fill the book. How many stamps does Jill have now? Write a number sentence for the problem. Use a for the missing number. Then solve. ________ Jill has __ stamps. The number of does Jill have now is 46 stamps. Explanation: Jill collects stamps and her stamp book has space for 64 stamps then she needs 18 more stamps to fill the book. So the number of does Jill have now is 64 – 18= 46 stamps. Question 14. Draw a quick picture to solve. Write the difference. Explain what you did to find the difference. _______________ _______________ The subtraction of 25 from 62 is 37. And there are three tens and seven ones. Explanation: In the above image, we can see that there are two tens and eight ones. So here we need to subtract 25 from 62, so we will perform regrouping of subtraction. Here Regrouping in subtraction is a process of exchanging one ten into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. So the subtraction of 25 from 62 is 62 – 25= 37 and there are three tens and seven ones. Conclusion: The information which was discussed in the above section which is Go Math Grade 2 Answer Key Chapter 5 2-Digit Subtraction is useful for you. Exchange these pdf links with your beloved ones and help them to acquire knowledge in maths. Stay connected with us to get the recent updates regarding the Go Math Grade 2 Answer Key for all the chapters.
# How to find the area of a rhombus A rhombus is the “jewel” of all parallelograms – after all, it is in the shape of a diamond. More technically, however, any shape that has four congruent sides is a rhombus. As you venture into the world of geometry, you will begin working with various shapes, including the rhombus. Getting to know more about it, its characteristics and how to work with it will help you along the way. To feel confident you have found a rhombus, look for the following characteristics that all rhombuses possess. The rhombus is an equal opportunity parallelogram. No side is bigger or better than the other. In fact, they are all equal. The rhombus is part of the bigger group of parallelograms. The altitude of any rhombus is the distance (at right angles) between two of the sides. When you create diagonal lines from each corner, they will bisect and create right angles. The Area of a Rhombus When you begin working with shapes in geometry, one of the first things you will do is find out all you can about it. One of these measurements is the area. To calculate the area of a rhombus, you follow the steps here. Step 1: Area = Altitude x s (s = side length) Step 2: The length of the side squared, or multiplied by sine of angle A or B. This would be written as Area = sin(A). Step 3: Multiply the diagonal lengths and then divide by 2. This would be written as Area = (p x q)/2. For example, what if you were given a rhombus with diagonals of 8 m and 10 m. What would the area of this rhombus be? To solve this problem, you simply use the equation above. Related text How to relieve sore throat Area = (8 m x 10 m)/2 = The Perimeter of a Rhombus The perimeter is defined as the distance around the edges of a rhombus. The perimeter is always equal to four times “s” (the length of the sides) since all of the sides are an equal length. This means that Perimeter = 4s. For example, the side length of a rhombus is 10 cm, the perimeter would be 4(10) = 40. Important Notes Squares are rhombuses. This is just a rhombus that has all right angles. While the most common term used to describe this shape in geometry is “rhombus” you may also hear the shape referred to as a rhomb or diamond. The plural for several of these is rhombuses or rhombi. The actual word rhombus is derived from “rhombos” which is Greek and means a piece of wood that is whirled on a string to create a roaring noise. Being able to work with a rhombus is essential. It will help you get through geometry and help you other times in life. Make sure to use the tips and equations here to ensure you can solve for the area and perimeter of any rhombus you are given.
# The Binomial Distribution: Mean and Variance ## Presentation on theme: "The Binomial Distribution: Mean and Variance"— Presentation transcript: The Binomial Distribution: Mean and Variance “Teach A Level Maths” Statistics 1 The Binomial Distribution: Mean and Variance © Christine Crisp Statistics 1 AQA MEI/OCR OCR "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages" There are very simple formulae that we can use to find the mean and variance of a Binomial Distribution. I’ll illustrate the formulae with an example. 3 2 1 x Suppose . The probability distribution is Can you find the mean using the formula for any discrete probability distribution? If you’ve forgotten how to calculate the mean, the results written in a table may remind you: 3 2 1 x mean, m = Can you spot the link to the parameters of the distribution ( n = 3 and p = 0·4 )? 3 2 1 x Suppose . The probability distribution is Can you find the mean using the formula for any discrete probability distribution? If you’ve forgotten how to calculate the mean, the results written in a table may remind you: 3 2 1 x mean, m = ANS: so, x2 1 4 9 3 2 1 x Now for the variance: 1 4 9 3 2 1 x The formula for the variance of any discrete probability distribution is The link with the parameters isn’t so easy to spot this time but it’s easy to remember. It is variance e.g.1 Find the mean and variance of the random variable X where Solution: mean, variance variance SUMMARY The mean of a Binomial Distribution is given by mean, The variance is given by variance Exercise 1. Find the mean and variance of the random variable X where Solutions: (a) mean variance mean (b) variance Exercise 1. Find the mean and variance of the random variable X where Solutions: (c) mean variance mean (d) variance The following slide contains repeats of information on earlier slides, shown without colour, so that it can be printed and photocopied. variance SUMMARY mean, The variance is given by The mean of a Binomial Distribution is given by Solution: e.g.1 Find the mean and variance of the random variable X where
# How do you define a variable and write an expression for each phrase: nine times the quotient of a number and three? Jul 28, 2016 $9 \times \left(x \div 3\right) = 9 \times \frac{x}{3} = \frac{9 \times x}{3}$ #### Explanation: In algebra we can work with numbers even if we do not know their exact value. WE can always just call a number $x , \mathmr{and} y , \mathmr{and} a , \mathmr{and} b$ This is called defining the variable. In this question where a number is being referred to, let's call the number "x". Let the number be $x$ The word "quotient" means the answer to a division and is always used with the word "AND". The QUOTIENT of $15 \mathmr{and} \text{3 is 5 because } \frac{15}{3} = 5$. "nine times the quotient of a number and three" means Muliply 9 by the answer to our number (x), divided by 3. $9 \times \left(x \div 3\right)$ OR $9 \times \frac{x}{3}$
NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.1 These NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.1 Question 1. Find the ratio of the following: (a) Speed of a cycle 15km per hour to the speed of scooter 30 km per hour. (b) 5 m to 10 km (c) 50 paise to ₹ 5 Solution: (a) Speed of the cycle = 15 km per hour. Speed of the scooter = 30 km per hour. ∴ Ratio = Speed of the cycle : Speed of the scooter = 15 : 30 = 1 : 2 (Divided by 15) (b) 10 km = 10 × 1000 m = 10000 m Ratio = 5 m : 10 km = 5 : 10000 = 1 : 2000 (c) ₹ 5 = 5 × 100 paise = 500 paise Ratio = 50 paise : ₹ 5 = 50 : 500 = 1 : 10 Question 2. Convert the following ratios to percentages. (a) 3 : 4 (b) 2 : 3 Solution: (a) 3 : 4 = $$\frac{3}{4}$$ = $$\frac{3}{4}$$ × 100% = 75% (b) 2 : 3 = $$\frac{2}{3}$$ = $$\frac{2}{3}$$ × 100 = $$\frac{200}{3}$$ = 66$$\frac{2}{3}$$ % Question 3. 72% of 25 students are good at mathematics. How many are not good in mathematics. Solution: Total number of students = 25 Number of students good in mathematics = 72% of 25 = $$\frac{72}{100}$$ × 25 = 18 Number of students who are not good in mathematics = 25 – 18 = 7 Percentage of students not good in mathematics = $$\frac{7}{25}$$ × 100 = 7 × 4 = 28% or = $$\frac{28}{100}$$ × 25 = 7 students are not good in Mathematics. Question 4. A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all? Solution: Number of matches won by the team = 10 Let the total number of matches be ‘x’ The team won 40% the total number matches. ∴ $$\frac{40}{100}$$ × x = 10 x = $$\frac{10 \times 100}{40}$$ = 25 ∴ Total number of matches played = 25. Question 5. If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning? Solution: Amount left with Chameli = (100 – 75)% = 25% Let the total amount with Chameli be ₹ ‘x’ ∴ 25% of total money = ₹ 600 $$\frac{25}{100}$$ × x = 600 x = $$\frac{600 \times 100}{25}$$ = ₹ 2400 Amount in the beginning = ₹ 2400 Question 6. If 60% of people in a city like a cricket 30% like football and the remaining like other games, then what percent of the people like other games? If the total number of people is 50 lakh, find the exact number who like each type of game. Solution: People who like cricket = 60% People who like football = 30% People who like other games = 100 – (60 + 30) = 100 – 90 = 10% Total number of people = 50,00,000 No. of people who like cricket = $$\frac{60}{100}$$ × 50,00,000 = 30,00,000 No. of people who like football = $$\frac{30}{100}$$ × 50,00,000 = 15,00,000 No. of people who like other games = $$\frac{10}{100}$$ × 50,00,000 = 5,00,000 error: Content is protected !!
# 4.5: Multiply and Divide Mixed Numbers and Complex Fractions (Part 1) $$\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 • Multiply and divide mixed numbers • Translate phrases to expressions with fractions • Simplify complex fractions • Simplify expressions written with a fraction bar ##### be prepared! Before you get started, take this readiness quiz. 1. Divide and reduce, if possible: $$(4 + 5) ÷ (10 − 7)$$. If you missed this problem, review Example 3.2.8. 2. Multiply and write the answer in simplified form: $$\dfrac{1}{8} \cdot \dfrac{2}{3}$$. If you missed this problem, review Example 4.2.7. 3. Convert $$2 \dfrac{3}{5}$$ into an improper fraction. If you missed this problem, review Example 4.1.11. ## Multiply and Divide Mixed Numbers In the previous section, you learned how to multiply and divide fractions. All of the examples there used either proper or improper fractions. What happens when you are asked to multiply or divide mixed numbers? Remember that we can convert a mixed number to an improper fraction. And you learned how to do that in Visualize Fractions. ##### Example $$\PageIndex{1}$$: multiply Multiply: $$3 \dfrac{1}{3} \cdot \dfrac{5}{8}$$ Solution Convert $$3 \dfrac{1}{3}$$ to an improper fraction. $$\dfrac{10}{3} \cdot \dfrac{5}{8}$$ Multiply. $$\dfrac{10 \cdot 5}{3 \cdot 8}$$ Look for common factors. $$\dfrac{\cancel{2} \cdot 5 \cdot 5}{3 \cdot \cancel{2} \cdot 4}$$ Remove common factors. $$\dfrac{5 \cdot 5}{3 \cdot 4}$$ Simplify. $$\dfrac{25}{12}$$ Notice that we left the answer as an improper fraction, $$\dfrac{25}{12}$$, and did not convert it to a mixed number. In algebra, it is preferable to write answers as improper fractions instead of mixed numbers. This avoids any possible confusion between $$2 \dfrac{1}{12}$$ and $$2 \cdot \dfrac{1}{12}$$. ##### Exercise $$\PageIndex{1}$$ Multiply, and write your answer in simplified form: $$5 \dfrac{2}{3} \cdot \dfrac{6}{17}$$. $$2$$ ##### Exercise $$\PageIndex{2}$$ Multiply, and write your answer in simplified form: $$\dfrac{3}{7} \cdot 5 \dfrac{1}{4}$$. $$\dfrac{9}{4}$$ ##### HOW TO: MULTIPLY OR DIVIDE MIXED NUMBERS Step 1. Convert the mixed numbers to improper fractions. Step 2. Follow the rules for fraction multiplication or division. Step 3. Simplify if possible. ##### Example $$\PageIndex{2}$$: Multiply, and write your answer in simplified form: $$2 \dfrac{4}{5} \left(− 1 \dfrac{7}{8}\right)$$. Solution Convert mixed numbers to improper fractions. $$\dfrac{14}{5} \left(-1 \dfrac{7}{8}\right)$$ Multiply. $$- \dfrac{14 \cdot 15}{5 \cdot 8}$$ Look for common factors. $$- \dfrac{\cancel{2} \cdot 7 \cdot \cancel{5} \cdot 3}{\cancel{5} \cdot \cancel{2} \cdot 4}$$ Remove common factors. $$- \dfrac{7 \cdot 3}{4}$$ Simplify. $$- \dfrac{21}{4}$$ ##### Exercise $$\PageIndex{3}$$ Multiply, and write your answer in simplified form. $$5 \dfrac{5}{7} \left(− 2 \dfrac{5}{8}\right)$$. $$-15$$ ##### Exercise $$\PageIndex{4}$$ Multiply, and write your answer in simplified form. $$-3 \dfrac{2}{5} \cdot 4 \dfrac{1}{6}$$. $$-\dfrac{85}{6}$$ ##### Example $$\PageIndex{3}$$: divide Divide, and write your answer in simplified form: $$3 \dfrac{4}{7} ÷ 5$$. Solution Convert mixed numbers to improper fractions. $$\dfrac{25}{7} \div \dfrac{5}{1}$$ Multiply the first fraction by the reciprocal of the second. $$\dfrac{25}{7} \cdot \dfrac{1}{5}$$ Multiply. $$\dfrac{25 \cdot 1}{7 \cdot 5}$$ Look for common factors. $$\dfrac{\cancel{5} \cdot 5 \cdot 1}{7 \cdot \cancel{5}}$$ Remove common factors. $$\dfrac{5 \cdot 1}{7}$$ Simplify. $$\dfrac{5}{7}$$ ##### Exercise $$\PageIndex{5}$$ Divide, and write your answer in simplified form: $$4 \dfrac{3}{8} ÷ 7$$. $$\dfrac{5}{8}$$ ##### Exercise $$\PageIndex{6}$$ Divide, and write your answer in simplified form: $$2 \dfrac{5}{8} ÷ 3$$. $$\dfrac{7}{8}$$ ##### Example $$\PageIndex{4}$$: divide Divide: $$2 \dfrac{1}{2} \div 1 \dfrac{1}{4}$$. Solution Convert mixed numbers to improper fractions. $$\dfrac{5}{2} \div \dfrac{5}{4}$$ Multiply the first fraction by the reciprocal of the second. $$\dfrac{5}{2} \cdot \dfrac{4}{5}$$ Multiply. $$\dfrac{5 \cdot 4}{2 \cdot 5}$$ Look for common factors. $$\dfrac{\cancel{5} \cdot \cancel{2} \cdot 2}{\cancel{2} \cdot 1 \cdot \cancel{5}}$$ Remove common factors. $$\dfrac{2}{1}$$ Simplify. $$2$$ ##### Exercise $$\PageIndex{7}$$ Divide, and write your answer in simplified form: $$2 \dfrac{2}{3} \div 1 \dfrac{1}{3}$$. $$2$$ ##### Exercise $$\PageIndex{8}$$ Divide, and write your answer in simplified form: $$3 \dfrac{3}{4} \div 1 \dfrac{1}{2}$$. $$\dfrac{5}{2}$$ ## Translate Phrases to Expressions with Fractions The words quotient and ratio are often used to describe fractions. In Subtract Whole Numbers, we defined quotient as the result of division. The quotient of $$a$$ and $$b$$ is the result you get from dividing $$a$$ by $$b$$, or $$\dfrac{a}{b}$$. Let’s practice translating some phrases into algebraic expressions using these terms. ##### Example $$\PageIndex{5}$$: translate Translate the phrase into an algebraic expression: “the quotient of $$3x$$ and $$8$$.” Solution The keyword is quotient; it tells us that the operation is division. Look for the words of and and to find the numbers to divide. The quotient of $$3x$$ and $$8$$. This tells us that we need to divide $$3x$$ by $$8$$. $$\dfrac{3x}{8}$$ ##### Exercise $$\PageIndex{9}$$ Translate the phrase into an algebraic expression: the quotient of $$9s$$ and $$14$$. $$\dfrac{9s}{14}$$ ##### Exercise $$\PageIndex{10}$$ Translate the phrase into an algebraic expression: the quotient of $$5y$$ and $$6$$. $$\dfrac{5y}{6}$$ ##### Example $$\PageIndex{6}$$: Translate the phrase into an algebraic expression: the quotient of the difference of $$m$$ and $$n$$, and $$p$$. Solution We are looking for the quotient of the difference of $$m$$ and $$n$$, and $$p$$. This means we want to divide the difference of $$m$$ and $$n$$ by $$p$$. $\dfrac{m − n}{p} \nonumber$ ##### Exercise $$\PageIndex{11}$$ Translate the phrase into an algebraic expression: the quotient of the difference of $$a$$ and $$b$$, and $$cd$$. $$\dfrac{a-b}{cd}$$ ##### Exercise $$\PageIndex{12}$$ Translate the phrase into an algebraic expression: the quotient of the sum of $$p$$ and $$q$$, and $$r$$. $$\dfrac{p+q}{r}$$ ## Simplify Complex Fractions Our work with fractions so far has included proper fractions, improper fractions, and mixed numbers. Another kind of fraction is called complex fraction, which is a fraction in which the numerator or the denominator contains a fraction. Some examples of complex fractions are: $\dfrac{\dfrac{6}{7}}{3} \quad \dfrac{\dfrac{3}{4}}{\dfrac{5}{8}} \quad \dfrac{\dfrac{x}{2}}{\dfrac{5}{6}} \nonumber$ To simplify a complex fraction, remember that the fraction bar means division. So the complex fraction $$\dfrac{\dfrac{3}{4}}{\dfrac{5}{8}}$$ can be written as $$\dfrac{3}{4} \div \dfrac{5}{8}$$. ##### Example $$\PageIndex{7}$$: simplify Simplify: $$\dfrac{\dfrac{3}{4}}{\dfrac{5}{8}}$$. Solution Rewrite as division. $$\dfrac{3}{4} \div \dfrac{5}{8}$$ Multiply the first fraction by the reciprocal of the second. $$\dfrac{3}{4} \cdot \dfrac{8}{5}$$ Multiply. $$\dfrac{3 \cdot 8}{4 \cdot 5}$$ Look for common factors. $$\dfrac{3 \cdot \cancel{4} \cdot 2}{\cancel{4} \cdot 5}$$ Remove common factors and simplify. $$\dfrac{6}{5}$$ ##### Exercise $$\PageIndex{13}$$ Simplify: $$\dfrac{\dfrac{2}{3}}{\dfrac{5}{6}}$$. $$\dfrac{4}{5}$$ ##### Exercise $$\PageIndex{14}$$ Simplify: $$\dfrac{\dfrac{3}{7}}{\dfrac{6}{11}}$$. $$\dfrac{11}{14}$$ ##### HOW TO: SIMPLIFY A COMPLEX FRACTION Step 1. Rewrite the complex fraction as a division problem. Step 2. Follow the rules for dividing fractions. Step 3. Simplify if possible. ##### Example $$\PageIndex{8}$$: simplify Simplify: $$\dfrac{− \dfrac{6}{7}}{3}$$. Solution Rewrite as division. $$- \dfrac{6}{7} \div 3$$ Multiply the first fraction by the reciprocal of the second. $$- \dfrac{6}{7} \cdot \dfrac{1}{3}$$ Multiply; the product will be negative. $$- \dfrac{6 \cdot 1}{7 \cdot 3}$$ Look for common factors. $$- \dfrac{\cancel{3} \cdot 2 \cdot 1}{7 \cdot \cancel{3}}$$ Remove common factors and simplify. $$- \dfrac{2}{7}$$ ##### Exercise $$\PageIndex{15}$$ Simplify: $$\dfrac{− \dfrac{8}{7}}{4}$$. $$-\dfrac{2}{7}$$ ##### Exercise $$\PageIndex{16}$$ Simplify: $$− \dfrac{3}{\dfrac{9}{10}}$$. $$-\dfrac{10}{3}$$ ##### Example $$\PageIndex{9}$$: simplify Simplify: $$\dfrac{\dfrac{x}{2}}{\dfrac{xy}{6}}$$. Solution Rewrite as division. $$\dfrac{x}{2} \div \dfrac{xy}{6}$$ Multiply the first fraction by the reciprocal of the second. $$\dfrac{x}{2} \cdot \dfrac{6}{xy}$$ Multiply. $$\dfrac{x \cdot 6}{2 \cdot xy}$$ Look for common factors. $$\dfrac{\cancel{x} \cdot 3 \cdot \cancel{2}}{\cancel{2} \cdot \cancel{x} \cdot y}$$ Remove common factors and simplify. $$\dfrac{3}{y}$$ ##### Exercise $$\PageIndex{17}$$ Simplify: $$\dfrac{\dfrac{a}{8}}{\dfrac{ab}{6}}$$. $$\dfrac{3}{4b}$$ ##### Exercise $$\PageIndex{18}$$ Simplify: $$\dfrac{\dfrac{p}{2}}{\dfrac{pq}{8}}$$. $$\dfrac{4}{q}$$ ##### Example $$\PageIndex{10}$$: simplify Simplify: $$\dfrac{2 \dfrac{3}{4}}{\dfrac{1}{8}}$$. Solution Rewrite as division. $$2 \dfrac{3}{4} \div \dfrac{1}{8}$$ Change the mixed number to an improper fraction. $$\dfrac{11}{4} \div \dfrac{1}{8}$$ Multiply the first fraction by the reciprocal of the second. $$\dfrac{11}{4} \cdot \dfrac{8}{1}$$ Multiply. $$\dfrac{11 \cdot 8}{4 \cdot 1}$$ Look for common factors. $$\dfrac{11 \cdot \cancel{4} \cdot 2}{\cancel{4} \cdot 1}$$ Remove common factors and simplify. $$22$$ ##### Exercise $$\PageIndex{19}$$ Simplify: $$\dfrac{\dfrac{5}{7}}{1 \dfrac{2}{5}}$$. $$\dfrac{25}{49}$$ ##### Exercise $$\PageIndex{20}$$ Simplify: $$\dfrac{\dfrac{8}{5}}{3 \dfrac{1}{5}}$$. $$\dfrac{1}{2}$$
# Thread: Prove that the limit exists using eps-delta definition 1. ## Prove that the limit exists using eps-delta definition Question 1 ) limx->4 (7-3 x) = -5 I was able to solve it and found that d= -E/3 - But my problem is that how do I verify/show that my delta works? Please explain. And there is another question where I am not sure how to approach it after simplifying it Question 2) lim x->2 (x^3)=8 \|x-2\|\|x^2+2x+4\|<E \|x-2\|<d 2. Originally Posted by dennosan Question 1 ) limx->4 (7-3 x) = -5 I was able to solve it and found that d= -E/3 Surely, you mean $\displaystyle \delta = \varepsilon/3$. Delta needs to be positive. But my problem is that how do I verify/show that my delta works? Well, how then did you find it? In order to prove that $\displaystyle \lim_{x\to 4}(7-3x)=-5$, you need to show that for any $\displaystyle \varepsilon>0$ there exists a $\displaystyle \delta>0$ such that for all x with $\displaystyle \|x-4\|<\delta$ the following inequality holds: $\displaystyle \|(7-3x)-(-5)\| < \varepsilon$ Now, simplify the left side to get $\displaystyle \|12-3x\|< \varepsilon$ Pull out a positive factor of 3 like this $\displaystyle 3\cdot \|4-x\|<\varepsilon$ and divide by 3 to get $\displaystyle \|4-x\|<\varepsilon/3$ which is equivalent to $\displaystyle \|x-4\|<\varepsilon/3$ Now all these inequalities are equivalent. So if the last inequality is satisfied, which is the case if you set $\displaystyle \delta := \varepsilon/3$ and require that $\displaystyle \|x-4\|<\delta$, then the first of these inequalities, the one that you have to show, also holds. And there is another question where I am not sure how to approach it after simplifying it Question 2) lim x->2 (x^3)=8 \|x-2\|\|x^2+2x+4\|<E \|x-2\|<d $\displaystyle \|x^3-8\|<\varepsilon$ and try to figure out how requiring $\displaystyle \|x-2\|<\delta$, for a suitably chosen $\displaystyle \delta>0$, guarantees that this inequality holds. $\displaystyle \|x-2\|\cdot \|x^2+2x+4\|<\varepsilon$ Well, $\displaystyle x^2+2x+4=(x+1)^2+3$ by completion of the square. Thus the quadratic polynomial and its absolute value are the same (because the polynomial is always positive) and it assumes its smallest value 3 at x=-1. So lets therefore already limit x to satisfy \|x-2\|<1, i.e. x must be from the interval (1,3). For such values of x, the quadratic factor assumes its largest value at x=3, which gives us the required upper bound $\displaystyle M := 3^2+2*3+4=19$. Finally, divide the above inequality by M on both sides to get your $\displaystyle \delta :=\min(1,\varepsilon/19)$. (We must make sure that $\displaystyle \delta < 1$, because otherwies our choice of M as an upper bound for the quadratic factor on the left side of the inequality would not be valid.)
# Order Of Operations Basic People have come up with a common set of basic rules for performing computations. Many years back, mathematicians came up with the standard “Order of Operations.” This is telling you which calculations should be made first in expressions that have more than just one operation. Recommended😍 ### Online GED Classes – Fast and Easy Learn Just 1 Hour A Day To Get Your Diploma in 2 Months. Get Quickly Prepared To Pass The GED Test. Without standard procedures for carrying out calculations, different people might get different answers when solving the same problem. 1. Solve: $$-12 + 6(-4)$$ A. B. C. D. Question 1 of 3 2. Solve: $$\frac{-24}{(-6)(-1)}$$ A. B. C. D. Question 2 of 3 3. Solve: $$48 \div 4(6)$$ A. B. C. D. Question 3 of 3 This lesson is provided by Onsego GED Prep. ### Video Transcription We explained parentheses earlier. You must do the math part inside of the parentheses first. When you’ve got ready with all of the stuff inside, you can move on to the operations that are in the problem’s open areas. Also, when they’re located inside the parentheses, you first need to complete the operations of multiplication and division. When the problem to solve is “3 + 5 x 6”, you need to do the part “5 x 6” of the problem first and add the three (3) later to the product. The same principle is true for divisions. A problem like, for example, “16 – 4 ÷ 2” needs to be done in two steps. First comes division, which results in the answer of 2. In the next step, you will subtract 2 from 16. And when you have to deal with a variety of operations, addition and subtraction are the last steps. So, begin with the parentheses, then move on to multiplication and division, and lastly, do addition and subtraction. Here’s an example: 6 ÷ 2 x 5 – 5 + (11 – 3) = ? 1: Parentheses first: 6 : 2 x 5 – 5+ (8) = ? 2: Multiplication and Division: In which order you need to be computing, multiplication, and division are always determined by which of the two comes first (from left to right). 6 ÷ 2 x 5 – 5+ (8) = ? 3 x 5 – 5 +(8) = ? 15 – 5 + 8= The order in which addition and subtraction must be dealt with first is determined by which one comes first as well (from left to right). 15-5+8= 10+8=18 Order of Operations Summary 1) First, carry out all operations within our grouping symbols. Grouping symbols include parentheses ( ), brackets [ ], braces { }, and fraction bars. 2) Then multiply and divide (always from left to right). 3) Then, add and subtract (from left to right). Summary The order in which we evaluate expressions can be ambiguous. The best way for avoiding ambiguities when evaluating expressions is establishing the correct order in which our operations must be performed. That’s crucial! The guidelines described here should be strictly enforced at all times when we evaluate expressions.
# Skip Counting Nines Skip counting nines may seem difficult at first. As the numbers we use get larger, it is easy to get overwhelmed, but honestly, once you get past eight, nine through twelve are easier--many of the multiplication problems have been covered in other units already--so there's not a lot of new material to learn. There is a little trick your child can use to remember the multiples of nine. ## Counting By Nines The trick is that whatever the answer is, it will always add up to nine. For example: 9 X 2 = 18 1 + 8 = 9 From 18, your child can then work his way through factor families by increasing the first digit (in this case "1") and decreasing the second digit (in this case "8"): 1 + 8 = 9 2 + 7 = 9 3 + 6 = 9 4 + 5 = 9 5 + 4 = 9 6 + 3 = 9 7 + 2 = 9 8 + 1 = 9 9 + 0 = 9 An easier way is to drop a digit from whatever factor nine is paired with and then add it up to nine. For example: 9 X 7 = ? In the example above, nine is paired with a factor of seven. Drop a digit, you now have the number 6. Six plus 3 equals nine so the answer is 63. Let's try another one. 9 X 5 = ? In this example, nine is paired with five. Drop down to 4. Four plus 5 equals nine. The answer is 45. It may seem tricky at first, but once your child gets this quick method down, he or she will be the fastest kid in class! It's not entirely useful for larger numbers, but it is a good way to check your work. Example: 9 X 26 = 234 2 + 3 + 4 = 9 ## Skip Counting Nines Worksheets In the first worksheet, skip count by nines to fill in the missing numbers on the baseball jerseys. The second worksheet will put your child's adding skills to the test. Remember how we added answers together to see if they equaled nine? She'll need to do that in order to circle the mice with multiples of nines on their money sacks. The final worksheet is a bit of a review. It's a color-by-multiples worksheet. Use the legend at the top to determine which color should be used in each box. If done correctly, your child should color an ice cream cone. Move on to the next skip counting lesson, review the previous one, or learn more about numbers and counting: › Skip Counting Nines
# Math: Place Value (Whole Number) ### Suggested Time Frame: 25 Instructional Days ##### Focus TEKS 3rd Grade – Using Place Value to Represent Numbers to 100,000 • 3.2A compose and decompose numbers up to 100,000 as a sum of so many ten thousands, so many thousands, so many hundreds, so many tens, and so many ones using objects, pictorial models, and numbers, including expanded notation as appropriate; – R RC1 • 3.2B describe the mathematical relationships found in the base-10 place value system through the hundred thousands place; – S RC1 • 3.2C represent a number on a number line as being between two consecutive multiples of 10; 100; 1,000; or 10,000 and use words to describe relative size of numbers in order to round whole numbers; and – S RC1 3rd Grade – Using Place Value to Compare Numbers up to 100,000 • 3.2D compare and order whole numbers up to 100,000 and represent comparisons using the symbols >, <, or =. – R RC1 Whole Number Place Value Through 1 Billion • 4.2B represent the value of the digit in whole numbers through 1,000,000,000 and decimals to the hundredths using expanded notation and numerals; – R RC1 • 4.2A interpret the value of each place-value position as 10 times the position to the right and as one-tenth of the value of the place to its left; – S RC1 • 4.2C compare and order whole numbers to 1,000,000,000 and represent comparisons using the symbols >, <, or =; – S RC1 • 4.2D round whole numbers to a given place value through the hundred thousands place; – S RC1 ##### Numeracy TEKS 3rd Grade – Building Fluency with Multiplication and Division Facts (Use-Ten and Use a Rule Strategies for Multiplication; Including the Think Multiplication Strategy for Division) (Suggested: 10 days) • 3.4F recall facts [zeros, ones, fives, and tens facts] to multiply up to 10 by 10 with automaticity and recall the corresponding division facts;- S RC2 [Including the Use-Ten and Use a Rule Strategies for Multiplication; Including the Think Multiplication Strategy for Division] • 3.4D determine the total number of objects when equally-sized groups of objects are combined or arranged in arrays up to 10 by 10 [arrays representing zeros, ones, fives, and tens facts]; – S RC2 • 3.4E represent multiplication facts [zeros, ones, fives, and tens facts] by using a variety of approaches such as repeated addition, equal-sized groups, arrays, area models, equal jumps on a number line, and skip counting; – S RC2 • 3.4H determine the number of objects in each group when a set of objects is partitioned into equal shares or a set of objects is shared equally; – S RC2 3rd Grade – Building Fluency with Multiplication and Division Facts (Doubling Strategy for Multiplication; Including the Think Multiplication Strategy for Division) (Suggested: 10 days) • 3.4F recall facts [twos, fours, and eights facts in this unit] to multiply up to 10 by 10 with automaticity and recall the corresponding division facts;- S RC2 [Including the Doubling Strategy for Multiplication; Including the Think Multiplication Strategy for Division] • 3.4D determine the total number of objects when equally-sized groups of objects are combined or arranged in arrays up to 10 by 10 [arrays representing twos, fours, and eights facts]; – S RC2 • 3.4E represent multiplication facts [twos, fours, and eights facts] by using a variety of approaches such as repeated addition, equal-sized groups, arrays, area models, equal jumps on a number line, and skip counting; – S RC2 • 3.4H determine the number of objects in each group when a set of objects is partitioned into equal shares or a set of objects is shared equally; – S RC2 Counting Patterns (Suggested: 3 days) • RRISD 4.5 Skip count by whole numbers starting from 0 and other starting numbers and describe patterns observed while counting; [Skip count by twos, fives, and tens within 1,200.] ##### Spiral Review TEKS Grade 3 Representing and Solving Story Problems (Suggested: 18 days) • Computation and Problem Solving • 3.4A solve with fluency one-step and two-step problems involving addition and subtraction within 1,000 using strategies based on place value, properties of operations, and the relationship between addition and subtraction; – R RC2 • 3.4B round to the nearest 10 or 100 or use compatible numbers to estimate solutions to addition and subtraction problems; – S RC2 • 3.4K solve one-step and two-step problems involving multiplication and division within 100 using strategies based on objects; pictorial models, including arrays, area models, and equal groups; properties of operations; or recall of facts. – R RC2 • 3.8B solve one- and two-step problems using categorical data represented with a frequency table, dot plot, pictograph, or bar graph with scaled intervals. – S RC4 • Representing • 3.5A represent one- and two-step problems involving addition and subtraction of whole numbers to 1,000 using pictorial models, number lines, and equations; – R RC2 • 3.5B represent and solve one- and two-step multiplication and division problems within 100 using arrays, strip diagrams, and equations; – R RC2 • 3.5E represent real-world relationships using number pairs in a table and verbal descriptions. – R RC2 Grade 3 Time Intervals (Suggested: 5 days) • 3.7C determine the solutions to problems involving addition and subtraction of time intervals in minutes using pictorial models or tools such as a 15-minute event plus a 30-minute event equals 45 minutes; – S RC3 [including relationships between minutes and hours] Counting Quantities (Suggested: 2 days, using both numeracy and spiral review time) • RRISD 4.1 Organize and count a collection of objects and create a recording of how the objects were counted • 4.1C select tools, including real objects, manipulatives, paper and pencil, and technology as appropriate, and techniques, including mental math, estimation, and number sense as appropriate, to solve problems; • 4.1E create and use representations to organize, record, and communicate mathematical ideas; • Counting Coins and Bills (Make this the focus on 1 day of Counting Collections) • 3.4C determine the value of a collection of coins and bills; – S RC4 Telling Time (Ongoing) • 2.9G read and write time to the nearest one-minute increment using analog and digital clocks and distinguish between a.m. and p.m.
Advertisement Remove all ads # Maximize Z = 50x + 30y Subject to 2 X + Y ≤ 18 3 X + 2 Y ≤ 34 X , Y ≥ 0 - Mathematics Advertisement Remove all ads Advertisement Remove all ads Advertisement Remove all ads Sum Maximize Z = 50x + 30y Subject to $2x + y \leq 18$ $3x + 2y \leq 34$ $x, y \geq 0$ Advertisement Remove all ads #### Solution First, we will convert the given inequations into equations, we obtain the following equations: 2x y = 18, 3x + 2y = 34 Region represented by 2x ≥ 18: The line 2x y = 18 meets the coordinate axes at A(9, 0) and B(0, 18) respectively. By joining these points we obtain the line 2x y = 18. Clearly (0,0) does not satisfies the inequation 2x ≥ 18. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 2x ≥ 18. Region represented by 3x + 2y ≤ 34: The line 3x + 2y = 34 meets the coordinate axes at $C\left( \frac{34}{3}, 0 \right)$ and D(0, 17) respectively. By joining these points we obtain the line 3x + 2y= 34. Clearly (0,0) satisfies the inequation 3x + 2y ≤ 34. So,the region containing the origin represents the solution set of the inequation 3x + 2y ≤ 34. The corner points of the feasible region are A(9, 0), $C\left( \frac{34}{3}, 0 \right)$ and E(2, 14). The values of Z at these corner points are as follows. Corner point Z = 50x + 30y A(9, 0) 50 × 9 + 3 × 0 = 450 $C\left( \frac{34}{3}, 0 \right)$ 50 x $\frac{34}{3}$ + 30 x 0= $\frac{1700}{3}$ E(2, 14) 50 × 2 + 30 × 14 = 520 Therefore, the maximum value of Z is $\frac{1700}{3}\text{ at the point } \left( \frac{34}{3}, 0 \right)$ .Hence, x = $\frac{34}{3}$ and y =0 is the optimal solution of the given LPP. Thus, the optimal value of Z is $\frac{1700}{3}$ . Concept: Graphical Method of Solving Linear Programming Problems Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 12 Maths Chapter 30 Linear programming Exercise 30.2 \| Q 4 \| Page 32 #### Video TutorialsVIEW ALL [1] Advertisement Remove all ads Share Notifications View all notifications Forgot password?
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 1.3: Shifting and Reflecting [ "article:topic", "reflection", "authorname:green", "horizontal shift", "Vertical Shift", "stretching", "compressing" ] $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ ### 1. Six Basic Functions Below are six basic functions: 1. 2. Memorize the shapes of these functions. ### 2. Horizontal Shifting Consider the graphs $$y =$$ • $$(x+0)^2$$ • $$(x+1)^2$$ • $$(x+2)^2$$ • $$(x+3)^2$$ Exercise Use the list features of a calculator to sketch the graph of $$y = \dfrac{1}{ [x - \{0,1,2,3\}] }$$ Horizontal Shifting Rules • Rule 1: $$f(x - a) = f(x)$$ shifted $$a$$ units to the right. • Rule 2: $$f(x + a) = f(x)$$ shifted $$a$$ units to the left. ### 3. Vertical Shifting Consider the graphs $$y =$$ • $$x^3$$ • $$x^3+ 1$$ • $$x^3 + 2$$ • $$x^3 + 3$$ Exercise Use the list features of a calculator to sketch the graph of $$y = x^3 - \{0,1,2,3\}$$ Vertical Shifting Rules • Rule 3: $$f(x ) + a = f(x)$$ shifted a units up. • Rule 4: $$f(x) - a = f(x)$$ shifted a units down. ### 4. Reflecting About the x-axis Consider the graphs of $$y = x^2$$ and $$y = -x^2$$. x-Axis Reflection Rule Rule 5: $$-f(x) = f(x)$$ reflected about the x-axis. ### 5. Reflecting About the y-axis Exercise Use the calculator to graph $$y=\sqrt{x}$$ and $$y=\sqrt{-x}$$ y-Axis Reflection Rule Rule 6: $$f(-x ) = f(x)$$ reflected about the y-axis. ### 6. Stretching and Compressing Exercise Graph the following: $$y = \{1,2,3,4\}x^3$$ $$y = {1/2,1/3,1/4,1/5}x^3$$ Stretching and Compression rules: • Rule 7: $$cf(x ) = f(x)$$ (for $$c > 1$$) stretched vertically. • Rule 8: $$cf(x ) = f(x)$$ (for $$c < 1$$) compressed vertically. Exercise Graph the following 1. $$y = x^2 - 10$$ 2. $$y = \sqrt{x - 2}$$ 3. $$y = -\|x - 5\| + 3$$ We will do some examples (including the graph of the winnings for the gambler and for the casino). ### 7. Increasing and Decreasing Functions Definition A function is called increasing if as an object moves from left to right, it is moving upwards along the graph. Or equivalently, If $$x < y$$, then $$f(x) < f(y)$$. Example 1 The curve $y = x^2$ is increasing on $$(0,\infty)$$ and decreasing on $$(-\infty,0)$$.
# Chapter 7 Extra Topics Crater Lake, Oregon ```Chapter 7 Extra Topics Crater Lake, Oregon Photo by Vickie Kelly, 1998 Greg Kelly, Hanford High School, Richland, Washington Fg Centers of Mass: d Torque is a function of force and distance. Lake Superior, Washburn, WI Photo by Vickie Kelly, 2004 (Torque is the tendency of a system to rotate about a point.)  If the forces are all gravitational, then torque  m1 g  mgx m2 g If the net torque is zero, then the system will balance. Since gravity is the same throughout the system, we could factor g out of the equation. M O   mk xk This is called the  If we divide Mo by the total mass, we can find the center of mass (balance point.) M O   mk xk MO x  M x m k k m k  x MO  M x m k For a thin rod or strip: k d = density per unit length m (d is the Greek letter delta.) k M O   x  d  x  dx b a M   d  x  dx b mass: center of mass: a MO x M For a rod of uniform density and thickness, the center of mass is in the middle.  For a two dimensional shape, we need two distances to locate the center of mass. y strip of mass dm x  distance from the y axis to the center of the strip x y  distance from the x axis to y x the center of the strip   y(pronounced dm Moment about x-axis: M xx tilde Center mass: ecksoftilda)  Moment about y-axis: M y  x dm  Mass: M  dm x My M Mx y M  For a two dimensional shape, we need two distances to locate the center of mass. y For a plate of uniform thickness and density, the density drops out of the equation when finding the center of mass. x y x Vocabulary: center of mass = center of gravity = centroid constant density d = homogeneous = uniform  9 8 yx 2 Mx   3 Mx   3 0 7 6 0 1 2 2 x  x dx 2 1 4 x dx 2 3 M y   x  x 2 dx 0 3 M y   x3dx 0 5 4 3 2 x 1 .5x 0 1 2 2 3 xx coordinate of 1 2 centroid y  =x (2.25, 2.7) 2 1 53 Mx  x 10 0 1 43 My  x 4 0 243 Mx  10 81 My  4 1 33 M   x dx  x  9 0 3 0 243 81 Mx 27 My 10 9 4 y   x   M 9 10 M 9 4 3 2  Note: The centroid does not have to be on the object.  If the center of mass is obvious, use a shortcut:  square  rectangle  circle  h 3 b 3 right triangle  Theorems of Pappus: When a two dimensional shape is rotated about an axis: Volume = area . distance traveled by the centroid. Surface Area = perimeter . distance traveled by the centroid of the arc. Consider an 8 cm diameter donut with a 3 cm diameter cross section: 1.5 V  2 r  area V  2  2.5    1.5  V  11.25 2 2.5 2 V  111cm 3  We can find the centroid of a semi-circular surface by using the Theorems of Pappus and working back to get the centroid. 1 2 A  r 2 y 4 3 V=  r 3 1 2 4 3 2 y   r   r 2 3 4r y 3  ```
# 2022 AMC 8 Problems/Problem 25 ## Problem A cricket randomly hops between $4$ leaves, on each turn hopping to one of the other $3$ leaves with equal probability. After $4$ hops what is the probability that the cricket has returned to the leaf where it started? $\textbf{(A) }\frac{2}{9}\qquad\textbf{(B) }\frac{19}{80}\qquad\textbf{(C) }\frac{20}{81}\qquad\textbf{(D) }\frac{1}{4}\qquad\textbf{(E) }\frac{7}{27}$ ## Solution 1 (Casework) Let $A$ denote the leaf where the cricket starts and $B$ denote one of the other $3$ leaves. Note that: • If the cricket is at $A,$ then the probability that it hops to $B$ next is $1.$ • If the cricket is at $B,$ then the probability that it hops to $A$ next is $\frac13.$ • If the cricket is at $B,$ then the probability that it hops to $B$ next is $\frac23.$ We apply casework to the possible paths of the cricket: 1. $A \rightarrow B \rightarrow A \rightarrow B \rightarrow A$ The probability for this case is $1\cdot\frac13\cdot1\cdot\frac13=\frac19.$ 2. $A \rightarrow B \rightarrow B \rightarrow B \rightarrow A$ The probability for this case is $1\cdot\frac23\cdot\frac23\cdot\frac13=\frac{4}{27}.$ Together, the probability that the cricket returns to $A$ after $4$ hops is $\frac19+\frac{4}{27}=\boxed{\textbf{(E) }\frac{7}{27}}.$ ~MRENTHUSIASM ## Solution 2 (Casework) We can label the leaves as shown below: Carefully counting cases, we see that there are $7$ ways for the cricket to return to leaf $A$ after four hops if its first hop was to leaf $B$: 1. $A \rightarrow B \rightarrow A \rightarrow B \rightarrow A$ 2. $A \rightarrow B \rightarrow A \rightarrow C \rightarrow A$ 3. $A \rightarrow B \rightarrow A \rightarrow D \rightarrow A$ 4. $A \rightarrow B \rightarrow C \rightarrow B \rightarrow A$ 5. $A \rightarrow B \rightarrow C \rightarrow D \rightarrow A$ 6. $A \rightarrow B \rightarrow D \rightarrow B \rightarrow A$ 7. $A \rightarrow B \rightarrow D \rightarrow C \rightarrow A$ By symmetry, we know that there are $7$ ways if the cricket's first hop was to leaf $C$, and there are $7$ ways if the cricket's first hop was to leaf $D$. So, there are $21$ ways in total for the cricket to return to leaf $A$ after four hops. Since there are $3^4 = 81$ possible ways altogether for the cricket to hop to any other leaf four times, the answer is $\frac{21}{81} = \boxed{\textbf{(E) }\frac{7}{27}}$. ~mahaler ## Solution 3 (Complement) There are always three possible leaves to jump to every time the cricket decides to jump, so there is a total number of $3^4$ routes. Let $A$ denote the leaf cricket starts at, and $B, C, D$ be the other leaves. If we want the cricket to move to leaf $A$ for its last jump, the cricket cannot jump to leaf $A$ for its third jump. Also, considering that the cricket starts at leaf $A$, he cannot jump to leaf $A$ for its first jump. Note that there are $3\cdot2=6$ paths if the cricket moves to leaf $A$ for its third jump. Therefore, we can conclude that the total number of possible paths for the cricket to return to leaf $A$ after four jumps is $3^3 - 6 = 21$, so the answer is $\frac{21}{3^4} = \frac{21}{81}=\boxed{\textbf{(E) }\frac{7}{27}}$. ## Solution 4 (Recursion) Denote $P_n$ to be the probability that the cricket would return back to the first point after $n$ hops. Then, we get the recursive formula $$P_n = \frac13(1-P_{n-1})$$ because if the leaf is not on the target leaf, then there is a $\frac13$ probability that it will make it back. With this formula and the fact that $P_1=0$ (After one hop, the cricket can never be back to the target leaf.), we have $$P_2 = \frac13, P_3 = \frac29, P_4 = \frac7{27},$$ so our answer is $\boxed{\textbf{(E) }\frac{7}{27}}$. ~wamofan ## Solution 5 (Dynamic Programming) Let $A$ denote the leaf cricket starts at, and $B, C, D$ be the other leaves, similar to Solution 2. Let $A(n)$ be the probability the cricket lands on $A$ after $n$ hops, $B(n)$ be the probability the cricket lands on $B$ after crawling $n$ hops, etc. Note that $A(1)=0$ and $B(1)=C(1)=D(1)=\frac13.$ For $n\geq2,$ the probability that the cricket land on each leaf after $n$ hops is $\frac13$ the sum of the probability the cricket land on other leaves after $n-1$ hops. So, we have \begin{align} A(n) &= \frac13 \cdot [B(n-1) + C(n-1) + D(n-1)], \\ B(n) &= \frac13 \cdot [A(n-1) + C(n-1) + D(n-1)], \\ C(n) &= \frac13 \cdot [A(n-1) + B(n-1) + D(n-1)], \\ D(n) &= \frac13 \cdot [A(n-1) + B(n-1) + C(n-1)]. \end{align} It follows that $A(n) = B(n-1) = C(n-1) = D(n-1).$ We construct the following table: $$\begin{array}{c\|cccc} & & & & \\ [-2ex] n & A(n) & B(n) & C(n) & D(n) \\ [1ex] \hline & & & & \\ [-1ex] 1 & 0 & \frac13 & \frac13 & \frac13 \\ & & & & \\ 2 & \frac13 & \frac29 & \frac29 & \frac29 \\ & & & & \\ 3 & \frac29 & \frac{7}{27} & \frac{7}{27} & \frac{7}{27} \\ & & & & \\ 4 & \frac{7}{27} & \frac{20}{81} & \frac{20}{81} & \frac{20}{81} \\ [1ex] \end{array}$$ Therefore, the answer is $A(4)=\boxed{\textbf{(E) }\frac{7}{27}}$. ## Solution 6 (Generating Function) Assign the leaves to $0, 1, 2,$ and $3$ modulo $4,$ and let $0$ be the starting leaf. We then use generating functions with relation to the change of leaves. For example, from $3$ to $1$ would be a change of $2,$ and from $1$ to $2$ would be a change of $1.$ This generating function is equal to $(x+x^2+x^3)^4.$ It is clear that we want the coefficients in the form of $x^{4n},$ where $n$ is a positive integer. One application of roots of unity filter gives us a successful case count of $\frac{81+1+1+1}{4} = 21.$ Therefore, the answer is $\frac{21}{3^4}=\boxed{\textbf{(E) }\frac{7}{27}}.$ ~sigma ## Solution 7 (Also Generating Functions) Let the leaves be $(0,0), (0,1), (1,0),$ and $(1,1)$ on the coordinate plane, with the cricket starting at $(0,0)$. Then we write a generating function. We denote $x$ a change in the x-value of the cricket, and similarly for $y$. Then our generating function is $(x+y+xy)^4,$ and we wish to compute the number of terms in which the exponents of both x and y are even. To do this, we first square to get $(x^2 + y^2 + x^2y^2 + 2xy + 2x^2y + 2xy^2)^2$. Note that every term squared will give even powers for x and y, so that gives us $3 + 4\cdot3 = 15.$ Then every combination of $x^2, y^2,$ and $x^2y^2$ will also give us even powers for x and y, so that yields $6$ more terms, for a total of $21.$ Now in total there $3^4 = 81$ possible sequences, so $21/81$ gives us the answer of $\boxed{\textbf{(E) }\frac{7}{27}}.$ ~littlefox_amc ## Remark This problem is a reduced version of 1985 AIME Problem 12, changing $7$ steps into $4$ steps. This problem is also similar to 2003 AIME II Problem 13. ~Math-X ## Video Solution(🚀Under 2 min🚀 Easy logic with all paths color-coded ✨) ~Education, the Study of Everything ~ pi_is_3.14 ## Video Solution ~Mathematical Dexterity ~Interstigation ~David ~STEMbreezy ~savannahsolver
# sin 30° ## Definition The value of sine when the angle of the right angled triangle equals to $30^\circ$, is called $\sin 30^\circ$. The value of $\sin 30^\circ$ can possibly find in mathematics by calculating the ratio of length of opposite side to length of hypotenuse of right angled triangle when the angle of right angled triangle is $30^\circ$. ### Proof The value of $\sin 30^\circ$ can be derived in two different approaches but the system of development is common and it is geometric system. 1 #### Fundamental approach In the case of right angled triangle whose angle is $30^\circ$, the length of hypotenuse is twice the length of the opposite side of the triangle. The value of sine of angle $30$ degrees can be derived on the basis of this principle. $Length \, of \, Hypotenuse$ $=$ $2 \times Length \, of \, Opposite \, side$ $$\implies \frac{1}{2} = \frac{Length \, of \, Opposite \, side}{Length \, of \, Hypotenuse}$$ $$\implies \frac{Length \, of \, Opposite \, side}{Length \, of \, Hypotenuse} = \frac{1}{2}$$ The ratio of length of the opposite side to length of hypotenuse is called sine of an angle but the angle of the right angled triangle here is $30^\circ$. So, the ratio between two sides is sine of angle $30^\circ$. $$\therefore \,\, \sin 30^\circ = \frac{1}{2}$$ The value of sine of angle $30^\circ$ is exactly real because the fundamental approach is derived by the system of geometric properties of the right angled triangle when the angle of the triangle is $30^\circ$. 2 #### Practical approach The value of $\sin 30^\circ$ can also be derived practically by the direct geometric approach. 1. Use ruler and draw a straight line horizontally. Call the left side point of this line as point $B$. 2. Take protractor, coincide point $B$ with centre of the protractor and also coincide horizontal line with right side base line of the protractor. Then mark a point at $30$ degrees angle by considering the bottom scale of the protractor. 3. Take ruler and draw a straight line from point $B$ through $30$ degrees angle point. 4. Take compass and set it to have $8$ centimetres length between pencil’s lead and needle point by considering the ruler. Then, draw an arc from point $B$ on $30$ degrees angle line and call the intersecting point of arc and $30^\circ$ angle line as point $C$. 5. Take set square, and draw a straight line from $C$ to horizontal line but the line should intersect it perpendicularly. Call the intersecting point as point $D$. In this way, a right angled triangle ($\Delta CBD$) is constructed with $30^\circ$ angle and $8$ centimeters long line. The $8$ centimetres long line is actually hypotenuse of this triangle but the length of opposite side is unknown. So, take ruler and measure the length of opposite side. You observe that the length of the opposite side is $4$ centimetres exactly. $$\sin 30^\circ = \frac{Length \, of \, Opposite \, side}{Length \, of \, Hypotenuse}$$ $$\implies \sin 30^\circ = \frac{CD}{BC}$$ $$\implies \sin 30^\circ = \frac{4}{8}$$ $$\implies \sin 30^\circ = \frac{1}{2} = 0.5$$ ##### Result The value of sine of angle $30$ degrees is derived in two methods and they both have given same value. So, it can be used anywhere without any doubt. ###### Representation It is usually expressed in three different forms in mathematics. In sexagesimal system, it is usually written in mathematical form as follows. $$\sin 30^° = \frac{1}{2}$$ In circular system, it is also usually written in mathematics as follows. $$\sin \Bigg(\frac{\pi}{6}\Bigg) = \frac{1}{2}$$ In centesimal system, it is also written in mathematical form as follows. $$\sin {33\frac{1}{3}}^g = \frac{1}{2}$$
# Factoring a linear binomial To factor a number means to write it as a product of its factors. A linear binomial has two terms and highest degree of one For example: 2x + 1; 9y + 43; 34p + 17q are linear binomials. To factor a linear binomial means to write it as a product of its factors. Rules to factor a linear binomial • At first, we find the highest common factor of the terms of the linear binomial • The HCF is factored out and the sum/difference of remaining factors is written in a pair of parentheses. • This is like reversing the distributive property of multiplication. Factor the following linear binomial: 28n + 63n2 ### Solution Step 1: The HCF of 28n and 63n2 is 7n Step 2: Factoring the linear binomial 28n + 63n2 = 7n (4 + 9n) Factor the following linear binomial: 65z – 52z4 ### Solution Step 1: The HCF of 65z and 52z4 is 13z Step 2: Factoring the linear binomial 65z – 52z4 = 13z (5 – 4z3) Factor the following linear binomial: 24x + 84x3 ### Solution Step 1: The HCF of 24x and 84x3 is 12x Step 2: Factoring the linear binomial 24x + 84x3 = 12x (2 + 7x2)
# A-level Mathematics/OCR/C1/Equations An equation consists of two expressions joined by the equals sign (${\displaystyle =}$). Everything on the left-hand side is equal to everything on the right-hand side, for example ${\displaystyle 2+3=4+1}$. Some equations contain a variable, usually denoted by ${\displaystyle x}$ or ${\displaystyle y}$ , though any symbol can be used. ## Manipulating expressions Sometimes, expressions will be messier than they need to be, and they can be represented in an easier to understand form. The skills here are essential to the rest of the A-level course, although it is very likely that you have already covered them at GCSE. ### Collecting like terms When collecting like terms, you simply add all the terms in ${\displaystyle x}$ together, all the terms in ${\displaystyle y}$ together, and all the terms in ${\displaystyle z}$ together. The same applies for any other letter that represents a variable. For example, ${\displaystyle 2x+4y+8z-3x-7y-2z+4x}$ becomes: ${\displaystyle 2x-3x+4x=3x}$ ${\displaystyle 4y-7y=-3y}$ ${\displaystyle 8z-2z=6z}$ So, by adding all the answers, ${\displaystyle 2x+4y+8z-3x-7y-2z+4x}$ simplified is ${\displaystyle 3x-3y+6z}$. ### Multiplication Multiplication of different variables such as ${\displaystyle a\times b}$ becomes ${\displaystyle ab}$. Single variables become indices, so ${\displaystyle x\times x}$ is ${\displaystyle x^{2}}$. Like addition and subtraction, you keep like terms together. So, for example: ${\displaystyle 2x^{2}z\times 3yz^{2}\times 4xy^{3}}$ becomes: ${\displaystyle 24{x^{3}}{y^{4}}{z^{3}}}$ ### Fractions It is quite often that fractions are encountered. Therefore we need to learn how to handle them properly. When working with fractions, the rule is to make all of the denominators equal, and then write the expression as one fraction. You need to multiply both the top and bottom by the same amount to keep the meaning of the fraction the same. For example, for ${\displaystyle {\frac {3x}{2}}+{\frac {2y}{5}}-{\frac {z}{10}}}$, the common denomimator is ${\displaystyle 10}$. Multiply both parts by ${\displaystyle 5}$: ${\displaystyle {\frac {15x}{10}}}$ Multiply both parts by ${\displaystyle 2}$: ${\displaystyle {\frac {4y}{10}}}$ Leave this as it is: ${\displaystyle {\frac {z}{10}}}$ You now have ${\displaystyle {\frac {15x}{10}}+{\frac {4y}{10}}-{\frac {z}{10}}}$, which becomes ${\displaystyle {\frac {15x+4y-z}{10}}}$. ## Solving equations Often, to solve an equation you must rearrange it so that the unknown term is on its own side of the equals sign. By rearranging ${\displaystyle 2+x=5}$ to ${\displaystyle x=5-2}$, ${\displaystyle x}$ has been made the subject of the equation. Now by simplifying the equation, you can find that the solution is ${\displaystyle x=3}$. An equation with a variable will only hold true for certain values of that variable. For example ${\displaystyle 2+x=5}$ is only true for ${\displaystyle x=3}$. The values that the variables have when the equation is true are called the solutions of the equation. Therefore ${\displaystyle x=3}$ is the solution of the equation ${\displaystyle 2+x=5}$. ### Changing the subject of an equation You will usually be given equations that are more complex than the example above. To move a term from one side of the equals sign to the other, you have to do the same thing on both sides of the equals sign. For example, to make ${\displaystyle x}$ the subject of ${\displaystyle y={\frac {4a(x^{2}+b)}{3}}}$: Multiply both sides by ${\displaystyle 3}$ ${\displaystyle 3y=4a(x^{2}+b)}$ Divide both sides by ${\displaystyle 4a}$ ${\displaystyle {\frac {3y}{4a}}=x^{2}+b}$ Subtract ${\displaystyle b}$ from both sides ${\displaystyle {\frac {3y}{4a}}-b=x^{2}}$ Square root both sides ${\displaystyle {\sqrt {{\frac {3y}{4a}}-b}}=x}$ or ${\displaystyle -{\sqrt {{\frac {3y}{4a}}-b}}=x}$ Quadratic equations are equations where the variable is raised to the power of 2 and, unlike linear equations, there are a maximum of two roots. A root is one value of the variable where the equation is true, and to fully solve an equation, you must find all of the roots. For a quadratic equation you can factorise it and then easily find which values make the equation valid. The example above is quite a simple case. You will usually be given a more complicated equation such as ${\displaystyle 2x^{2}+5x+3=0}$. If the equation isn't already in the form ${\displaystyle ax^{2}+bx+c=0}$, rearrange it so that it is. The steps needed to factorise ${\displaystyle 2x^{2}+5x+3=0}$ are: Multiply ${\displaystyle 2}$ by ${\displaystyle 3}$ (coefficient of ${\displaystyle x^{2}}$ multiplied by the constant term) ${\displaystyle 2\times 3=6}$ Find two numbers that add to give ${\displaystyle 5}$ (coefficient of ${\displaystyle x}$) and multiply to give ${\displaystyle 6}$ (answer from previous step) ${\displaystyle 2\times 3=6}$, ${\displaystyle 2+3=5}$ Split ${\displaystyle 5x}$ to ${\displaystyle 2x+3x}$ (from the results of the previous step) ${\displaystyle 2x^{2}+2x+3x+3=0}$ Simplify ${\displaystyle 2x(x+1)+3(x+1)=0}$ Simplify further ${\displaystyle (2x+3)(x+1)=0}$ So ${\displaystyle (2x+3)(x+1)=0}$ is ${\displaystyle 2x^{2}+5x+3=0}$ in factorised form. You can now use the fact that any number multiplied by ${\displaystyle 0}$ is ${\displaystyle 0}$ to find the roots of the equation. The numbers that make one bracket equal to ${\displaystyle 0}$ are the roots of the equation. In the example, the roots are ${\displaystyle -1.5}$ and ${\displaystyle -1}$. It is also possible to solve a quadratic equation using the quadratic formula or by completing the square. ## Simultaneous equations Simultaneous equations are useful in solving two or more variables at once. Basic simultaneous equations consist of two linear expressions and can be solved by three different methods: elimination, substitution or by plotting the graph. ### Elimination method The basic principle of the elimination method is to manipulate one or more of the expressions in order to cancel out one of the variables, and then solve for the correct solution. An example of this: ${\displaystyle 2x+3y=10}$ (1) (Assigning the number (1) to this equation) ${\displaystyle 2x+6y=6}$ (2) (Assigning the number (2) to this equation) From this, we can see that by multiplying equation (1) by a factor of 2 and then subtracting this new equation from (2), the ${\displaystyle y}$-variable will be eliminated. (1) ${\displaystyle \times 2\rightarrow 4x+6y=20}$ (1a) (Assigning the number (1a) to this equation) Now subtracting (2) from (1a): ${\displaystyle 4x+6y=20}$ (1a) ${\displaystyle -}$ ${\displaystyle 2x+6y=6}$ (2) ${\displaystyle =}$ ${\displaystyle 2x+0y=14}$ Now that we have ${\displaystyle 2x=14}$, we can solve for ${\displaystyle x}$, which in this case is ${\displaystyle 7}$. ${\displaystyle x=7}$. Substitute the newly found ${\displaystyle x}$ into (1): ${\displaystyle 2\times 7+3y=10}$ ${\displaystyle 14+3y=10}$ And we find that ${\displaystyle y=-4/3}$ So, the solution to the two equations (1) and (2) are: ${\displaystyle x=7}$ ${\displaystyle y=-4/3}$ ### Substitution method The substitution method relies on being able to rearrange the expressions to isolate a single variable, in the form variable = expression. From this result this new expression can then be substituted for the variable itself, and the solutions evaluated. An example of this: ${\displaystyle 2x+3y=12}$ (1) (Assigning the number (1) to this equation) ${\displaystyle x+y=6}$ (2) (Assigning the number (2) to this equation) From this expression, it is possible to see that (2) is the most simplistic expression, and thus will be the better choice to rearrange. Taking (2), and rearranging this into ${\displaystyle x=6-y}$. (2a) Subbing (2a) into (1) we get ${\displaystyle 2(6-y)+3y=12}$ Solving this, we get that ${\displaystyle y=0}$ Again we can sub this result into one of the original equations to solve for ${\displaystyle x}$. In this case ${\displaystyle x=6}$. Note that for situations in which one of the equations is non-linear, you must isolate one variable in the linear equation and substitute it into the non-linear one. Then you can solve the quadratic equation with one of the methods above. Another form of substitution is if you've got a similar expression in both equations, like in this case: ${\displaystyle 2x+3y=10}$ (1) (Assigning the number (1) to this equation) ${\displaystyle 2x+6y=6}$ (2) (Assigning the number (2) to this equation) Here, ${\displaystyle 2x}$ is found in both equations, so: ${\displaystyle 2x=10-3y}$ (1) ${\displaystyle 2x=6-6y}$ (2) And since ${\displaystyle 2x=2x}$, you could do: ${\displaystyle 10-3y=6-6y}$ ${\displaystyle 6y-3y=6-10}$ ${\displaystyle 3y=-4}$ ${\displaystyle y=-4/3}$ Now you've got ${\displaystyle y}$, and finding ${\displaystyle x}$ will be the same as above. Graph by first solving for y. 2x+3y=12 ### Solving problems with simultaneous equations Often, you will be given problems which you must be able to write out as a pair of simultaneous equations. You will need to recognise such problems, and write them out correctly before solving them. Most problems will be similar to these examples with some differences. Example At a record store, 2 albums and 1 single costs £10. 1 album and 2 singles cost £8. Find the cost of an album and the cost of a single. Taking an album as ${\displaystyle a}$ and a single as ${\displaystyle s}$, the two equations would be: ${\displaystyle 2a+s=10}$ ${\displaystyle a+2s=8}$ This is part of the C1 (Core Mathematics 1) module of the A-level Mathematics text.
PUMPA - SMART LEARNING எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம் Book Free Demo Statement: If a line touches a circle and from the point of contact a chord is drawn, the angles between the tangent and the chord are respectively equal to the angles in the corresponding alternate segments. Explanation: In the figure, the chord $$PQ$$ divides the circle into two segments. Then, the tangent $$AB$$ is drawn such that it touches the circle at $$P$$. Thus, the angle in the alternate segment for $$\angle QPB$$ is $$\angle QSP$$ and that for $$\angle QPA$$ is $$\angle PTQ$$ are equal. Then, $$\angle QPB$$ $$=$$ $$\angle QSP$$ and $$\angle QPA$$ $$=$$ $$\angle PTQ$$. Proof for the theorem: Given: Let $$O$$ be the centre of the circle. The tangent $$AB$$ touches the circle at $$P$$, and $$PQ$$ is a chord. Let $$S$$ and $$T$$ be the two points on the circle on the opposite sides of chord $$PQ$$. To prove: (i) $$\angle QPB$$ $$=$$ $$\angle QSP$$ and (ii) $$\angle QPA$$ $$=$$ $$\angle PTQ$$ Construction: Draw the diameter $$POR$$ and draw $$QR$$, $$QS$$ and $$PS$$. Proof: The diameter $$RP$$ is perpendicular to the tangent $$AB$$. So, $$\angle RPB$$ $$=$$ $$90^{\circ}$$. $$\Rightarrow$$ $$\angle RPQ$$ $$+$$ $$\angle QPB$$ $$=$$ $$90^{\circ}$$ …… $$(1)$$ Consider the triangle $$RPQ$$. We know that: Angle in a semicircle is $$90^{\circ}$$. By the theorem, we have $$\angle RQP$$ $$=$$ $$90^{\circ}$$. …… $$(2)$$ Thus, $$RPQ$$ is a right-angled triangle. The sum of the two acute angles in the right-angled triangle, is $$90^{\circ}$$. $$\Rightarrow$$ $$\angle QRP$$ $$+$$ $$\angle RPQ$$ $$=$$ $$90^{\circ}$$ …… $$(3)$$ From equations $$(1)$$ and $$(3)$$, we have: $$\angle RPQ$$ $$+$$ $$\angle QPB$$ $$=$$ $$\angle QRP$$ $$+$$ $$\angle RPQ$$ $$\Rightarrow$$ $$\angle QPB$$ $$=$$ $$\angle QRP$$ …… $$(4)$$ We know that: Angles in the same segment are equal. By the theorem, we have $$\angle QRP$$ $$=$$ $$\angle PSQ$$. …… $$(5)$$ From equations $$(4)$$ and $$(5)$$, we have: $$\angle QPB$$ $$=$$ $$\angle PSQ$$. …… $$(6)$$ Therefore, statement (i) is proved. It is observed that $$\angle QPA$$ and $$\angle QPB$$ are linear angles. We know that: The linear pair of angle are always supplementary. $$\Rightarrow$$ $$\angle QPA$$ $$+$$ $$\angle QPB$$ $$=$$ $$180^{\circ}$$ …… $$(7)$$ Consider the cyclic quadrilateral $$PSQT$$. By the theorem, we have: The sum of opposite angles of a cyclic quadrilateral is $$180^{\circ}$$. $$\angle PSQ$$ $$+$$ $$\angle PTQ$$ $$=$$ $$180^{\circ}$$. …… $$(8)$$ From equations $$(7)$$ and $$(8)$$, we have: $$\angle QPA$$ $$+$$ $$\angle QPB$$ $$=$$ $$\angle PSQ$$ $$+$$ $$\angle PTQ$$ Substitute equation $$(6)$$ in the above equation. $$\Rightarrow$$ $$\angle QPA$$ $$+$$ $$\angle QPB$$ $$=$$ $$\angle QPB$$ $$+$$ $$\angle PTQ$$ $$\Rightarrow$$ $$\angle QPA$$ $$=$$ $$\angle PTQ$$ Therefore, statement (ii) is proved. Example: In the figure, $$AB$$ is the tangent, and $$CF$$ is the chord of the given circle. Next, find the value of $$x$$. Solution: Given, $$\angle BCF$$ $$=$$ $$57^{\circ}$$. By the Alternate Segment Theorem, $$\angle BCF$$ $$=$$ $$\angle CDE$$. Thus, $$x$$ $$=$$ $$57^{\circ}$$. Therefore, the value of $$x$$ $$=$$ $$57^{\circ}$$. Important! This theorem is also known as Tangent - chord theorem.
# Calculating Instantaneous Rates of Change When calculating instantaneous rates of change students need to visualise the properties of the gradient for a straight line graph. I use the starter activity to see if they can match four graphs with their corresponding equations. The only clue is the direction and steepness of the red lines in relation to the blue line y = x. All the graphs have an intercept of zero, to allow students to focus on the gradient. At the end of this activity I feedback to make sure students understand that lines which go downwards have a negative gradient and those that are steeper than y = x has a gradient greater than one. ##### Calculating Instantaneous Rates of Change To introduce how to calculate an instantaneous rate of change on a curve we discuss how the steepness of the graph changes depending on the x value. I like to use the Geogebra applet below to demonstrate how the gradient of the tangent changes along the curve. The teacher can change the function depending on the point they are trying to make. I provide the students with a print out of the next few slides for them to write on as we progress through the examples. This saves time and helps students develop a clear written method. I encourage the students to use integer coordinate pairs for calculating the change in horizontal and vertical whenever possible. This video demonstrates the written method I teach the students. ##### Practical Reasons for Calculating Instantaneous Rates of Change As we progress through the lesson, I emphasise the practical reasons for calculating instantaneous rates of change. First, we calculate the rate of leakage from a water tank using a distance-time graph. Second, we use a velocity-time graph to estimate the acceleration of a ball as it travels through the air. In this example we discuss gravity as a force slowing the acceleration towards the turning-point then increasing as it begins to fall. ##### Assessing Progress and Feeding Back I use the plenary to check student’s progress and understanding in two key areas. 1. How well can students find the gradient of a tangent to estimate an instantaneous rate of change? 2. Can students interpret the practical meaning of the gradient in the context of the two variables? Students attempt this on mini-whiteboards and present their working to me for assessment. Estimating an instantaneous rate of change typically takes two, one hour lessons, During this time students work through the worksheet and several examination questions. It is important for students to become confident performing the calculations and understanding them within the context of the problems. Mr Mathematics members can access this lesson online and download the worksheet by clicking here. ## Revising Area Under a Curve Students revise how to estimate the distance travelled in a... ## Revising Function Notation and Composite Functions Students revise how to substitute known values and expressions into... ## Estimating the Area Under a Curve Students learn about estimating the area under a curve using... ## Instantaneous Rates of Change Students learn how to calculate and interpret instantaneous rates of... This site uses Akismet to reduce spam. Learn how your comment data is processed. ### Mr Mathematics Blog #### Angles in Polygons There are two key learning points when solving problems with angles in polygons. The first is to understand why all the exterior angles of a polygon have a sum of 360°. The second is to understand the interior and exterior angles appear on the same straight line. Students can be told these two facts and […] #### Getting Ready for a New School Year When getting ready for a new school year I have a list of priorities to work through. Knowing my team have all the information and resources they need to teach their students gives me confidence we will start the term in the best possible way. Mathematics Teaching and Learning Folder All teachers receive a folder […] #### Mathematics OFSTED Inspection – The Deep Dive Earlier this week, my school took part in a trial OFSTED inspection as part of getting ready for the new inspection framework in September 2019. This involved three Lead Inspectors visiting our school over the course of two days. The first day involved a ‘deep dive’ by each of the Lead Inspectors into Mathematics, English […]
# Surface Area Of Cuboid & Cube: Formula, Derivation, Volume & Examples In this blog, we will introduce the surface area of cuboid and cubes. We will begin with the surface area of the cuboid. Let us take a cuboid. We can see in each face of the cuboid there is a rectangle. So let us find out how many rectangles does a cuboid has? 1, 2, 3, 4, 5, and 6. So there are 6 rectangles as there are 6 faces of a cuboid. Now open this cuboid and mark its dimensions. On opening, we have this. So we have used six rectangular pieces to cover the complete outer surface of the cuboid. This shows us that the outer surface of a cuboid is made up of six rectangles (in fact, rectangular regions, called the faces of the cuboid), whose areas can be found by multiplying the length by breadth for each of them separately and then adding the six areas together. So, the sum of the areas of the six rectangles is: Area of rectangle 1= (l × h) + Area of rectangle 2= (l × b) + Area of rectangle 3= (l × h) + Area of rectangle 4= (l × b) + Area of rectangle 5= (b × h) + Area of rectangle 6 = (b × h) That is = 2(l × b) + 2(b × h) + 2(l × h) taking 2 common we have = 2(lb + bh + hl) This gives us: Surface Area of a Cuboid = 2(lb + bh + hl) What about a cube? Let us see in the case of the cube. For this also open up a cube. A cuboid, whose length, breadth, and height are all equal, is called a cube. If each edge of the cube is a, then the surface area of this cube would be 2(a × a + a × a + a × a) i.e., 6a2 The surface area of a cuboid (or a cube) is sometimes also referred to as the total surface area. Solved Examples 1. Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden box covered with colored paper with a picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth, and height as 80 cm, 40 cm, and 20 cm respectively how many square sheets of paper on the side 40 cm would she require? Solution: Since Mary wants to paste the paper on the outer surface of the box; the quantity of paper required would be equal to the surface area of the box which is of the shape of a cuboid. The dimensions of the box are: Length =80 cm, Breadth = 40 cm, Height = 20 cm. so we need to go for surface area of box so the surface area of the box = 2(lb + bh + hl) = 2[(80 × 40) + (40 × 20) + (20 × 80)] cm2 = 2[3200 + 800 + 1600] cm2 = 2 × 5600 cm2 = 11200 cm2 Now let us go for the area of the sheet. The area of each sheet of the paper = 40 × 40 cm2= 1600 cm2 Therefore, the number of sheets required = surface area of box divided by area of the sheet. That is 11200 upon 1600 gives 7. Number of sheets required = 7 sheets 2. Hameed has built a cubical water tank with a lid for his house, with each outer edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of the side 25 cm. Find how much he would spend for the tiles if the cost of the tiles is Rs 360 per dozen. Solution: Since Hameed is getting the five outer faces of the tank covered with tiles, he would need to know the surface area of the tank, to decide on the number of tiles required. Edge of the cubical tank = 1.5 m = 150 cm (= a) So, surface area of the tank = 5 × 150 × 150 cm2 Area of each square tile = side × side = 25 × 25 cm2 So, the number of tiles required = surface area of tank upon area of tile. That is 5 x 150 x 150 upon 25 x 25. Cost of one dozen tiles i.e. cost of 12 tiles = Rs 360 Cost of one tile is = 360/12 =Rs30 So the cost of 180 tiles is = 180 x 30 = Rs 5400 Curved Surface Area Of Cuboid Formula: The curved surface area of a cuboid is a total of four rectangle planes, except the top (upper) and bottom (lower) surfaces. Mathematically, the lateral or curved surface of cuboid formula is given as: Curved surface area of cuboid formula = 2 (lh + wh) = 2 h (l + w) square units. Volume Of Cuboid and Cube Let’s discuss the volume of solids. If an object is solid, then the space occupied by such an object is measured and is termed the Volume of the object. On the other hand, if the object is hollow, then the interior is empty and can be filled with air, or some liquid that will take the shape of its container. In this case, the volume of the substance that can fill the interior is called the capacity of the container. In short, the volume of an object is the measure of the space it occupies, and the capacity of an object is the volume of substance its interior can accommodate. Hence, the unit of measurement of either of the two is cubic unit. First, we will consider the volume of the cuboid. Let us consider some rectangles stacked on top of each other. Suppose we say that the area of each rectangle is A, the height up to which the rectangles are stacked is h and the volume of the cuboid is V. Can you tell what would be the relationship between V, A, and h? The area of the plane region occupied by each rectangle × height = Measure of the space occupied by the cuboid So, we get A × h = V That is, Volume of a Cuboid = base area × height = length × breadth × height or l × b × h, where l, b and h are respectively the length, breadth and height of the cuboid. Similarly in cube volume is area into height. And we also know that all three dimensions of cube are equal so we have Volume of a Cube = edge × edge × edge = a cube where a is the edge of the cube. Solved Examples 1. A wall of length 10 m was to be built across open ground. The height of the wall is 4 m and the thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm × 12 cm × 8 cm, how many bricks would be required? Solution: Since the wall with all its bricks makes up the space occupied by it, we need to find the volume of the wall, which is nothing but a cuboid. Here, Length = 10 m = 1000 cm Thickness = 24 cm Height = 4 m = 400 cm Therefore, Volume of the wall = length × thickness × height = 1000 × 24 × 400 cm3 Now, each brick is a cuboid with length = 24 cm, breadth = 12 cm and height = 8 cm So, volume of each brick = length × breadth × height = 24 × 12 × 8 cm3 2. A child playing with building blocks, which are of the shape of cubes, has built a structure. If the edge of each cube is 3 cm, find the volume of the structure built by the child. Solution: Volume of each cube = edge × edge × edge = 3 × 3 × 3 cm cube = 27 cm3 Number of cubes in the structure = 15 Therefore, volume of the structure = 27 × 15 cm3 = 405 cm3
# Question Video: Finding the Equation of the Normal to a Curve Defined Implicitly at a Given Point Involving Using the Product Rule Mathematics • Higher Education Find the equation of the normal to the curve π₯Β²π¦Β² β 4π₯ + 2π¦ β 20 = 0 at the point (1, 4). 05:12 ### Video Transcript Find the equation of the normal to the curve π₯ squared times π¦ squared minus four π₯ plus two π¦ minus 20 is equal to zero at the point one, four. The question wants us to find the equation of a normal to a curve defined implicitly at the point one, four. To find the equation of a normal, we recall if the tangent to the curve at the point π has a slope of π. Then the normal to the curve at the point π has a slope negative one divided by π. This is because the tangent and the normal lines are perpendicular to each other. In fact, this tells us if the tangent line is horizontal, then the normal line will be vertical, and vice versa. So we can use the slope of the tangent to find the slope of our normal. To find the slope of our tangent at the point one, four, weβre going to use differentiation. We differentiate both sides of our equation with respect to π₯ to help us find an expression for the slope dπ¦ by dπ₯. This gives us the derivative of π₯ squared π¦ squared minus four π₯ plus two π¦ minus 20 with respect to π₯ is equal to the derivative of zero with respect to π₯. We can differentiate each term separately. We know the derivative of zero with respect to π₯ is just equal to zero. The derivative of negative 20 with respect to π₯ is also equal to zero. And the derivative of negative four π₯ with respect to π₯ is equal to negative four. To differentiate the remaining two terms, we recall that π¦ is a function of π₯. So we can differentiate these by using the chain rule. The chain rule tells us since π¦ is a function of π₯, the derivative of π of π¦ with respect to π₯ is equal to the derivative of π with respect to π¦ multiplied by the derivative of π¦ with respect to π₯. We can use this to find the derivative of two π¦ with respect to π₯. Itβs equal to the derivative of two π¦ with respect to π¦ multiplied by dπ¦ by dπ₯. And we know the derivative of two π¦ with respect to π¦ is just equal to two. We see the final term we need to differentiate is the product of two functions. So weβll need to use the product rule. The product rule tells us the derivative of the product of two functions π’ and π£ with respect to π₯ is equal to π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. So we can use the product rule and the chain rule together to differentiate π₯ squared π¦ squared with respect to π₯. Applying the product rule gives us π₯ squared times the derivative of π¦ squared with respect to π₯ plus π¦ squared times the derivative of π₯ squared with respect to π₯. We can differentiate π¦ squared with respect to π₯ by using the chain rule. Itβs the derivative of π¦ squared with respect to π¦. Thatβs two π¦ multiplied by dπ¦ by dπ₯. And we can evaluate the derivative of π₯ squared with respect to π₯ as two π₯. So we have two π₯ squared π¦ dπ¦ by dπ₯ plus two π₯ π¦ squared minus four plus two dπ¦ by dπ₯ plus zero is equal to zero. And we want to find an expression for the slope of our tangents. So we need to rewrite this expression to have dπ¦ by dπ₯ as the subject. First, weβll subtract two π₯π¦ squared and add four to both sides of the equation. Then weβll take out the shared factor of dπ¦ by dπ₯ from the remaining two terms. This gives us two π₯ squared π¦ plus two multiplied by dπ¦ by dπ₯ is equal to four minus two π₯π¦ squared. Finally, we can divide both sides of our equation by two π₯ squared π¦ plus two to get dπ¦ by dπ₯ is equal to four minus two π₯π¦ squared all divided by two π₯ squared π¦ plus two. We want to find the equation of the normal to the curve at the point one, four. Thatβs when π₯ is equal to one and π¦ is equal to four. So we can substitute π₯ is equal to one and π¦ is equal to four into our expression for dπ¦ by dπ₯ to find the slope of the tangent to the curve at the point one, four. Itβs equal to four minus two times one times four squared all divided by two times one squared times four plus two. We can evaluate this to get an answer of negative 14 divided by five. However, this is the slope of the tangent to the curve at this point. We want the slope of the normal to the curve at this point. We can do this by finding negative one times the reciprocal of this value. The reciprocal of negative 14 divided by five is negative five divided by 14. We then multiply this by negative one to just get five divided by 14. Since weβve found the slope of our normal line to be five divided by 14, we can write it in the standard form for a line. π¦ is equal to five divided by 14π₯ plus π. And from the question, we know that our normal line must pass through the point one, four. Since our line passes through the point one, four, we can substitute π₯ is equal to one and π¦ is equal to four to find the value of π. We subtract five divided by 14 from both sides. And we see that π is equal to 51 divided by 14. Therefore, by using the fact π is equal to 51 over 14 and then rearranging our equation. We have the equation for the normal to the curve π₯ squared π¦ squared minus four π₯ plus two π¦ minus 20 is equal to zero at the point one, four is π¦ minus five π₯ over 14 minus 51 over 14 is equal to zero.
# The positive Predictive Value ## Overview Teaching: 20 min Exercises: 40 min Questions • What is the positive Predictive Value (PPV) ? Objectives • After this lesson, you should understand what is the PPV ## Definitions: remiders • $$H_0$$ : null hypothesis: The hypothesis that the effect we are testing for is null • $$H_A$$ : alternative hypothesis : Not $$H_0$$, so there is some signal • $$T$$ : The random variable that takes value “significant” or “not significant” • $$T_S$$ : Value of T when test is significant (eg $$T = T_S$$) • $$T_N$$ : Value of T when test is not significant (eg $$T = T_N$$) • $$\alpha$$ : false positive rate - probability to reject $$H_0$$ when $$H_0$$ is true ($$H_A$$ is false) $$= P(T_S \mid H_0)$$ • $$\beta$$ : false negative rate - probability to accept $$H_0$$ when $$H_A$$ is true ($$H_0$$ is false) $$= P(T_N \mid H_A)$$ • power = $$W = 1-\beta$$ ## PPV : definition PPV = $$P(H_A \mid T_S)$$ ## PPV : How do I compute it ? where does it come from ? ### Let’s do some basic probability We consider that the hypotheses H_0 and H_A are random, i.e. they have associated probabilities. For instance, the probability of $$H_0$$ to be true could be 20%. We have either $$H_A$$ is true, or else $$H_0$$ is true. Therefore: $$P(H_A = True) + P(H_0 = True) = 1$$. We simply note $$P(H_A = True)$$ as $$P(H_A)$$ and $$P(T = T_S)$$ as $$P(T_S)$$ We are interested in the probability of a significant test, we can write $P(T_S) = P(T_S, H_A) + P(T_S, H_0)$ because $T_S$ occurs either under the null or the alternative (mutually exclusive) and their represent all the possibilities. ## Bayes theorem The famous Bayes theorem states: $P(A, B) = P(A \mid B) P(B)$ and therefore $P(A \mid B) = \frac{P(A, B)}{P(B)} = \frac{P(B \mid A) P(A)}{P(B)}$ ## Apply this to our question: Now, apply this to the probability of the test results $$T$$. The probability of a significant result of the test $$T=T_S$$ is : $P(T_S) = P(T_S, H_A) + P(T_S, H_0)$ $P(T_S) = P(T_S \mid H_A) P(H_A) + P(T_S \mid H_0) P(H_0)$ What is the probability of $$H_A$$ given that the test is significant (eg, PPV) ? $P(H_A \mid T_S) = \frac{P(T_S \mid H_A) P(H_A)}{P(T_S)} = \frac{P(T_S \mid H_A) P(H_A)}{P(T_S \mid H_A) Pr(H_A) + Pr(T_S \mid H_0) Pr(H_0)}$ And we know that $P(T_S \mid H_A) = 1 - P(T_N \mid H_A) = 1 - \beta = W$ Substituting: $P(H_A \mid T_S) = \frac{W P(H_A)}{W P(H_A) + \alpha P(H_0)}$ Defining: $$R = \frac{P(H_A)}{P(H_0)}$$ the odd ratio of the alternative over the null $P(H_A \mid T_S) = \frac{W R}{W R + \alpha}$ ## Task: Play with the PPV - understand the impact of the parameters Pick a recent study that you have done in fMRI or using anatomical data. try to propose values for power, alpha, and prior Vary power from .1 to .9 and print or plot results First define a function to compute the PPV from power, odd ratio and alpha The solution below is in Python, but feel free to do it in your favorite scripting language ## Solution def PPV_OR(odd_ratio, power, alpha, verbose=True): """ returns PPV from odd_ratio, power and alpha parameters: ----------- odd_ratio: float P(H_A)/(1-P(H_A)) power: float Power for this study alpha: float type I risk of error Returns: ---------- float The positive predicted value """ ppv = (powerodd_ratio)/(powerodd_ratio + alpha) if verbose: print("With odd ratio=%3.2f, " "Power=%3.2f, alpha=%3.2f, " "We have PPV=%3.2f" %(odd_ratio,power,alpha,ppv)) return ppv Second define a function to display easily the results ## Solution def plot_ppv(xvalues, yvalues, xlabel, ylabel, title): ''' simply plot yvalues against xvalues, with labels and title Parameters: ----------- xvalues, yvalues : iterables of numbers labels and title : string ''' fig = plt.figure(); axis.plot(xvalues, yvalues, color='red', marker='o', linestyle='dashed', linewidth=2, markersize=14); axis.set_xlabel(xlabel,fontsize=20); axis.set_ylabel(ylabel,fontsize=20); axis.set_title(figure_title, fontsize=20); return fig, axis Last play with parameters : first, let's vary power ## Solution #----------------------------------------------------------------- # An example R = 1./5. Pw = .5 alph = .05 ppv = PPV_OR(R, Pw, alph) #----------------------------------------------------------------- # Vary power: Pw = np.arange(.1,1,.2) ppvs = [PPV_OR(R, pw, alph, verbose = False) for pw in Pw] xlabel = 'Power' ylabel = 'PPV' figure_title = 'With an odd ratio of {odd_ratio}'.format(odd_ratio=R) #----------------------------------------------------------------- # print plot_ppv(Pw, ppvs, xlabel, ylabel, figure_title); Then, let's vary odd ratio ## Solution #----------------------------------------------------------------- # Vary odd ratio: Pw = .5 alph = .05 odd_ratios = np.arange(.05,.5,.05) ppvs = [PPV_OR(R, Pw, alph, verbose = False) for R in odd_ratios] #----------------------------------------------------------------- # print figure_title = 'With a power of {power}'.format(power=Pw) xlabel = 'odd_ratios' ylabel = 'PPV' plot_ppv(odd_ratios, ppvs, xlabel, ylabel, figure_title); Last, let's vary alpha, remember that p-hacking may give us large type I risk of errors ## Solution #----------------------------------------------------------------- # Vary alpha: Pw = .5 R = 1/5 alphas = np.arange(0, .2, 0.01)# [0.001, .005, 0.01, 0.05, 0.1] #, 0.2, 0.3, 0.4, 0.5] ppvs = [PPV_OR(R, Pw, alph, verbose = False) for alph in alphas] #----------------------------------------------------------------- # print xlabel = 'alpha' ylabel = 'PPV' figure_title = 'With a power of {power} and odd ratio of {odd_ratio}'.format( power=Pw, odd_ratio=R) plot_ppv(alphas, ppvs, xlabel, ylabel, figure_title); #----------------------------------------------------------------- ## This is also in a python notebook The above discussed points can also be explored within this jupyter notebook ### How to work with the notebooks ? There are two cases. 1. You do not really know python, and how to install the jupyter notebook. You can still read the notebook, skipping the code sections. The notebook will introduce some definitions, and then play with different settings. But the true benefit comes if you can install the jupyter project see here and actually play with the code, for instance changing the sample size or effect size to understand better what power is. 2. Download the notebook, and try to understand the concepts and the code. If the code is unclear, please make an issue on the repronim github site. 3. Run it interactively on binder. ## Key Points • A significant (say at the 0.05 level) may have a low chance of replication. • The PPV estimates the probability that the alternative hypothesis $$H_A$$ is true given that the test is significant at some $$\alpha$$ level. • This probability depends on several factors such as power $$\beta$$, $$\alpha$$ level, but also the prior chance that $$H_A$$ is true.
Math Unit 4 homework # Few factors that charlie should factor into his This preview shows pages 1–3. Sign up to view the full content. few factors that Charlie should factor into his buying decision are shipping costs and set up costs. Will he have to pay someone to train his employees and himself how to use it? Will it speed up production and increase his income allowing him to pay it off even quicker. How expensive are the replacement parts? How much does a routine upkeep visit cost? 1. The local bank will loan Charlie \$25,000 for 1 year at an interest rate of 12% with only one payment due at the end of the year. If Charlie borrows the full \$25,000 for the new engraver, what will the total cost of the loan be? The total cost of the loan would be \$25,300. (\$25,000 x.012 = \$300). This preview has intentionally blurred sections. Sign up to view the full version. View Full Document 2. Calculate the total amount of revenue (gross profit) that will be lost if the engraver breaks. The total amount of revenue lost if the engraver breaks for 18 days is \$17,750. (\$97518 days = \$17, 550) 3. If the engraving business makes \$975 per day in revenue and generates a net profit of 25%, how much profit is generated per day? The amount of profit generated per day is \$243.75. (\$97525% = \$243.75 per day profit). 4. What is the total net profit lost if the engraver is out of commission for the full 18 days? The amount of profit if the engraver breaks is \$4,387.50. (\$17,550*25% = \$4,387.50). 5. If the engraver is kept busy 269 full days per year, how much revenue (gross profit) will be generated? If the engraver is kept running for 269 days @ \$975 in revenue per day the total amount would be \$262,275.00. 6. This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]} ### Page1 / 3 few factors that Charlie should factor into his buying... This preview shows document pages 1 - 3. Sign up to view the full document. View Full Document Ask a homework question - tutors are online
# How do you differentiate f(x)=(1+arctanx)/(2-3arctanx)? Mar 24, 2017 $\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{5}{\left(1 + {x}^{2}\right) {\left(2 - 3 \arctan x\right)}^{2}}$ #### Explanation: Here we can use quotient rule. Also note that derivative of $\arctan x$, which is also known as ${\tan}^{- 1} x$ is $\frac{1}{1 + {x}^{2}}$ i.e. $\frac{d}{\mathrm{dx}} \arctan x = \frac{1}{1 + {x}^{2}}$ Now according to quotient rule if $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$ then $\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{h \left(x\right) \cdot \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} - g \left(x\right) \cdot \frac{\mathrm{dh} \left(x\right)}{\mathrm{dx}}}{{\left(h \left(x\right)\right)}^{2}}$ Here $g \left(x\right) = 1 + \arctan x$ hence $\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = \frac{1}{1 + {x}^{2}}$ and $h \left(x\right) = 2 - 3 \arctan x$ hence $\frac{\mathrm{dh} \left(x\right)}{\mathrm{dx}} = - \frac{3}{1 + {x}^{2}}$ Hence $\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{\left(2 - 3 \arctan x\right) \cdot \frac{1}{1 + {x}^{2}} - \left(1 + \arctan x\right) \left(- \frac{3}{1 + {x}^{2}}\right)}{{\left(2 - 3 \arctan x\right)}^{2}}$ = $\frac{\left(\frac{2}{1 + {x}^{2}} - \frac{3 \arctan x}{1 + {x}^{2}} + \frac{3}{1 + {x}^{2}} + \frac{3 \arctan x}{1 + {x}^{2}}\right)}{{\left(2 - 3 \arctan x\right)}^{2}}$ = $\frac{\frac{5}{1 + {x}^{2}}}{{\left(2 - 3 \arctan x\right)}^{2}}$ = $\frac{5}{\left(1 + {x}^{2}\right) {\left(2 - 3 \arctan x\right)}^{2}}$
# How do find the derivative of y = x^2 sinx? Aug 5, 2015 $\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \sin \left(x\right) + {x}^{2} \cos \left(x\right)$ #### Explanation: You can use the product rule to find the derivative. The product rule says the following: If $h \left(x\right) = f \left(x\right) g \left(x\right)$, then $h ' \left(x\right) = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$ In our case, $h \left(x\right) = {x}^{2} \sin x$ $f \left(x\right) = {x}^{2}$ $g \left(x\right) = \sin x$ $f ' \left(x\right) = 2 x$ $g ' \left(x\right) = \cos x$ Plug in those values into our definition for the product rule to get $\frac{\mathrm{dy}}{\mathrm{dx}} = h ' \left(x\right) = 2 x \sin \left(x\right) + {x}^{2} \cos \left(x\right)$
# Section 1.1 A Preview of Calculus 42 ```42 CHAPTER 1 Limits and Their Properties Section 1.1 As you progress through this course, remember that learning calculus is just one of your goals. Your most important goal is to learn how to use calculus to model and solve real-life • Be sure you understand the question. What is given? What are you asked to find? • Outline a plan. There are many approaches you could use: look for a pattern, solve a simpler problem, work backwards, draw a diagram, use technology, or any of many other approaches. • Complete your plan. Be sure to writing the answer as x ! 4.6, it would be better to write the answer as “The area of the region is 4.6 square meters.” • Look back at your work. Does your answer make sense? Is there a way you can check the reasonableness of STUDY TIP A Preview of Calculus • Understand what calculus is and how it compares with precalculus. • Understand that the tangent line problem is basic to calculus. • Understand that the area problem is also basic to calculus. What Is Calculus? Calculus is the mathematics of change—velocities and accelerations. Calculus is also the mathematics of tangent lines, slopes, areas, volumes, arc lengths, centroids, curvatures, and a variety of other concepts that have enabled scientists, engineers, and economists to model real-life situations. Although precalculus mathematics also deals with velocities, accelerations, tangent lines, slopes, and so on, there is a fundamental difference between precalculus mathematics and calculus. Precalculus mathematics is more static, whereas calculus is more dynamic. Here are some examples. • An object traveling at a constant velocity can be analyzed with precalculus mathematics. To analyze the velocity of an accelerating object, you need calculus. • The slope of a line can be analyzed with precalculus mathematics. To analyze the slope of a curve, you need calculus. • A tangent line to a circle can be analyzed with precalculus mathematics. To analyze a tangent line to a general graph, you need calculus. • The area of a rectangle can be analyzed with precalculus mathematics. To analyze the area under a general curve, you need calculus. Each of these situations involves the same general strategy—the reformulation of precalculus mathematics through the use of a limit process. So, one way to answer the question “What is calculus?” is to say that calculus is a “limit machine” that involves three stages. The first stage is precalculus mathematics, such as the slope of a line or the area of a rectangle. The second stage is the limit process, and the third stage is a new calculus formulation, such as a derivative or integral. Precalculus mathematics GRACE CHISHOLM YOUNG (1868–1944) Grace Chisholm Young received her degree in mathematics from Girton College in Cambridge, England. Her early work was published under the name of William Young, her husband. Between 1914 and 1916, Grace Young published work on the foundations of calculus that won her the Gamble Prize from Girton College. Limit process Calculus Some students try to learn calculus as if it were simply a collection of new formulas. This is unfortunate. If you reduce calculus to the memorization of differentiation and integration formulas, you will miss a great deal of understanding, self-confidence, and satisfaction. On the following two pages some familiar precalculus concepts coupled with their calculus counterparts are listed. Throughout the text, your goal should be to learn how precalculus formulas and techniques are used as building blocks to produce the more general calculus formulas and techniques. Don’t worry if you are unfamiliar with some of the “old formulas” listed on the following two pages—you will be reviewing all of them. As you proceed through this text, come back to this discussion repeatedly. Try to keep track of where you are relative to the three stages involved in the study of calculus. For example, the first three chapters break down as shown. Chapter P: Preparation for Calculus Chapter 1: Limits and Their Properties Chapter 2: Differentiation Precalculus Limit process Calculus SECTION 1.1 Without Calculus 43 With Differential Calculus y y y = f (x) Value of f !x" when x ! c x c ∆y Slope of a line y = f (x) Limit of f !x" as x approaches c Slope of a curve dy dx Secant line to a curve Tangent line to a curve Average rate of change between t ! a and t ! b Instantaneous rate of change at t ! c t=a x c ∆x t=b Curvature of a circle t=c Curvature of a curve y y Height of a curve when x!c A Preview of Calculus c x Maximum height of a curve on an interval Tangent plane to a sphere Tangent plane to a surface Direction of motion along a line Direction of motion along a curve a b x 44 CHAPTER 1 Limits and Their Properties Without Calculus With Integral Calculus y Area of a rectangle Area under a curve Work done by a constant force Work done by a variable force x y Center of a rectangle Centroid of a region x Length of a line segment Length of an arc Surface area of a cylinder Surface area of a solid of revolution Mass of a solid of constant density Mass of a solid of variable density Volume of a rectangular solid Volume of a region under a surface Sum of a finite number of terms a1 " a2 " . . . " an ! S Sum of an infinite number of terms a1 " a2 " a3 " . . . ! S SECTION 1.1 45 A Preview of Calculus The Tangent Line Problem y y = f(x) Tangent line P x . The tangent line to the graph of f at P Figure 1.1 Video The notion of a limit is fundamental to the study of calculus. The following brief descriptions of two classic problems in calculus—the tangent line problem and the area problem—should give you some idea of the way limits are used in calculus. In the tangent line problem, you are given a function f and a point P on its graph and are asked to find an equation of the tangent line to the graph at point P, as shown in Figure 1.1. Except for cases involving a vertical tangent line, the problem of finding the tangent line at a point P is equivalent to finding the slope of the tangent line at P. You can approximate this slope by using a line through the point of tangency and a second point on the curve, as shown in Figure 1.2(a). Such a line is called a secant line. If P!c, f !c"" is the point of tangency and Q!c " #x, f !c " #x"" is a second point on the graph of f, the slope of the secant line through these two points is given by msec ! f !c " #x" \$ f !c" f !c " #x" \$ f !c" ! . c " #x \$ c #x y y Q Q(c + ∆ x, f(c + ∆ x)) Secant lines P(c, f(c)) f (c + ∆ x) − f(c) P Tangent line ∆x x . x (a) The secant line through !c, f !c"" and !c " #x, f !c " #x"" (b) As Q approaches P, the secant lines approach the tangent line. Figure 1.2 Animation As point Q approaches point P, the slope of the secant line approaches the slope of the tangent line, as shown in Figure 1.2(b). When such a “limiting position” exists, the slope of the tangent line is said to be the limit of the slope of the secant line. E X P L O R AT I O N The following points lie on the graph of f !x" ! x2. Q1!1.5, f !1.5"", Q2!1.1, f !1.1"", Q3!1.01, f !1.01"", Q4!1.001, f !1.001"", Q5!1.0001, f !1.0001"" Each successive point gets closer to the point P!1, 1". Find the slope of the secant line through Q1 and P, Q2 and P, and so on. Graph these secant lines on a graphing utility. Then use your results to estimate the slope of the tangent line to the graph of f at the point P. 46 CHAPTER 1 Limits and Their Properties The Area Problem y y = f (x) a b In the tangent line problem, you saw how the limit process can be applied to the slope of a line to find the slope of a general curve. A second classic problem in calculus is finding the area of a plane region that is bounded by the graphs of functions. This problem can also be solved with a limit process. In this case, the limit process is applied to the area of a rectangle to find the area of a general region. As a simple example, consider the region bounded by the graph of the function y ! f !x", the x-axis, and the vertical lines x ! a and x ! b, as shown in Figure 1.3. You can approximate the area of the region with several rectangular regions, as shown in Figure 1.4. As you increase the number of rectangles, the approximation tends to become better and better because the amount of area missed by the rectangles decreases. Your goal is to determine the limit of the sum of the areas of the rectangles as the number of rectangles increases without bound. x Area . under a curve Figure 1.3 y y Video y = f(x) y = f(x) HISTORICAL NOTE In one of the most astounding events ever to occur in mathematics, it was discovered that the tangent line problem and the area problem are closely related. This discovery led to the birth of calculus. You will learn about the relationship between these two problems when you study the Fundamental Theorem of Calculus in Chapter 4. a b x a Approximation using four rectangles . x b Approximation using eight rectangles Figure 1.4 Animation E X P L O R AT I O N Consider the region bounded by the graphs of f !x" ! x2, y ! 0, and x ! 1, as shown in part (a) of the figure. The area of the region can be approximated by two sets of rectangles—one set inscribed within the region and the other set circumscribed over the region, as shown in parts (b) and (c). Find the sum of the areas of each set of rectangles. Then use your results to approximate the area of the region. y y y f (x) = x2 f (x) = 1 1 1 x x 1 (a) Bounded region f (x) = x 2 x2 x 1 (b) Inscribed rectangles 1 (c) Circumscribed rectangles SECTION 1.1 47 A Preview of Calculus Exercises for Section 1.1 The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. Click on to view the complete solution of the exercise. Click on to print an enlarged copy of the graph. In Exercises 1– 6, decide whether the problem can be solved using precalculus, or whether calculus is required. If the problem can be solved using precalculus, solve it. If the problem seems to require calculus, explain your reasoning and use a graphical or numerical approach to estimate the solution. (c) Use the results of part (b) to estimate the slope of the tangent line of f at P "4, 2#. Describe how to improve your approximation of the slope. 9. (a) Use the rectangles in each graph to approximate the area of the region bounded by y ! 5!x, y ! 0, x ! 1, and x ! 5. y 1. Find the distance traveled in 15 seconds by an object traveling at a constant velocity of 20 feet per second. y 5 5 2. Find the distance traveled in 15 seconds by an object moving with a velocity of v"t# ! 20 \$ 7 cos t feet per second. 4 4 3 3 3. A bicyclist is riding on a path modeled by the function f "x# ! 0.04"8x # x2#, where x and f "x# are measured in miles. Find the rate of change of elevation when x ! 2. 2 2 y y 3 f(x) = 0.04 (8x x 2 ) 2 1 x 1 2 3 4 5 6 Figure for 3 1 x 1 1 2 3 4 5 1 3 2 4 5 x y 1 1 Figure for 4 U 2 5. Find the area of the shaded region. y x U U 2 U x (b) Describe how you could continue this process to obtain a more accurate approximation of the area. y 5 1 5 4 y 6 4. A bicyclist is riding on a path modeled by the function f "x# ! 0.08x, where x and f "x# are measured in miles. Find the rate of change of elevation when x ! 2. 4 3 2 1 3 2 10. (a) Use the rectangles in each graph to approximate the area of the region bounded by y ! sin x, y ! 0, x ! 0, and x ! ". f(x) = 0.08x 1 1 1 x (b) Describe how you could continue this process to obtain a more accurate approximation of the area. 3 2 1 3 (2, 3) (5, 0) (0, 0) 3 4 Figure for 5 1 x 5 6 2 1 x 11. Consider the length of the graph of f "x# ! 5!x from "1, 5# to "5, 1#. 1 Figure for 6 6. Find the area of the shaded region. 7. Secant Lines Consider the function f "x# ! 4x # x2 and the point P "1, 3# on the graph of f. (a) Graph f and the secant lines passing through P "1, 3# and Q "x, f "x## for x-values of 2, 1.5, and 0.5. (b) Find the slope of each secant line. (c) Use the results of part (b) to estimate the slope of the tangent line of f at P "1, 3#. Describe how to improve your approximation of the slope. y y 5 (1, 5) 5 4 4 3 3 2 (5, 1) 1 x 1 2 3 4 5 (1, 5) 2 (5, 1) 1 x 1 2 3 4 5 (a) Approximate the length of the curve by finding the distance between its two endpoints, as shown in the first figure. 8. Secant Lines Consider the function f "x# ! \$x and the point P "4, 2# on the graph of f. (b) Approximate the length of the curve by finding the sum of the lengths of four line segments, as shown in the second figure. (a) Graph f and the secant lines passing through P "4, 2# and Q "x, f "x## for x-values of 1, 3, and 5. (c) Describe how you could continue this process to obtain a more accurate approximation of the length of the curve. (b) Find the slope of each secant line. 48 CHAPTER 1 Limits and Their Properties Section 1.2 Finding Limits Graphically and Numerically • Estimate a limit using a numerical or graphical approach. • Learn different ways that a limit can fail to exist. • Study and use a formal definition of limit. . An Introduction to Limits Video Suppose you are asked to sketch the graph of the function f given by f !x" ! x " 1. For all values other than x ! 1, you can use standard curve-sketching techniques. However, at x ! 1, it is not clear what to expect. To get an idea of the behavior of the graph of f near x ! 1, you can use two sets of x-values—one set that approaches 1 from the left and one set that approaches 1 from the right, as shown in the table. lim f(x) = 3 x→1 x3 # 1 , x#1 (1, 3) y x approaches 1 from the left. 3 2 x approaches 1 from the right. x 0.75 0.9 0.99 0.999 1 1.001 1.01 1.1 1.25 f #x\$ 2.313 2.710 2.970 2.997 ? 3.003 3.030 3.310 3.813 . f !x" approaches 3. 3 f(x) = x − 1 x −1 −2 −1 x 1 The . limit of f !x" as x approaches 1 is 3. Figure 1.5 f !x" approaches 3. Animation The graph of f is a parabola that has a gap at the point !1, 3", as shown in Figure 1.5. Although x cannot equal 1, you can move arbitrarily close to 1, and as a result f !x" moves arbitrarily close to 3. Using limit notation, you can write lim f !x" ! 3. Animation This is read as “the limit of f !x" as x approaches 1 is 3.” x→1 This discussion leads to an informal description of a limit. If f !x" becomes arbitrarily close to a single number L as x approaches c from either side, the limit of f !x", as x approaches c, is L. This limit is written as lim f !x" ! L. x→c E X P L O R AT I O N The discussion above gives an example of how you can estimate a limit numerically by constructing a table and graphically by drawing a graph. Estimate the following limit numerically by completing the table. lim x→2 x f #x\$ x2 # 3x \$ 2 x#2 1.75 1.9 1.99 1.999 2 2.001 2.01 2.1 2.25 ? ? ? ? ? ? ? ? ? Then use a graphing utility to estimate the limit graphically. SECTION 1.2 EXAMPLE 1 Finding Limits Graphically and Numerically 49 Estimating a Limit Numerically Evaluate the function f !x" ! x'!&x \$ 1 # 1" at several points near x ! 0 and use the results to estimate the limit lim x→0 x . &x \$ 1 # 1 y Solution The table lists the values of f !x" for several x-values near 0. f is undefined at x = 0. f(x) = x approaches 0 from the left. x x+1−1 x 1 f !x" #0.01 #0.001 #0.0001 0 0.0001 0.001 0.01 1.99499 1.99950 1.99995 ? 2.00005 2.00050 2.00499 x −1 f !x" approaches 2. 1 . The limit of f !x" as x approaches 0 is 2. Figure 1.6 Editable Graph x approaches 0 from the right. f !x" approaches 2. From the results shown in the table, you can estimate the limit to be 2. This limit is reinforced by the graph of f (see Figure 1.6). Exploration A Try It Exploration B In Example 1, note that the function is undefined at x ! 0 and yet f (x) appears to be approaching a limit as x approaches 0. This often happens, and it is important to realize that the existence or nonexistence of f !x" at x ! c has no bearing on the existence of the limit of f !x" as x approaches c. EXAMPLE 2 Finding a Limit Find the limit of f !x" as x approaches 2 where f is defined as f !x" ! x"2 x ! 2. Solution Because f !x" ! 1 for all x other than x ! 2, you can conclude that the limit is 1, as shown in Figure 1.7. So, you can write y 2 %1,0, f (x) = 1, x 2 0, x 2 lim f !x" ! 1. x→2 The fact that f !2" ! 0 has no bearing on the existence or value of the limit as x approaches 2. For instance, if the function were defined as 1 2 3 x .. The limit of f !x" as x approaches 2 is 1. Figure 1.7 Editable Graph f !x" ! %1,2, x"2 x!2 the limit would be the same. Try It Exploration A So far in this section, you have been estimating limits numerically and graphically. Each of these approaches produces an estimate of the limit. In Section 1.3, you will study analytic techniques for evaluating limits. Throughout the course, try to develop a habit of using this three-pronged approach to problem solving. 1. Numerical approach Construct a table of values. 2. Graphical approach 3. Analytic approach Draw a graph by hand or using technology. Use algebra or calculus. 50 CHAPTER 1 Limits and Their Properties Limits That Fail to Exist In the next three examples you will examine some limits that fail to exist. EXAMPLE 3 y Behavior That Differs from the Right and Left Show that the limit does not exist. x  f(x) = x lim 1 x→0 (x( x f (x) = 1 (( x −1 −δ 1 δ Solution Consider the graph of the function f !x" ! x 'x. From Figure 1.8, you can see that for positive x-values (x( ! 1, x x > 0 and for negative x-values f(x) = −1 (x( ! #1, lim f !x" does not exist. x < 0. x x→ 0 . This means that no matter how close x gets to 0, there will be both positive and negative x-values that yield f !x" ! 1 and f !x" ! #1. Specifically, if % (the lowercase Greek letter delta) is a positive number, then for x-values satisfying the inequality 0 < x < %, you can classify the values of x 'x as shown. Figure 1.8 Editable Graph (( (( !# %, 0" !0, %" Negative x-values yield x 'x ! #1. Positive x-values yield x 'x ! 1. (( (( . This implies that the limit does not exist. Exploration A Try It EXAMPLE 4 Exploration B Unbounded Behavior Discuss the existence of the limit lim x→0 Solution Let f !x" ! 1'x 2. In Figure 1.9, you can see that as x approaches 0 from either the right or the left, f !x" increases without bound. This means that by choosing x close enough to 0, you can force f !x" to be as large as you want. For instance, f !x) 1 will be larger than 100 if you choose x that is within 10 of 0. That is, y f (x) = 1 x2 4 3 (( 0 < x < 2 −1 1 lim . f !x" does not exist. . x→ 0 Figure 1.9 Editable Graph 1 10 f !x" ! 1 > 100. x2 Similarly, you can force f !x" to be larger than 1,000,000, as follows. 1 −2 1 . x2 2 x (( 0 < x < 1 1000 f !x" ! 1 > 1,000,000 x2 Because f !x" is not approaching a real number L as x approaches 0, you can conclude that the limit does not exist. Try It Exploration A Exploration B SECTION 1.2 1 Discuss the existence of the limit lim sin . x→0 x 1 f (x) = sin x 1 x −1 Solution Let f !x" ! sin!1'x". In Figure 1.10, you can see that as x approaches 0, f !x" oscillates between #1 and 1. So, the limit does not exist because no matter how small you choose %, it is possible to choose x1 and x2 within % units of 0 such that sin!1'x1" ! 1 and sin!1'x2 " ! #1, as shown in the table. 1 x . . 51 Oscillating Behavior EXAMPLE 5 y Finding Limits Graphically and Numerically −1 lim f !x" does not exist. x→ 0 Figure 1.10 Editable Graph 2'& 2'3& 2'5& 2'7& 2'9& 2'11& x→0 1 #1 1 #1 1 #1 Limit does not exist. sin #1/x\$ Try It Exploration A Open Exploration Common Types of Behavior Associated with Nonexistence of a Limit 1. f !x" approaches a different number from the right side of c than it approaches from the left side. 2. f !x" increases or decreases without bound as x approaches c. 3. f !x" oscillates between two fixed values as x approaches c. There are many other interesting functions that have unusual limit behavior. An often cited one is the Dirichlet function 0, if x is rational. f !x" ! 1, if x is irrational. % Because this function has no limit at any real number c, it is not continuous at any real number c. You will study continuity more closely in Section 1.4. TECHNOLOGY PITFALL When you use a graphing utility to investigate the behavior of a function near the x-value at which you are trying to evaluate a limit, remember that you can’t always trust the pictures that graphing utilities draw. If you use a graphing utility to graph the function in Example 5 over an interval containing 0, you will most likely obtain an incorrect graph such as that shown in Figure 1.11. The reason that a graphing utility can’t show the correct graph is that the graph has infinitely many oscillations over any interval that contains 0. 1.2 PETER GUSTAV DIRICHLET (1805–1859) . In the early development of calculus, the definition of a function was much more restricted than it is today, and “functions” such as the Dirichlet function would not have been considered. The modern definition of function was given by the German mathematician Peter Gustav Dirichlet. −0.25 0.25 −1.2 MathBio Incorrect graph of f !x" ! sin!1'x". Figure 1.11 52 CHAPTER 1 Limits and Their Properties A Formal Definition of Limit Let’s take another look at the informal description of a limit. If f !x" becomes arbitrarily close to a single number L as x approaches c from either side, then the limit of f !x" as x approaches c is L, written as lim f !x" ! L. x→c At first glance, this description looks fairly technical. Even so, it is informal because exact meanings have not yet been given to the two phrases “ f !x" becomes arbitrarily close to L” and “x approaches c.” The first person to assign mathematically rigorous meanings to these two phrases was Augustin-Louis Cauchy. His ' -% definition of limit is the standard used today. In Figure 1.12, let ' (the lowercase Greek letter epsilon) represent a (small) positive number. Then the phrase “f !x" becomes arbitrarily close to L” means that f !x" lies in the interval !L # ', L \$ '". Using absolute value, you can write this as L +ε L (c, L) ( f !x" # L( < '. L−ε Similarly, the phrase “x approaches c” means that there exists a positive number % such that x lies in either the interval !c # %, c" or the interval !c, c \$ %". This fact can be concisely expressed by the double inequality c +δ c c−δ The '-% definition of the limit of f !x" as x approaches c Figure 1.12 ( ( 0 < x # c < %. The first inequality ( ( 0 < x#c The distance between x and c is more than 0. expresses the fact that x " c. The second inequality (x # c( < % x is within % units of c. says that x is within a distance % of c. Definition of Limit Let f be a function defined on an open interval containing c (except possibly at c) and let L be a real number. The statement lim f !x" ! L x→c means that for each ' > 0 there exists a % > 0 such that if ( ( 0 < x # c < %, then ( f !x" # L( < '. FOR FURTHER INFORMATION For more on the introduction of rigor to calculus, see “Who Gave You the Epsilon? Cauchy and the Origins of Rigorous Calculus” by Judith V. . Grabiner in The American Mathematical Monthly. MathArticle NOTE Throughout this text, the expression lim f !x" ! L x→c implies two statements—the limit exists and the limit is L. Some functions do not have limits as x → c, but those that do cannot have two different limits as x → c. That is, if the limit of a function exists, it is unique (see Exercise 69). SECTION 1.2 Finding Limits Graphically and Numerically 53 The next three examples should help you develop a better understanding of the '-% definition of limit. Finding a ! for a Given " EXAMPLE 6 y = 1.01 y=1 y = 0.99 Given the limit lim !2x # 5" ! 1 y 2 x→3 x = 2.995 x=3 x = 3.005 ( ( ( ( find % such that !2x # 5" # 1 < 0.01 whenever 0 < x # 3 < %. Solution In this problem, you are working with a given value of '—namely, ' ! 0.01. To find an appropriate %, notice that 1 x 1 2 3 4 −1 ( ( is equivalent to 2 x # 3 < 0.01, you can choose % ! 2!0.01" ! 0.005. This choice works because ( ( 0 < x # 3 < 0.005 f (x) = 2x − 5 −2 (!2x # 5" # 1( ! (2x # 6( ! 2(x # 3(. Because the inequality (!2x # 5" # 1( < 0.01 1 implies that (!2x # 5" # 1( ! 2(x # 3( < 2!0.005" ! 0.01 . The limit of f !x" as x approaches 3 is 1. as shown in Figure 1.13. Figure 1.13 Exploration B Exploration A Try It NOTE In Example 6, note that 0.005 is the largest value of % that will guarantee !2x # 5" # 1 < 0.01 whenever 0 < x # 3 < %. Any smaller positive value of % would also work. ( ( ( ( In Example 6, you found a %-value for a given '. This does not prove the existence of the limit. To do that, you must prove that you can find a % for any ', as shown in the next example. y=4+ε Using the "-! Definition of Limit EXAMPLE 7 y=4 Use the '-% definition of limit to prove that y=4−ε lim !3x # 2" ! 4. x→2 x=2+δ x=2 x=2−δ y Solution You must show that for each ' > 0, there exists a % > 0 such that !3x # 2" # 4 < ' whenever 0 < x # 2 < %. Because your choice of % depends on ', you need to establish a connection between the absolute values !3x # 2" # 4 and x # 2 . ( ( ( ( ( ( (!3x # 2" # 4( ! (3x # 6( ! 3(x # 2( 4 3 ( So, for a given ' > 0 you can choose % ! ''3. This choice works because 2 ( ( 0 < x#2 < %! 1 f (x) = 3x − 2 implies that x 1 2 3 4 . The limit of f !x" as x approaches 2 is 4. Figure 1.14 ' 3 (!3x # 2" # 4( ! 3(x # 2( < 3)3* ! ' ' as shown in Figure 1.14. Try It Exploration A ( 54 CHAPTER 1 Limits and Their Properties EXAMPLE 8 Using the "-! Definition of Limit Use the '-% definition of limit to prove that f(x) = x 2 lim x 2 ! 4. 4+ε x→2 (2 + δ )2 Solution You must show that for each ' > 0, there exists a % > 0 such that 4 (x 2 # 4( < ' (2 − δ )2 4−ε 2+δ 2 2−δ The . limit of f !x" as x approaches 2 is 4. Figure 1.15 ( ( To find an appropriate %, begin by writing (x2 # 4( ! (x # 2((x \$ 2(. For all x in the interval !1, 3", you know that (x \$ 2( < 5. So, letting % be the minimum of ''5 and 1, it follows that, whenever 0 < (x # 2( < %, you have whenever 0 < x # 2 < %. (x2 # 4( ! (x # 2((x \$ 2( < )5!5" ! ' ' as shown in Figure 1.15. Try It Exploration A Throughout this chapter you will use the '-% definition of limit primarily to prove theorems about limits and to establish the existence or nonexistence of particular types of limits. For finding limits, you will learn techniques that are easier to use than the '-% definition of limit. 54 CHAPTER 1 Limits and Their Properties Exercises for Section 1.2 The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. Click on to view the complete solution of the exercise. Click on to print an enlarged copy of the graph. In Exercises 1–8, complete the table and use the result to estimate the limit. Use a graphing utility to graph the function xq2 x 1.9 1.99 1.999 2.001 2.01 2.1 6. lim xq4 2. lim x!2 x2 ! 4 x 1.9 1.99 1.999 2.001 2.01 2.1 !0.1 2.999 3.001 3.01 3.1 4.001 4.01 4.1 !x""x # 1#\$ ! "4"5# x!4 3.9 3.99 3.999 7. lim sinx x xq0 x !0.01 !0.1 !0.01 !0.001 0.001 0.01 0.1 !0.01 !0.001 0.001 0.01 0.1 f &x' x x 2.99 f &x' %x # 3 ! %3 xq0 2.9 x f &x' 3. lim !1""x # 1#\$ ! "1"4# x!3 f &x' f &x' xq2 xq3 x x!2 x2 ! x ! 2 1. lim 5. lim !0.001 0.001 0.01 0.1 8. lim cos x ! 1 x x !0.1 xq0 f &x' 4. lim %1 ! x ! 2 xq!3 x#3 x !3.1 f &x' !3.01 f &x' !3.001 !2.999 !2.99 !2.9 SECTION 1.2 In Exercises 9–18, use the graph to find the limit (if it exists). If the limit does not exist, explain why. 9. lim "4 ! x# 10. lim "x 2 # 2# xq3 y 4 4 3 3 y (b) lim f "x# 6 5 xq1 (c) f "4# 3 2 1 (d) lim f "x# 2 xq4 1 1 x 1 2 3 11. lim f "x# 1 20. (a) f "!2# x&2 x\$2 f "x# \$ 2 (c) f "0# x&1 x\$1 (d) lim f "x# xq0 y 4 4 3 3 (e) f "2# 1 2 3 )x ! 5) 13. lim xq5 (h) lim f "x# xq4 x 1 2 x!5 In Exercises 21 and 22, use the graph of f to identify the values of c for which lim f &x' exists. xqc 1 xq3 x ! 3 14. lim y 21. y 6 y 4 3 2 1 2 (g) f "4# 2 1 4 x 1 2 3 4 5 xq2 1 x 2 1 (f ) lim f "x# 2 1 4 3 2 xq!2 (x1, # 2, y y (b) lim f "x# xq1 (40,! x, 1 2 3 4 5 6 2 12. lim f "x# xq2 f "x# \$ x 1 x 2 1 4 55 In Exercises 19 and 20, use the graph of the function f to decide whether the value of the given quantity exists. If it does, find it. If not, explain why. 19. (a) f "1# xq1 y Finding Limits Graphically and Numerically 4 2 1 x 6 7 8 9 2 3 4 x 2 1 2 15. lim sin % x 2 4 2 22. y 4 xq0 y 2 y x 4 2 1 4 6 16. lim sec x xq1 x 2 2 4 6 x 1 2 U 2 17. lim cos xq0 1 x x U 2 xqc 23. f "x# \$ 8 ! 2x, 4, xq %"2 y 1 2 1 x 1 1 1 U 2 U 2 ( ( x2, 18. lim tan x y In Exercises 23 and 24, sketch the graph of f. Then identify the values of c for which lim f &x' exists. U 3U 2 x sin x, 24. f "x# \$ 1 ! cos x, cos x, x f 2 2 < x < 4 x v 4 x < 0 0 f x f % x > % 56 CHAPTER 1 Limits and Their Properties In Exercises 25 and 26, sketch a graph of a function f that satisfies the given values. (There are many correct answers.) 25. f "0# is undefined. f "x# \$ 26. f "!2# \$ 0 lim f "x# \$ 4 f "2# \$ 0 f "2# \$ 6 lim f "x# \$ 0 xq0 lim f "x# does not exist. xq2 1 x!1 xq2 27. Modeling Data The cost of a telephone call between two cities is \$0.75 for the first minute and \$0.50 for each additional minute or fraction thereof. A formula for the cost is given by ) f "x# ! 1) < 0.01. 2.0 1.0 0.5 f "x# \$ 2 ! C 3.6 3.7 y 2.5 2.9 C 3 f x 1 3.1 3.5 4 ? Does the limit of C"t# as t approaches 3 exist? Explain. 32. The graph of f "x# \$ x 2 ! 1 is shown in the figure. Find ' such that if 0 < x ! 2 < ' then f "x# ! 3 < 0.2. ) ) y f 4 2 C"t# \$ 0.35 ! 0.12,!"t ! 1#-. y = 3.2 y=3 y = 2.8 1 29. The graph of f "x# \$ x # 1 is shown in the figure. Find ' such that if 0 < x ! 2 < ' then f "x# ! 3 < 0.4. ) ) 1 2 3 4 In Exercises 33–36, find the limit L. Then find " > 0 such that ) f &x' ! L) < 0.01 whenever 0 < )x ! c) < ". 5 2.6 x ) y 3.4 2 3 28. Repeat Exercise 27 for ) ) y = 1.1 y=1 y = 0.9 ? 2 ) is shown in the figure. Find ' such that if 0 < x ! 1 < ' then ) f "x# ! 1) < 0.1. 4 (c) Use the graph to complete the table and observe the behavior of the function as t approaches 3. t 4 1 x 1 3.5 3 31. The graph of 2 lim C "t#. 2 tq3.5 3.4 ) 201 2 199 101 99 x "Note: ,x- \$ greatest integer n such that n f x. For example, ,3.2- \$ 3 and ,!1.6- \$ !2.# (a) Use a graphing utility to graph the cost function for 0 < t f 5. (b) Use the graph to complete the table and observe the behavior of the function as t approaches 3.5. Use the graph and the table to find 3.3 ) 1.01 1.00 0.99 1.5 1 where t is the time in minutes. 3 ) y C"t# \$ 0.75 ! 0.50 ,! "t ! 1#- t ) is shown in the figure. Find ' such that if 0 < x ! 2 < ' then xq!2 lim f "x# \$ 3 30. The graph of 4 33. lim "3x # 2# 3 xq2 2 34. lim 4 ! xq4 x 0.5 1.0 1.5 2.0 2.5 1.6 2.4 3.0 x 2 + 35. lim "x 2 ! 3# xq2 36. lim "x 2 # 4# xq5 SECTION 1.2 In Exercises 37–48, find the limit L. Then use the ( -" definition to prove that the limit is L. 37. lim "x # 3# xq2 38. lim "2x # 5# Finding Limits Graphically and Numerically 57 56. Identify three types of behavior associated with the nonexistence of a limit. Illustrate each type with a graph of a function. xq!3 "12x ! 1# 2 lim "3x # 9# xq1 39. lim xq!4 40. 57. Jewelry A jeweler resizes a ring so that its inner circumference is 6 centimeters. 41. lim 3 (a) What is the radius of the ring? 42. lim "!1# (b) If the ring’s inner circumference can vary between 5.5 centimeters and 6.5 centimeters, how can the radius vary? xq6 xq2 3 x 43. lim % xq0 44. lim %x xq4 ) ) 45. lim x ! 2 xq!2 ) xq3 ) 46. lim x ! 3 (c) Use the (-' definition of limit to describe this situation. Identify ( and '. 58. Sports A sporting goods manufacturer designs a golf ball having a volume of 2.48 cubic inches. 47. lim "x 2 # 1# (a) What is the radius of the golf ball? 48. lim "x 2 # 3x# (b) If the ball’s volume can vary between 2.45 cubic inches and 2.51 cubic inches, how can the radius vary? xq1 xq!3 Writing In Exercises 49–52, use a graphing utility to graph the function and estimate the limit (if it exists). What is the domain of the function? Can you detect a possible error in determining the domain of a function solely by analyzing the graph generated by a graphing utility? Write a short paragraph about the importance of examining a function analytically as well as graphically. 49. f "x# \$ %x # 5 ! 3 x!4 lim f "x) xq4 x!3 50. f "x# \$ 2 x ! 4x # 3 lim f "x# xq3 x!9 51. f "x# \$ %x ! 3 lim f "x# xq9 52. f "x# \$ x!3 x2 ! 9 lim f "x# xq3 53. Write a brief description of the meaning of the notation lim f "x# \$ 25. xq8 54. If f "2# \$ 4, can you conclude anything about the limit of f "x# as x approaches 2? Explain your reasoning. 55. If the limit of f "x# as x approaches 2 is 4, can you conclude (c) Use the (-' definition of limit to describe this situation. Identify ( and '. 59. Consider the function f "x# \$ "1 # x#1"x. Estimate the limit lim "1 # x#1"x xq0 by evaluating f at x-values near 0. Sketch the graph of f. 60. Consider the function f "x# \$ )x # 1) ! )x ! 1). x Estimate lim )x # 1) ! )x ! 1) x xq0 by evaluating f at x-values near 0. Sketch the graph of f. 61. Graphical Analysis lim xq2 The statement x2 !4 \$4 x!2 means that for each ( > 0 there corresponds a ' > 0 such that if 0 < x ! 2 < ', then ) ) ) ) x2 ! 4 ! 4 < (. x!2 ) x2 ! 4 ! 4 < 0.001. x!2 If ( \$ 0.001, then ) Use a graphing utility to graph each side of this inequality. Use the zoom feature to find an interval "2 ! ', 2 # '# such that the graph of the left side is below the graph of the right side of the inequality. 58 CHAPTER 1 62. Graphical Analysis Limits and Their Properties 72. (a) Given that The statement lim "3x # 1#"3x ! 1#x2 # 0.01 \$ 0.01 x 2 ! 3x xq3 x ! 3 lim xq0 means that for each ( > 0 there corresponds a ' > 0 such that if 0 < x ! 3 < ', then prove that there exists an open interval "a, b# containing 0 such that "3x # 1#"3x ! 1#x2 # 0.01 > 0 for all x & 0 in "a, b#. ) ) ) x2 ! 3x ! 3 < (. x!3 ) ) x2 ! 3x ! 3 < 0.001. x!3 (b) Given that lim g "x# \$ L, where L > 0, prove that there xqc exists an open interval "a, b# containing c such that g"x# > 0 for all x & c in "a, b#. If ( \$ 0.001, then ) 73. Programming Use the programming capabilities of a graphing utility to write a program for approximating lim f "x#. xqc Use a graphing utility to graph each side of this inequality. Use the zoom feature to find an interval "3 ! ', 3 # '# such that the graph of the left side is below the graph of the right side of the inequality. True or False? In Exercises 63–66, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 63. If f is undefined at x \$ c, then the limit of f "x# as x approaches c does not exist. Assume the program will be applied only to functions whose limits exist as x approaches c. Let y1 \$ f "x# and generate two lists whose entries form the ordered pairs "c ± !0.1\$ n , f "c ± !0.1\$ n ## for n \$ 0, 1, 2, 3, and 4. 74. Programming Use the program you created in Exercise 73 to approximate the limit x 2 ! x ! 12 . xq4 x!4 lim 64. If the limit of f "x# as x approaches c is 0, then there must exist a number k such that f "k# < 0.001. Putnam Exam Challenge 65. If f "c# \$ L, then lim f "x# \$ L. xqc 75. Inscribe a rectangle of base b and height h and an isosceles triangle of base b in a circle of radius one as shown. For what value of h do the rectangle and triangle have the same area? 66. If lim f "x# \$ L, then f "c# \$ L. xqc 67. Consider the function f "x# \$ %x. (a) Is lim %x \$ 0.5 a true statement? Explain. xq0.25 (b) Is lim %x \$ 0 a true statement? Explain. xq0 68. Writing The definition of limit on page 52 requires that f is a function defined on an open interval containing c, except possibly at c. Why is this requirement necessary? 69. Prove that if the limit of f "x# as x q c exists, then the limit must be unique. [Hint: Let lim f "x# \$ L1 and xqc lim f "x# \$ L 2 xqc and prove that L1 \$ L2.] 70. Consider the line f "x# \$ mx # b, where m & 0. Use the (-' definition of limit to prove that lim f "x# \$ mc # b. xqc 71. Prove that lim f "x# \$ L is equivalent to lim ! f "x# ! L\$ \$ 0. xqc xqc h b 76. A right circular cone has base of radius 1 and height 3. A cube is inscribed in the cone so that one face of the cube is contained in the base of the cone. What is the side-length of the cube? These problems were composed by the Committee on the Putnam Prize Competition. SECTION 1.3 Section 1.3 Evaluating Limits Analytically 59 Evaluating Limits Analytically • • • • . Evaluate a limit using properties of limits. Develop and use a strategy for finding limits. Evaluate a limit using dividing out and rationalizing techniques. Evaluate a limit using the Squeeze Theorem. Properties of Limits Video In Section 1.2, you learned that the limit of f "x# as x approaches c does not depend on the value of f at x ! c. It may happen, however, that the limit is precisely f "c#. In such cases, the limit can be evaluated by direct substitution. That is, lim f "x# ! f "c#. Substitute c for x. x→c Such well-behaved functions are continuous at c. You will examine this concept more closely in Section 1.4. y f (c) = x THEOREM 1.1 Some Basic Limits Let b and c be real numbers and let n be a positive integer. c+ ε ε =δ 1. lim b ! b f(c) = c 2. lim x ! c x→c x→c 3. lim x n ! c n x→c ε =δ c−ε c−δ c c+δ x Figure 1.16 Proof To prove Property 2 of Theorem 1.1, you need to show that for each \$ > 0 there exists a # > 0 such that x " c < \$ whenever 0 < x " c < #. To do this, choose # ! \$. The second inequality then implies the first, as shown in Figure 1.16. This completes the proof. (Proofs of the other properties of limits in this section are listed in Appendix A or are discussed in the exercises.) ! ! ! ! Evaluating Basic Limits NOTE When you encounter new notations or symbols in mathematics, be sure . you know how the notations are read. For instance, the limit in Example 1(c) is read as “the limit of x 2 as x approaches 2 is 4.” EXAMPLE 1 . The editable graph feature allows you to edit the graph of a function to visually evaluate the limit as x approaches c. a. lim 3 ! 3 x→2 b. lim x ! "4 Try It a. x→"4 x→2 Exploration A Editable Graph THEOREM 1.2 c. lim x 2 ! 2 2 ! 4 b. Editable Graph c. Editable Graph Properties of Limits Let b and c be real numbers, let n be a positive integer, and let f and g be functions with the following limits. lim f "x# ! L x→c and 1. Scalar multiple: 2. Sum or difference: 3. Product: 4. Quotient: 5. Power: lim g "x# ! K x→c lim \$b f "x#% ! bL x→c lim \$ f "x# ± g"x#% ! L ± K x→c lim \$ f "x#g"x#% ! LK x→c lim x→c f "x# L ! , g"x# K lim \$ f "x#%n ! Ln x→c provided K % 0 60 CHAPTER 1 Limits and Their Properties EXAMPLE 2 The Limit of a Polynomial lim "4x 2 & 3# ! lim 4x 2 & lim 3 x→2 x→2 & . ' Property 2 ! 4 lim x 2 & lim 3 Property 1 ! 4"22# & 3 Example 1 ! 19 Simplify. x→2 x→2 Exploration A Try It . x→2 The editiable graph feature allows you to edit the graph of a function to visually evaluate the limit as x approaches c. Editable Graph In Example 2, note that the limit (as x → 2) of the polynomial function p"x# ! 4x 2 & 3 is simply the value of p at x ! 2. lim p"x# ! p"2# ! 4"22# & 3 ! 19 x→2 This direct substitution property is valid for all polynomial and rational functions with nonzero denominators. THEOREM 1.3 Limits of Polynomial and Rational Functions If p is a polynomial function and c is a real number, then lim p"x# ! p"c#. x→c If r is a rational function given by r "x# ! p"x#(q"x# and c is a real number such that q"c# % 0, then lim r "x# ! r "c# ! x→c EXAMPLE 3 p"c# . q"c# The Limit of a Rational Function 2 Find the limit: lim x & x & 2 . x→1 x&1 Solution Because the denominator is not 0 when x ! 1, you can apply Theorem 1.3 to obtain . x 2 & x & 2 12 & 1 & 2 4 ! ! ! 2. x→1 x&1 1&1 2 lim Try It . Exploration A The editiable graph feature allows you to edit the graph of a function to visually evaluate the limit as x approaches c. Editable Graph THE SQUARE ROOT SYMBOL The first use of a symbol to denote the square root can be traced to the sixteenth century. Mathematicians first used the symbol ) , which had only two strokes. This symbol was chosen because it resembled a lowercase r, to .stand for the Latin word radix, meaning root. Video Video Polynomial functions and rational functions are two of the three basic types of algebraic functions. The following theorem deals with the limit of the third type of algebraic function—one that involves a radical. See Appendix A for a proof of this theorem. THEOREM 1.4 The Limit of a Function Involving a Radical Let n be a positive integer. The following limit is valid for all c if n is odd, and is valid for c > 0 if n is even. n x !) n c lim ) x→c SECTION 1.3 Evaluating Limits Analytically 61 The following theorem greatly expands your ability to evaluate limits because it shows how to analyze the limit of a composite function. See Appendix A for a proof of this theorem. THEOREM 1.5 The Limit of a Composite Function If f and g are functions such that lim g"x# ! L and lim f "x# ! f "L#, then x→c & x→L ' lim f "g "x## ! f lim g"x# ! f "L#. x→c EXAMPLE 4 x→c The Limit of a Composite Function a. Because lim "x 2 & 4# ! 0 2 & 4 ! 4 x→0 and lim )x ! 2 x→4 it follows that lim )x2 & 4 ! )4 ! 2. x→0 b. Because lim "2x 2 " 10# ! 2"32# " 10 ! 8 and x→3 3 x ! 2 lim ) x→8 it follows that . 3 2x 2 " 10 ! ) 3 8 ! 2. lim ) x→3 Exploration A Try It . Open Exploration The editable graph feature allows you to edit the graph of a function to visually evaluate the limit as x approaches c. a. Editable Graph b. Editable Graph You have seen that the limits of many algebraic functions can be evaluated by direct substitution. The six basic trigonometric functions also exhibit this desirable quality, as shown in the next theorem (presented without proof). THEOREM 1.6 Limits of Trigonometric Functions Let c be a real number in the domain of the given trigonometric function. 1. lim sin x ! sin c 2. lim cos x ! cos c 3. lim tan x ! tan c 4. lim cot x ! cot c 5. lim sec x ! sec c 6. lim csc x ! csc c x→c x→c x→c x→c x→c EXAMPLE 5 x→c Limits of Trigonometric Functions a. lim tan x ! tan"0# ! 0 x→0 & b. lim "x cos x# ! lim x . x→ ' x→ ' '& lim cos x' ! ' cos"'# ! " ' x→ ' c. lim sin2 x ! lim "sin x#2 ! 02 ! 0 x→0 Try It x→0 Exploration A 62 CHAPTER 1 Limits and Their Properties A Strategy for Finding Limits On the previous three pages, you studied several types of functions whose limits can be evaluated by direct substitution. This knowledge, together with the following theorem, can be used to develop a strategy for finding limits. A proof of this theorem is given in Appendix A. y 3 f(x) = x − 1 x−1 THEOREM 1.7 3 Functions That Agree at All But One Point Let c be a real number and let f "x# ! g"x# for all x % c in an open interval containing c. If the limit of g"x# as x approaches c exists, then the limit of f "x# also exists and 2 lim f "x# ! lim g"x#. x→c . −2 x −1 1 Editable Graph x→c EXAMPLE 6 Finding the Limit of a Function Find the limit: lim x→1 x3 " 1 . x"1 Solution Let f "x# ! "x3 " 1#("x " 1#. By factoring and dividing out like factors, you can rewrite f as y 3 f "x# ! "x " 1#"x2 & x & 1# ! x2 & x & 1 ! g"x#, "x " 1# x % 1. So, for all x-values other than x ! 1, the functions f and g agree, as shown in Figure 1.17. Because lim g"x# exists, you can apply Theorem 1.7 to conclude that f and g 2 x→1 have the same limit at x ! 1. g (x) = x 2 + x + 1 . −2 −1 x 1 f and g agree at all but one point. Editable Graph . lim x→1 x3 " 1 "x " 1#"x 2 & x & 1# ! lim x"1 x→1 x"1 "x " 1#"x2 & x & 1# ! lim x"1 x→1 ! lim "x 2 & x & 1# x→1 ! 12 & 1 & 1 !3 Figure 1.17 Try It . STUDY TIP When applying this strategy for finding a limit, remember that some functions do not have a limit (as x approaches c). For instance, the following limit does not exist. x3 & 1 x→1 x " 1 lim Factor. Divide out like factors. Apply Theorem 1.7. Use direct substitution. Simplify. Exploration A Exploration B Exploration C Exploration D A Strategy for Finding Limits 1. Learn to recognize which limits can be evaluated by direct substitution. (These limits are listed in Theorems 1.1 through 1.6.) 2. If the limit of f "x# as x approaches c cannot be evaluated by direct substitution, try to find a function g that agrees with f for all x other than x ! c. [Choose g such that the limit of g"x# can be evaluated by direct substitution.] 3. Apply Theorem 1.7 to conclude analytically that lim f "x# ! lim g"x# ! g"c#. x→c x→c 4. Use a graph or table to reinforce your conclusion. SECTION 1.3 Evaluating Limits Analytically 63 Dividing Out and Rationalizing Techniques Two techniques for finding limits analytically are shown in Examples 7 and 8. The first technique involves dividing out common factors, and the second technique involves rationalizing the numerator of a fractional expression. Dividing Out Technique EXAMPLE 7 x2 & x " 6 . x→"3 x&3 Find the limit: lim Solution Although you are taking the limit of a rational function, you cannot apply Theorem 1.3 because the limit of the denominator is 0. y −2 −1 1 2 x→"3 −1 f (x) = x2 + x − 6 x+3 −4 (−3, −5) x2 & x " 6 x→"3 x&3 lim −2 −3 lim "x 2 & x " 6# ! 0 x Direct substitution fails. lim "x & 3# ! 0 x→"3 Because the limit of the numerator is also 0, the numerator and denominator have a common factor of "x & 3#. So, for all x % "3, you can divide out this factor to obtain −5 . f is undefined when x ! " 3. f "x# ! Figure 1.18 x 2 & x " 6 "x & 3#"x " 2# ! ! x " 2 ! g"x#, x&3 x&3 Using Theorem 1.7, it follows that Editable Graph x2 & x " 6 ! lim "x " 2# x→"3 x&3 x→"3 ! "5. lim NOTE In the solution of Example 7, . be sure you see the usefulness of the Factor Theorem of Algebra. This theorem states that if c is a zero of a polynomial function, "x " c# is a factor of the polynomial. So, if you apply direct substitution to a rational function and obtain r "c# ! p"c# 0 ! q"c# 0 you can conclude that "x " c# must be a common factor to both p"x# and q"x#. −3 − δ −5 + ε −3 + δ Glitch near (−3, −5) −5 − ε Incorrect graph of f Figure 1.19 x % "3. Apply Theorem 1.7. Use direct substitution. This result is shown graphically in Figure 1.18. Note that the graph of the function f coincides with the graph of the function g"x# ! x " 2, except that the graph of f has a gap at the point ""3, "5#. Try It Exploration A Open Exploration In Example 7, direct substitution produced the meaningless fractional form 0(0. An expression such as 0(0 is called an indeterminate form because you cannot (from the form alone) determine the limit. When you try to evaluate a limit and encounter this form, remember that you must rewrite the fraction so that the new denominator does not have 0 as its limit. One way to do this is to divide out like factors, as shown in Example 7. A second way is to rationalize the numerator, as shown in Example 8. TECHNOLOGY PITFALL f "x# ! x2 & x " 6 x&3 and Because the graphs of g"x# ! x " 2 differ only at the point ""3, "5#, a standard graphing utility setting may not distinguish clearly between these graphs. However, because of the pixel configuration and rounding error of a graphing utility, it may be possible to find screen settings that distinguish between the graphs. Specifically, by repeatedly zooming in near the point ""3, "5# on the graph of f, your graphing utility may show glitches or irregularities that do not exist on the actual graph. (See Figure 1.19.) By changing the screen settings on your graphing utility you may obtain the correct graph of f. 64 CHAPTER 1 Limits and Their Properties Rationalizing Technique EXAMPLE 8 Find the limit: lim )x & 1 " 1 x x→0 . Solution By direct substitution, you obtain the indeterminate form 0(0. lim ")x & 1 " 1# ! 0 x→0 lim )x & 1 " 1 Direct substitution fails. x x→0 lim x ! 0 x→0 In this case, you can rewrite the fraction by rationalizing the numerator. )x & 1 " 1 x f(x) = )x & 1 " 1 '& )x & 1 & 1 x )x & 1 & 1 "x & 1# " 1 ! x")x & 1 & 1# x ! x")x & 1 & 1# 1 ! , x%0 )x & 1 & 1 y 1 & ! x +1−1 x ' Now, using Theorem 1.7, you can evaluate the limit as shown. x −1 lim 1 )x & 1 " 1 x x→0 x→0 Figure 1.20 1 )x & 1 & 1 1 1&1 1 ! 2 ! −1 1 The . limit of f "x# as x approaches 0 is 2 . ! lim A table or a graph can reinforce your conclusion that the limit is 12. (See Figure 1.20.) Editable Graph x approaches 0 from the left. x approaches 0 from the right. x "0.25 f "x# 0.5359 0.5132 0.5013 0.5001 ? "0.1 "0.01 "0.001 0 f "x# approaches 0.5. 0.001 0.01 0.1 0.25 0.4999 0.4988 0.4881 0.4721 f "x# approaches 0.5. . Exploration A Try It Exploration B Exploration C NOTE The rationalizing technique for evaluating limits is based on multiplication by a convenient form of 1. In Example 8, the convenient form is 1! )x & 1 & 1 )x & 1 & 1 . SECTION 1.3 Evaluating Limits Analytically 65 The Squeeze Theorem h (x) ≤ f (x) ≤ g(x) The next theorem concerns the limit of a function that is squeezed between two other functions, each of which has the same limit at a given x-value, as shown in Figure 1.21. (The proof of this theorem is given in Appendix A.) y f lies in here. g g f THEOREM 1.8 f If h"x# ≤ f "x# ≤ g"x# for all x in an open interval containing c, except possibly at c itself, and if h h lim h"x# ! L ! lim g"x# x→c x c . The Squeeze Theorem x→c then lim f "x# exists and is equal to L. x→c The Squeeze Theorem Figure 1.21 Video You can see the usefulness of the Squeeze Theorem in the proof of Theorem 1.9. THEOREM 1.9 1. lim x→0 y (cos θ , sin θ ) (1, tan θ ) θ (1, 0) Two Special Trigonometric Limits sin x !1 x 2. lim x→0 1 " cos x !0 x Proof To avoid the confusion of two different uses of x, the proof is presented using the variable (, where ( is an acute positive angle measured in radians. Figure 1.22 shows a circular sector that is squeezed between two triangles. x tan θ 1 sin θ θ θ A circular sector is used to prove Theorem 1.9. Figure 1.22 θ 1 1 Area of triangle tan ( 2 ≥ ≥ Area of sector ( 2 1 ≥ ≥ Area of triangle sin ( 2 Multiplying each expression by 2(sin ( produces 1 ( ≥ ≥ 1 cos ( sin ( and taking reciprocals and reversing the inequalities yields FOR FURTHER INFORMATION f "x# ! "sin x#(x, see the article “The Function "sin x#(x” by William B. . Gearhart and Harris S. Shultz in The College Mathematics Journal. MathArticle cos ( ≤ sin ( ≤ 1. ( Because cos ( ! cos ""(# and "sin (#(( ! \$sin"" (#%("" (#, you can conclude that this inequality is valid for all nonzero ( in the open interval "" '(2, '(2#. Finally, because lim cos ( ! 1 and lim 1 ! 1, you can apply the Squeeze Theorem to ( →0 ( →0 conclude that lim "sin (#(( ! 1. The proof of the second limit is left as an exercise (see ( →0 Exercise 120). 66 CHAPTER 1 Limits and Their Properties A Limit Involving a Trigonometric Function EXAMPLE 9 Find the limit: lim x→0 tan x . x Solution Direct substitution yields the indeterminate form 0(0. To solve this problem, you can write tan x as "sin x#("cos x# and obtain lim x→0 f(x) = tan x x & tan x sin x ! lim x x→0 x '&cos1 x'. Now, because 4 lim x→0 sin x !1 x and lim x→0 1 !1 cos x you can obtain −' 2 ' 2 lim x→0 & tan x sin x ! lim x→0 x x Figure 1.23 Editable Graph x→0 ! "1#"1# ! 1. −2 The . limit of f "x# as x approaches 0 is 1. '& lim cos1 x' (See Figure 1.23.) Exploration A Try It A Limit Involving a Trigonometric Function EXAMPLE 10 Find the limit: lim x→0 sin 4x . x Solution Direct substitution yields the indeterminate form 0(0. To solve this problem, you can rewrite the limit as g(x) = lim x→0 sin 4x x lim x→0 ' 2 −2 The . limit of g"x# as x approaches 0 is 4. Figure 1.24 Editable Graph ' Multiply and divide by 4. Now, by letting y ! 4x and observing that x → 0 if and only if y → 0, you can write 6 −' 2 & sin 4x sin 4x ! 4 lim . x x→0 4x & & sin 4x sin 4x ! 4 lim x→0 x 4x sin y ! 4 lim y→0 y ! 4"1# ! 4. ' ' (See Figure 1.24.) Exploration A Try It Use a graphing utility to confirm the limits in the examples and exercise set. For instance, Figures 1.23 and 1.24 show the graphs of TECHNOLOGY f "x# ! tan x x and g"x# ! sin 4x . x Note that the first graph appears to contain the point "0, 1# and the second graph appears to contain the point "0, 4#, which lends support to the conclusions obtained in Examples 9 and 10. SECTION 1.3 67 Evaluating Limits Analytically Exercises for Section 1.3 The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. Click on to view the complete solution of the exercise. Click on to print an enlarged copy of the graph. In Exercises 1–4, use a graphing utility to graph the function and visually estimate the limits. 1. h!x" " x 2 ! 5x 12!&x ! 3" x!9 2. g!x" " (a) lim h!x" (a) lim g!x" (b) lim h!x" (b) lim g!x" xq5 xq0 ) ) (a) lim f !x" (b) lim f !x" '\$6x( In Exercises 37–40, use the information to evaluate the limits. 3 38. lim f !x" " 2 xqc 1 lim g!x" " 2 xqc xqc (a) lim #4f !x"\$ (a) lim f !t" (b) lim # f !x" # g!x"\$ (b) lim # f !x" # g!x"\$ (b) lim f !t" (c) lim # f !x" g!x"\$ (c) lim # f !x" g!x"\$ tq!1 xqc xqc 6. lim x3 7. lim !2x ! 1" 8. lim !3x # 2" xq!3 f !x" g!x" (d) lim xqc 39. lim f !x" " 4 xq!2 xqc xqc f !x" g!x" xqc 5. lim x 4 xqc xqc (d) lim In Exercises 5–22, find the limit. xq0 xq7 (a) lim #5g!x"\$ tq4 xq2 36. lim sec lim g!x" " 3 4. f !t" " t t ! 4 xq \$%3 '\$4x( xqc 3. f !x" " x cos x xq0 xq3 37. lim f !x" " 2 xq4 xq!1 35. lim tan 40. lim f !x" " 27 xqc xqc (a) lim # f !x"\$3 3 f !x" (a) lim & xqc xqc 11. lim !2x 2 # 4x # 1" 12. lim !3x 3 ! 2x 2 # 4" (c) lim #3f !x"\$ f !x" 18 (c) lim # f !x"\$ 2 1 13. lim xq2 x 2 14. lim xq!3 x # 2 (d) lim # f !x"\$3%2 (d) lim # f !x"\$ 2%3 9. lim !x 2 # 3x" 10. lim !!x 2 # 1" xq!3 xq1 xq!3 xq1 15. lim x!3 x2 # 4 16. lim 17. lim 5x &x # 2 18. lim xq1 xq7 xq3 xq3 2x ! 3 x#5 &x # 1 x!4 3 x # 4 20. lim & xq4 21. lim !x # 3" 2 22. lim !2x ! 1"3 xq!4 (b) lim xqc xqc xqc 19. lim &x # 1 xq3 (b) lim &f !x" xqc xqc xqc In Exercises 41–44, use the graph to determine the limit visually (if it exists). Write a simpler function that agrees with the given function at all but one point. 41. g!x" " !2x 2 # x x 42. h!x" " y y xq0 3 (b) lim g!x" 24. f !x" " x # 7, g!x" " (a) lim f !x" xq!3 (c) lim g! f !x"" xq4 x2 xq1 (b) lim g!x" (c) lim g! f !x"" xq4 xq!3 25. f !x" " 4 ! x 2, g!x" " &x # 1 (a) lim f !x" xq1 xq4 2 x 1 (b) lim g!x" (c) lim g! f !x"" xq3 xq1 (b) lim g!x" (a) lim g!x" (a) lim h!x" (b) lim g!x" (b) lim h!x" xq0 xq!2 xq0 x3 ! x 43. g!x" " x!1 44. f !x" " y (c) lim g! f !x"" xq21 5 1 xq4 x x2 ! x y 2 3 1 In Exercises 27– 36, find the limit of the trigonometric function. 2 27. lim sin x 28. lim tan x 1 \$x 29. lim cos xq2 3 30. lim sin 31. lim sec 2x 32. lim cos 3x (a) lim g!x" (a) lim f !x" 33. 34. (b) lim g!x" (b) lim f !x" xq \$%2 xq0 lim sin x xq5\$%6 3 3 xq!1 3 x#6 26. f !x" " 2x 2 ! 3x # 1, g!x" " & (a) lim f !x" 1 2 1 23. f !x" " 5 ! x, g!x" " x3 xq1 x 2 1 In Exercises 23–26, find the limits. (a) lim f !x" x 2 ! 3x x x 2 xq \$ xq1 \$x 2 xq \$ lim cos x xq5\$%3 2 1 xq1 xq!1 x 1 2 xq1 xq0 3 68 CHAPTER 1 Limits and Their Properties In Exercises 45–48, find the limit of the function (if it exists). Write a simpler function that agrees with the given function at all but one point. Use a graphing utility to confirm your result. 45. lim xq!1 47. lim xq2 x2 ! 1 x#1 46. lim xq!1 x3 !8 x!2 48. lim 2x 2 ! x ! 3 x#1 x3 xq!1 #1 x#1 Graphical, Numerical, and Analytic Analysis In Exercises 79–82, use a graphing utility to graph the function and estimate the limit. Use a table to reinforce your conclusion. Then find the limit by analytic methods. sin 3t t 80. lim cos x ! 1 2x2 sin x 2 x xq0 82. lim sin x 3 & x 79. lim tq0 xq0 81. lim xq0 In Exercises 49–62, find the limit (if it exists). x!5 2 xq5 x ! 25 50. lim x2 # x ! 6 xq!3 x2 ! 9 52. lim 51. lim 53. lim &x # 5 ! &5 x xq0 55. lim 59. 61. 62. #xq0 x2 ! 5x # 4 xq4 x2 ! 2x ! 8 54. lim &2 # x ! &2 x xq0 &x # 5 ! 3 56. lim &x # 1 ! 2 x!4 x!3 xq3 #1%!3 # x"\$ ! !1%3" #1%!x # 4"\$ ! !1%4" 58. lim lim x x xq0 xq0 2!x # 'x" ! 2x !x # 'x"2 ! x 2 60. lim lim 'x 'x 'xq0 'xq0 2 2 !x # 'x" ! 2!x # ' x" # 1 ! !x ! 2x # 1" lim 'x 'xq0 !x # 'x"3 ! x3 lim 'x 'xq0 xq4 57. In Exercises 83–86, find lim 2!x 2 xq2 x ! 4 49. lim Graphical, Numerical, and Analytic Analysis In Exercises 63–66, use a graphing utility to graph the function and estimate the limit. Use a table to reinforce your conclusion. Then find the limit by analytic methods. 63. lim &x # 2 ! &2 x #1%!2 # x"\$ ! !1%2" 65. lim x xq0 xq0 66. lim xq2 x5 ! 32 x!2 68. lim 3!1 ! cos x" x 69. lim 70. lim cos & tan & & sin2 x 71. lim x xq0 72. lim tan2 x x xq0 sin x 5x xq0 sin x!1 ! cos x" xq0 2x2 & q0 xq0 !1 ! cos h"2 74. lim % sec % h hq0 %q\$ cos x 1 ! tan x 75. lim 76. lim xq \$%2 cot x xq \$%4 sin x ! cos x sin 3t 77. lim tq0 2t sin 2x 2 sin 2x 3x Hint: Find lim . 78. lim 2x 3 sin 3x xq0 sin 3x xq0 ' (' 4 x 86. f !x" " x 2 ! 4x In Exercises 87 and 88, use the Squeeze Theorem to find lim f x+. xqc 87. c " 0 4 ! x 2 f f !x" f 4 # x 2 88. c " a ) ) ) ) b ! x ! a f f !x" f b # x ! a In Exercises 89–94, use a graphing utility to graph the given function and the equations y ! x and y ! " x in the same viewing window. Using the graphs to observe the Squeeze Theorem visually, find lim f x+. )) )) xq0 ) )) ) 89. f !x" " x cos x 90. f !x" " x sin x 91. f !x" " x sin x 92. f !x" " x cos x )) 1 x 94. h!x" " x cos 1 x 95. In the context of finding limits, discuss what is meant by two functions that agree at all but one point. 96. Give an example of two functions that agree at all but one point. 97. What is meant by an indeterminate form? 98. In your own words, explain the Squeeze Theorem. 99. Writing 73. lim , 85. f !x" " 84. f !x" " &x In Exercises 67–78, determine the limit of the trigonometric function (if it exists). 67. lim 83. f !x" " 2x # 3 93. f !x" " x sin 4 ! &x 64. lim xq16 x ! 16 f x \$ #x+ " f x+ . #x (- Use a graphing utility to graph f !x" " x, g!x" " sin x, and h!x" " sin x x in the same viewing window. Compare the magnitudes of f !x" and g!x" when x is close to 0. Use the comparison to write a short paragraph explaining why lim h!x" " 1. xq0 SECTION 1.3 100. Writing sin2 x x in the same viewing window. Compare the magnitudes of f !x" and g!x" when x is close to 0. Use the comparison to write a short paragraph explaining why lim h!x" " 0. xq0 )x) " 1 x sin x "1 x 114. lim 115. If f !x" " g!x" for all real numbers other than x " 0, and Free-Falling Object In Exercises 101 and 102, use the position function st + ! "16t 2 \$ 1000, which gives the height (in feet) of an object that has fallen for t seconds from a height of 1000 feet. The velocity at time t ! a seconds is given by tqa 113. lim xq\$ xq0 lim 69 True or False? In Exercises 113–118, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. Use a graphing utility to graph f !x" " x, g!x" " sin2 x, and h!x" " Evaluating Limits Analytically lim f !x" " L, then lim g!x" " L. xq0 xq0 116. If lim f !x" " L, then f !c" " L. xqc 117. lim f !x" " 3, where f !x" " sa+ " st+ . a"t xq2 101. If a construction worker drops a wrench from a height of 1000 feet, how fast will the wrench be falling after 5 seconds? /3,0, x f 2 x > 2 118. If f !x" < g!x" for all x ( a, then lim f !x" < lim g!x". xqa xqa 102. If a construction worker drops a wrench from a height of 1000 feet, when will the wrench hit the ground? At what velocity will the wrench impact the ground? 119. Think About It Find a function f to show that the converse of Exercise 112(b) is not true. [Hint: Find a function f such that lim f !x" " L but lim f !x" does not exist.] Free-Falling Object In Exercises 103 and 104, use the position function st+ ! "4.9t 2 \$ 150, which gives the height (in meters) of an object that has fallen from a height of 150 meters. The velocity at time t ! a seconds is given by 120. Prove the second part of Theorem 1.9 by proving that xqc lim tqa sa+ " st+ . a"t lim xq0 ) /0,1, 104. At what velocity will the object impact the ground? 105. Find two functions f and g such that lim f !x" and lim g!x" do xq0 not exist, but lim # f !x" # g!x"\$ does exist. g!x" " /0,x, if x is rational if x is irrational. Find (if possible) lim f !x" and lim g!x". xq0 xq0 106. Prove that if lim f !x" exists and lim # f !x" # g!x"\$ does not xqc exist, then lim g!x" does not exist. 108. Prove Property 3 of Theorem 1.1. (You may use Property 3 of Theorem 1.2.) ) ) ) 110. Prove that if lim f !x" " 0, then lim f !x" " 0. ) 111. Prove that if lim f !x" " 0 and g!x" f M for a fixed number xqc xqc ) 112. (a) Prove that if lim f !x" " 0, then lim f !x" " 0. xqc (Note: This is the converse of Exercise 110.) ) ) )) Use the inequality 0 f !x") ! )L0 f ) f !x" ! L).\$ (b) Prove that if lim f !x" " L, then lim f !x" " L . #Hint: xqc sec x ! 1 . x2 (b) Use a graphing utility to graph f. Is the domain of f obvious from the graph? If not, explain. (c) Use the graph of f to approximate lim f !x". xqc (d) Confirm the answer in part (c) analytically. 123. Approximation (a) Find lim xq0 M and all x ( c, then lim f !x"g!x" " 0. ) xqc Consider f !x" " xq0 109. Prove Property 1 of Theorem 1.2. xqc 122. Graphical Reasoning xq0 (a) Find the domain of f. xqc 107. Prove Property 1 of Theorem 1.1. xqc if x is rational if x is irrational and 103. Find the velocity of the object when t " 3. xqc xqc 1 ! cos x " 0. x 121. Let f !x" " xq0 ) )) 1 ! cos x . x2 (b) Use the result in part (a) to derive the approximation 1 cos x . 1 ! 2x 2 for x near 0. (c) Use the result in part (b) to approximate cos!0.1". (d) Use a calculator to approximate cos!0.1" to four decimal places. Compare the result with part (c). 124. Think About It When using a graphing utility to generate a table to approximate lim #!sin x"%x\$, a student concluded that xq0 the limit was 0.01745 rather than 1. Determine the probable cause of the error. 70 CHAPTER 1 Limits and Their Properties Section 1.4 Continuity and One-Sided Limits • • • • Determine continuity at a point and continuity on an open interval. Determine one-sided limits and continuity on a closed interval. Use properties of continuity. Understand and use the Intermediate Value Theorem. Continuity at a Point and on an Open Interval E X P L O R AT I O N Informally, you might say that a function is continuous on an open interval if its graph can be drawn . with a pencil without lifting the pencil from the paper. Use a graphing utility to graph each function on the given interval. From the graphs, which functions would you say are continuous on the interval? Do you think you can trust the results you obtained graphically? Explain your reasoning. Function Interval a. y ! x2 \$ 1 !"3, 3" b. y ! 1 x"2 !"3, 3" c. y ! sin x x !" %, %" d. y ! x2 " 4 x\$2 !"3, 3" e. y ! %x \$ 1, !"3, 3" 2x " 4, x ≤ 0 x > 0 In mathematics, the term continuous has much the same meaning as it has in everyday usage. Informally, to say that a function f is continuous at x ! c means that there is no interruption in the graph of f at c. That is, its graph is unbroken at c and there are no holes, jumps, or gaps. Figure 1.25 identifies three values of x at which the graph of f is not continuous. At all other points in the interval !a, b", the graph of f is uninterrupted and continuous. Animation y y y lim f(x) f(c) is not defined. x→c does not exist. lim f (x) ≠ f (c) x→c x a c x b a c b x a c b Three conditions exist for which the graph of f is not continuous at x ! c. Figure 1.25 In Figure 1.25, it appears that continuity at x ! c can be destroyed by any one of the following conditions. 1. The function is not defined at x ! c. 2. The limit of f !x" does not exist at x ! c. 3. The limit of f !x" exists at x ! c, but it is not equal to f !c". If none of the above three conditions is true, the function f is called continuous at c, as indicated in the following important definition. FOR FURTHER INFORMATION For continuity, see the article “Leibniz and the Spell of the Continuous” by Hardy . in The College Mathematics Grant Journal. Definition of Continuity Continuity at a Point: conditions are met. A function f is continuous at c if the following three 1. f !c" is defined. 2. lim f !x" exists. x→c MathArticle 3. lim f !x" ! f !c". . Continuity on an Open Interval: A function is continuous on an open interval #a, b\$ if it is continuous at each point in the interval. A function that is continuous on the entire real line !" #, #" is everywhere continuous. x→c Video SECTION 1.4 y Continuity and One-Sided Limits 71 Consider an open interval I that contains a real number c. If a function f is defined on I (except possibly at c), and f is not continuous at c, then f is said to have a discontinuity at c. Discontinuities fall into two categories: removable and nonremovable. A discontinuity at c is called removable if f can be made continuous by appropriately defining (or redefining) f !c". For instance, the functions shown in Figure 1.26(a) and (c) have removable discontinuities at c and the function shown in Figure 1.26(b) has a nonremovable discontinuity at c. x a c b EXAMPLE 1 Continuity of a Function Discuss the continuity of each function. (a) Removable discontinuity a. f !x" ! y 1 x b. g!x" ! x2 " 1 x"1 c. h!x" ! %x x \$ 1, x ≤ 0 2 \$ 1, x > 0 d. y ! sin x Solution x a c b (b) Nonremovable discontinuity y a. The domain of f is all nonzero real numbers. From Theorem 1.3, you can conclude that f is continuous at every x-value in its domain. At x ! 0, f has a nonremovable discontinuity, as shown in Figure 1.27(a). In other words, there is no way to define f !0" so as to make the function continuous at x ! 0. b. The domain of g is all real numbers except x ! 1. From Theorem 1.3, you can conclude that g is continuous at every x-value in its domain. At x ! 1, the function has a removable discontinuity, as shown in Figure 1.27(b). If g!1" is defined as 2, the “newly defined” function is continuous for all real numbers. c. The domain of h is all real numbers. The function h is continuous on !" #, 0" and !0, #", and, because lim h!x" ! 1, h is continuous on the entire real line, as shown x→0 in Figure 1.27(c). d. The domain of y is all real numbers. From Theorem 1.6, you can conclude that the function is continuous on its entire domain, !" #, #", as shown in Figure 1.27(d). y y 3 x a c 3 1 f (x) = x 2 2 2 g(x) = x − 1 x −1 b Figure 1.26 1 1 (c) Removable discontinuity x −1 1 2 3 1 2 3 −1 (a) Nonremovable discontinuity at x ! 0 Editable Graph (b) Removable discontinuity at x ! 1 Editable Graph y y 3 h (x) = 1 x + 1, x ≤ 0 x 2 + 1, x > 0 π 2 x −1 y = sin x 1 2 Some people may refer to . the function in Example 1(a) as “discontinuous.” We have found that this terminology can be confusing. Rather than saying the function is discontinuous, we . prefer to say that it has a discontinuity at x ! 0. x −1 −1 . STUDY TIP (1, 2) 1 2 3 3π 2 x −1 −1 (c) Continuous on entire real line Editable Graph (d) Continuous on entire real line Editable Graph Figure 1.27 Try It Exploration A Exploration B Exploration C 72 CHAPTER 1 Limits and Their Properties One-Sided Limits and Continuity on a Closed Interval y To understand continuity on a closed interval, you first need to look at a different type of limit called a one-sided limit. For example, the limit from the right means that x approaches c from values greater than c [see Figure 1.28(a)]. This limit is denoted as x approaches c from the right. x c<x lim f !x" ! L. Limit from the right x→c \$ (a) Limit from right Similarly, the limit from the left means that x approaches c from values less than c [see Figure 1.28(b)]. This limit is denoted as y x approaches c from the left. lim f !x" ! L. Limit from the left x→c " x One-sided limits are useful in taking limits of functions involving radicals. For instance, if n is an even integer, c>x (b) Limit from left Figure 1.28 n x ! 0. lim ( x→0 \$ y EXAMPLE 2 Find the limit of f !x" ! (4 " x 2 as x approaches "2 from the right. 3 4 − x2 f (x) = A One-Sided Limit Solution As shown in Figure 1.29, the limit as x approaches "2 from the right is . lim (4 " x2 ! 0. 1 −2 x→"2\$ 1 2 −1 The limit of f !x" as x approaches " 2 from the. right is 0. Figure 1.29 One-sided limits can be used to investigate the behavior of step functions. One common type of step function is the greatest integer function &x', defined by &x' ! greatest integer n such that n ≤ x. Greatest integer function For instance, &2.5' ! 2 and &"2.5' ! "3. Editable Graph EXAMPLE 3 y Exploration A Try It x −1 The Greatest Integer Function Find the limit of the greatest integer function f !x" ! &x' as x approaches 0 from the left and from the right. f(x) = [[x]] 2 Solution As shown in Figure 1.30, the limit as x approaches 0 from the left is given by 1 lim &x' ! "1 −2 x −1 1 2 3 x→0" and the limit as x approaches 0 from the right is given by lim &x' ! 0. −2 . Greatest integer function Figure 1.30 Editable Graph x→0\$ The greatest integer function has a discontinuity at zero because the left and right limits at zero are different. By similar reasoning, you can see that the greatest integer function has a discontinuity at any integer n. Try It Exploration A Exploration B SECTION 1.4 Continuity and One-Sided Limits 73 When the limit from the left is not equal to the limit from the right, the (twosided) limit does not exist. The next theorem makes this more explicit. The proof of this theorem follows directly from the definition of a one-sided limit. THEOREM 1.10 The Existence of a Limit Let f be a function and let c and L be real numbers. The limit of f !x" as x approaches c is L if and only if lim f !x" ! L x→c" and lim f !x" ! L. x→c\$ The concept of a one-sided limit allows you to extend the definition of continuity to closed intervals. Basically, a function is continuous on a closed interval if it is continuous in the interior of the interval and exhibits one-sided continuity at the endpoints. This is stated formally as follows. y Definition of Continuity on a Closed Interval A function f is continuous on the closed interval [a, b] if it is continuous on the open interval !a, b" and lim f !x" ! f !a" x a x→a\$ b Continuous function on a closed interval Figure 1.31 and lim f !x" ! f !b". x→b" The function f is continuous from the right at a and continuous from the left at b (see Figure 1.31). Similar definitions can be made to cover continuity on intervals of the form !a, b* and )a, b" that are neither open nor closed, or on infinite intervals. For example, the function f !x" ! (x is continuous on the infinite interval )0, #", and the function g!x" ! (2 " x is continuous on the infinite interval !" #, 2. EXAMPLE 4 Continuity on a Closed Interval Discuss the continuity of f !x" ! (1 " x 2. Solution The domain of f is the closed interval )"1, 1. At all points in the open interval !"1, 1", the continuity of f follows from Theorems 1.4 and 1.5. Moreover, because y 1 f (x) = 1 − x2 lim (1 " x 2 ! 0 ! f !"1" x→"1\$ −1 x 1 and lim (1 " x 2 ! 0 ! f !1" x→1" . f is continuous on )" 1, 1. Figure 1.32 Editable Graph Continuous from the right Continuous from the left you can conclude that f is continuous on the closed interval )"1, 1, as shown in Figure 1.32. Try It Exploration A 74 CHAPTER 1 Limits and Their Properties The next example shows how a one-sided limit can be used to determine the value of absolute zero on the Kelvin scale. EXAMPLE 5 Charles’s Law and Absolute Zero On the Kelvin scale, absolute zero is the temperature 0 K. Although temperatures of approximately 0.0001 K have been produced in laboratories, absolute zero has never been attained. In fact, evidence suggests that absolute zero cannot be attained. How did scientists determine that 0 K is the “lower limit” of the temperature of matter? What is absolute zero on the Celsius scale? V Solution The determination of absolute zero stems from the work of the French physicist Jacques Charles (1746–1823). Charles discovered that the volume of gas at a constant pressure increases linearly with the temperature of the gas. The table illustrates this relationship between volume and temperature. In the table, one mole of hydrogen is held at a constant pressure of one atmosphere. The volume V is measured in liters and the temperature T is measured in degrees Celsius. 30 25 V = 0.08213T + 22.4334 15 10 (−273.15, 0) −300 −200 5 −100 100 T The volume of hydrogen gas depends on its. temperature. Figure 1.33 T "40 "20 0 20 40 60 80 V 19.1482 20.7908 22.4334 24.0760 25.7186 27.3612 29.0038 The points represented by the table are shown in Figure 1.33. Moreover, by using the points in the table, you can determine that T and V are related by the linear equation or V ! 0.08213T \$ 22.4334 Editable Graph T! V " 22.4334 . 0.08213 By reasoning that the volume of the gas can approach 0 (but never equal or go below 0) you can determine that the “least possible temperature” is given by V " 22.4334 V→0 0.08213 0 " 22.4334 ! 0.08213 + "273.15. lim\$T ! lim\$ V→0 . Use direct substitution. So, absolute zero on the Kelvin scale !0 K" is approximately "273.15& on the Celsius scale. Exploration A Try It The following table shows the temperatures in Example 5, converted to the Fahrenheit scale. Try repeating the solution shown in Example 5 using these temperatures and volumes. Use the result to find the value of absolute zero on the Fahrenheit scale. In 1995, physicists Carl Wieman and Eric Cornell of the University of Colorado at Boulder used lasers and evaporation to produce a supercold gas in which atoms overlap. This gas is called a Bose-Einstein condensate. “We get to within a billionth of a degree of absolute zero,”reported Wieman. (Source: Time magazine, April 10, 2000) T "40 "4 32 68 104 140 176 V 19.1482 20.7908 22.4334 24.0760 25.7186 27.3612 29.0038 NOTE Charles’s Law for gases (assuming constant pressure) can be stated as V ! RT Charles’s Law where V is volume, R is constant, and T is temperature. In the statement of this law, what property must the temperature scale have? SECTION 1.4 AUGUSTIN-LOUIS CAUCHY (1789–1857) . The concept of a continuous function was first introduced by Augustin-Louis Cauchy in 1821. The definition given in his text Cours d’Analyse stated that indefinite small changes in y were the result of indefinite small changes in x. “… f !x" will be called a continuous function if … the numerical values of the difference f !x \$ ( " " f !x" decrease indefinitely with those of ( ….” MathBio Continuity and One-Sided Limits 75 Properties of Continuity In Section 1.3, you studied several properties of limits. Each of those properties yields a corresponding property pertaining to the continuity of a function. For instance, Theorem 1.11 follows directly from Theorem 1.2. THEOREM 1.11 Properties of Continuity If b is a real number and f and g are continuous at x ! c, then the following functions are also continuous at c. 1. Scalar multiple: bf 2. Sum and difference: f ± g 3. Product: fg 4. Quotient: f , g if g!c" ' 0 The following types of functions are continuous at every point in their domains. p!x" ! anxn \$ an"1xn"1 \$ . . . \$ a1x \$ a0 p!x" 2. Rational functions: r!x" ! , q!x" ' 0 q!x" n x f !x" ! ( 4. Trigonometric functions: sin x, cos x, tan x, cot x, sec x, csc x 1. Polynomial functions: By combining Theorem 1.11 with this summary, you can conclude that a wide variety of elementary functions are continuous at every point in their domains. EXAMPLE 6 Applying Properties of Continuity By Theorem 1.11, it follows that each of the following functions is continuous at every point in its domain. . f !x" ! x \$ sin x, Try It f !x" ! 3 tan x, Exploration A f !x" ! x2 \$ 1 cos x Open Exploration The next theorem, which is a consequence of Theorem 1.5, allows you to determine the continuity of composite functions such as f !x" ! sin 3x, THEOREM 1.12 f !x" ! (x2 \$ 1, 1 f !x" ! tan . x Continuity of a Composite Function If g is continuous at c and f is continuous at g!c", then the composite function given by ! f & g"!x" ! f !g!x"" is continuous at c. One consequence of Theorem 1.12 is that if f and g satisfy the given conditions, you can determine the limit of f !g!x"" as x approaches c to be . lim f !g!x"" ! f !g!c"". x→c Technology 76 CHAPTER 1 Limits and Their Properties Testing for Continuity EXAMPLE 7 Describe the interval(s) on which each function is continuous. a. f !x" ! tan x b. g!x" ! sin 1 , x ' 0 x 0, x!0 % c. h!x" ! % x sin 1 , x ' 0 x 0, x!0 Solution a. The tangent function f !x" ! tan x is undefined at x! % \$ n%, 2 n is an integer. At all other points it is continuous. So, f !x" ! tan x is continuous on the open intervals . . . ., " /. /. / 3% % % % % 3% ," , " , , , ,. . . 2 2 2 2 2 2 as shown in Figure 1.34(a). b. Because y ! 1-x is continuous except at x ! 0 and the sine function is continuous for all real values of x, it follows that y ! sin !1-x" is continuous at all real values except x ! 0. At x ! 0, the limit of g!x" does not exist (see Example 5, Section 1.2). So, g is continuous on the intervals !" #, 0" and !0, #", as shown in Figure 1.34(b). c. This function is similar to that in part (b) except that the oscillations are damped by the factor x. Using the Squeeze Theorem, you obtain ,, " x ≤ x sin 1 ≤ x, x ,, x'0 and you can conclude that lim h!x" ! 0. x→0 y So, h is continuous on the entire real line, as shown in Figure 1.34(c). 4 3 y y 2 y = x  1 −π 1 π 1 x −3 x −1 1 x −1 1 −4 f(x) = tan x . f is continuous on each open interval in its (a) domain. . Editable Graph Figure 1.34 . −1 −1 g (x) = 1 sin x , x ≠ 0 x=0 0, (b) g is continuous on !" #, 0" and !0, #". Editable Graph Try It y = − x  h (x) = 0, x=0 (c) h is continuous on the entire real line. Editable Graph Exploration A x sin 1x , x ≠ 0 SECTION 1.4 Continuity and One-Sided Limits 77 The Intermediate Value Theorem Theorem 1.13 is an important theorem concerning the behavior of functions that are continuous on a closed interval. THEOREM 1.13 Intermediate Value Theorem If f is continuous on the closed interval )a, b* and k is any number between f !a" and f !b), then there is at least one number c in )a, b* such that . f !c" ! k. Video NOTE The Intermediate Value Theorem tells you that at least one c exists, but it does not give a method for finding c. Such theorems are called existence theorems. By referring to a text on advanced calculus, you will find that a proof of this theorem is based on a property of real numbers called completeness. The Intermediate Value Theorem states that for a continuous function f, if x takes on all values between a and b, f !x" must take on all values between f !a" and f !b". As a simple example of this theorem, consider a person’s height. Suppose that a girl is 5 feet tall on her thirteenth birthday and 5 feet 7 inches tall on her fourteenth birthday. Then, for any height h between 5 feet and 5 feet 7 inches, there must have been a time t when her height was exactly h. This seems reasonable because human growth is continuous and a person’s height does not abruptly change from one value to another. The Intermediate Value Theorem guarantees the existence of at least one number c in the closed interval )a, b. There may, of course, be more than one number c such that f !c" ! k, as shown in Figure 1.35. A function that is not continuous does not necessarily exhibit the intermediate value property. For example, the graph of the function shown in Figure 1.36 jumps over the horizontal line given by y ! k, and for this function there is no value of c in )a, b such that f !c" ! k. y y f (a) f (a) k k f (b) f (b) a c1 c2 c3 x b f is continuous on )a, b. [There exist three c’s such that f !c" ! k.] Figure 1.35 x a b f is not continuous on )a, b. [There are no c’s such that f !c" ! k.] Figure 1.36 The Intermediate Value Theorem often can be used to locate the zeros of a function that is continuous on a closed interval. Specifically, if f is continuous on )a, b* and f !a" and f !b" differ in sign, the Intermediate Value Theorem guarantees the existence of at least one zero of f in the closed interval )a, b. 78 CHAPTER 1 y Limits and Their Properties f (x) = x 3 + 2x − 1 An Application of the Intermediate Value Theorem Use the Intermediate Value Theorem to show that the polynomial function f !x" ! x 3 \$ 2x " 1 has a zero in the interval )0, 1. (1, 2) 2 EXAMPLE 8 Solution Note that f is continuous on the closed interval )0, 1. Because f !0" ! 0 3 \$ 2!0" " 1 ! "1 and 1 f !1" ! 13 \$ 2!1" " 1 ! 2 it follows that f !0" < 0 and f !1" > 0. You can therefore apply the Intermediate Value Theorem to conclude that there must be some c in )0, 1 such that . (c, 0) −1 x 1 f !c" ! 0 f has a zero in the closed interval )0, 1. as shown in Figure 1.37. −1 (0, −1) f is continuous on )0, 1 with f !0" < 0 and f !.1" > 0. Figure 1.37 Editable Graph Exploration A Try It The bisection method for approximating the real zeros of a continuous function is similar to the method used in Example 8. If you know that a zero exists in the closed interval )a, b, the zero must lie in the interval )a, !a \$ b"-2 or )!a \$ b"-2, b. From the sign of f !)a \$ b-2", you can determine which interval contains the zero. By repeatedly bisecting the interval, you can “close in” on the zero of the function. You can also use the zoom feature of a graphing utility to approximate the real zeros of a continuous function. By repeatedly zooming in on the point where the graph crosses the x-axis, and adjusting the x-axis scale, you can approximate the zero of the function to any desired accuracy. The zero of x3 \$ 2x " 1 is approximately 0.453, as shown in Figure 1.38. TECHNOLOGY 0.2 −0.2 0.013 1 0.4 −0.2 Figure 1.38 −0.012 Zooming in on the zero of f !x" ! x \$ 2x " 1 3 0.5 78 CHAPTER 1 Limits and Their Properties Exercises for Section 1.4 The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. Click on to view the complete solution of the exercise. Click on to print an enlarged copy of the graph. In Exercises 1–6, use the graph to determine the limit, and discuss the continuity of the function. (a) lim" f !x" (b) lim! f !x" xqc 1. xqc 2 x 1 2 2 3 4 (3, 1) (2, 2) 1 x 2 5. 4 (3, 0) c=3 2 2 1 3 (2, 2) 2 x 2 c=3 4 (2, 3) y 2 y c = 2 4 xqc c = 2 (3, 1) 1 4. y (c) lim f !x" 2. y 3. 6. y (4, 2) 4 3 c=4 x 1 3 (1, 2) 1 2 3 4 5 6 (4, 2) x 4 3 2 1 y 3 2 1 1 6 3 c = 1 2 (1, 0) x 1 SECTION 1.4 In Exercises 7–24, find the limit (if it exists). If it does not exist, explain why. 7. lim# x"5 x2 " 25 8. lim# 2"x x2 " 4 xq5 xq2 9. 10. lim" 16. 17. 18. 3 &x " 2& x"2 30. f #t\$ ! 3 " )9 " t % 31. f #x\$ ! %3 # 32. g#x\$ ! 1 x2 " 4 3 " x, 1 2 x, 2 x f 0 x > 0 33. f #x\$ ! x 2 " 2x # 1 34. f #x\$ ! 1 x2 # 1 35. f #x\$ ! 3x " cos x 36. f #x\$ ! cos 19. lim cot x xq \$ \$x 2 20. lim sec x 37. f #x\$ ! x x2 " x 21. lim" #3x+ " 5\$ 38. f #x\$ ! x x2 " 1 39. f #x\$ ! x x2 # 1 40. f #x\$ ! x"3 x2 " 9 41. f #x\$ ! x#2 x 2 " 3x " 10 42. f #x\$ ! x"1 x2 # x " 2 xq \$02 xq4 22. lim##2x " x+\$ xq2 23. lim #2 " "x+ \$ xq3 - ./ x 24. lim 1 " " xq1 2 In Exercises 25–28, discuss the continuity of each function. x2 " 1 26. f #x\$ ! x#1 1 25. f #x\$ ! 2 x "4 y y 3 2 1 3 2 1 x 1 3 3 2 1 3 x 1 2 '"5, 5( '"3, 3( '"1, 4( '"1, 2( In Exercises 33–54, find the x-values (if any) at which f is not continuous. Which of the discontinuities are removable? % % % 1 2 3 3 Interval 29. g#x\$ ! )25 " x 2 #x # % x\$2 # x # % x " #x 2 # x\$ %x %xq0 x#2 , x f 3 2 lim" f #x\$, where f #x\$ ! 12 " 2x xq3 , x > 3 3 x2 " 4x # 6, x < 2 lim f #x\$, where f #x\$ ! "x2 # 4x " 2, x v 2 xq2 x3 # 1, x < 1 lim f #x\$, where f #x\$ ! x # 1, x v 1 xq1 x, x f 1 lim f #x\$, where f #x\$ ! 1 " x, x > 1 xq1# 3 1 2 2 3 Function lim # , x 3 2 In Exercises 29–32, discuss the continuity of the function on the closed interval. 1 1 " x # %x x 13. lim " %x %xq0 15. 1 2 3 && 14. x 3 2 1 )x2 " 9 x"4 xq2 3 2 1 3 2 1 x 11. lim" x xq0 12. lim# y y )x " 2 xq4 % x, x < 1 x!1 28. f #x\$ ! 2, 2x " 1, x > 1 1 27. f #x\$ ! 2x+ # x x lim xq"3" 79 Continuity and One-Sided Limits 3 43. f #x\$ ! &x # 2& 44. f #x\$ ! &x " 3& x#2 x"3 %x,x , xx >f 11 "2x # 3, x < 1 46. f #x\$ ! % x , x v 1 45. f #x\$ ! 2 2 80 CHAPTER 1 47. f #x\$ ! % 1 2x Limits and Their Properties In Exercises 65–68, use a graphing utility to graph the function. Use the graph to determine any x-values at which the function is not continuous. # 1, x f 2 3 " x, x > 2 % "2x, x f 2 48. f #x\$ ! 2 x " 4x # 1, x > 2 65. f #x\$ ! x+ " x tan \$ x, 4 49. f #x\$ ! x, &x& < 1 &x& v 1 67. g#x\$ ! %x csc \$ x , 6 50. f #x\$ ! 2, &x " 3& f 2 &x " 3& > 2 68. f #x\$ ! % % % 51. f #x\$ ! csc 2x 69. f #x\$ ! 53. f #x\$ ! x " 1+ 54. f #x\$ ! 3 " x+ x#2 x 2 # 4x #x # 2\$ 56. f #x\$ ! x#4 % 2 2 (3, 0) x 4 2 2 4 3 4 4 \$x 4 72. f #x\$ ! x#1 )x y 4 4 3 x 2 2 1 x 4 % % % 1 2 Writing In Exercises 73 and 74, use a graphing utility to graph the function on the interval [!4, 4]. Does the graph of the function appear continuous on this interval? Is the function continuous on [!4, 4]? Write a short paragraph about the importance of examining a function analytically as well as graphically. 2, x f "1 59. f #x\$ ! ax # b, "1 < x < 3 "2, x v 3 x2 " a2 , x&a 60. g #x\$ ! x " a 8, x!a 73. f #x\$ ! In Exercises 61– 64, discuss the continuity of the composite function h!x" # f ! g!x"". g#x\$ ! x 2 # 5 1 1 2 2 4 sin x , x < 0 x a " 2x, x v 0 1 63. f #x\$ ! x"6 4 y x3, x f 2 57. f #x\$ ! ax 2, x > 2 g #x\$ ! x " 1 y 1 71. f #x\$ ! sec & 61. f #x\$ ! x 2 70. f #x\$ ! x)x # 3 2 In Exercises 57–60, find the constant a, or the constants a and b, such that the function is continuous on the entire real line. 58. g#x\$ ! x x2 # 1 x xq0! x 2 " 4&x f #x\$ ! & & cos x " 1 , x < 0 x 5x, x v 0 lim f !x". and Is the function continuous on the entire real line? Explain. 55. 2x " 4, x f 3 2 " 2x, x > 3 y In Exercises 55 and 56, use a graphing utility to graph the function. From the graph, estimate lim f !x" 1 x2 " x " 2 In Exercises 69–72, describe the interval(s) on which the function is continuous. \$x 52. f #x\$ ! tan 2 xq0" 66. h#x\$ ! 62. f #x\$ ! 1 )x g #x\$ ! x " 1 64. f #x\$ ! sin x g #x\$ ! x2 sin x x 74. f #x\$ ! x3 " 8 x"2 Writing In Exercises 75–78, explain why the function has a zero in the given interval. Interval Function 75. f #x\$ ! 1 4 16 x 3 "x #3 76. f #x\$ ! x3 # 3x " 2 77. f #x\$ ! x 2 " 2 " cos x , / 4 \$x 78. f #x\$ ! " # tan x 8 '1, 2( '0, 1( '0, \$( '1, 3( SECTION 1.4 In Exercises 79–82, use the Intermediate Value Theorem and a graphing utility to approximate the zero of the function in the interval [0, 1]. Repeatedly “zoom in” on the graph of the function to approximate the zero accurate to two decimal places. Use the zero or root feature of the graphing utility to approximate the zero accurate to four decimal places. Continuity and One-Sided Limits 81 89. Sketch the graph of any function f such that lim f #x\$ ! 1 lim f #x\$ ! 0. and xq3# xq3" Is the function continuous at x ! 3? Explain. 79. f #x\$ ! x3 # x " 1 90. If the functions f and g are continuous for all real x, is f # g always continuous for all real x? Is f0g always continuous for all real x? If either is not continuous, give an example to 80. f #x\$ ! x3 # 3x " 2 81. g#t\$ ! 2 cos t " 3t 82. h#(\$ ! 1 # ( " 3 tan ( In Exercises 83–86, verify that the Intermediate Value Theorem applies to the indicated interval and find the value of c guaranteed by the theorem. 83. f #x\$ ! x 2 # x " 1, f #c\$ ! 11 '0, 5(, 84. f #x\$ ! " 6x # 8, '0, 3(, f #c\$ ! 0 85. f #x\$ ! x3 " x 2 # x " 2, f #c\$ ! 4 '0, 3(, 2 x #x 5 86. f #x\$ ! f #c\$ ! 6 , ,4 , x"1 2 x2 True or False? In Exercises 91–94, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 91. If lim f #x\$ ! L and f #c\$ ! L, then f is continuous at c. xqc 92. If f #x\$ ! g#x\$ for x & c and f #c\$ & g#c\$, then either f or g is not continuous at c. 1 2 93. A rational function can have infinitely many x-values at which it is not continuous. & 87. State how continuity is destroyed at x ! c for each of the following graphs. (a) y (b) y & 94. The function f #x\$ ! x " 1 0#x " 1\$ is continuous on #" ', '\$. 95. Swimming Pool Every day you dissolve 28 ounces of chlorine in a swimming pool. The graph shows the amount of chlorine f #t\$ in the pool after t days. y 140 112 84 c x c x 56 28 t (c) (d) y 1 y 2 3 4 5 6 7 lim" f #t\$ and lim# f #t\$. Estimate and interpret tq4 tq4 Describe how the functions f #x\$ ! 3 # x+ and c x c x 88. Describe the difference between a discontinuity that is removable and one that is nonremovable. In your explanation, give examples of the following descriptions. (a) A function with a nonremovable discontinuity at x ! 2 (b) A function with a removable discontinuity at x ! "2 (c) A function that has both of the characteristics described in parts (a) and (b) g#x\$ ! 3 " "x+ differ. 97. Telephone Charges A dial-direct long distance call between two cities costs \$1.04 for the first 2 minutes and \$0.36 for each additional minute or fraction thereof. Use the greatest integer function to write the cost C of a call in terms of time t (in minutes). Sketch the graph of this function and discuss its continuity. 82 CHAPTER 1 Limits and Their Properties 98. Inventory Management The number of units in inventory in a small company is given by , -t #2 2. " t/ N#t\$ ! 25 2 where t is the time in months. Sketch the graph of this function and discuss its continuity. How often must this company replenish its inventory? 99. Déjà Vu At 8:00 A.M. on Saturday a man begins running up the side of a mountain to his weekend campsite (see figure). On Sunday morning at 8:00 A.M. he runs back down the mountain. It takes him 20 minutes to run up, but only 10 minutes to run down. At some point on the way down, he realizes that he passed the same place at exactly the same time on Saturday. Prove that he is correct. [Hint: Let s#t\$ and r #t\$ be the position functions for the runs up and down, and apply the Intermediate Value Theorem to the function f #t\$ ! s#t\$ " r #t\$.] 105. Modeling Data After an object falls for t seconds, the speed S (in feet per second) of the object is recorded in the table. t 0 5 10 15 20 25 30 S 0 48.2 53.5 55.2 55.9 56.2 56.3 (a) Create a line graph of the data. (b) Does there appear to be a limiting speed of the object? If there is a limiting speed, identify a possible cause. 106. Creating Models A swimmer crosses a pool of width b by swimming in a straight line from #0, 0\$ to #2b, b\$. (See figure.) (a) Let f be a function defined as the y-coordinate of the point on the long side of the pool that is nearest the swimmer at any given time during the swimmer’s path across the pool. Determine the function f and sketch its graph. Is it continuous? Explain. (b) Let g be the minimum distance between the swimmer and the long sides of the pool. Determine the function g and sketch its graph. Is it continuous? Explain. y (2b, b) b Not drawn to scale Saturday 8:00 A.M. Sunday 8:00 A.M. 100. Volume Use the Intermediate Value Theorem to show that for all spheres with radii in the interval '1, 5(, there is one with a volume of 275 cubic centimeters. 101. Prove that if f is continuous and has no zeros on 'a, b(, then either f #x\$ > 0 for all x in 'a, b( or f #x\$ < 0 for all x in 'a, b(. %0,1, if x is rational if x is irrational 103. Show that the function % is continuous only at x ! 0. (Assume that k is any nonzero real number.) 104. The signum function is defined by % "1, x < 0 sgn#x\$ ! 0, x!0 1, x > 0. 2 x f c x > c 108. Prove that for any real number y there exists x in #" \$02, \$02\$ such that tan x ! y. (b) lim# sgn#x\$ xq0 112. (a) Let f1#x\$ and f2#x\$ be continuous on the closed interval 'a, b(. If f1#a\$ < f2#a\$ and f1#b\$ > f2#b\$, prove that there exists c between a and b such that f1#c\$ ! f2#c\$. (b) Show that there exists c in '0, \$2( such that cos x ! x. Use a graphing utility to approximate c to three decimal places. Putnam Exam Challenge Sketch a graph of sgn#x\$ and find the following (if possible). xq0 %x,1 " x , 111. Discuss the continuity of the function h#x\$ ! x x+. if x is rational if x is irrational (a) lim" sgn#x\$ f #x\$ ! 110. Prove that if lim f #c # % x\$ ! f #c\$, then f is continuous %xq0 at c. is not continuous at any real number. 0, f #x\$ ! kx, 107. Find all values of c such that f is continuous on #" ', '\$. 109. Let f #x\$ ! #)x # c2 " c\$0x, c > 0. What is the domain of f ? How can you define f at x ! 0 in order for f to be continuous there? 102. Show that the Dirichlet function f #x\$ ! x (0, 0) (c) lim sgn#x\$ xq0 113. Prove or disprove: if x and y are real numbers with y v 0 and y# y # 1\$ f #x # 1\$2, then y# y " 1\$ f x2. 114. Determine all polynomials P#x\$ such that P#x2 # 1\$ ! #P#x\$\$2 # 1 and P#0\$ ! 0. These problems were composed by the Committee on the Putnam Prize Competition. SECTION 1.5 Section 1.5 83 Infinite Limits Infinite Limits • Determine infinite limits from the left and from the right. • Find and sketch the vertical asymptotes of the graph of a function. Infinite Limits y Let f be the function given by 3 → ∞, x−2 as x → 2+ 6 4 2 x −6 −4 4 6 f !x" ! From Figure 1.39 and the table, you can see that f !x" decreases without bound as x approaches 2 from the left, and f !x" increases without bound as x approaches 2 from the right. This behavior is denoted as −2 3 → −∞, −4 x−2 as x → 2− −6 lim\$ 3 ! \$" x\$2 f !x" decreases without bound as x approaches 2 from the left. lim 3 ! x\$2 " f !x" increases without bound as x approaches 2 from the right. x→2 3 f(x) = x−2 3 . x\$2 and f !x" increases and decreases without bound as x approaches 2. x→2 # Figure 1.39 x approaches 2 from the right. x approaches 2 from the left. x 1.5 1.9 1.99 1.999 2 2.001 2.01 2.1 2.5 f #x\$ \$6 \$30 \$300 \$3000 ? 3000 300 30 6 f !x" decreases without bound. f !x" increases without bound. A limit in which f !x" increases or decreases without bound as x approaches c is called an infinite limit. Definition of Infinite Limits Let f be a function that is defined at every real number in some open interval containing c (except possibly at c itself). The statement lim f !x" ! " x→c means that for each M > 0 there exists a % > 0 such that f !x" > M whenever 0 < x \$ c < % (see Figure 1.40). Similarly, the statement % y lim f !x" ! \$ " x→c lim f (x) = ∞ means that for each N < 0 there exists a % > 0 such that f !x" < N whenever 0 < x \$ c < %. To define the infinite limit from the left, replace 0 < x \$ c < % by c \$ % < x < c. To define the infinite limit from the right, replace 0 < x \$ c < % by c < x < c # %. x→c % % M δ δ . c Infinite limits Figure 1.40 % x % % % % Video Be sure you see that the equal sign in the statement lim f !x" ! " does not mean that the limit exists! On the contrary, it tells you how the limit fails to exist by denoting the unbounded behavior of f !x" as x approaches c. 84 CHAPTER 1 Limits and Their Properties E X P L O R AT I O N Use a graphing utility to graph each function. For each function, analytically find the single real number c that is not in the domain. Then graphically find the limit of f !x" as x approaches c from the left and from the right. 3 x\$4 2 c. f !x" ! !x \$ 3" 2 1 2\$x \$3 d. f !x" ! !x # 2" 2 a. f !x" ! EXAMPLE 1 b. f !x" ! Determining Infinite Limits from a Graph Use Figure 1.41 to determine the limit of each function as x approaches 1 from the left and from the right. y y 2 3 1 2 x −1 2 . −3 (a) f(x) = 1 x−1 −1 −2 x 2 −1 f (x) = −2 −1 3 1 (x − 1) 2 (b) Editable Graph y 2 1 3 −2 −2 y f (x) = 2 −1 x−1 −1 (x − 1) 2 1 x 2 −1 −2 −1 x −1 −2 −2 −3 −3 (c) Editable Graph f(x) = 2 (d) Editable Graph Editable Graph Figure 1.41 Each graph has an asymptote at x ! 1. Solution 1 ! \$" x\$1 1 b. lim !" x→1 !x \$ 1" 2 \$1 c. lim\$ !" x→1 x \$ 1 a. lim\$ and x→1 . d. lim x→1 1 ! x\$1 " Limit from each side is ". and \$1 ! \$" !x \$ 1" 2 Try It lim x→1 # lim x→1 # \$1 ! \$" x\$1 Limit from each side is \$ ". Exploration A Vertical Asymptotes If it were possible to extend the graphs in Figure 1.41 toward positive and negative infinity, you would see that each graph becomes arbitrarily close to the vertical line x ! 1. This line is a vertical asymptote of the graph of f. (You will study other types of asymptotes in Sections 3.5 and 3.6.) NOTE If the graph of a function f has a vertical asymptote at x ! c, then f is not continuous at c. Definition of Vertical Asymptote If f !x" approaches infinity (or negative infinity) as x approaches c from the right or the left, then the line x ! c is a vertical asymptote of the graph of f. SECTION 1.5 Infinite Limits 85 In Example 1, note that each of the functions is a quotient and that the vertical asymptote occurs at a number where the denominator is 0 (and the numerator is not 0). The next theorem generalizes this observation. (A proof of this theorem is given in Appendix A.) THEOREM 1.14 Vertical Asymptotes Let f and g be continuous on an open interval containing c. If f !c" ' 0, g!c" ! 0, and there exists an open interval containing c such that g!x" ' 0 for all x ' c in the interval, then the graph of the function given by h !x" ! . has a vertical asymptote at x ! c. y f (x) = 1 2(x + 1) 2 Video 1 x −1 EXAMPLE 2 1 −1 a. f !x" ! (a) Editable Graph f ((x) = x +1 x2 − 1 f !x" ! 4 x −2 2 4 (b) Editable Graph y f (x) = cot x 6 2 π 2π x −4 −6 . (c) Editable Graph Functions with vertical asymptotes Figure 1.42 c. f !x" ! cot x 1 2!x # 1" x2 # 1 x2 # 1 ! x 2 \$ 1 !x \$ 1"!x # 1" you can see that the denominator is 0 at x ! \$1 and x ! 1. Moreover, because the numerator is not 0 at these two points, you can apply Theorem 1.14 to conclude that the graph of f has two vertical asymptotes, as shown in Figure 1.42(b). c. By writing the cotangent function in the form f !x" ! cot x ! 4 −2π x2 # 1 x2 \$ 1 is 0 and the numerator is not 0. So, by Theorem 1.14, you can conclude that x ! \$1 is a vertical asymptote, as shown in Figure 1.42(a). b. By factoring the denominator as f !x" ! . . b. f !x" ! a. When x ! \$1, the denominator of 2 −4 1 2!x # 1" Solution y 2 Finding Vertical Asymptotes Determine all vertical asymptotes of the graph of each function. −2 . f !x" g!x" cos x sin x you can apply Theorem 1.14 to conclude that vertical asymptotes occur at all values of x such that sin x ! 0 and cos x ' 0, as shown in Figure 1.42(c). So, the graph of this function has infinitely many vertical asymptotes. These asymptotes occur when x ! n&, where n is an integer. Try It Exploration A Exploration B Open Exploration Theorem 1.14 requires that the value of the numerator at x ! c be nonzero. If both the numerator and the denominator are 0 at x ! c, you obtain the indeterminate form 0&0, and you cannot determine the limit behavior at x ! c without further investigation, as illustrated in Example 3. 86 CHAPTER 1 Limits and Their Properties EXAMPLE 3 A Rational Function with Common Factors Determine all vertical asymptotes of the graph of f (x) = f !x" ! x 2 + 2x − 8 x2 − 4 Solution Begin by simplifying the expression, as shown. y 4 x 2 # 2x \$ 8 x2 \$ 4 !x # 4"!x \$ 2" ! !x # 2"!x \$ 2" x#4 ! , x'2 x#2 f !x" ! Undefined when x = 2 2 −4 2 −2 x 2 # 2x \$ 8 . x2 \$ 4 x Vertical asymptote at x = − 2 f !x" increases and decreases without bound as.x approaches \$ 2. Figure 1.43 At all x-values other than x ! 2, the graph of f coincides with the graph of g!x" ! !x # 4"&!x # 2". So, you can apply Theorem 1.14 to g to conclude that there is a vertical asymptote at x ! \$2, as shown in Figure 1.43. From the graph, you can see that Editable Graph . lim \$ x→\$2 x 2 # 2x \$ 8 ! \$" x2 \$ 4 and lim # x→\$2 x 2 # 2x \$ 8 ! ". x2 \$ 4 Note that x ! 2 is not a vertical asymptote. Exploration A Try It EXAMPLE 4 Exploration B Determining Infinite Limits Find each limit. f(x) = 6 −4 lim\$ x→1 x 2 − 3x x−1 and lim# x→1 x 2 \$ 3x x\$1 Solution Because the denominator is 0 when x ! 1 (and the numerator is not zero), you know that the graph of 6 −6 f .has a vertical asymptote at x ! 1. Figure 1.44 x 2 \$ 3x x\$1 f !x" ! x 2 \$ 3x x\$1 has a vertical asymptote at x ! 1. This means that each of the given limits is either " or \$ ". A graphing utility can help determine the result. From the graph of f shown in Figure 1.44, you can see that the graph approaches " from the left of x ! 1 and approaches \$ " from the right of x ! 1. So, you can conclude that Editable Graph lim\$ x 2 \$ 3x !" x\$1 The limit from the left is infinity. lim# x2 \$ 3x ! \$". x\$1 The limit from the right is negative infinity. x→1 and . x→1 Try It Exploration A TECHNOLOGY PITFALL When using a graphing calculator or graphing software, be careful to interpret correctly the graph of a function with a vertical asymptote—graphing utilities often have difficulty drawing this type of graph. SECTION 1.5 THEOREM 1.15 Infinite Limits 87 Properties of Infinite Limits Let c and L be real numbers and let f and g be functions such that lim f !x" ! " lim g!x" ! L. and x→c x→c 1. Sum or difference: lim ) f !x" ± g!x"* ! " x→c lim ) f !x"g!x"* ! ", 2. Product: L > 0 x→c lim ) f !x"g!x"* ! \$ ", x→c L < 0 g!x" !0 f !x" Similar properties hold for one-sided limits and for functions for which the limit of f !x" as x approaches c is \$ ". 3. Quotient: lim x→c Proof To show that the limit of f !x" # g!x" is infinite, choose M > 0. You then need to find % > 0 such that ) f !x" # g!x"* > M % % % % whenever 0 < x \$ c < %. For simplicity’s sake, you can assume L is positive. Let M1 ! M # 1. Because the limit of f !x" is infinite, there exists %1 such that f !x" > M1 whenever 0 < x \$ c < %1. Also, because the limit of g!x" is L, there exists % 2 such that g!x" \$ L < 1 whenever 0 < x \$ c < %2. By letting % be the smaller of %1 and % 2, you can conclude that 0 < x \$ c < % implies f !x" > M # 1 and g!x" \$ L < 1. The second of these two inequalities implies that g!x" > L \$ 1, and, adding this to the first inequality, you can write % % % % % % % % f !x" # g!x" > !M # 1" # !L \$ 1" ! M # L > M. So, you can conclude that lim ) f !x" # g!x"* ! ". x→c The proofs of the remaining properties are left as exercises (see Exercise 72). Determining Limits EXAMPLE 5 a. Because lim 1 ! 1 and lim x→0 ' lim 1 # x→0 x→0 1 ! ", you can write x2 ( 1 ! ". x2 Property 1, Theorem 1.15 b. Because lim\$ !x 2 # 1" ! 2 and lim\$ !cot & x" ! \$ ", you can write x→1 lim\$ x→1 x→1 2 x #1 ! 0. cot & x Property 3, Theorem 1.15 c. Because lim# 3 ! 3 and lim# cot x ! ", you can write x→0 . x→0 lim 3 cot x ! ". x→0 # Try It Property 2, Theorem 1.15 Exploration A 88 CHAPTER 1 Limits and Their Properties Exercises for Section 1.5 The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. Click on to view the complete solution of the exercise. Click on to print an enlarged copy of the graph. In Exercises 1–4, determine whether f 'x( approaches # or "# as x approaches "2 from the left and from the right. ) ) 1. f !x" # 2 x2 x "4 2. f !x" # 1 x!2 y 2 2 1 x 2 2 3. f !x" # tan \$x 4 4. f !x" # sec 25. f !x" # 27. s!t" # \$x 4 1 x 2 2 6 6 2 x 2 6 Numerical and Graphical Analysis In Exercises 5–8, determine whether f 'x( approaches # or "# as x approaches "3 from the left and from the right by completing the table. Use a x "3.5 "3.1 "3.01 "3.001 "2.99 "2.9 t sin t x2 x "9 2 "2.5 x 6. f !x" # 2 x "9 8. f !x" # sec \$x 6 In Exercises 9–28, find the vertical asymptotes (if any) of the graph of the function. 1 9. f !x" # 2 x 4 10. f !x" # !x " 2"3 x2 " 2 11. h!x" # 2 x "x"2 x2 x2 "4 2!x 12. g!x" # 2 x !1 " x" 14. f !x" # "4x x2 ! 4 t"1 15. g!t" # 2 t !1 2s " 3 16. h!s" # 2 s " 25 17. f !x" # tan 2x 18. f !x" # sec \$ x x2 " 4 x 3 ! 2x 2 ! x ! 2 26. h!t" # t 2 " 2t t 4 " 16 28. g!%" # tan % % 30. f !x" # x 2 " 6x " 7 x!1 31. f !x" # x2 ! 1 x!1 32. f !x" # sin!x ! 1" x!1 In Exercises 33–48, find the limit. 33. lim! x"3 x"2 34. lim! 2!x 1"x 35. lim! x2 x2 " 9 36. lim" x2 x 2 ! 16 xq2 lim " xq"3 xq1 x 2 ! 2x " 3 x2 ! x " 6 % 41. lim" 1 ! 1 5. f !x" # 2 x "9 13. f !x" # x2 " 2x " 15 " 5x2 ! x " 5 x2 " x 39. lim 2 xq1 !x ! 1"!x " 1" f 'x( 7. f !x" # x3 24. h!x" # x2 " 1 x!1 37. "2.999 " x 2 " 4x " 6x " 24 3x 2 29. f !x" # xq3 f 'x( x 1 3 2x In Exercises 29–32, determine whether the graph of the function has a vertical asymptote or a removable discontinuity at x ! "1. Graph the function using a graphing utility to confirm your y 3 2 1 x3 ! 1 x!1 x 1 20. g!x" # 4x 2 ! 4x " 24 " 2x 3 " 9x 2 ! 18x x4 23. g!x" # 2 3 4 y 6 22. f !x" # 3 2 4 4 t2 x x2 ! x " 2 21. f !x" # y 6 19. T !t" # 1 " xq0 1 x & xq4 38. lim xq !"1#2" 40. lim xq3 44. 45. lim \$x csc x 46. lim 47. lim x sec \$ x xq1#2 x"2 x2 xq0 2 sin x xq \$ 6x 2 ! x " 1 4x 2 " 4x " 3 % 42. lim" x 2 " 43. lim! xq0 ! lim xq !\$#2" ! xq0 1 x & "2 cos x x!2 cot x 48. lim x 2 tan \$ x xq1#2 In Exercises 49–52, use a graphing utility to graph the function and determine the one-sided limit. 49. f !x" # x2 ! x ! 1 x3 " 1 lim f !x" xq1 ! 51. f !x" # 1 x 2 " 25 lim f !x" xq5 " 50. f !x" # x3 " 1 x !x!1 2 lim f !x" xq1 " 52. f !x" # sec lim f !x" xq3 ! \$x 6 SECTION 1.5 53. In your own words, describe the meaning of an infinite limit. Is & a real number? 54. In your own words, describe what is meant by an asymptote of a graph. 55. Write a rational function with vertical asymptotes at x # 6 and x # "2, and with a zero at x # 3. 56. Does the graph of every rational function have a vertical asymptote? Explain. 57. Use the graph of the function f (see figure) to sketch the graph of g!x" # 1#f !x" on the interval "2, 3+. To print an enlarged copy of the graph, select the MathGraph button. 89 61. Relativity According to the theory of relativity, the mass m of a particle depends on its velocity v. That is, m# m0 \$1 " !v2#c2" where m0 is the mass when the particle is at rest and c is the speed of light. Find the limit of the mass as v approaches c " . 62. Rate of Change A 25-foot ladder is leaning against a house (see figure). If the base of the ladder is pulled away from the house at a rate of 2 feet per second, the top will move down the wall at a rate of r# y 2x \$625 " x2 ft/sec where x is the distance between the base of the ladder and the house. 2 f x 2 1 1 1 2 3 (a) Find the rate r when x is 7 feet. (b) Find the rate r when x is 15 feet. (c) Find the limit of r as x q 25 " . 58. Boyle’s Law For a quantity of gas at a constant temperature, the pressure P is inversely proportional to the volume V. Find the limit of P as V q 0 ! . 59. Rate of Change A patrol car is parked 50 feet from a long warehouse (see figure). The revolving light on top of the car turns at a rate of 12 revolution per second. The rate at which the light beam moves along the wall is r # 50\$ Infinite Limits sec2 % ft/sec. (a) Find the rate r when % is \$#6. (b) Find the rate r when % is \$#3. (c) Find the limit of r as % q !\$#2" " . 25 ft r ft 2 sec 63. Average Speed On a trip of d miles to another city, a truck driver’s average speed was x miles per hour. On the return trip the average speed was y miles per hour. The average speed for the round trip was 50 miles per hour. (a) Verify that y # 25x . What is the domain? x " 25 (b) Complete the table. x V 50 ft 40 50 60 y x 60. Illegal Drugs The cost in millions of dollars for a governmental agency to seize x% of an illegal drug is C# 30 528x , 0 f x < 100. 100 " x Are the values of y different than you expected? Explain. (c) Find the limit of y as x q 25 ! and interpret its meaning. 64. Numerical and Graphical Analysis Use a graphing utility to complete the table for each function and graph each function to estimate the limit. What is the value of the limit when the power on x in the denominator is greater than 3? (a) Find the cost of seizing 25% of the drug. x (b) Find the cost of seizing 50% of the drug. f 'x( 1 (c) Find the cost of seizing 75% of the drug. (d) Find the limit of C as x q 100 " and interpret its meaning. 0.5 0.2 0.1 0.01 0.001 (a) lim! x " sin x x (b) lim! x " sin x x2 (c) lim! x " sin x x3 (d) lim! x " sin x x4 xq0 xq0 xq0 xq0 0.0001 90 CHAPTER 1 Limits and Their Properties 65. Numerical and Graphical Analysis Consider the shaded region outside the sector of a circle of radius 10 meters and inside a right triangle (see figure). (a) Write the area A # f !% " of the region as a function of %. Determine the domain of the function. (b) Use a graphing utility to complete the table and graph the function over the appropriate domain. 0.3 % 0.6 0.9 1.2 (d) Use a graphing utility to complete the table. & 0.3 0.9 1.2 1.5 L (e) Use a graphing utility to graph the function over the appropriate domain. (f) Find 1.5 0.6 f '%( lim ) q !\$#2" " L. Use a geometric argument as the basis of a second method of finding this limit. (g) Find lim! L. (c) Find the limit of A as % q !\$#2"". ) q0 True or False? In Exercises 67–70, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. V 10 m 66. Numerical and Graphical Reasoning A crossed belt connects a 20-centimeter pulley (10-cm radius) on an electric motor with a 40-centimeter pulley (20-cm radius) on a saw arbor (see figure). The electric motor runs at 1700 revolutions per minute. 20 cm 10 cm K 67. If p!x" is a polynomial, then the graph of the function given by p!x" has a vertical asymptote at x # 1. f !x" # x"1 68. The graph of a rational function has at least one vertical asymptote. 69. The graphs of polynomial functions have no vertical asymptotes. 70. If f has a vertical asymptote at x # 0, then f is undefined at x # 0. 71. Find functions f and g such that lim f !x" # & and xqc lim g!x" # & but lim f !x" " g!x"+ ( 0. xqc xqc 72. Prove the remaining properties of Theorem 1.15. 73. Prove that if lim f !x" # &, then lim xqc (a) Determine the number of revolutions per minute of the saw. (b) How does crossing the belt affect the saw in relation to the motor? (c) Let L be the total length of the belt. Write L as a function of ), where ) is measured in radians. What is the domain of the function? (Hint: Add the lengths of the straight sections of the belt and the length of the belt around each pulley.) 74. Prove that if lim xqc xqc 1 # 0. f !x" 1 # 0, then lim f 'x( does not exist. f !x" xqc Infinite Limits In Exercises 75 and 76, use the '-\$ definition of infinite limits to prove the statement. 75. lim! xq3 1 # x"3 & 76. lim" xq4 1 # "& x"4 REVIEW EXERCISES 91 Review Exercises for Chapter 1 The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. Click on to view the complete solution of the exercise. Click on to print an enlarged copy of the graph. In Exercises 1 and 2, determine whether the problem can be solved using precalculus or if calculus is required. If the problem can be solved using precalculus, solve it. If the problem seems to require calculus, explain your reasoning. Use a graphical or numerical approach to estimate the solution. x 3 # 125 x#5 19. lim xq"5 x2 " 4 3 xq"2 x # 8 20. lim 1 " cos x sin x 1. Find the distance between the points #1, 1\$ and #3, 9\$ along the curve y ! x 2. 21. lim 2. Find the distance between the points #1, 1\$ and #3, 9\$ along the line y ! 4x " 3. 22. lim 4x tan x 23. lim sin&#'(6\$ # &x% " #1(2\$ &x In Exercises 3 and 4, complete the table and use the result to estimate the limit. Use a graphing utility to graph the function xq0 xq '(4 &xq0 [Hint: sin#\$ # %\$ ! sin \$ cos % # cos \$ sin %] 24. lim &xq0 x "0.1 "0.01 "0.001 0.001 0.01 cos#' # &x\$ # 1 &x [Hint: cos#\$ # %\$ ! cos \$ cos % " sin \$ sin %] 0.1 f !x" 3 In Exercises 25 and 26, evaluate the limit given lim f !x" " ! 4 xqc 2 and lim g!x" " 3. xqc &4(#x # 2\$% " 2 xq0 x 4#'x # 2 " '2 \$ 4. lim x xq0 3. lim 25. lim & f #x\$g#x\$% xqc 26. lim & f #x\$ # 2g#x\$% xqc In Exercises 5 and 6, use the graph to determine each limit. 5. h#x\$ ! x 2 " 2x x 6. g#x\$ ! y y 8 h x 2 lim f !x". xq1 # g (b) Use a graphing utility to graph the function and use the graph to estimate the limit. 4 4 4 4 x 4 4 8 (c) Rationalize the numerator to find the exact value of the limit analytically. x (a) lim h#x\$ (b) lim h#x\$ xq0 xq"1 (a) lim g#x\$ (b) lim g#x\$ xq2 7. lim #3 " x\$ xq1 9. lim #x 2 " 3\$ xq2 8. lim 'x 10. lim 9 tq4 tq"2 15. lim xq4 t#2 t2 " 4 'x " 2 x"4 &1(#x # 1\$% " 1 x xq0 17. lim 14. lim tq3 16. lim xq0 18. lim sq0 1.001 1.0001 '2x # 1 " '3 x"1 1 x"1 3 x "' t2 Free-Falling Object In Exercises 29 and 30, use the position function s!t" " !4.9t 2 # 200, which gives the height (in meters) of an object that has fallen from a height of 200 meters. The velocity at time t " a seconds is given by '4 # x " 2 lim ) ) 12. lim 3 y " 1 yq4 1.01 &Hint: a3 " b3 ! #a " b\$#a 2 # ab # b2\$% xq5 In Exercises 11–24, find the limit (if it exists). 11. lim 't # 2 27. f #x\$ ! 28. f #x\$ ! xq9 1.1 f !x" xq0 In Exercises 7–10, find the limit L. Then use the (-\$ definition to prove that the limit is L. 13. lim In Exercises 27 (a) Complete the table to estimate the limit. 2 2 2 3x x"2 Numerical, Graphical, and Analytic Analysis and 28, consider "9 t"3 x #1('1 # s \$ " 1 s tqa s!a" ! s!t" . a!t 29. Find the velocity of the object when t ! 4. 30. At what velocity will the object impact the ground? 92 CHAPTER 1 Limits and Their Properties In Exercises 31–36, find the limit (if it exists). If the limit does not exist, explain why. 31. lim" xq3 51. Let f #x\$ ! )x " 3) x2 " 4 . Find each limit (if possible). x"2 ) ) (a) lim" f #x\$ xq2 x"3 (b) lim# f #x\$ 32. lim -x " 1. xq2 xq4 #x " 2\$2, x f 2 ,2 " x, x > 2 1 " x, x f 1 34. lim g#x\$, where g#x\$ ! , x # 1, x > 1 t # 1, t < 1 35. lim h#t\$, where h#t\$ ! , #t # 1\$, t v 1 "s " 4s " 2, s f "2 36. lim f #s\$, where f #s\$ ! , s # 4s # 6, s > "2 33. lim f #x\$, where f #x\$ ! xq2 (c) lim f #x\$ xq2 52. Let f #x\$ ! 'x#x " 1\$ . (a) Find the domain of f. ' (b) Find lim" f #x\$. xq1# xq0 3 (c) Find lim# f #x\$. 1 2 tq1 xq1 2 2 sq"2 In Exercises 53–56, find the vertical asymptotes (if any) of the graphs of the function. In Exercises 37–46, determine the intervals on which the function is continuous. 53. g#x\$ ! 1 # 37. f #x\$ ! -x # 3. 55. f #x\$ ! 38. f #x\$ ! 3x 2 "x"2 x"1 , 57. 59. , 5 " x, x f 2 2x " 3, x > 2 41. f #x\$ ! 1 #x " 2\$ 2 43. f #x\$ ! 3 x#1 45. f #x\$ ! csc 'x 2 42. f #x\$ ! 'x #x 1 44. f #x\$ ! x#1 2x # 2 46. f #x\$ ! tan 2x 47. Determine the value of c such that the function is continuous on the entire real line. f #x\$ ! ,xcx##3,6, x f 2 x > 2 48. Determine the values of b and c such that the function is continuous on the entire real line. f #x\$ ! , x # 1, x 2 # bx # c, 4x 4 " x2 54. h#x\$ ! 8 #x " 10\$ 2 56. f #x\$ ! csc ' x In Exercises 57–68, find the one-sided limit. 3x 2 " x " 2 , x ) 1 x"1 39. f #x\$ ! 0, x!1 40. f #x\$ ! 2 x 1 < x < 3 x"2 v 1 ) ) 49. Use the Intermediate Value Theorem to show that f #x\$ ! 2x 3 " 3 has a zero in the interval &1, 2%. 50. Delivery Charges The cost of sending an overnight package from New York to Atlanta is \$9.80 for the first pound and \$2.50 for each additional pound or fraction thereof. Use the greatest integer function to create a model for the cost C of overnight delivery of a package weighing x pounds. Use a graphing utility to graph the function and discuss its continuity. lim " 2x 2 # x # 1 x#2 58. lim x#1 x3 # 1 60. xq"2 xq"1 # 61. lim" xq1 lim # x 2 " 2x # 1 x#1 * 64. lim" xq"1 xq2 sin 4x 65. lim# 5x xq0 xq0 x#1 x4 " 1 62. + x 2x " 1 lim xq"1 " x 2 # 2x # 1 x"1 1 63. lim# x " 3 x xq0 67. lim# lim xq #1(2\$ # 1 3 x2 ' "4 sec x 66. lim# x xq0 csc 2x x 68. lim" xq0 cos 2 x x 69. Environment A utility company burns coal to generate electricity. The cost C in dollars of removing p% of the air pollutants in the stack emissions is C! 80,000p , 100 " p 0 f p < 100. Find the cost of removing (a) 15%, (b) 50%, and (c) 90% of the pollutants. (d) Find the limit of C as p q 100". 70. The function f is defined as shown. f #x\$ ! tan 2x , x (a) Find lim xq0 x)0 tan 2x (if it exists). x (b) Can the function f be defined at x ! 0 such that it is continuous at x ! 0? P.S. P.S. 93 Problem Solving Problem Solving The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. Click on to view the complete solution of the exercise. Click on to print an enlarged copy of the graph. 1. Let P"x, y# be a point on the parabola y # x 2 in the first quadrant. Consider the triangle !PAO formed by P, A"0, 1#, and the origin O"0, 0#, and the triangle !PBO formed by P, B"1, 0#, and the origin. y 3. (a) Find the area of a regular hexagon inscribed in a circle of radius 1. How close is this area to that of the circle? (b) Find the area An of an n-sided regular polygon inscribed in (c) Complete the table. P A n 1 6 12 24 48 96 An B O x 1 (d) What number does An approach as n gets larger and larger? y (a) Write the perimeter of each triangle in terms of x. 6 (b) Let r"x# be the ratio of the perimeters of the two triangles, Perimeter !PAO r"x# # . Perimeter !PBO 2 Complete the table. 6 x P(3, 4) 1 4 2 1 0.1 0.01 Perimeter !PAO Q x 2 6 6 Figure for 3 Perimeter !PBO 2 O Figure for 4 4. Let P"3, 4# be a point on the circle x 2 ! y 2 # 25. r \$x% (a) What is the slope of the line joining P and O"0, 0#? (b) Find an equation of the tangent line to the circle at P. (c) Calculate lim! r"x#. xq0 2. Let P"x, y# be a point on the parabola y # x 2 in the first quadrant. Consider the triangle !PAO formed by P, A"0, 1#, and the origin O"0, 0#, and the triangle !PBO formed by P, B"1, 0#, and the origin. y (c) Let Q"x, y# be another point on the circle in the first quadrant. Find the slope mx of the line joining P and Q in terms of x. (d) Calculate lim mx. How does this number relate to your xq3 5. Let P"5, "12# be a point on the circle x 2 ! y 2 # 169. y P A 1 15 B O 5 x 1 15 5 O (a) Write the area of each triangle in terms of x. (a) What is the slope of the line joining P and O"0, 0#? Area !PBO . Area !PAO (b) Find an equation of the tangent line to the circle at P. Complete the table. x 4 Area !PAO Area !PBO a\$x% 2 1 0.1 0.01 (c) Let Q"x, y# be another point on the circle in the fourth quadrant. Find the slope mx of the line joining P and Q in terms of x. (d) Calculate lim mx. How does this number relate to your xq5 6. Find the values of the constants a and b such that lim (c) Calculate lim! a"x#. xq0 Q 15 P(5, 12) (b) Let a"x# be the ratio of the areas of the two triangles, a"x# # x 5 xq0 !a ! bx " !3 x # !3. 94 CHAPTER 1 Limits and Their Properties 7. Consider the function f "x# # !3 ! x1,3 " 2 x"1 12. To escape Earth’s gravitational field, a rocket must be launched with an initial velocity called the escape velocity. A rocket launched from the surface of Earth has velocity v (in miles per second) given by . (a) Find the domain of f. (b) Use a graphing utility to graph the function. f "x#. (c) Calculate lim xq"27! v# (d) Calculate lim f "x#. xq1 8. Determine all values of the constant a such that the following function is continuous for all real numbers. ax , f "x# # tan x a 2 " 2, & !v !2GM r 2 0 x < 0 9. Consider the graphs of the four functions g1, g2, g3, and g4. y v# g2 2 1 2 3 1 y 2 x 3 v# y 3 g3 2 x 2 " 2.17. x 3 !v !10,600 r 2 0 " 6.99. 13. For positive numbers a < b, the pulse function is defined as 1 1 2 0 Find the escape velocity for this planet. Is the mass of this planet larger or smaller than that of Earth? (Assume that the mean density of this planet is the same as that of Earth.) g4 2 1 !v !1920 r (c) A rocket launched from the surface of a planet has velocity v (in miles per second) given by x 3 1 2 3 For each given condition of the function f, which of the graphs could be the graph of f ? (a) lim f "x# # 3 xq2 & 0, Pa,b"x# # H"x " a# " H"x " b# # 1, 0, where H"x# # &1,0, x < a a f x < b x v b x v 0 is the Heaviside function. x < 0 (a) Sketch the graph of the pulse function. (b) f is continuous at 2. (b) Find the following limits: (c) lim" f "x# # 3 xq2 (i) + 1 10. Sketch the graph of the function f "x# # . x lim Pa,b"x# xqa! (iii) lim! Pa,b"x# xqb (ii) lim Pa,b"x# xqa" (iv) lim" Pa,b"x# xqb (a) Evaluate f " #, f "3#, and f "1#. (c) Discuss the continuity of the pulse function. (b) Evaluate the limits lim" f "x#, lim! f "x#, lim" f "x#, and xq1 xq1 xq0 lim! f "x#. (d) Why is 1 4 xq0 U"x# # (c) Discuss the continuity of the function. 11. Sketch the graph of the function f "x# # (x) ! ("x). 1 (a) Evaluate f "1#, f "0#, f "2 #, and f ""2.7#. (b) Evaluate the limits lim" f "x#, lim! f "x#, and lim1 f "x#. xq1 " 48 Find the escape velocity for the moon. 1 1 2 0 (b) A rocket launched from the surface of the moon has velocity v (in miles per second) given by 3 g1 !v !192,000 r where v0 is the initial velocity, r is the distance from the rocket to the center of Earth, G is the gravitational constant, M is the mass of Earth, and R is the radius of Earth (approximately 4000 miles). y 2 2GM ' R (a) Find the value of v0 for which you obtain an infinite limit for r as v tends to zero. This value of v0 is the escape velocity for Earth. x v 0 3 " xq1 (c) Discuss the continuity of the function. xq 2 1 P "x# b " a a,b called the unit pulse function? lim f "x# # L, then 14. Let a be a nonzero constant. Prove that if xq0 lim f "ax# # L. Show by means of an example that a must be xq0 nonzero. 96 CHAPTER 2 Differentiation Section 2.1 The Derivative and the Tangent Line Problem • Find the slope of the tangent line to a curve at a point. • Use the limit definition to find the derivative of a function. • Understand the relationship between differentiability and continuity. ISAAC NEWTON (1642–1727) In addition to his work in calculus, Newton including the Law of Universal Gravitation .and his three laws of motion. The Tangent Line Problem Calculus grew out of four major problems that European mathematicians were working on during the seventeenth century. MathBio 1. 2. 3. 4. y P x The tangent line problem (Section 1.1 and this section) The velocity and acceleration problem (Sections 2.2 and 2.3) The minimum and maximum problem (Section 3.1) The area problem (Sections 1.1 and 4.2) Each problem involves the notion of a limit, and calculus can be introduced with any of the four problems. A brief introduction to the tangent line problem is given in Section 1.1. Although partial solutions to this problem were given by Pierre de Fermat (1601–1665), René Descartes (1596–1650), Christian Huygens (1629–1695), and Isaac Barrow (1630 –1677), credit for the first general solution is usually given to Isaac Newton (1642–1727) and Gottfried Leibniz (1646–1716). Newton’s work on this problem stemmed from his interest in optics and light refraction. What does it mean to say that a line is tangent to a curve at a point? For a circle, the tangent line at a point P is the line that is perpendicular to the radial line at point P, as shown in Figure 2.1. For a general curve, however, the problem is more difficult. For example, how would you define the tangent lines shown in Figure 2.2? You might say that a line is tangent to a curve at a point P if it touches, but does not cross, the curve at point P. This definition would work for the first curve shown in Figure 2.2, but not for the second. Or you might say that a line is tangent to a curve if the line touches or intersects the curve at exactly one point. This definition would work for a circle but not for more general curves, as the third curve in Figure 2.2 shows. y y y y = f(x) Tangent line to a circle Figure 2.1 P P P x FOR FURTHER INFORMATION For mathematical discoveries to the first “discoverer,” see the article “Mathematical Firsts—Who Done It?” by. Richard H. Williams and Roy D. Mazzagatti in Mathematics Teacher. MathArticle y = f(x) y = f (x) x Tangent line to a curve at a point Figure 2.2 E X P L O R AT I O N Identifying a Tangent Line Use a graphing utility to graph the function f !x" " 2x 3 ! 4x 2 # 3x ! 5. On the same screen, graph y " x ! 5, y " 2x ! 5, and y " 3x ! 5. Which of these lines, if any, appears to be tangent to the graph of f at the point !0, !5"? Explain your reasoning. x SECTION 2.1 y (c + ∆ x , f(c + ∆ x)) f (c + ∆ x) − f (c) = ∆y (c, f(c)) ∆x x The Derivative and the Tangent Line Problem 97 Essentially, the problem of finding the tangent line at a point P boils down to the problem of finding the slope of the tangent line at point P. You can approximate this slope using a secant line through the point of tangency and a second point on the curve, as shown in Figure 2.3. If !c, f !c"" is the point of tangency and !c # \$ x, f !c # \$ x"" is a second point on the graph of f, the slope of the secant line through the two points is given by substitution into the slope formula y 2 ! y1 x 2 ! x1 f !c # \$x" ! f !c" msec " !c # \$x" ! c m" The secant line through !c, f !c"" and !c # \$x, f !c # \$x"" msec " Figure 2.3 f !c # \$x" ! f !c" . \$x Change in y Change in x Slope of secant line The right-hand side of this equation is a difference quotient. The denominator \$x is the change in x, and the numerator \$y " f !c # \$x" ! f !c" is the change in y. The beauty of this procedure is that you can obtain more and more accurate approximations of the slope of the tangent line by choosing points closer and closer to the point of tangency, as shown in Figure 2.4. THE TANGENT LINE PROBLEM In 1637, mathematician René Descartes stated this about the tangent line problem: “And I dare say that this is not only the most useful and general problem in geometry that I know, but even that I ever desire to know.” (c, f (c)) ∆x ∆x (c, f(c)) ∆y (c, f (c)) ∆y ∆x ∆x → 0 ∆y (c, f (c)) ∆y ∆x (c, f (c)) ∆y ∆x ∆x → 0 (c, f(c)) (c, f(c)) ∆y ∆x (c, f (c)) Tangent line Tangent line Tangent line approximations Figure 2.4 . To view a sequence of secant lines approaching a tangent line, select the Animation button. Animation Definition of Tangent Line with Slope m If f is defined on an open interval containing c, and if the limit lim \$x→0 . \$y f !c # \$x" ! f !c" " lim "m \$x \$x→0 \$x exists, then the line passing through !c, f !c"" with slope m is the tangent line to the graph of f at the point !c, f !c"". Video Video The slope of the tangent line to the graph of f at the point !c, f !c"" is also called the slope of the graph of f at x ! c. * This use of the word secant comes from the Latin secare, meaning to cut, and is not a reference to the trigonometric function of the same name. 98 CHAPTER 2 Differentiation EXAMPLE 1 The Slope of the Graph of a Linear Function Find the slope of the graph of f !x" " 2x ! 3 at the point !2, 1". f (x) = 2x − 3 y Solution To find the slope of the graph of f when c " 2, you can apply the definition of the slope of a tangent line, as shown. ∆x = 1 3 lim \$x→0 ∆y = 2 2 m=2 1 (2, 1) x 1 2 3 f !2 # \$x" ! f !2" #2!2 # \$x" ! 3\$ ! #2!2" ! 3\$ " lim \$x→0 \$x \$x 4 # 2\$x ! 3 ! 4 # 3 " lim \$x→0 \$x 2\$x " lim \$x→0 \$x " lim 2 \$x→0 "2 . slope of f at !2, 1" is m " 2. The The slope of f at !c, f !c"" " !2, 1" is m " 2, as shown in Figure 2.5. Figure 2.5 .Editable Graph NOTE In Example 1, the limit definition of the slope of f agrees with the definition of the slope of a line as discussed in Section P.2. Exploration A Try It The graph of a linear function has the same slope at any point. This is not true of nonlinear functions, as shown in the following example. EXAMPLE 2 y Find the slopes of the tangent lines to the graph of 4 3 Tangent line at (−1,2 ) −2 f (x) = x 2 + 1 2 −1 Tangent line at (0, 1) 1 2 Editable Graph f !x" " x 2 # 1 at the points !0, 1" and !!1, 2", as shown in Figure 2.6. Solution Let !c, f !c"" represent an arbitrary point on the graph of f. Then the slope of the tangent line at !c, f !c"" is given by x The slope of f at any point !c, f !c"" is m ." 2c. Figure 2.6 Tangent Lines to the Graph of a Nonlinear Function lim \$x→0 f !c # \$x" ! f !c" #!c # \$x" 2 # 1\$ ! !c 2 # 1" " lim \$x→0 \$x \$x c 2 # 2c!\$x" # !\$x" 2 # 1 ! c 2 ! 1 " lim \$x→0 \$x 2c!\$x" # !\$x" 2 " lim \$x→0 \$x " lim !2c # \$x" \$x→0 " 2c. So, the slope at any point !c, f !c"" on the graph of f is m " 2c. At the point !0, 1", the slope is m " 2!0" " 0, and at !!1, 2", the slope is m " 2!!1" " !2. . NOTE In Example 2, note that c is held constant in the limit process !as \$ x → 0". Try It Exploration A SECTION 2.1 y lim \$x→0 (c, f(c)) x The graph of f has a vertical tangent line at !c, f !c"". Figure 2.7 99 The definition of a tangent line to a curve does not cover the possibility of a vertical tangent line. For vertical tangent lines, you can use the following definition. If f is continuous at c and Vertical tangent line c The Derivative and the Tangent Line Problem f !c # \$x" ! f !c" "& \$x or lim \$x→0 f !c # \$x" ! f !c" " !& \$x the vertical line x " c passing through !c, f !c"" is a vertical tangent line to the graph of f. For example, the function shown in Figure 2.7 has a vertical tangent line at !c, f !c"". If the domain of f is the closed interval #a, b\$, you can extend the definition of a vertical tangent line to include the endpoints by considering continuity and limits from the right !for x " a" and from the left !for x " b". The Derivative of a Function You have now arrived at a crucial point in the study of calculus. The limit used to define the slope of a tangent line is also used to define one of the two fundamental operations of calculus—differentiation. Definition of the Derivative of a Function The derivative of f at x is given by f%!x" " lim \$x→0 . f !x # \$x" ! f !x" \$x provided the limit exists. For all x for which this limit exists, f % is a function of x. Video Be sure you see that the derivative of a function of x is also a function of x. This “new” function gives the slope of the tangent line to the graph of f at the point !x, f !x"", provided that the graph has a tangent line at this point. The process of finding the derivative of a function is called differentiation. A function is differentiable at x if its derivative exists at x and is differentiable on an open interval &a, b' if it is differentiable at every point in the interval. In addition to f%!x", which is read as “ f prime of x,” other notations are used to denote the derivative of y " f !x". The most common are f%!x", dy , dx y%, d # f !x"\$, dx Dx # y\$. Notation for derivatives The notation dy%dx is read as “the derivative of y with respect to x” or simply “dy ! dx”. Using limit notation, you can write dy \$y " lim \$x→0 dx \$x f !x # \$x" ! f !x" " lim \$x→0 \$x . " f%!x". History 100 CHAPTER 2 Differentiation EXAMPLE 3 Finding the Derivative by the Limit Process Find the derivative of f !x" " x 3 # 2x. Solution f%!x" " lim \$x→0 " lim \$x→0 When using the definition to find a derivative of a function, the key is to rewrite the difference quotient so that \$x does not occur as a factor of the denominator. STUDY TIP " lim \$x→0 " lim \$x→0 " lim \$x→0 " lim \$x→0 . f !x # \$x" ! f !x" Definition of derivative \$x !x # \$x"3 # 2!x # \$x" ! !x3 # 2x" \$x x3 # 3x2\$x # 3x!\$x" 2 # !\$x"3 # 2x # 2\$x ! x3 ! 2x \$x 3x 2\$x # 3x!\$x" 2 # !\$x"3 # 2\$x \$x \$x #3x 2 # 3x\$x # !\$x" 2 # 2\$ \$x #3x 2 # 3x\$x # !\$x" 2 # 2\$ " 3x 2 # 2 . Exploration A Try It Exploration C . Exploration B Open Exploration The editable graph feature below allows you to edit the graph of a function and its derivative. Editable Graph Remember that the derivative of a function f is itself a function, which can be used to find the slope of the tangent line at the point !x, f !x"" on the graph of f. EXAMPLE 4 Using the Derivative to Find the Slope at a Point Find f%!x" for f !x" " (x. Then find the slope of the graph of f at the points !1, 1" and !4, 2". Discuss the behavior of f at !0, 0". Solution Use the procedure for rationalizing numerators, as discussed in Section 1.3. f !x # \$x" ! f !x" f%!x" " lim Definition of derivative \$x→0 \$x (x # \$x ! (x " lim \$x→0 \$x \$x→0 y 3 (4, 2) 2 (1, 1) m= m= 1 2 (0, 0) 1 f(x) = x 3 4 The slope of f at !x, f !x"", x > 0, is m .." 1+! 2(x ". Figure 2.8 Editable Graph 1 4 x 2 ) " 1 , 2(x (x # \$x ! (x ) (x # \$x # (x (x # \$x # (x \$x !x # \$x" ! x " lim \$x→0 \$x !(x # \$x # (x " \$x " lim \$x→0 \$x !(x # \$x # (x " 1 " lim \$x→0 (x # \$x # (x " lim x > 0 At the point !1, 1", the slope is f%!1" " 12. At the point !4, 2", the slope is f%!4" " 14. See Figure 2.8. At the point !0, 0", the slope is undefined. Moreover, the graph of f has a vertical tangent line at !0, 0". Try It Exploration A Exploration B Exploration C SECTION 2.1 The Derivative and the Tangent Line Problem 101 In many applications, it is convenient to use a variable other than x as the independent variable, as shown in Example 5. EXAMPLE 5 Finding the Derivative of a Function Find the derivative with respect to t for the function y " 2%t. Solution Considering y " f !t", you obtain dy f !t # \$t" ! f !t" " lim \$t→0 dt \$t 2 2 ! t # \$t t " lim \$t→0 \$t 2t ! 2!t # \$t" t!t # \$t" " lim \$t→0 \$t !2\$t " lim \$t→0 \$t!t"!t # \$t" !2 " lim \$t→0 t !t # \$t" 2 " ! 2. t . 4 . y= f !t # \$t" " 2%!t # \$t" and f !t" " 2%t Combine fractions in numerator. Divide out common factor of \$t. Simplify. Evaluate limit as \$t → 0. Exploration A Try It 2 t Definition of derivative Open Exploration The editable graph feature below allows you to edit the graph of a function and its derivative. Editable Graph (1, 2) A graphing utility can be used to reinforce the result given in Example 5. For instance, using the formula dy%dt " !2%t 2, you know that the slope of the graph of y " 2%t at the point !1, 2" is m " !2. This implies that an equation of the tangent line to the graph at !1, 2" is TECHNOLOGY 0 6 0 y = −2t + 4 At the point !1, 2" the line y " ! 2t # 4 is tangent to the graph of y " 2% t. Figure 2.9 y ! 2 " !2!t ! 1" or y " !2t # 4 as shown in Figure 2.9. Differentiability and Continuity The following alternative limit form of the derivative is useful in investigating the relationship between differentiability and continuity. The derivative of f at c is y (x, f(x)) (c, f (c)) f%!c" " lim x→c x−c f(x) − f (c) lim x x As x approaches c, the secant line approaches the tangent line. Figure 2.10 Alternative form of derivative provided this limit exists (see Figure 2.10). (A proof of the equivalence of this form is given in Appendix A.) Note that the existence of the limit in this alternative form requires that the one-sided limits x→c! c f !x" ! f !c" x!c f !x" ! f !c" x!c and lim x→c# f !x" ! f !c" x!c exist and are equal. These one-sided limits are called the derivatives from the left and from the right, respectively. It follows that f is differentiable on the closed interval [a, b] if it is differentiable on !a, b" and if the derivative from the right at a and the derivative from the left at b both exist. 102 CHAPTER 2 Differentiation y If a function is not continuous at x " c, it is also not differentiable at x " c. For instance, the greatest integer function 2 f !x" " -x. 1 −2 is not continuous at x " 0, and so it is not differentiable at x " 0 (see Figure 2.11). You can verify this by observing that x −1 1 3 2 f(x) = [[x]] lim! f !x" ! f !0" -x. ! 0 " lim! "& x→0 x!0 x Derivative from the left lim f !x" ! f !0" -x. ! 0 " lim# " 0. x→0 x!0 x Derivative from the right x→0 −2 The greatest integer function is not differentiable at x " 0, because it is not continuous at x " 0. and x→0 # Figure 2.11 Although it is true that differentiability implies continuity (as shown in Theorem 2.1 on the next page), the converse is not true. That is, it is possible for a function to be continuous at x " c and not differentiable at x " c. Examples 6 and 7 illustrate this possibility. EXAMPLE 6 The function y , shown in Figure 2.12 is continuous at x " 2. But, the one-sided limits m = −1 1 2 3 4 , , Derivative from the left , , Derivative from the right lim! x!2 !0 f !x" ! f !2" " lim! " !1 x→2 x!2 x!2 lim x!2 !0 f !x" ! f !2" " lim# "1 x→2 x!2 x!2 x→2 m=1 1 , f !x" " x ! 2 f (x) =x − 2 3 2 A Graph with a Sharp Turn and x f is not differentiable at x " 2, because the derivatives from the left and from the right are. not equal. Figure 2.12 Editable Graph x→2# are not equal. So, f is not differentiable at x " 2 and the graph of f does not have a tangent line at the point !2, 0". Exploration A Try It EXAMPLE 7 Open Exploration A Graph with a Vertical Tangent Line y f(x) = x 1/3 The function f !x" " x1%3 1 is continuous at x " 0, as shown in Figure 2.13. But, because the limit −2 x −1 1 2 lim x→0 −1 f is not differentiable at x " 0, because f has. a vertical tangent at x " 0. Figure 2.13 Editable Graph f !x" ! f !0" x1%3 ! 0 " lim x→0 x!0 x 1 " lim 2%3 x→0 x "& is infinite, you can conclude that the tangent line is vertical at x " 0. So, f is not differentiable at x " 0. Try It Exploration A Exploration B Exploration C From Examples 6 and 7, you can see that a function is not differentiable at a point at which its graph has a sharp turn or a vertical tangent. SECTION 2.1 TECHNOLOGY Some graphing utilities, such as Derive, Maple, perform symbolic differentiation. Others perform numerical differentiation by finding values of derivatives using the formula f %!x" 1 f !x # \$x" ! f !x ! \$x" 2\$x THEOREM 2.1 The Derivative and the Tangent Line Problem 103 Differentiability Implies Continuity If f is differentiable at x " c, then f is continuous at x " c. Proof You can prove that f is continuous at x " c by showing that f !x" approaches f !c" as x → c. To do this, use the differentiability of f at x " c and consider the following limit. ) f !xx" !! cf !c"0 f !x" ! f !c" " / lim !x ! c"0/ lim x!c 0 / lim # f !x" ! f !c"\$ " lim !x ! c" where \$x is a small number such as 0.001. Can you see any problems with this definition? For instance, using this definition, what is the value of the derivative of f !x" " x when x " 0? x→c x→c x→c ,, x→c " !0"# f %!c"\$ "0 Because the difference f !x" ! f !c" approaches zero as x → c, you can conclude that lim f !x" " f !c". So, f is continuous at x " c. x→c The following statements summarize the relationship between continuity and differentiability. 1. If a function is differentiable at x " c, then it is continuous at x " c. So, differentiability implies continuity. 2. It is possible for a function to be continuous at x " c and not be differentiable at x " c. So, continuity does not imply differentiability. SECTION 2.1 The Derivative and the Tangent Line Problem 103 Exercises for Section 2.1 The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. Click on to view the complete solution of the exercise. Click on to print an enlarged copy of the graph. In Exercises 1 and 2, estimate the slope of the graph at the points #x1, y1\$ and #x2, y2\$. 1. (a) y y (b) y (x1, y1) (x2, y2) In Exercises 3 and 4, use the graph shown in the figure. To print an enlarged copy of the graph, select the MathGraph button. (x2, y2) (x1, y1) x x 6 5 4 3 2 1 (4, 5) f (1, 2) x 1 2 3 4 5 6 2. (a) y 3. Identify or sketch each of the quantities on the figure. y (b) (a) f !1" and f !4" (x1, y1) (c) y # (x2, y2) x x (x1, y1) (x2, y2) (b) f !4" ! f !1" f !4" ! f !1" !x ! 1" \$ f !1" 4!1 4. Insert the proper inequality symbol !< or >" between the given quantities. (a) f !4" ! f !1" f !4" ! f !3" 4!1 ! 4!3 (b) f !4" ! f !1" f "!1" 4!1 ! 104 CHAPTER 2 Differentiation In Exercises 5 –10, find the slope of the tangent line to the graph of the function at the given point. 3 2x 5. f !x" # 3 ! 2x, !!1, 5" 6. g!x" # 7. g!x" # x 2 ! 4, !1, !3" 8. g!x" # 5 ! x 2, 9. f !t" # 3t ! t 2, !0, 0" 39. 5 4 3 2 1 \$ 1, !!2, !2" !2, 1" 2 10. h!t" # t \$ 3, !!2, 7" 12. g!x" # !5 13. f !x" # !5x 14. f !x" # 3x \$ 2 2 15. h!s" # 3 \$ 3 s 1 16. f !x" # 9 ! 2x 17. f !x" # 2x 2 \$ x ! 1 18. f !x" # 1 ! x 2 19. f !x" # x3 20. f !x" # x3 21. f !x" # 1 x!1 22. f !x" # 1 x2 ! 12x 23. f !x" # %x \$ 1 24. f !x" # \$ x x2 4 3 2 1 2 3 4 5 27. f !x" # !2, 8" !1, 1" 4 31. f !x" # x \$ , !4, 5" x 28. f !x" # 29. f !x" # %x, x3 \$ 1, !1, 2" 30. f !x" # %x ! 1, 32. f !x" # !5, 2" 1 , !0, 1" x\$1 Line 33. f !x" # x 3 3x ! y \$ 1 # 0 34. f !x" # x 3 \$ 2 3x ! y ! 4 # 0 35. f !x" # 1 1 36. f !x" # %x ! 1 x 3 2 1 5 4 3 2 1 f 1 2 3 2 3 42. The tangent line to the graph of y # h!x" at the point !!1, 4" passes through the point !3, 6". Find h!!1" and h"!!1". In Exercises 43– 46, sketch the graph of f!. Explain how you 44. y 2 1 1 2 2 3 4 y 4 x 2 y x 3 2 3 4 5 6 3 2 1 f x 1 2 3 45. 1 2 4 f f 6 46. y 7 6 5 4 3 2 1 x 2 2 6 38. x 1 2 3 41. The tangent line to the graph of y # g!x" at the point !5, 2" passes through the point !9, 0". Find g!5" and g" !5". x \$ 2y \$ 7 # 0 y fe 3 2 1 1 2 3 43. In Exercises 37–40, the graph of f is given. Select the graph of f!. 37. fe 2 3 x \$ 2y ! 6 # 0 %x 3 2 1 In Exercises 33–36, find an equation of the line that is tangent to the graph of f and parallel to the given line. Function y (d) 3 2 1 3 2 1 2 3 2 y (c) x 3 2 1 x 25. f !x" # x 2 \$ 1, !2, 5" x 3, fe fe %x 26. f !x" # x 2 \$ 2x \$ 1, !!3, 4" x 1 2 3 y (b) 1 In Exercises 25–32, (a) find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. f 3 2 1 y 5 4 3 2 1 4 5 4 3 2 1 2 3 4 5 (a) y f 1 In Exercises 11–24, find the derivative by the limit process. 11. f !x" # 3 40. y y 7 6 f 4 3 2 1 f x 1 2 3 4 5 6 7 x 1 2 3 4 5 6 7 8 47. Sketch a graph of a function whose derivative is always negative. SECTION 2.1 48. Sketch a graph of a function whose derivative is always positive. 60. Graphical Reasoning Use a graphing utility to graph each function and its tangent lines at x # !1, x # 0, and x # 1. Based on the results, determine whether the slopes of tangent lines to the graph of a function at different values of x are always distinct. In Exercises 49–52, the limit represents f! #c\$ for a function f and a number c. Find f and c. &5 ! 3!1 \$ 'x"' ! 2 'xq0 'x 2 !x \$ 36 51. lim xq6 x!6 49. lim !!2 \$ 'x"3 \$ 8 'xq0 'x 2%x ! 6 52. lim xq9 x!9 50. lim In Exercises 53 –55, identify a function f that has the following characteristics. Then sketch the function. 53. f !0" # 2; & f" !x" < 0 for x < 0; 55. f !0" # 0; f" !0" # 0; f" !x" > 0 if x % 0 56. Assume that f" !c" # 3. Find f " !!c" if (a) f is an odd function and if (b) f is an even function. In Exercises 57 and 58, find equations of the two tangent lines to the graph of f that pass through the indicated point. 58. f !x" # x 2 y 10 8 6 4 (2, 5) 4 3 x 1 3 2 5 59. Graphical Reasoning 6 4 2 4 !2 !0.5 0 0.5 1 1.5 2 62. f !x" # 12x 2 Graphical Reasoning In Exercises 63 and 64, use a graphing utility to graph the functions f and g in the same viewing window where g#x\$ # f #x \$ 0.01\$ % f #x\$ . 0.01 Label the graphs and describe the relationship between them. 64. f !x" # 3%x 66. f !x" # 14 x 3 x 2 4 (1, 3) 6 The figure shows the graph of g". Graphical Reasoning In Exercises 67 and 68, use a graphing utility to graph the function and its derivative in the same viewing window. Label the graphs and describe the relationship between them. ge 1 %x 68. f !x" # x3 ! 3x 4 Writing In Exercises 69 and 70, consider the functions f and S"x where x 4 6 4 6 (a) g"!0" # ! !1 f #x\$ 67. f !x" # 6 4 2 !1.5 65. f !x" # x!4 ! x" y 6 4 x In Exercises 65 and 66, evaluate f #2\$ and f #2.1\$ and use the results to approximate f!#2\$. 2 1 Graphical, Numerical, and Analytic Analysis In Exercises 61 and 62, use a graphing utility to graph f on the interval [%2, 2]. Complete the table by graphically estimating the slopes of the graph at the indicated points. Then evaluate the slopes analytically and compare your results with those obtained graphically. 63. f !x" # 2x ! x 2 y 5 (b) g !x" # x 3 61. f !x" # 14 x 3 f" !x" > 0 for x > 0 57. f !x" # 4x ! x 2 (a) f !x" # x 2 f !#x\$ 54. f !0" # 4; f" !0" # 0; f " !x" # !3, ! & < x < 105 The Derivative and the Tangent Line Problem S"x #x\$ # (b) g"!3" # ! (c) What can you conclude about the graph of g knowing that g" !1" # ! 83? (d) What can you conclude about the graph of g knowing that g" !!4" # 73? (e) Is g!6" ! g!4" positive or negative? Explain. (f) Is it possible to find g !2" from the graph? Explain. f #2 \$ "x\$ % f #2\$ #x % 2\$ \$ f #2\$. "x (a) Use a graphing utility to graph f and S"x in the same viewing window for "x # 1, 0.5, and 0.1. (b) Give a written description of the graphs of S for the different values of "x in part (a). 69. f !x" # 4 ! !x ! 3" 2 70. f !x" # x \$ 1 x 106 CHAPTER 2 Differentiation In Exercises 71–80, use the alternative form of the derivative to find the derivative at x # c (if it exists). 71. f !x" # x 2 ! 1, c # 2 72. g!x" # x!x ! 1", c # 1 73. f !x" # x 3 \$ 2x 2 \$ 1, c # !2 74. f !x" # x 3 \$ 2x, c # 1 (( c#3 78. g!x" # !x \$ 3"13, c # !3 ( ( ( ( 80. f !x" # x ! 4 , c # 4 In Exercises 81– 86, describe the x-values at which f is differentiable. 81. f !x" # 1 x\$1 ( 12 10 1 x 4 2 2 x 4 x2 x2 ! 4 5 4 3 2 5 4 3 1 x 4 x 3 4 1 2 3 4 5 6 3 86. f !x" # )x4 !!x4,, x f 0 x > 0 2 2 y y 3 4 2 2 1 x 4 4 Graphical Analysis In Exercises 87–90, use a graphing utility to find the x-values at which f is differentiable. ( ( 87. f !x" # x \$ 3 88. f !x" # 89. f !x" # x25 90. f !x" # )xx !! 3x2x, \$ 3x, 3 2 2 x f 1 x > 1 2 x f 1 x > 1 95. f #x\$ # ) x 2 \$ 1, 4x ! 3, x f 2 x > 2 96. f !x" # 1 2x ) \$ 1, %2x , x < 2 x v 2 97. Graphical Reasoning A line with slope m passes through the point !0, 4" and has the equation y # mx \$ 4. 2x x!1 (b) Graph g and g" on the same set of axes. True or False? In Exercises 99–102, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 99. The slope of the tangent line to the differentiable function f at f !2 \$ ' x" ! f !2" . the point !2, f !2"" is 'x 100. If a function is continuous at a point, then it is differentiable at that point. 101. If a function has derivatives from both the right and the left at a point, then it is differentiable at that point. x 4 4 3 )x,x , (d) Find f "!x" if f !x" # x 4. Compare the result with the conjecture in part (c). Is this a proof of your conjecture? Explain. y 2 94. f !x" # (c) Identify a pattern between f and g and their respective derivatives. Use the pattern to make a conjecture about h"!x" if h !x" # x n, where n is an integer and n v 2. 4 y 1 x f 1 x > 1 (a) Graph f and f " on the same set of axes. 2 84. f !x" # 85. f !x" # %x ! 1 2 98. Conjecture Consider the functions f !x" # x 2 and g!x" # x3. 6 4 2 1 83. f !x" # !x ! 3" 23 )!!xx !! 11"" , (b) Use a graphing utility to graph the function d in part (a). Based on the graph, is the function differentiable at every value of m? If not, where is it not differentiable? y 1 92. f !x" # %1 ! x 2 3, (a) Write the distance d between the line and the point !3, 1" as a function of m. ( 82. f !x" # x 2 ! 9 y 2 ( In Exercises 95 and 96, determine whether the function is differentiable at x # 2. 77. f !x" # !x ! 6"23, c # 6 79. h!x" # x \$ 5 , c # !5 ( 91. f !x" # x ! 1 93. f !x" # 75. g!x" # % x , c # 0 76. f !x" # 1x, In Exercises 91–94, find the derivatives from the left and from the right at x # 1 (if they exist). Is the function differentiable at x # 1? 102. If a function is differentiable at a point, then it is continuous at that point. 103. Let f !x" # 1 1 x sin , x % 0 x 2 sin , x % 0 x x . and g !x" # 0, 0, x#0 x#0 ) ) Show that f is continuous, but not differentiable, at x # 0. Show that g is differentiable at 0, and find g"!0". 104. Writing Use a graphing utility to graph the two functions f !x" # x 2 \$ 1 and g!x" # x \$ 1 in the same viewing window. Use the zoom and trace features to analyze the graphs near the point !0, 1". What do you observe? Which function is differentiable at this point? Write a short paragraph describing the geometric significance of differentiability at a point. (( SECTION 2.2 . Section 2.2 . Basic Differentiation Rules and Rates of Change Basic Differentiation Rules and Rates of Change Video Video Video Video • • • • • • Find the derivative of a function using the Constant Rule. Find the derivative of a function using the Power Rule. Find the derivative of a function using the Constant Multiple Rule. Find the derivative of a function using the Sum and Difference Rules. Find the derivatives of the sine function and of the cosine function. Use derivatives to find rates of change. The Constant Rule y In Section 2.1 you used the limit definition to find derivatives. In this and the next two sections you will be introduced to several “differentiation rules” that allow you to find derivatives without the direct use of the limit definition. The slope of a horizontal line is 0. THEOREM 2.2 f (x) = c The derivative of a constant function is 0. The Constant Rule The derivative of a constant function is 0. That is, if c is a real number, then d #c\$ " 0. dx x The Constant Rule Proof Figure 2.14 Let f !x" " c. Then, by the limit definition of the derivative, d #c\$ " f!!x" dx f !x & %x" \$ f !x" %x c\$c " lim %x→0 %x " lim %x→0 NOTE In Figure 2.14, note that the Constant Rule is equivalent to saying that the slope of a horizontal line is 0. This demonstrates the relationship between slope and derivative. " lim 0 %x→0 " 0. EXAMPLE 1 Using the Constant Rule Function Derivative dy "0 dx f!!x" " 0 s!!t" " 0 y! " 0 a. y " 7 . b. f !x" " 0 c. s!t" " \$3 d. y " k# 2, k is constant Try It . Exploration A The editable graph feature below allows you to edit the graph of a function. . 107 a. Editable Graph b. Editable Graph c. Editable Graph d. Editable Graph E X P L O R AT I O N Writing a Conjecture Use the definition of the derivative given in Section 2.1 to find the derivative of each function. What patterns do you see? Use your results to write a conjecture about the derivative of f !x" " x n. a. f !x" " x1 d. f !x" " x4 b. f !x" " x 2 e. f !x" " x1%2 c. f !x" " x 3 f. f !x" " x\$1 108 CHAPTER 2 Differentiation The Power Rule Before proving the next rule, it is important to review the procedure for expanding a binomial. !x & %x" 2 " x 2 & 2x%x & !%x" 2 !x & %x" 3 " x 3 & 3x 2%x & 3x!%x"2 & !%x"3 The general binomial expansion for a positive integer n is !x & %x" n " x n & nx n\$1 !%x" & n!n \$ 1"x n\$2 !%x" 2 & . . . & !%x" n. 2 !%x"2 is a factor of these terms. This binomial expansion is used in proving a special case of the Power Rule. THEOREM 2.3 The Power Rule If n is a rational number, then the function f !x" " x n is differentiable and d n #x \$ " nx n\$1. dx For f to be differentiable at x " 0, n must be a number such that x n\$1 is defined on an interval containing 0. Proof If n is a positive integer greater than 1, then the binomial expansion produces d n !x & %x"n \$ x n #x \$ " lim dx %x→0 %x n!n \$ 1"x n\$2 !%x" 2 & . . . & !%x" n \$ x n 2 " lim %x %x→0 n\$2 n!n \$ 1"x " lim nx n\$1 & !%x" & . . . & !%x" n\$1 2 %x→0 " nx n\$1 & 0 & . . . & 0 " nx n\$1. x n & nx n\$1!%x" & & ' This proves the case for which n is a positive integer greater than 1. You will prove the case for n " 1. Example 7 in Section 2.3 proves the case for which n is a negative integer. In Exercise 75 in Section 2.5 you are asked to prove the case for which n is rational. (In Section 5.5, the Power Rule will be extended to cover irrational values of n.) y 4 3 y=x When using the Power Rule, the case for which n " 1 is best thought of as a separate differentiation rule. That is, 2 1 x 1 2 3 Power Rule when n " 1 4 The slope of the line y " x is 1. Figure 2.15 d #x\$ " 1. dx This rule is consistent with the fact that the slope of the line y " x is 1, as shown in Figure 2.15. SECTION 2.2 EXAMPLE 2 Basic Differentiation Rules and Rates of Change Using the Power Rule Function Derivative a. f !x" " x 3 f!!x) " 3x 2 d 1%3 1 1 g!!x" " #x \$ " x\$2%3 " 2%3 dx 3 3x dy d \$2 2 " #x \$ " !\$2"x\$3 " \$ 3 dx dx x 3 x b. g!x" " ( . c. y " 109 1 x2 Exploration A Try It In Example 2(c), note that before differentiating, 1%x 2 was rewritten as x\$2. Rewriting is the first step in many differentiation problems. Rewrite: Given: 1 y" 2 x y f (x) = x 4 Simplify: dy 2 "\$ 3 dx x Differentiate: dy " !\$2"x\$3 dx y " x\$2 2 EXAMPLE 3 (−1, 1) 1 Finding the Slope of a Graph Find the slope of the graph of f !x" " x 4 when (1, 1) a. x " \$1 x (0, 0) −1 1 Note that the slope of the graph is negative at the point !\$1, 1", the slope is zero at the point !0, 0", and the slope is positive at the .. point !1, 1". Figure 2.16 a. When x " \$1, the slope is f!!\$1" " 4!\$1"3 " \$4. b. When x " 0, the slope is f!!0" " 4!0"3 " 0. c. When x " 1, the slope is f!!1" " 4!1"3 " 4. f (x) = x 2 Slope is positive. Exploration A Open Exploration Finding an Equation of a Tangent Line Solution To find the point on the graph of f, evaluate the original function at x " \$2. 4 3 !\$2, f !\$2"" " !\$2, 4" Point on graph To find the slope of the graph when x " \$2, evaluate the derivative, f!!x" " 2x, at x " \$2. 2 m " f!!\$2" " \$4 1 x 1 2 y = −4x − 4 The line y " \$ 4x \$ 4 is tangent to the .. graph of f !x" " x 2 at the point !\$ 2, 4". Editable Graph Slope is zero. Find an equation of the tangent line to the graph of f !x" " x 2 when x " \$2. y Figure 2.17 Slope is negative. See Figure 2.16. EXAMPLE 4 −2 c. x " 1. Solution The slope of a graph at a point is the value of the derivative at that point. The derivative of f is f!!x" " 4x3. Try It Editable Graph (−2, 4) b. x " 0 Slope of graph at !\$2, 4" Now, using the point-slope form of the equation of a line, you can write y \$ y1 " m!x \$ x1" y \$ 4 " \$4#x \$ !\$2"\$ y " \$4x \$ 4. Point-slope form Substitute for y1, m, and x1. Simplify. See Figure 2.17. Try It Exploration A Exploration B Open Exploration 110 CHAPTER 2 Differentiation The Constant Multiple Rule THEOREM 2.4 The Constant Multiple Rule If f is a differentiable function and c is a real number, then cf is also d differentiable and #cf !x"\$ " cf!!x". dx Proof d cf !x & %x" \$ cf !x" #cf !x"\$ " lim %x→0 dx %x f !x & %x" \$ f !x" " lim c %x→0 %x f !x & %x" \$ f !x" " c lim %x→0 %x " cf!!x" & Definition of derivative ' ' & Apply Theorem 1.2. Informally, the Constant Multiple Rule states that constants can be factored out of the differentiation process, even if the constants appear in the denominator. d d #cf !x"\$ " c # dx dx f !x"\$ " cf!!x" d f !x" d 1 " f !x" dx c dx c 1 d 1 " # f !x"\$ " f!!x" c dx c & ' EXAMPLE 5 Function a. y " 2 x b. f !t" " 4t 2 5 c. y " 2(x 1 3 x2 2( 3x e. y " \$ 2 d. y " . Try It &) * ' )* )* Using the Constant Multiple Rule Derivative dy d d 2 " #2x\$1\$ " 2 #x\$1\$ " 2!\$1"x\$2 " \$ 2 dx dx dx x d 4 2 4 d 2 4 8 f!!t" " t " #t \$ " !2t" " t dt 5 5 dt 5 5 dy d 1 1 " #2x1%2\$ " 2 x\$1%2 " x\$1%2 " dx dx 2 (x dy d 1 \$2%3 1 2 1 " x " \$ x\$5%3 " \$ 5%3 dx dx 2 2 3 3x d 3 3 3 y! " \$ x " \$ !1" " \$ dx 2 2 2 & ' ) & & ' * ) * ' Exploration A The Constant Multiple Rule and the Power Rule can be combined into one rule. The combination rule is Dx #cx n\$ " cnx n\$1. SECTION 2.2 EXAMPLE 6 111 Using Parentheses When Differentiating Original Function Rewrite Differentiate 5 2x 3 5 b. y " !2x"3 7 c. y " \$2 3x 7 d. y " !3x"\$2 5 y " !x\$3" 2 5 y " !x\$3" 8 7 y " !x 2" 3 5 y! " !\$3x\$4" 2 5 y! " !\$3x\$4" 8 7 y! " !2x" 3 y " 63!x 2" y! " 63!2x" Try It Exploration A a. y " . Basic Differentiation Rules and Rates of Change Simplify 15 2x 4 15 y! " \$ 4 8x 14x y! " 3 y! " \$ y! " 126x The Sum and Difference Rules THEOREM 2.5 The Sum and Difference Rules The sum (or difference) of two differentiable functions f and g is itself differentiable. Moreover, the derivative of f & g !or f \$ g" is the sum (or difference) of the derivatives of f and g. d # f !x" & g!x"\$ " f!!x" & g!!x" dx d # f !x" \$ g!x"\$ " f!!x" \$ g!!x" dx Sum Rule Difference Rule Proof A proof of the Sum Rule follows from Theorem 1.2. (The Difference Rule can be proved in a similar way.) d # f !x & %x" & g!x & %x"\$ \$ # f !x" & g!x"\$ # f !x" & g!x"\$ " lim %x→0 dx %x f !x & %x" & g!x & %x" \$ f !x" \$ g!x" " lim %x→0 %x f !x & %x" \$ f !x" g!x & %x" \$ g!x" " lim & %x→0 %x %x f !x & %x" \$ f !x" g!x & %x" \$ g!x" " lim & lim %x→0 %x→0 %x %x " f!!x" & g!!x" & ' The Sum and Difference Rules can be extended to any finite number of functions. For instance, if F!x" " f !x" & g!x" \$ h!x", then F!!x" " f!!x" & g!!x" \$ h!!x". EXAMPLE 7 Using the Sum and Difference Rules Function . a. f !x" " x 3 \$ 4x & 5 x4 b. g!x" " \$ & 3x 3 \$ 2x 2 Try It . Derivative f!!x" " 3x 2 \$ 4 g!!x" " \$2x 3 & 9x 2 \$ 2 Exploration A The editable graph feature below allows you to edit the graph of a function and its derivative. Editable Graph 112 CHAPTER 2 Differentiation FOR FURTHER INFORMATION For the outline of a geometric proof of the derivatives of the sine and cosine functions, see the article “The Spider’s Spacewalk Derivation of sin! and cos! ” by Tim . Hesterberg in The College Mathematics Journal. MathArticle Derivatives of Sine and Cosine Functions In Section 1.3, you studied the following limits. sin %x "1 %x→0 %x and lim 1 \$ cos %x "0 %x→0 %x lim These two limits can be used to prove differentiation rules for the sine and cosine functions. (The derivatives of the other four trigonometric functions are discussed in Section 2.3.) THEOREM 2.6 Derivatives of Sine and Cosine Functions d #cos x\$ " \$sin x dx d #sin x\$ " cos x dx y y′ = 0 1 y′ = −1 y′ = 1 π y′ = 1 π 2 −1 Proof y = sin x 2π x y′ = 0 y decreasing y increasing y increasing y′ positive y′ negative y ′ positive y π 2 −1 π 2π y ′ = cos x The derivative of the sine function is the . function. cosine Figure 2.18 Animation y = 2 sin x 2 −# x d sin!x & %x" \$ sin x Definition of derivative #sin x\$ " lim %x→0 dx %x sin x cos %x & cos x sin %x \$ sin x " lim %x→0 %x cos x sin %x \$ !sin x"!1 \$ cos %x" " lim %x→0 %x sin %x 1 \$ cos %x " lim !cos x" \$ !sin x" %x→0 %x %x sin %x 1 \$ cos %x " cos x lim \$ sin x lim %x→0 %x→0 %x %x " !cos x"!1" \$ !sin x"!0" " cos x & ) # * ) ) ' Derivatives Involving Sines and Cosines Function 3 sin x 2 * This differentiation rule is shown graphically in Figure 2.18. Note that for each x, the slope of the sine curve is equal to the value of the cosine. The proof of the second rule is left as an exercise (see Exercise 116). EXAMPLE 8 y= ) a. y " 2 sin x sin x 1 b. y " " sin x 2 2 c. y " x & cos x Derivative y! " 2 cos x 1 cos x y! " cos x " 2 2 y! " 1 \$ sin x A graphing utility can provide insight into the interpretation of a derivative. For instance, Figure 2.19 shows the graphs of TECHNOLOGY −2 y = sin x y= 1 sin x 2 d #a sin x\$ " a cos x dx. Figure 2.19 y " a sin x for a " 12, 1, 32, and 2. Estimate the slope of each graph at the point !0, 0". Then verify your estimates analytically by evaluating the derivative of each function when x " 0. Try It Exploration A Open Exploration SECTION 2.2 Basic Differentiation Rules and Rates of Change 113 Rates of Change You have seen how the derivative is used to determine slope. The derivative can also be used to determine the rate of change of one variable with respect to another. Applications involving rates of change occur in a wide variety of fields. A few examples are population growth rates, production rates, water flow rates, velocity, and acceleration. A common use for rate of change is to describe the motion of an object moving in a straight line. In such problems, it is customary to use either a horizontal or a vertical line with a designated origin to represent the line of motion. On such lines, movement to the right (or upward) is considered to be in the positive direction, and movement to the left (or downward) is considered to be in the negative direction. The function s that gives the position (relative to the origin) of an object as a function of time t is called a position function. If, over a period of time %t, the object changes its position by the amount %s " s!t & %t" \$ s!t", then, by the familiar formula Rate " distance time the average velocity is Change in distance %s " . Change in time %t EXAMPLE 9 Average velocity Finding Average Velocity of a Falling Object If a billiard ball is dropped from a height of 100 feet, its height s at time t is given by the position function s " \$16t 2 & 100 Position function where s is measured in feet and t is measured in seconds. Find the average velocity over each of the following time intervals. a. #1, 2\$ b. #1, 1.5\$ c. #1, 1.1\$ Solution a. For the interval #1, 2\$, the object falls from a height of s!1" " \$16!1"2 & 100 " 84 feet to a height of s!2" " \$16!2"2 & 100 " 36 feet. The average velocity is %s 36 \$ 84 \$48 " " " \$48 feet per second. %t 2\$1 1 b. For the interval #1, 1.5\$, the object falls from a height of 84 feet to a height of 64 feet. The average velocity is %s 64 \$ 84 \$20 " " " \$40 feet per second. %t 1.5 \$ 1 0.5 c. For the interval #1, 1.1\$, the object falls from a height of 84 feet to a height of 80.64 feet. The average velocity is %s 80.64 \$ 84 \$3.36 " " " \$33.6 feet per second. %t 1.1 \$ 1 0.1 . Note that the average velocities are negative, indicating that the object is moving downward. Try It Exploration A Exploration B 114 CHAPTER 2 Differentiation s Suppose that in Example 9 you wanted to find the instantaneous velocity (or simply the velocity) of the object when t " 1. Just as you can approximate the slope of the tangent line by calculating the slope of the secant line, you can approximate the velocity at t " 1 by calculating the average velocity over a small interval #1, 1 & %t\$ (see Figure 2.20). By taking the limit as %t approaches zero, you obtain the velocity when t " 1. Try doing this—you will find that the velocity when t " 1 is \$32 feet per second. In general, if s " s!t" is the position function for an object moving along a straight line, the velocity of the object at time t is Tangent line P Secant line t1 = 1 t2 t The average velocity between t1 and t2 is the slope of the secant line, and the instantaneous velocity at t1 is the slope of the. tangent line. Figure 2.20 Animation v!t" " lim %t→0 s!t & %t" \$ s!t" " s!!t". %t Velocity function In other words, the velocity function is the derivative of the position function. Velocity can be negative, zero, or positive. The speed of an object is the absolute value of its velocity. Speed cannot be negative. The position of a free-falling object (neglecting air resistance) under the influence of gravity can be represented by the equation s!t" " 1 2 gt & v0t & s0 2 Position function where s0 is the initial height of the object, v0 is the initial velocity of the object, and g is the acceleration due to gravity. On Earth, the value of g is approximately \$32 feet per second per second or \$9.8 meters per second per second. . History EXAMPLE 10 32 ft Using the Derivative to Find Velocity At time t " 0, a diver jumps from a platform diving board that is 32 feet above the water (see Figure 2.21). The position of the diver is given by s!t" " \$16t2 & 16t & 32 Position function where s is measured in feet and t is measured in seconds. a. When does the diver hit the water? b. What is the diver’s velocity at impact? Solution Velocity is positive when an object is rising, . is negative when an object is falling. and Figure 2.21 Animation NOTE In Figure 2.21, note that the diver moves upward for the first halfsecond because the velocity is positive for 0 < t < 12. When the velocity is 0, . diver has reached the maximum the height of the dive. a. To find the time t when the diver hits the water, let s " 0 and solve for t. \$16t 2 & 16t & 32 " 0 \$16!t & 1"!t \$ 2" " 0 t " \$1 or 2 Set position function equal to 0. Factor. Solve for t. Because t ≥ 0, choose the positive value to conclude that the diver hits the water at t " 2 seconds. b. The velocity at time t is given by the derivative s!!t" " \$32t & 16. So, the velocity at time t " 2 is s!!2" " \$32!2" & 16 " \$48 feet per second. Try It Exploration A SECTION 2.2 115 Basic Differentiation Rules and Rates of Change Exercises for Section 2.2 The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. Click on to view the complete solution of the exercise. Click on to print an enlarged copy of the graph. In Exercises 1 and 2, use the graph to estimate the slope of the tangent line to y ! xn at the point &1, 1'. Verify your answer analytically. To print an enlarged copy of the graph, select the MathGraph button. 1. (a) y ! x1%2 29. y ! 30. y ! (b) y ! x 3 y Original Function Rewrite Differentiate Simplify #x x 4 x#3 y 2 In Exercises 31–38, find the slope of the graph of the function at the given point. Use the derivative feature of a graphing utility 2 1 1 (1, 1) (1, 1) Function x 1 x 2 1 2. (a) y ! x#1\$2 2 y 2 2 (1, 1) 1 2 x 1 2 3. y ! 8 4. f !x" ! #2 5. y ! 6. y ! !0, # 12 " 34. y ! 3x 3 # 6 !2, 18" !0, 1" !5, 0" !0, 0" !\$, #1" 37. f !%" ! 4 sin % # % In Exercises 39–52, find the derivative of the function. 39. f !x" ! x 2 " 5 # 3x #2 x8 41. g!t" ! t 2 # 1 8. y ! 8 x 5 x 9. f !x" ! # 1 7 33. f !x" ! # 2 " 5x 3 38. g!t" ! 2 " 3 cos t In Exercises 3 –24, find the derivative of the function. 1 7. y ! 7 x !35, 2" 36. f !x" ! 3!5 # x"2 3 x6 3 5t 35. y ! !2x " 1" x 1 !1, 3" 2 (1, 1) 1 3 x2 32. f !t" ! 3 # (b) y ! x#1 y 31. f !x" ! Point 43. f !x" ! 4 x 10. g!x" ! # 4 t3 40. f !x" ! x 2 # 3x # 3x#2 42. f !x" ! x " x 3 # 3x 2 " 4 x2 44. h!x" ! 1 x2 2x 2 # 3x " 1 x 11. f !x" ! x " 1 12. g!x" ! 3x # 1 45. y ! x!x 2 " 1" 13. f !t" ! #2t 2 " 3t # 6 14. y ! t 2 " 2t # 3 3 x 47. f !x" ! #x # 6 # 3 x "# 5 x 48. f !x" ! # 15. g!x" ! 16. y ! 8 # 49. h!s" ! s 50. f !t" ! t 2\$3 # t1\$3 " 4 x2 " 4x 3 17. s!t" ! t 3 # 2t " 4 19. y ! 18. f !x" ! 2x 3 # x 2 " 3x \$ sin % # cos % 2 22. y ! 5 " sin x 1 # 3 sin x x 24. y ! 5 " 2 cos x !2x"3 In Exercises 25–30, complete the table. Original Function 25. y ! 5 2x 2 2 26. y ! 2 3x 3 27. y ! !2x" 3 28. y ! \$ !3x" 2 4\$5 #s 2\$3 51. f !x" ! 6#x " 5 cos x 52. f !x" ! 20. g!t" ! \$ cos t 1 21. y ! x 2 # 2 cos x 23. y ! x3 46. y ! 3x!6x # 5x 2" Rewrite Differentiate " 3 cos x In Exercises 53–56, (a) find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. Function Simplify 2 3 x # 53. y ! x 4 # 3x 2 " 2 54. y ! x 3 " x 55. f !x" ! 2 4 3 # x 56. y ! !x 2 " 2x"!x " 1" Point !1, 0" !#1, #2" !1, 2" !1, 6" 116 CHAPTER 2 Differentiation In Exercises 57–62, determine the point(s) (if any) at which the graph of the function has a horizontal tangent line. 57. y ! x 4 # 8x 2 " 2 In Exercises 71 and 72, the graphs of a function f and its derivative f" are shown on the same set of coordinate axes. Label the graphs as f or f" and write a short paragraph stating the criteria used in making the selection. To print an enlarged copy of the graph, select the MathGraph button. 58. y ! x 3 " x 59. y ! 1 x2 60. y ! x 2 " 1 71. 61. y ! x " sin x, 0 f x < 2\$ 72. y y 2 1 3 62. y ! #3x " 2 cos x, 0 f x < 2\$ 1 In Exercises 63–66, find k such that the line is tangent to the graph of the function. 63. f !x" ! 1 2 3 4 1 2 3 2 Line Function x2 3 2 1 x 2 1 x y ! 4x # 9 # kx 64. f !x" ! k # x 2 y ! #4x " 7 k 65. f !x" ! x 3 y!# x"3 4 66. f !x" ! k#x y!x"4 67. Use the graph of f to answer each question. To print an enlarged copy of the graph, select the MathGraph button. y 73. Sketch the graphs of y ! x 2 and y ! #x 2 " 6x # 5, and sketch the two lines that are tangent to both graphs. Find equations of these lines. 74. Show that the graphs of the two equations y ! x and y ! 1\$x have tangent lines that are perpendicular to each other at their point of intersection. 75. Show that the graph of the function f !x" ! 3x " sin x " 2 does not have a horizontal tangent line. f 76. Show that the graph of the function B C A D f !x" ! x5 " 3x3 " 5x E x (a) Between which two consecutive points is the average rate of change of the function greatest? (b) Is the average rate of change of the function between A and B greater than or less than the instantaneous rate of change at B? (c) Sketch a tangent line to the graph between C and D such that the slope of the tangent line is the same as the average rate of change of the function between C and D. 68. Sketch the graph of a function f such that f& > 0 for all x and the rate of change of the function is decreasing. In Exercises 69 and 70, the relationship between f and g is given. Explain the relationship between f" and g". does not have a tangent line with a slope of 3. In Exercises 77 and 78, find an equation of the tangent line to the graph of the function f through the point &x0, y0' not on the graph. To find the point of tangency &x, y' on the graph of f , solve the equation f"&x' ! y0 # y . x0 # x 77. f !x" ! #x !x0, y0" ! !#4, 0" 78. f !x" ! 2 x !x0, y0" ! !5, 0" 79. Linear Approximation Use a graphing utility, with a square window setting, to zoom in on the graph of 69. g!x" ! f !x" " 6 f !x" ! 4 # 12 x 2 70. g!x" ! #5 f !x" to approximate f& !1". Use the derivative to find f& !1". 80. Linear Approximation Use a graphing utility, with a square window setting, to zoom in on the graph of f !x" ! 4#x " 1 to approximate f& !4". Use the derivative to find f& !4". SECTION 2.2 81. Linear Approximation Consider the function f !x" ! x 3'2 with the solution point !4, 8". (a) Use a graphing utility to graph f. Use the zoom feature to obtain successive magnifications of the graph in the neighborhood of the point !4, 8". After zooming in a few times, the graph should appear nearly linear. Use the trace feature to determine the coordinates of a point near !4, 8". Find an equation of the secant line S!x" through the two points. 117 Basic Differentiation Rules and Rates of Change #1 , x 91. f !x" ! (1, 2) 92. f !x" ! sin x, 0, 6 + \$ Vertical Motion In Exercises 93 and 94, use the position function s!t" ! #16 t 2 \$ v0 t \$ s0 for free-falling objects. 93. A silver dollar is dropped from the top of a building that is 1362 feet tall. (a) Determine the position and velocity functions for the coin. (b) Find the equation of the line T !x" ! f&!4"!x # 4" " f !4" (b) Determine the average velocity on the interval (1, 2). tangent to the graph of f passing through the given point. Why are the linear functions S and T nearly the same? (d) Find the time required for the coin to reach ground level. #3 #2 #1 #0.5 #0.1 0 f &4 \$ %x' T&4 \$ %x' %x 0.1 0.5 1 2 3 f &4 \$ %x' T&4 \$ %x' 82. Linear Approximation Repeat Exercise 81 for the function f !x" ! x 3 where T!x" is the line tangent to the graph at the point !1, 1". Explain why the accuracy of the linear approximation decreases more rapidly than in Exercise 81. 95. A projectile is shot upward from the surface of Earth with an initial velocity of 120 meters per second. What is its velocity after 5 seconds? After 10 seconds? 96. To estimate the height of a building, a stone is dropped from the top of the building into a pool of water at ground level. How high is the building if the splash is seen 6.8 seconds after the stone is dropped? Think About It In Exercises 97 and 98, the graph of a position function is shown. It represents the distance in miles that a person drives during a 10-minute trip to work. Make a sketch of the corresponding velocity function. 97. True or False? In Exercises 83–88, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 83. If f&!x" ! g&!x", then f !x" ! g!x". 84. If f !x" ! g!x" " c, then f&!x" ! g&!x". 86. If y ! x\$\$, then dy\$dx ! 1\$\$. 87. If g!x" ! 3 f !x", then g& !x" ! 3f&!x". 88. If f !x" ! 1\$x n, then f& !x" ! 1\$!nx n#1". In Exercises 89–92, find the average rate of change of the function over the given interval. Compare this average rate of change with the instantaneous rates of change at the endpoints of the interval. 89. f !t" ! 2t " 7, (1, 2) 90. f !t" ! t2 # 3, (2, 2.1) 10 8 6 4 2 (10, 6) (4, 2) (6, 2) (0, 0) 2 4 6 8 10 Time (in minutes) t s 10 8 6 4 2 (6, 5) (10, 6) (8, 5) (0, 0) 2 4 6 8 10 Time (in minutes) t Think About It In Exercises 99 and 100, the graph of a velocity function is shown. It represents the velocity in miles per hour during a 10-minute drive to work. Make a sketch of the corresponding position function. 99. 100. v Velocity (in mph) 85. If y ! \$ 2, then dy\$dx ! 2\$. 98. s Distance (in miles) %x Vertical Motion In Exercises 95 and 96, use the position function s&t' ! #4.9t 2 \$ v0 t \$ s0 for free-falling objects. 60 50 40 30 20 10 t 2 4 6 8 10 Time (in minutes) Velocity (in mph) (d) Demonstrate the conclusion in part (c) by completing the table. (e) Find the velocity of the coin at impact. 94. A ball is thrown straight down from the top of a 220-foot building with an initial velocity of #22 feet per second. What is its velocity after 3 seconds? What is its velocity after falling 108 feet? Distance (in miles) (c) Use a graphing utility to graph f and T on the same set of coordinate axes. Note that T is a good approximation of f when x is close to 4. What happens to the accuracy of the approximation as you move farther away from the point of tangency? (c) Find the instantaneous velocities when t ! 1 and t ! 2. v 60 50 40 30 20 10 t 2 4 6 8 10 Time (in minutes) 118 CHAPTER 2 Differentiation 101. Modeling Data The stopping distance of an automobile, on dry, level pavement, traveling at a speed v (kilometers per hour) is the distance R (meters) the car travels during the reaction time of the driver plus the distance B (meters) the car travels after the brakes are applied (see figure). The table shows the results of an experiment. Reaction time Braking distance R B Driver sees obstacle Driver applies brakes Car stops 40 60 80 100 Reaction Time Distance, R 8.3 16.7 25.0 33.3 41.7 Braking Time Distance, B 2.3 The annual inventory cost C for a 1,008,000 " 6.3Q Q 107. Writing The number of gallons N of regular unleaded gasoline sold by a gasoline station at a price of p dollars per gallon is given by N ! f ! p". (a) Describe the meaning of f&!1.479". (b) Is f&!1.479" usually positive or negative? Explain. 9.0 20.2 35.8 55.9 (a) Use the regression capabilities of a graphing utility to find a linear model for reaction time distance. (b) Use the regression capabilities of a graphing utility to find a quadratic model for braking distance. (c) Determine the polynomial giving the total stopping distance T. (d) Use a graphing utility to graph the functions R, B, and T in the same viewing window. (e) Find the derivative of T and the rates of change of the total stopping distance for v ! 40, v ! 80, and v ! 100. (f) Use the results of this exercise to draw conclusions about the total stopping distance as speed increases. 102. Fuel Cost A car is driven 15,000 miles a year and gets x miles per gallon. Assume that the average fuel cost is \$1.55 per gallon. Find the annual cost of fuel C as a function of x and use this function to complete the table. 15 106. Inventory Management manufacturer is where Q is the order size when the inventory is replenished. Find the change in annual cost when Q is increased from 350 to 351, and compare this with the instantaneous rate of change when Q ! 350. 20 10 1 s!t" ! # 2at 2 " c. C! Speed, v x 105. Velocity Verify that the average velocity over the time interval (t0 # (t, t0 " (t) is the same as the instantaneous velocity at t ! t0 for the position function 20 25 30 35 40 C dC/dx Who would benefit more from a one-mile-per-gallon increase in fuel efficiency—the driver of a car that gets 15 miles per gallon or the driver of a car that gets 35 miles per gallon? Explain. 103. Volume The volume of a cube with sides of length s is given by V ! s3. Find the rate of change of the volume with respect to s when s ! 4 centimeters. 104. Area The area of a square with sides of length s is given by A ! s2. Find the rate of change of the area with respect to s when s ! 4 meters. 108. Newton’s Law of Cooling This law states that the rate of change of the temperature of an object is proportional to the difference between the object’s temperature T and the temperature Ta of the surrounding medium. Write an equation for this law. 109. Find an equation of the parabola y ! ax2 " bx " c that passes through !0, 1" and is tangent to the line y ! x # 1 at !1, 0". 110. Let !a, b" be an arbitrary point on the graph of y ! 1\$x, x > 0. Prove that the area of the triangle formed by the tangent line through !a, b" and the coordinate axes is 2. 111. Find the tangent line(s) to the curve y ! x3 # 9x through the point !1, #9". 112. Find the equation(s) of the tangent line(s) to the parabola y ! x 2 through the given point. (a) !0, a" (b) !a, 0" Are there any restrictions on the constant a? In Exercises 113 and 114, find a and b such that f is differentiable everywhere. -x " b, cos x, 114. f !x" ! ax " b, 113. f !x" ! ax3, 2 x f 2 x >2 x < 0 x v 0 , , ,, 115. Where are the functions f1!x" ! sin x and f2!x" ! sin x differentiable? 116. Prove that d (cos x) ! #sin x. dx FOR FURTHER INFORMATION For a geometric interpretation of the derivatives of trigonometric functions, see the article “Sines and Cosines of the Times” by Victor J. Katz in Math Horizons. MathArticle SECTION 2.3 Section 2.3 Product and Quotient Rules and Higher-Order Derivatives 119 Product and Quotient Rules and Higher-Order Derivatives • • • • Find the derivative of a function using the Product Rule. Find the derivative of a function using the Quotient Rule. Find the derivative of a trigonometric function. Find a higher-order derivative of a function. The Product Rule In Section 2.2 you learned that the derivative of the sum of two functions is simply the sum of their derivatives. The rules for the derivatives of the product and quotient of two functions are not as simple. THEOREM 2.7 NOTE A version of the Product Rule that some people prefer is d # f !x"g !x"\$ ! f\$ !x"g!x" " f !x"g\$!x". dx . The advantage of this form is that it generalizes easily to products involving three or more factors. The Product Rule The product of two differentiable functions f and g is itself differentiable. Moreover, the derivative of fg is the first function times the derivative of the second, plus the second function times the derivative of the first. d # f !x"g!x"\$ ! f !x"g\$!x" " g!x" f\$!x" dx Video Proof Some mathematical proofs, such as the proof of the Sum Rule, are straightforward. Others involve clever steps that may appear unmotivated to a reader. This proof involves such a step—subtracting and adding the same quantity—which is shown in color. d f !x " %x"g!x " %x" # f !x"g!x" # f !x"g!x"\$ ! lim dx %x→ 0 %x f !x " %x"g!x " %x" # f !x " %x"g!x" " f !x " %x"g!x" # f !x"g!x" ! lim %x→0 %x g!x " %x" # g!x" f !x " %x" # f !x" ! lim f !x " %x" " g!x" %x→ 0 %x %x g!x " %x" # g!x" f !x " %x" # f !x" ! lim f !x " %x" " lim g!x" %x→0 %x→0 %x %x g!x " %x" # g!x" f !x " %x" # f !x" ! lim f !x " %x" & lim " lim g!x" & lim %x→0 %x→0 %x→0 %x→0 %x %x ! f !x"g\$!x" " g!x"f\$!x" % % THE PRODUCT RULE When Leibniz originally wrote a formula for the Product Rule, he was motivated by the expression !x " dx"! y " dy" # xy from which he subtracted dx dy (as being negligible) and obtained the differential form x dy " y dx. This derivation resulted in the traditional form of the Product Rule. (Source:The History of Mathematics by David M. Burton) & & % & Note that lim f !x " %x" ! f !x" because f is given to be differentiable and therefore %x→ 0 is continuous. The Product Rule can be extended to cover products involving more than two factors. For example, if f, g, and h are differentiable functions of x, then d # f !x"g!x"h!x"\$ ! f\$!x"g!x"h!x" " f !x"g\$!x"h!x" " f !x"g!x"h\$!x". dx For instance, the derivative of y ! x2 sin x cos x is dy ! 2x sin x cos x " x2 cos x cos x " x2 sin x!#sin x" dx ! 2x sin x cos x " x2!cos2x # sin2x". 120 CHAPTER 2 Differentiation The derivative of a product of two functions is not (in general) given by the product of the derivatives of the two functions. To see this, try comparing the product of the derivatives of f !x" ! 3x # 2x 2 and g!x" ! 5 " 4x with the derivative in Example 1. Using the Product Rule EXAMPLE 1 Find the derivative of h!x" ! !3x # 2x2"!5 " 4x". Solution Derivative of second First Derivative of first Second d d #5 " 4x\$ " !5 " 4x" #3x # 2x2\$ dx dx ! !3x # 2x2"!4" " !5 " 4x"!3 # 4x" ! !12x # 8x2" " !15 # 8x # 16x2" ! #24x2 " 4x " 15 h\$!x" ! !3x # 2x2" Apply Product Rule. In Example 1, you have the option of finding the derivative with or without the Product Rule. To find the derivative without the Product Rule, you can write . Dx #!3x # 2x 2"!5 " 4x"\$ ! Dx ##8x 3 " 2x 2 " 15x\$ ! #24x 2 " 4x " 15. Exploration A Try It In the next example, you must use the Product Rule. Using the Product Rule EXAMPLE 2 Find the derivative of y ! 3x2 sin x. Solution . d d d #3x2 sin x\$ ! 3x2 #sin x\$ " sin x #3x2\$ dx dx dx ! 3x2 cos x " !sin x"!6x" ! 3x2 cos x " 6x sin x ! 3x!x cos x " 2 sin x" . Technology Exploration A Try It The editable graph feature below allows you to edit the graph of a function and its derivative. Editable Graph Using the Product Rule EXAMPLE 3 Find the derivative of y ! 2x cos x # 2 sin x. Solution Product Rule NOTE In Example 3, notice that you use the Product Rule when both factors of the product are variable, and you use the. Constant Multiple Rule when one of the factors is a constant. Apply Product Rule. ' ( Constant Multiple Rule ' ( dy d d d ! !2x" #cos x\$ " !cos x" #2x\$ # 2 #sin x\$ dx dx dx dx ! !2x"!#sin x" " !cos x"!2" # 2!cos x" ! #2x sin x Try It Exploration A Technology SECTION 2.3 121 Product and Quotient Rules and Higher-Order Derivatives The Quotient Rule THEOREM 2.8 The Quotient Rule The quotient f)g of two differentiable functions f and g is itself differentiable at all values of x for which g!x" ' 0. Moreover, the derivative of f)g is given by the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. d f !x" g!x" f\$!x" # f !x"g\$!x" ! , dx g!x" # g!x"\$ 2 % & . g!x" ' 0 Video Proof As with the proof of Theorem 2.7, the key to this proof is subtracting and f !x " %x" f !x" # d f !x" g!x " % x" g!x" Definition of derivative ! lim %x→ 0 dx g!x" %x g!x" f !x " %x" # f !x"g!x " %x" ! lim %x→ 0 %xg!x"g!x " %x" g!x"f !x " %x" # f !x"g!x" " f !x"g!x" # f !x"g!x " %x" ! lim %x→ 0 %xg!x"g !x " %x" g!x"# f !x " % x" # f !x"\$ f !x"# g!x " %x" # g!x"\$ lim # lim %x→ 0 %x→ 0 %x %x ! lim #g!x"g!x " %x"\$ % & %x→ 0 % g!x" lim TECHNOLOGY A graphing utility can be used to compare the graph of a function with the graph of its derivative. For instance, in Figure 2.22, the graph of the function in Example 4 appears to have two points that have horizontal tangent lines. What are the values of y\$ at these two points? y′ = −5x 2 + 4x + 5 (x 2 + 1) 2 . % & %x→0 Note that lim g!x " %x" ! g!x" because g is given to be differentiable and therefore %x→ 0 is continuous. EXAMPLE 4 Using the Quotient Rule 5x # 2 . x2 " 1 % & d d #5x # 2\$ # !5x # 2" #x 2 " 1\$ dx dx !x 2 " 1"2 !x 2 " 1"!5" # !5x # 2"!2x" ! !x 2 " 1" 2 !5x 2 " 5" # !10x 2 # 4x" ! !x 2 " 1" 2 #5x 2 " 4x " 5 ! !x 2 " 1"2 d 5x # 2 ! dx x 2 " 1 −4 Graphical comparison of a function and its derivative Figure 2.22 Try It . & Solution 6 8 5x − 2 x2 + 1 f !x " %x" # f !x" g!x " %x" # g!x" # f !x" lim %x→0 %x %x lim #g!x"g!x " %x"\$ g!x" f\$!x" # f !x"g\$!x" ! # g!x"\$ 2 Find the derivative of y ! −7 y= ! %x→0 !x 2 " 1" Exploration A Apply Quotient Rule. Exploration B The editable graph feature below allows you to edit the graph of a function and its derivative. Editable Graph 122 CHAPTER 2 Differentiation Note the use of parentheses in Example 4. A liberal use of parentheses is recommended for all types of differentiation problems. For instance, with the Quotient Rule, it is a good idea to enclose all factors and derivatives in parentheses, and to pay special attention to the subtraction required in the numerator. When differentiation rules were introduced in the preceding section, the need for rewriting before differentiating was emphasized. The next example illustrates this point with the Quotient Rule. EXAMPLE 5 Rewriting Before Differentiating Find an equation of the tangent line to the graph of f !x" ! Solution Begin by rewriting the function. 3 # !1)x" x"5 1 x 3# x ! x!x " 5" 3x # 1 ! 2 x " 5x !x 2 " 5x"!3" # !3x # 1"!2x " 5" f \$ !x" ! !x 2 " 5x"2 !3x 2 " 15x" # !6x 2 " 13x # 5" ! !x 2 " 5x" 2 #3x 2 " 2x " 5 ! !x 2 " 5x"2 f !x" ! f(x) = ' 1 3− x x+5 y 5 4 3 y=1 (−1, 1) −7 − 6 − 5 − 4 − 3 − 2 −1 x 1 2 3 −2 −3 −4 −5 The line y ! 1 is tangent to the graph of f !.x" at the point !# 1, 1". Figure 2.23 . ( 3 # !1)x" at !#1, 1". x"5 Write original function. Multiply numerator and denominator by x. Rewrite. Quotient Rule Simplify. To find the slope at !#1, 1", evaluate f \$ !#1". f \$ !#1" ! 0 Slope of graph at !#1, 1" Then, using the point-slope form of the equation of a line, you can determine that the equation of the tangent line at !#1, 1" is y ! 1. See Figure 2.23. Exploration A Try It The editable graph feature below allows you to edit the graph of a function. Editable Graph Not every quotient needs to be differentiated by the Quotient Rule. For example, each quotient in the next example can be considered as the product of a constant times a function of x. In such cases it is more convenient to use the Constant Multiple Rule. EXAMPLE 6 Using the Constant Multiple Rule Original Function x 2 " 3x 6 4 5x b. y ! 8 #3!3x # 2x 2" c. y ! 7x 9 d. y ! 2 5x a. y ! NOTE To see the benefit of using the Constant Multiple Rule for some quotients, try using the Quotient Rule to differentiate the functions in Example . 6—you should obtain the same results, but with more work. Try It Rewrite Differentiate 1 y ! !x 2 " 3x" 6 5 y ! x4 8 3 y ! # !3 # 2x" 7 9 y ! !x#2" 5 1 y\$ ! !2x " 3" 6 5 y\$ ! !4x 3" 8 3 y\$ ! # !#2" 7 9 y\$ ! !#2x#3" 5 Exploration A Simplify 2x " 3 6 5 y\$ ! x 3 2 6 y\$ ! 7 18 y\$ ! # 3 5x y\$ ! SECTION 2.3 Product and Quotient Rules and Higher-Order Derivatives 123 In Section 2.2, the Power Rule was proved only for the case where the exponent n is a positive integer greater than 1. The next example extends the proof to include negative integer exponents. EXAMPLE 7 Proof of the Power Rule (Negative Integer Exponents) If n is a negative integer, there exists a positive integer k such that n ! #k. So, by the Quotient Rule, you can write % & d n d 1 #x \$ ! dx dx x k x k !0" # !1"!kx k#1" ! !x k"2 0 # kx k#1 ! x 2k ! #kx#k#1 ! nx n#1. Quotient Rule and Power Rule n ! #k So, the Power Rule Dx #x n\$ ! nx n#1 . Power Rule is valid for any integer. In Exercise 75 in Section 2.5, you are asked to prove the case for which n is any rational number. Try It Exploration A Derivatives of Trigonometric Functions Knowing the derivatives of the sine and cosine functions, you can use the Quotient Rule to find the derivatives of the four remaining trigonometric functions. THEOREM 2.9 . Derivatives of Trigonometric Functions d #tan x\$ ! sec 2 x dx d #sec x\$ ! sec x tan x dx d #cot x\$ ! #csc2x dx d #csc x\$ ! #csc x cot x dx Video Proof Considering tan x ! !sin x")!cos x" and applying the Quotient Rule, you obtain !cos x"!cos x" # !sin x"!#sin x" d #tan x\$ ! dx cos 2 x cos2 x " sin2 x ! cos2 x 1 ! cos2 x ! sec2 x. Apply Quotient Rule. The proofs of the other three parts of the theorem are left as an exercise (see Exercise 89). 124 CHAPTER 2 Differentiation EXAMPLE 8 NOTE Because of trigonometric identities, the derivative of a trigonometric function can take many forms. This presents a challenge when you are trying . the back of the text. Differentiating Trigonometric Functions Function Derivative dy ! 1 # sec2 x dx y\$ ! x!sec x tan x" " !sec x"!1" ! !sec x"!1 " x tan x" a. y ! x # tan x b. y ! x sec x Exploration A Try It EXAMPLE 9 Open Exploration Different Forms of a Derivative Differentiate both forms of y ! 1 # cos x ! csc x # cot x. sin x Solution 1 # cos x sin x !sin x"!sin x" # !1 # cos x"!cos x" y\$ ! sin2 x sin2 x " cos2 x # cos x ! sin2 x 1 # cos x ! sin2 x First form: y ! Second form: y ! csc x # cot x y\$ ! #csc x cot x " csc2 x To show that the two derivatives are equal, you can write . ' (' 1 # cos x 1 1 cos x ! # sin 2 x sin 2 x sin x sin x ! csc 2 x # csc x cot x. Try It Exploration A ( Technology The summary below shows that much of the work in obtaining a simplified form of a derivative occurs after differentiating. Note that two characteristics of a simplified form are the absence of negative exponents and the combining of like terms. f! x+ After Differentiating f! x+ After Simplifying Example 1 !3x # 2x2"!4" " !5 " 4x"!3 # 4x" #24x2 " 4x " 15 Example 3 !2x"!#sin x" " !cos x"!2" # 2!cos x" #2x sin x Example 4 !x2 " 1"!5" # !5x # 2"!2x" !x2 " 1" 2 #5x2 " 4x " 5 !x2 " 1"2 Example 5 !x2 " 5x"!3" # !3x # 1"!2x " 5" !x2 " 5x"2 #3x2 " 2x " 5 !x2 " 5x"2 Example 9 !sin x"!sin x" # !1 # cos x"!cos x" sin2 x 1 # cos x sin2 x SECTION 2.3 Product and Quotient Rules and Higher-Order Derivatives 125 Higher-Order Derivatives Just as you can obtain a velocity function by differentiating a position function, you can obtain an acceleration function by differentiating a velocity function. Another way of looking at this is that you can obtain an acceleration function by differentiating a position function twice. s!t" v!t" ! s\$!t" a!t" ! v\$!t" ! s( !t" NOTE: The second derivative of f is the derivative of the first derivative of f. Position function Velocity function Acceleration function The function given by a!t" is the second derivative of s!t" and is denoted by s( !t". The second derivative is an example of a higher-order derivative. You can define derivatives of any positive integer order. For instance, the third derivative is the derivative of the second derivative. Higher-order derivatives are denoted as follows. y\$, f\$!x", Second derivative: y(, f ( !x", Third derivative: y\$\$\$, f\$\$\$!x", Fourth derivative: y !4", f !4"!x", First derivative: dy , dx d 2y , dx 2 d 3y , dx 3 d4y , dx 4 d # f !x"\$, dx d2 # f !x"\$, dx 2 d3 # f !x"\$, dx 3 d4 # f !x"\$, dx 4 dny , dx n dn # f !x"\$, dx n Dx # y\$ Dx2 # y\$ Dx3# y\$ Dx4 # y\$ ! nth derivative: EXAMPLE 10 f !n"!x", y!n", Dxn # y\$ Finding the Acceleration Due to Gravity Because the moon has no atmosphere, a falling object on the moon encounters no air resistance. In 1971, astronaut David Scott demonstrated that a feather and a hammer fall at the same rate on the moon. The position function for each of these falling objects is given by s!t" ! #0.81t 2 " 2 where s!t" is the height in meters and t is the time in seconds. What is the ratio of Earth’s gravitational force to the moon’s? THE MOON The moon’s mass is 7.349 ) 1022 kilograms, and Earth’s mass is 5.976 ) 1024 kilograms. The moon’s radius is 1737 kilometers, and Earth’s radius is 6378 kilometers. Because the gravitational force on the surface of a planet is directly proportional to its mass and inversely proportional to the square of its radius, the ratio of the gravitational force on Earth to the gravitational force on the moon is .. !5.976 ) 1024")63782 , 6.03. !7.349 ) 1022")17372 Video Solution To find the acceleration, differentiate the position function twice. s!t" ! #0.81t 2 " 2 s\$!t" ! #1.62t s( !t" ! #1.62 Position function Velocity function Acceleration function So, the acceleration due to gravity on the moon is #1.62 meters per second per second. Because the acceleration due to gravity on Earth is #9.8 meters per second per second, the ratio of Earth’s gravitational force to the moon’s is Earth’s gravitational force #9.8 ! Moon’s gravitational force #1.62 , 6.05. Try It Exploration A 126 CHAPTER 2 Differentiation Exercises for Section 2.3 The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. Click on to view the complete solution of the exercise. Click on to print an enlarged copy of the graph. In Exercises 1–6, use the Product Rule to differentiate the function. 1. g#x\$ " #x 2 # 1\$#x 2 \$ 2x\$ 2. f #x\$ " #6x # 5\$#x 3 \$ 2\$ 3. h#t\$ " 4. g#s\$ " %s#4 \$ # # 4\$ 3 5. f #x\$ " x cos x 3 t % t2 s2 \$ 6. g#x\$ " %x sin x In Exercises 7–12, use the Quotient Rule to differentiate the function. 7. f #x\$ " 9. h#x\$ " 11. g#x\$ " x x2 # 1 8. g#t\$ " 3 x % x3 t2 # 2 2t \$ 7 s %s \$ 1 cos t 12. f #t\$ " 3 t 10. h#s\$ " #1 sin x x2 13. f #x\$ " # x3 Value of c \$ 3x\$# 2x 2 # 3x # 5\$ 14. f #x\$ " #x 2 \$ 2x # 1\$#x 3 \$ 1\$ 15. f #x\$ " x2 \$ 4 x\$3 36. f #x\$ " #x 2 \$ x\$#x 2 # 1\$#x 2 # x # 1\$ 37. f #x\$ " x2 # c2 , c is a constant x2 \$ c2 38. f #x\$ " c2 \$ x 2 , c is a constant c2 # x 2 46. h#s\$ " 47. y " Differentiate sin x x 1 \$ 10 csc s s sec x 48. y " x 3#1 \$ sin x\$ 2 cos x 49. y " \$csc x \$ sin x 50. y " x sin x # cos x 51. f #x\$ " x 2 tan x 52. f #x\$ " sin x cos x 53. y " 2x sin x # x2 54. h#%\$ " 5% sec % # % tan % cos x In Exercises 55–58, use a computer algebra system to differentiate the function. Simplify !xx ## 12"#2x \$ 5\$ x \$x\$3 f #x\$ " ! #x # x # 1\$ x #1 " 55. g#x\$ " 56. 2 2 57. g#%\$ " 7 21. y " 3 3x 2 % 1 \$ sin % 58. f #%\$ " sin % 1 \$ cos % In Exercises 59–62, evaluate the derivative of the function at the given point. Use a graphing utility to verify your result. Function 59. y " 1 # csc x 1 \$ csc x 60. f #x\$ " tan x cot x In Exercises 25–38, find the derivative of the algebraic function. 3 \$ 2x \$ x 2 x2 \$ 1 35. f #x\$ " #3x3 # 4x\$#x \$ 5\$#x # 1\$ 4 t # 8 sec t 45. g#t\$ " % 5x 2 \$ 3 20. y " 4 25. f #x\$ " !2x \$ x #1 1" 44. y " x # cot x x 2 # 2x 19. y " 3 3x 2 \$ 5 7 34. g#x\$ " x 2 43. f #x\$ " \$x # tan x In Exercises 19–24, complete the table without using the Quotient Rule. 24. y " 1 x 33. f #x\$ " x\$3 2\$ c"1 ! c" 6 4x 3&2 x 32. h#x\$ " #x2 \$ 1\$2 42. f #x\$ " sin x 18. f #x\$ " x 23. y " " 3 x#%x # 3\$ 30. f #x\$ " % 29. f #x\$ " c"1 17. f #x\$ " x cos x 4 5x 2 2x # 5 %x 31. h#s\$ " #s3 \$ 2\$2 2 x#1 40. f #%\$ " #% # 1\$ cos % ! c" 4 22. y " ! 28. f #x\$ " x 4 1 \$ cos t 41. f #t\$ " t c"2 Rewrite " 39. f #t\$ " t 2 sin t c"0 x#1 16. f #x\$ " x\$1 Function 4 x#3 In Exercises 39–54, find the derivative of the trigonometric function. In Exercises 13–18, find f !'x( and f !'c(. Function ! 27. f #x\$ " x 1 \$ 26. f #x\$ " x 3 # 3x # 2 x2 \$ 1 61. h#t\$ " sec t t 62. f #x\$ " sin x#sin x # cos x\$ Point !!6 , \$3" #1, 1\$ !!, \$ !1 " !!4 , 1" SECTION 2.3 127 Product and Quotient Rules and Higher-Order Derivatives In Exercises 63–68, (a) find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. In Exercises 79 and 80, verify that f! 'x( " g!'x(, and explain the relationship between f and g. 79. f #x\$ " 3x 5x # 4 , g#x\$ " x#2 x#2 63. f #x\$ " #x3 \$ 3x # 1\$#x # 2\$, 80. f #x\$ " sin x \$ 3x sin x # 2x , g#x\$ " x x 64. f #x\$ " #x \$ 1\$#x 2 \$ 2\$, x , #2, 2\$ x\$1 65. f #x\$ " #0, 2\$ 66. f #x\$ " !!4 , 1" 67. f #x\$ " tan x, #1, \$3\$ #x \$ 1\$ , #x # 1\$ !2, 13" !!3 , 2" 68. f #x\$ " sec x, In Exercises 81 and 82, use the graphs of f and g. Let f 'x( p 'x( " f 'x(g'x( and q'x( " . g'x( Famous Curves In Exercises 69–72, find an equation of the tangent line to the graph at the given point. (The graphs in Exercises 69 and 70 are called witches of Agnesi. The graphs in Exercises 71 and 72 are called serpentines.) 69. 70. y 6 4 4 3 2 x 2 2 4 4 f (x) = 2 72. y 4 3 2 1 4 (2, ) 8 5 (2, ) 4 5 x 1 2 3 4 f (x) = 8 4x x2 + 6 In Exercises 73–76, determine the point(s) at which the graph of the function has a horizontal tangent line. 73. f #x\$ " 75. f #x\$ " x2 x\$1 4x \$ 2 x2 74. f #x\$ " 76. f #x\$ " 8 f g 4 g 2 x 2 4 6 8 2 10 x 2 4 6 8 10 83. Area The length of a rectangle is given by 2t # 1 and its height is %t, where t is time in seconds and the dimensions are in centimeters. Find the rate of change of the area with respect to time. 4 x 8 10 f 8 2 y 16x f(x) = 2 x + 16 4 10 27 x2 + 9 x 2 y 6 2 8 (b) Find q&#7\$. y 2 2 71. (b) Find q&#4\$. 4 (3, ) (2, 1) 82. (a) Find p&#4\$. y 6 8 f(x) = 2 x +4 81. (a) Find p&#1\$. x2 x2 # 1 x\$4 x2 \$ 7 77. Tangent Lines Find equations of the tangent lines to the x#1 graph of f #x\$ " that are parallel to the line 2y # x " 6. x\$1 Then graph the function and the tangent lines. 78. Tangent Lines Find equations of the tangent lines to the x graph of f #x\$ " that pass through the point #\$1, 5\$. x\$1 Then graph the function and the tangent lines. 84. Volume The radius of a right circular cylinder is given by 1 %t # 2 and its height is 2 %t, where t is time in seconds and the dimensions are in inches. Find the rate of change of the volume with respect to time. 85. Inventory Replenishment The ordering and transportation cost C for the components used in manufacturing a product is C " 100 x # , !200 x x # 30 " 2 x v 1 where C is measured in thousands of dollars and x is the order size in hundreds. Find the rate of change of C with respect to x when (a) x " 10, (b) x " 15, and (c) x " 20. What do these rates of change imply about increasing order size? 86. Boyle’s Law This law states that if the temperature of a gas remains constant, its pressure is inversely proportional to its volume. Use the derivative to show that the rate of change of the pressure is inversely proportional to the square of the volume. 87. Population Growth A population of 500 bacteria is introduced into a culture and grows in number according to the equation ! P#t\$ " 500 1 # 4t 50 # t 2 " where t is measured in hours. Find the rate at which the population is growing when t " 2. 128 CHAPTER 2 Differentiation 88. Gravitational Force Newton’s Law of Universal Gravitation states that the force F between two masses, m1 and m2, is F" Gm1m2 d2 93. f #x\$ " 4x3&2 89. Prove the following differentiation rules. (a) d )sec x " sec x tan x dx (c) d )cot x* " \$csc2 x dx (b) d )csc x* " \$csc x cot x dx 94. f #x\$ " x # 32x\$2 x x\$1 95. f #x\$ " where G is a constant and d is the distance between the masses. Find an equation that gives an instantaneous rate of change of F with respect to d. (Assume m1 and m2 represent moving points.) 96. f #x\$ " 97. f #x\$ " 3 sin x x 2 # 2x \$ 1 x 98. f #x\$ " sec x In Exercises 99–102, find the given higher-order derivative. 99. f&#x\$ " x 2, 2 100. f ' #x\$ " 2 \$ , x f ' #x\$ 101. f&&&#x\$ " 2%x, f #4\$#x\$ f&&&#x\$ 102. f #4\$#x\$ " 2x # 1, f #6\$#x\$ 90. Rate of Change Determine whether there exist any values of x in the interval )0, 2! \$ such that the rate of change of f #x\$ " sec x and the rate of change of g#x\$ " csc x are equal. 91. Modeling Data The table shows the numbers n (in thousands) of motor homes sold in the United States and the retail values v (in billions of dollars) of these motor homes for the years 1996 through 2001. The year is represented by t, with t " 6 corresponding to 1996. (Source: Recreation Vehicle Industry Association) Year, t In Exercises 93–98, find the second derivative of the function. 6 7 8 9 10 11 n 247.5 254.5 292.7 321.2 300.1 256.8 v 6.3 6.9 8.4 10.4 9.5 8.6 (a) Use a graphing utility to find cubic models for the number of motor homes sold n#t\$ and the total retail value v#t\$ of the motor homes. (b) Graph each model found in part (a). (c) Find A " v#t\$&n#t\$, then graph A. What does this function represent? 103. Sketch the graph of a differentiable function f such that f #2\$ " 0, f& < 0 for \$ ) < x < 2, and f& > 0 for 2 < x < ). 104. Sketch the graph of a differentiable function f such that f > 0 and f& < 0 for all real numbers x. In Exercises 105–108, use the given information to find f&'2(. g'2( " 3 and h'2( " \$1 g&'2( " \$2 h!'2( " 4 and 105. f #x\$ " 2g#x\$ # h#x\$ 106. f #x\$ " 4 \$ h#x\$ g#x\$ 107. f #x\$ " h#x\$ 108. f #x\$ " g#x\$h#x\$ In Exercises 109 and 110, the graphs of f, f!, and f # are shown on the same set of coordinate axes. Which is which? Explain your reasoning. To print an enlarged copy of the graph, select the MathGraph button. 109. 110. y y 2 (d) Interpret A&#t\$ in the context of these data. 92. Satellites When satellites observe Earth, they can scan only part of Earth’s surface. Some satellites have sensors that can measure the angle % shown in the figure. Let h represent the satellite’s distance from Earth’s surface and let r represent r r h V 2 (b) Find the rate at which h is changing with respect to % when % " 30(. (Assume r " 3960 miles.) 1 2 x 3 1 2 In Exercises 111–114, the graph of f is shown. Sketch the graphs of f! and f # . To print an enlarged copy of the graph, select the MathGraph button. 111. (a) Show that h " r #csc % \$ 1\$. x 1 112. y 4 y 8 f 4 2 4 2 2 x 4 8 x f 4 4 SECTION 2.3 113. 114. y f 4 3 2 1 (a) Use the Product Rule to generate rules for finding f ' #x\$, f&&&#x\$, and f #4\$#x\$. f 2 U 2 3U 2 (b) Use the results in part (a) to write a general rule for f #n\$#x\$. 1 1 2 4 U 2 U 3U 2 2U x 115. Acceleration The velocity of an object in meters per second is v#t\$ " 36 \$ t 2, 0 f t f 6. Find the velocity and acceleration of the object when t " 3. What can be said about the speed of the object when the velocity and acceleration have opposite signs? 116. Acceleration An automobile’s velocity starting from rest is 100t v #t\$ " 2t # 15 117. Stopping Distance A car is traveling at a rate of 66 feet per second (45 miles per hour) when the brakes are applied. The position function for the car is s#t\$ " \$8.25t 2 # 66t, where s is measured in feet and t is measured in seconds. Use this function to complete the table, and find the average velocity during each time interval. 0 1 2 3 4 In Exercises 123 and 124, find the derivatives of the function f for n " 1, 2, 3, and 4. Use the results to write a general rule for f!'x( in terms of n. 123. f #x\$ " x n sin x 124. f #x\$ " cos x xn Differential Equations In Exercises 125–128, verify that the function satisfies the differential equation. Differential Equation 1 125. y " , x > 0 x x3 y' # 2x2 y& " 0 126. y " 2x3 \$ 6x # 10 \$y'& \$ xy' \$ 2y& " \$24x2 127. y " 2 sin x # 3 y' # y " 3 128. y " 3 cos x # sin x y' # y " 0 True or False? In Exercises 129–134, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 129. If y " f #x\$g#x\$, then dy&dx " f&#x\$g&#x\$. s't( 130. If y " #x # 1\$#x # 2\$#x # 3\$#x # 4\$, then d 5y&dx 5 " 0. v't( 131. If f&#c\$ and g&#c\$ are zero and h#x\$ " f #x\$g#x\$, then h&#c\$ " 0. a't( 132. If f #x\$ is an nth-degree polynomial, then f #n#1\$#x\$ " 0. 118. Particle Motion The figure shows the graphs of the position, velocity, and acceleration functions of a particle. y 16 12 8 4 1 122. Finding a Pattern Develop a general rule for )x f #x\$#n\$ where f is a differentiable function of x. Function where v is measured in feet per second. Find the acceleration at (a) 5 seconds, (b) 10 seconds, and (c) 20 seconds. t 129 121. Finding a Pattern Consider the function f #x\$ " g#x\$h#x\$. y 4 x Product and Quotient Rules and Higher-Order Derivatives 134. If the velocity of an object is constant, then its acceleration is zero. 135. Find a second-degree polynomial f #x\$ " ax2 # bx # c such that its graph has a tangent line with slope 10 at the point #2, 7\$ and an x-intercept at #1, 0\$. t 1 133. The second derivative represents the rate of change of the first derivative. 4 5 6 7 136. Consider the third-degree polynomial f #x\$ " ax3 # bx2 # cx # d, a 0. (a) Copy the graphs of the functions shown. Identify each graph. Explain your reasoning. To print an enlarged copy of the graph, select the MathGraph button. (b) On your sketch, identify when the particle speeds up and when it slows down. Explain your reasoning. Finding a Pattern In Exercises 119 and 120, develop a general rule for f 'n('x( given f 'x(. 119. f #x\$ " x n 1 120. f #x\$ " x Determine conditions for a, b, c, and d if the graph of f has (a) no horizontal tangents, (b) exactly one horizontal tangent, and (c) exactly two horizontal tangents. Give an example for each case. ++ 137. Find the derivative of f #x\$ " x x . Does f ' #0\$ exist? 138. Think About It Let f and g be functions whose first and second derivatives exist on an interval I. Which of the following formulas is (are) true? (a) fg' \$ f 'g " # fg& \$ f&g\$& (b) fg' # f 'g " # fg\$' 130 CHAPTER 2 Differentiation Section 2.4 The Chain Rule • • • • Find the derivative of a composite function using the Chain Rule. Find the derivative of a function using the General Power Rule. Simplify the derivative of a function using algebra. Find the derivative of a trigonometric function using the Chain Rule. The Chain Rule This text has yet to discuss one of the most powerful differentiation rules—the Chain Rule. This rule deals with composite functions and adds a surprising versatility to the rules discussed in the two previous sections. For example, compare the functions shown below. Those on the left can be differentiated without the Chain Rule, and those on the right are best done with the Chain Rule. Without the Chain Rule With the Chain Rule y ! x2 # 1 y ! sin x y ! \$x 2 # 1 y ! sin 6x y ! 3x # 2 y ! x # tan x y ! "3x # 2#5 y ! x # tan x2 Basically, the Chain Rule states that if y changes dy!du times as fast as u, and u changes du!dx times as fast as x, then y changes "dy!du#"du!dx# times as fast as x. . Video The Derivative of a Composite Function EXAMPLE 1 3 Gear 2 Gear 1 Axle 2 Gear 4 1 Axle 1 dy dy ! dx du Gear 3 1 Animation Axle 3 Solution Because the circumference of the second gear is three times that of the first, the first axle must make three revolutions to turn the second axle once. Similarly, the second axle must make two revolutions to turn the third axle once, and you can write dy !3 du and du ! 2. dx Combining these two results, you know that the first axle must make six revolutions to turn the third axle once. So, you can write dy ! dx ! ! . du " dx . 2 Axle 1: y revolutions per minute Axle 2: u revolutions per minute . 3: x revolutions per minute Axle Figure 2.24 A set of gears is constructed, as shown in Figure 2.24, such that the second and third gears are on the same axle. As the first axle revolves, it drives the second axle, which in turn drives the third axle. Let y, u, and x represent the numbers of revolutions per minute of the first, second, and third axles. Find dy!du, du!dx, and dy!dx, and show that Rate of change of first axle with respect to second axle dy du " Rate of change of second axle with respect to third axle du " dx ! 3 " 2 ! 6 Rate of change of first axle with respect to third axle . In other words, the rate of change of y with respect to x is the product of the rate of change of y with respect to u and the rate of change of u with respect to x. Try It Exploration A SECTION 2.4 E X P L O R AT I O N Using the Chain Rule Each of the following functions can be differentiated using rules that you studied in Sections 2.2 and 2.3. For each function, find the derivative using those rules. Then find the derivative using the Chain Rule. Compare your results. Which method is simpler? 2 a. 3x # 1 b. "x # 2#3 c. sin 2x The Chain Rule 131 Example 1 illustrates a simple case of the Chain Rule. The general rule is stated below. THEOREM 2.10 The Chain Rule If y ! f "u# is a differentiable function of u and u ! g"x# is a differentiable function of x, then y ! f "g"x## is a differentiable function of x and dy dy ! dx du du " dx or, equivalently, d ' f "g"x##( ! f\$"g"x##g\$ "x#. dx Proof Let h"x# ! f "g"x##. Then, using the alternative form of the derivative, you need to show that, for x ! c, h\$"c# ! f\$"g"c##g\$"c#. An important consideration in this proof is the behavior of g as x approaches c. A problem occurs if there are values of x, other than c, such that g"x# ! g"c#. Appendix A shows how to use the differentiability of f and g to overcome this problem. For now, assume that g"x# ' g"c# for values of x other than c. In the proofs of the Product Rule and the Quotient Rule, the same quantity was added and subtracted to obtain the desired form. This proof uses a similar technique—multiplying and dividing by the same (nonzero) quantity. Note that because g is differentiable, it is also continuous, and it follows that g"x# → g"c# as x → c. f "g"x## & f "g"c## x→c x&c f "g"x## & f "g"c## ! lim x→c g"x# & g"c# f "g"x## & f "g"c## ! lim x→c g"x# & g"c# ! f\$"g"c##g\$"c# h\$"c# ! lim % % " &% g"x# & g"c# , g"x# ' g"c# x&c g"x# & g"c# lim x→c x&c & & When applying the Chain Rule, it is helpful to think of the composite function f % g as having two parts—an inner part and an outer part. Outer function y ! f "g"x## ! f "u# Inner function The derivative of y ! f "u# is the derivative of the outer function (at the inner function u) times the derivative of the inner function. y\$ ! f\$"u# " u\$ 132 CHAPTER 2 Differentiation Decomposition of a Composite Function EXAMPLE 2 y ! f "g"x## 1 x#1 b. y ! sin 2x c. y ! \$3x2 & x # 1 d. y ! tan 2 x a. y ! . y ! x 6 # 3x 4 # 3x 2 # 1 and . y\$ ! 6x5 # 12x3 # 6x. Verify that this is the same as the derivative in Example 3. Which method would you use to find . d 2 "x # 1#50? dx u!x#1 y! u ! 2x u ! 3x 2 & x # 1 u ! tan x 1 u y ! sin u y ! \$u y ! u2 Using the Chain Rule EXAMPLE 3 You could also solve the problem in Example 3 without using the Chain Rule by observing that y ! f "u# Exploration A Try It STUDY TIP u ! g"x# Find dy!dx for y ! "x 2 # 1#3. Solution For this function, you can consider the inside function to be u ! x 2 # 1. By the Chain Rule, you obtain dy ! 3"x 2 # 1#2"2x# ! 6x"x 2 # 1# 2. dx dy du du dx Exploration A Try It Exploration B The editable graph feature below allows you to edit the graph of a function and its derivative. Editable Graph The General Power Rule The function in Example 3 is an example of one of the most common types of composite functions, y ! 'u"x#(n. The rule for differentiating such functions is called the General Power Rule, and it is a special case of the Chain Rule. THEOREM 2.11 The General Power Rule If y ! 'u"x#( where u is a differentiable function of x and n is a rational number, then n, dy du ! n'u"x#(n&1 dx dx or, equivalently, d n 'u ( ! nu n&1 u\$. dx . Video Proof Because y ! un, you apply the Chain Rule to obtain ) ) dy dy du ! dx du dx d n du ! 'u ( . du dx By the (Simple) Power Rule in Section 2.2, you have Du 'un( ! nu n&1, and it follows that dy du ! n ' u"x#(n&1 . dx dx SECTION 2.4 . EXAMPLE 4 Video The Chain Rule 133 Applying the General Power Rule Find the derivative of f "x# ! "3x & 2x 2#3. Solution Let u ! 3x & 2x2. Then f "x# ! "3x & 2x2#3 ! u3 and, by the General Power Rule, the derivative is n un&1 u\$ d '3x & 2x 2( dx ! 3"3x & 2x 2# 2"3 & 4x#. f\$"x# ! 3"3x & 2x 2#2 . Differentiate 3x & 2x 2. Exploration A Try It . Apply General Power Rule. The editable graph feature below allows you to edit the graph of a function. f(x) = 3 (x 2 − 1) 2 Editable Graph y EXAMPLE 5 2 3 Find all points on the graph of f "x# ! \$ "x 2 & 1# 2 for which f\$"x# ! 0 and those for which f\$"x# does not exist. Solution Begin by rewriting the function as −2 x −1 1 2 −1 f "x# ! "x 2 & 1#2!3. Then, applying the General Power Rule (with u ! x2 & 1# produces n −2 un&1 u\$ 2 2 "x & 1#&1!3 "2x# 3 4x ! 3 2 . 3\$x & 1 f\$"x# ! f ′(x) = 4x 3 3 x2 − 1 The derivative of f is 0 at x ! 0 and is .. undefined at x ! ± 1. Figure 2.25 Editable Graph Apply General Power Rule. So, f\$"x# ! 0 when x ! 0 and f\$"x# does not exist when x ! ± 1, as shown in Figure 2.25. Exploration A Try It EXAMPLE 6 Differentiating Quotients with Constant Numerators Differentiate g"t# ! &7 . "2t & 3# 2 Solution Begin by rewriting the function as g"t# ! &7"2t & 3#&2. NOTE Try differentiating the function in Example 6 using the Quotient Rule. You should obtain the same result, but using the Quotient Rule is less efficient than using the General Power Rule. Then, applying the General Power Rule produces un&1 n u\$ g\$"t# ! "&7#"&2#"2t & 3#&3"2# Apply General Power Rule. Constant Multiple Rule . ! 28"2t & 3#&3 28 ! . "2t & 3#3 Try It Exploration A Simplify. Write with positive exponent. Exploration B 134 CHAPTER 2 Differentiation Simplifying Derivatives The next three examples illustrate some techniques for simplifying the “raw derivatives” of functions involving products, quotients, and composites. Simplifying by Factoring Out the Least Powers EXAMPLE 7 f "x# ! x2\$1 & x2 ! x 2"1 & x 2#1!2 d d f\$"x# ! x 2 '"1 & x 2#1!2( # "1 & x 2#1!2 'x 2( dx dx 1 ! x 2 "1 & x 2#&1!2"&2x# # "1 & x 2#1!2"2x# 2 3 ! &x "1 & x 2#&1!2 # 2x"1 & x 2#1!2 ! x"1 & x 2#&1!2'&x 2"1# # 2"1 & x 2#( x"2 & 3x 2# ! \$1 & x 2 % . & x x "x 2 # 4#1!3 "x 2 # 4#1!3"1# & x"1!3#"x 2 # 4#&2!3"2x# f\$"x# ! "x 2 # 4#2!3 1 3"x 2 # 4# & "2x 2#"1# ! "x 2 # 4#&2!3 3 "x 2 # 4#2!3 x 2 # 12 ! 3"x2 # 4#4!3 % Simplify. Factor. Simplify. & Rewrite. Quotient Rule Factor. Simplify. Exploration A Try It Simplifying the Derivative of a Power EXAMPLE 9 )3xx #& 31* 2 Original function 2 un&1 n u\$ )3xx #& 31* dxd % 3xx #& 31& 2"3x & 1# "x # 3#"3# & "3x & 1#"2x# !% & x # 3 &% "x # 3# y\$ ! 2 General Power Rule Original function 3 x2 # 4 \$ ! y! Product Rule Simplifying the Derivative of a Quotient EXAMPLE 8 f "x# ! Rewrite. Exploration A Try It TECHNOLOGY Symbolic differentiation utilities are capable of differentiating very complicated functions. Often, however, the result is given in unsimplified form. If you find the derivatives of the functions given in Examples 7, 8, and 9. Then compare the results with those given Original function 2 2 General Power Rule 2 2 2 2"3x & 1#"3x 2 # 9 & 6x 2 # 2x# "x 2 # 3#3 2"3x & 1#"&3x 2 # 2x # 9# ! "x 2 # 3#3 ! . 2 Try It Exploration A Quotient Rule Multiply. Simplify. Open Exploration SECTION 2.4 The Chain Rule 135 Trigonometric Functions and the Chain Rule The “Chain Rule versions” of the derivatives of the six trigonometric functions are as shown. d 'sin u( ! "cos u# u\$ dx d 'tan u( ! "sec 2 u# u\$ dx d 'sec u( ! "sec u tan u# u\$ dx . d 'cos u( ! & "sin u# u\$ dx d 'cot u( ! & "csc 2 u# u\$ dx d 'csc u( ! & "csc u cot u# u\$ dx Technology EXAMPLE 10 Applying the Chain Rule to Trigonometric Functions cos u u a. y ! sin 2x . u\$ d '2x( ! "cos 2x#"2# ! 2 cos 2x dx y\$ ! &sin"x & 1# y\$ ! 3 sec 2 3x y\$ ! cos 2x b. y ! cos"x & 1# c. y ! tan 3x Exploration A Try It Be sure that you understand the mathematical conventions regarding parentheses and trigonometric functions. For instance, in Example 10(a), sin 2x is written to mean sin"2x#. EXAMPLE 11 a. b. c. d. . Parentheses and Trigonometric Functions y ! cos 3x 2 ! cos"3x 2# y ! "cos 3#x 2 y ! cos"3x#2 ! cos"9x 2# y ! cos 2 x ! "cos x# 2 e. y ! \$cos x ! "cos x#1!2 ! "&sin 3x 2#"6x# ! &6x sin 3x 2 ! "cos 3#"2x# ! 2x cos 3 ! "&sin 9x 2#"18x# ! &18x sin 9x 2 ! 2"cos x#"&sin x# ! &2 cos x sin x 1 sin x y\$ ! "cos x#&1!2"&sin x# ! & 2 2\$cos x y\$ y\$ y\$ y\$ Exploration A Try It To find the derivative of a function of the form k"x# ! f "g"h"x###, you need to apply the Chain Rule twice, as shown in Example 12. EXAMPLE 12 Repeated Application of the Chain Rule f "t# ! sin3 4t ! "sin 4t#3 Original function Rewrite. d 'sin 4t( dt d ! 3"sin 4t#2"cos 4t# '4t( dt ! 3"sin 4t#2"cos 4t#"4# ! 12 sin 2 4t cos 4t f\$"t# ! 3"sin 4t#2 . Try It Exploration A Apply Chain Rule once. Apply Chain Rule a second time. Simplify. 136 CHAPTER 2 Differentiation EXAMPLE 13 y Tangent Line of a Trigonometric Function Find an equation of the tangent line to the graph of f(x) = 2 sin x + cos 2x f "x# ! 2 sin x # cos 2x 2 at the point "(, 1#, as shown in Figure 2.26. Then determine all values of x in the interval "0, 2(# at which the graph of f has a horizontal tangent. ( π , 1) 1 π 2 π 3π 2 2π x Solution Begin by finding f\$"x#. f "x# ! 2 sin x # cos 2x f\$"x# ! 2 cos x # "&sin 2x#"2# ! 2 cos x & 2 sin 2 x −2 −3 −4 Write original function. Apply Chain Rule to cos 2x. Simplify. To find the equation of the tangent line at "(, 1#, evaluate f\$"(#. Figure 2.26 f \$ "(# ! 2 cos ( & 2 sin 2( ! &2 Substitute. Slope of graph at "(, 1# Now, using the point-slope form of the equation of a line, you can write y & y1 ! m"x & x1# y & 1 ! &2"x & (# y ! 1 & 2x # 2(. Point-slope form Substitute for y1, m, and x1. Equation of tangent line at "(, 1# ( ( 5( 3( , and . So, f has a You can then determine that f\$"x# ! 0 when x ! , , 6 2 6 2 ( ( 5( 3( horizontal tangent at x ! , , , and . 6 2 6 2 STUDY TIP To become skilled at . differentiation, you should memorize each rule. As an aid to memorization, note that the cofunctions (cosine, cotangent, and cosecant) require a negative sign as part of their derivatives. Try It Exploration A This section concludes with a summary of the differentiation rules studied so far. Summary of Differentiation Rules General Differentiation Rules Let f, g, and u be differentiable functions of x. Constant Multiple Rule: Sum or Difference Rule: d 'cf ( ! cf \$ dx d ' f ± g( ! f \$ ± g\$ dx Product Rule: Quotient Rule: d ' fg( ! fg\$ # gf\$ dx d f gf\$ & fg\$ ! dx g g2 Derivatives of Algebraic Functions Constant Rule: "Simple# Power Rule: d 'c( ! 0 dx d n 'x ( ! nxn&1, dx Derivatives of Trigonometric Functions d 'sin x( ! cos x dx d 'cos x( ! &sin x dx d 'tan x( ! sec 2 x dx d 'cot x( ! &csc 2 x dx Chain Rule Chain Rule: General Power Rule: d ' f "u#( ! f \$"u# u\$ dx d n 'u ( ! nu n&1 u\$ dx %& d 'x( ! 1 dx d 'sec x( ! sec x tan x dx d 'csc x( ! &csc x cot x dx SECTION 2.4 137 The Chain Rule Exercises for Section 2.4 The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. Click on to view the complete solution of the exercise. Click on to print an enlarged copy of the graph. In Exercises 1–6, complete the table. y " f \$g\$x%% u " g\$x% In Exercises 39 and 40, find the slope of the tangent line to the sine function at the origin. Compare this value with the number of complete cycles in the interval [0, 2!]. What can you conclude about the slope of the sine function sin ax at the origin? y " f \$u% 1. y " \$6x \$ 5%4 2. y " 1 39. (a) #x # 1 y y = sin x 2 3. y " #x2 \$ 1 1 U 2 5. y " csc 3x 3x 2 7. y " \$2x \$ 7%3 9. g\$x% " 3\$4 \$ 9x% 14. g\$x% " #x 2 \$ 2x # 1 15. y " 4 4 2# 4 2 \$ 9x 16. f \$x% " \$3 # 17. y " 1 x\$2 \$ x2 18. s\$t% " ! 1 19. f \$t% " t\$3 " 5 20. y " \$ \$t # 3%3 2 1 #x # 2 23. f \$x% " x 2\$x \$ 2%4 22. g\$t% " 21. y " 26. y " x 28. y " #x 2 # 1 ! " t 30. h\$t% " ! t # 2" 1 \$ 2v 31. f \$v% " ! 1#v" 3x \$ 2 32. g\$x% " ! 2x # 3 " # 1 t2 \$ 2 24. f \$x% " x\$3x \$ 9%3 25. y " x#1 \$ x 2 1 2 2 x #16 \$ x2 x #x 4 # 4 2 x#5 x2 # 2 2 1 t 2 # 3t \$ 1 x y = sin 2x 2 2U x 1 2 U 2 U 3U 2 2U x 2 In Exercises 41–58, find the derivative of the function. 41. y " cos 3x 42. y " sin ! x 43. g\$x% " 3 tan 4x 44. h\$x% " sec x 2 45. y " sin\$!x%2 46. y " cos\$1 \$ 2x%2 47. h\$x% " sin 2x cos 2x 49. f \$x% " cot x sin x 51. y " 4 sec2 x 1 4 1 1 48. g\$%% " sec\$2 %% tan\$2%% 50. g\$v% " cos v csc v 52. g\$t% " 5 cos 2 ! t 53. f \$%% " sin 2 2% 54. h\$t% " 2 cot2\$! t # 2% 55. f \$t% " 3 56. y " 3x \$ 5 cos\$! x%2 \$! t \$ 1% 57. y " #x # sin\$2x%2 sec2 3 x # # 3 sin x 58. y " sin # 2 In Exercises 59–66, evaluate the derivative of the function at the given point. Use a graphing utility to verify your result. 3 2 Function 3 59. s\$t% " #x # 1 x2 # 1 x#1 x cos ! x # 1 x #t 2 Point # 2t # 8 5 3x 3 # 4x 60. y " # In Exercises 33–38, use a computer algebra system to find the derivative of the function. Then use the utility to graph the function and its derivative on the same set of coordinate axes. Describe the behavior of the function that corresponds to any zeros of the graph of the derivative. 37. y " U 1 4 3 # 2U 1 10. f \$t% " \$9t # 2% 3 9x 2 # 4 13. y " # U y (b) 1 12. g\$x% " #5 \$ 3x 35. y " U 2 y = sin 3x 2&3 11. f \$t% " #1 \$ t 33. y " x y 2 8. y " 3\$4 \$ x 2%5 4 29. g\$x% " 2U 2 40. (a) 27. y " U 2 In Exercises 7–32, find the derivative of the function. y = sin 2x 2 1 4. y " 3 tan\$! x 2% 6. y " cos y (b) 34. y " # 2x x#1 36. g\$x% " #x \$ 1 # #x # 1 38. y " x 2 tan 1 x 3 x3 \$ 4 1 f \$x% " 2 \$x \$ 3x%2 3t # 2 f \$t% " t\$1 x#1 f \$x% " 2x \$ 3 y " 37 \$ sec 3\$2x% 61. f \$x% " 62. 63. 64. 65. 1 66. y " # #cos x x \$2, 4% \$2, 2% !\$1, \$ 53" !4, 161 " \$0, \$2% \$2, 3% \$0, 36% ! 2 , 2 ! ! " 138 CHAPTER 2 Differentiation In Exercises 67–74, (a) find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of the graphing utility to confirm your results. Function Point 67. f \$x% " #3x 2 \$ 2 69. y " \$2x3 # 1%2 70. f \$x% " \$9 \$ x2%2&3 71. f \$x% " sin 2x 73. f \$x% " tan x 2 74. y " 2 tan3 x 75. g\$t% " 3t #t2 # 2t \$ 1 85. f \$x% " sin x 2 86. f \$x% " sec 2! x ! " 89. f \$x% " cos\$x2%, 91. 25 x2 y 8 2 x2 4 3 2 x 3 (1, 1) 1 4 6 3 2 1 x 1 2 3 2 81. Horizontal Tangent Line Determine the point(s) in the interval \$0, 2!% at which the graph of f \$x% " 2 cos x # sin 2x has a horizontal tangent. 82. Horizontal Tangent Line Determine the point(s) at which the x graph of f \$x% " has a horizontal tangent. #2x \$ 1 2 x 2 3 4 4 In Exercises 95 and 96, the relationship between f and g is given. Explain the relationship between f" and g". 95. g\$x% " f \$3x% 2 y 3 x 3 (3, 4) 2 4 f (x) = 4 2 94. y 80. Bullet-nose curve y 6 4 2 1 2 3 4 2 3 Famous Curves In Exercises 79 and 80, find an equation of the tangent line to the graph at the given point. Then use a graphing utility to graph the function and its tangent line in the same viewing window. x x x 3 93. 4 y 4 3 2 2 ! " 6 92. y 3 2 \$4, 8% \$4 \$ 2t%#1 # t 4 77. s \$t% " , 0, 3 3 2 # 78. y " \$t \$ 9% t # 2, \$2, \$10% f(x) = " In Exercises 91–94, the graphs of a function f and its derivative f" are shown. Label the graphs as f or f" and write a short paragraph stating the criteria used in making the selection. To print an enlarged copy of the graph, select the MathGraph button. 76. f \$x% " #x \$2 \$ x%2, 79. Top half of circle \$0, 1% ! , #3 6 !12, 32" , ! 90. g\$t% " tan 2t, In Exercises 75–78, (a) use a graphing utility to find the derivative of the function at the given point, (b) find an equation of the tangent line to the graph of the function at the given point, and (c) use the utility to graph the function and its tangent line in the same viewing window. 2 84. f \$x% " 87. h\$x% " 19 \$3x # 1%3, \$1, 64 9% 1 1 , 0, 88. f \$x% " #x # 4 2 ! " ! " ! " 72. y " cos 3x 1 x\$2 83. f \$x% " 2\$x 2 \$ 1%3 In Exercises 87–90, evaluate the second derivative of the function at the given point. Use a computer algebra system to verify \$3, 5% \$2, 2% \$\$1, 1% \$1, 4% \$!, 0% ! #2 ,\$ 4 2 ! ,1 4 ! ,2 4 68. f \$x% " 13x#x 2 # 5 In Exercises 83–86, find the second derivative of the function. 96. g\$x% " f \$x 2% 97. Given that g\$5% " \$3, g&\$5% " 6, h\$5% " 3, and h&\$5% " \$2, find f&\$5% (if possible) for each of the following. If it is not possible, state what additional information is required. (a) f \$x% " g\$x%h\$x% (c) f \$x% " g\$x% h\$x% (b) f \$x% " g\$h\$x%% (d) f \$x% " 'g\$x%( 3 SECTION 2.4 98. Think About It The table shows some values of the derivative of an unknown function f. Complete the table by finding (if possible) the derivative of each transformation of f. (a) g\$x% " f \$x% \$ 2 (b) h\$x% " 2 f \$x% (c) r\$x% " f \$\$3x% (d) s\$x% " f \$x # 2% x f" )x* \$2 \$1 0 1 2 3 4 2 3 \$ 13 \$1 \$2 \$4 102. Harmonic Motion The displacement from equilibrium of an object in harmonic motion on the end of a spring is y " 13 cos 12t \$ 14 sin 12t 103. Pendulum A 15-centimeter pendulum moves according to the equation % " 0.2 cos 8t, where % is the angular displacement from the vertical in radians and t is the time in seconds. Determine the maximum angular displacement and the rate of change of % when t " 3 seconds. h" )x* 104. Wave Motion A buoy oscillates in simple harmonic motion y " A cos (t as waves move past it. The buoy moves a total of 3.5 feet (vertically) from its low point to its high point. It returns to its high point every 10 seconds. r" )x* s" \$x% In Exercises 99 and 100, the graphs of f and g are shown. Let h)x* # f )g)x** and s)x* # g) f )x*. Find each derivative, if it exists. If the derivative does not exist, explain why. 99. (a) Find h&\$1%. 100. (a) Find h&\$3%. (b) Find s&\$5%. (b) Find s&\$9%. y 10 8 f 4 g 6 g 2 2 x 2 4 6 8 10 x 2 4 6 8 10 101. Doppler Effect The frequency F of a fire truck siren heard by a stationary observer is 132,400 331 ± v where ± v represents the velocity of the accelerating fire truck in meters per second (see figure). Find the rate of change of F with respect to v when (a) the fire truck is approaching at a velocity of 30 meters per second (use \$v). (b) the fire truck is moving away at a velocity of 30 meters per second (use #v ). 132,400 331 + v F= (b) Determine the velocity of the buoy as a function of t. S " C\$R 2 \$ r 2% 10 f 8 (a) Write an equation describing the motion of the buoy if it is at its high point at t " 0. 105. Circulatory System The speed S of blood that is r centimeters from the center of an artery is y F= 139 where y is measured in feet and t is the time in seconds. Determine the position and velocity of the object when t " !&8. g" )x F" The Chain Rule 132,400 331 v where C is a constant, R is the radius of the artery, and S is measured in centimeters per second. Suppose a drug is administered and the artery begins to dilate at a rate of dR&dt. At a constant distance r, find the rate at which S changes with respect to t for C " 1.76 ' 105, R " 1.2 ' 10\$2, and dR&dt " 10\$5. 106. Modeling Data The normal daily maximum temperatures T (in degrees Fahrenheit) for Denver, Colorado, are shown in the table. (Source: National Oceanic and Atmospheric Month Jan Feb Mar Apr May Jun Temperature 43.2 47.2 53.7 60.9 70.5 82.1 Month Jul Aug Sep Oct Nov Dec Temperature 88.0 86.0 77.4 66.0 51.5 44.1 (a) Use a graphing utility to plot the data and find a model for the data of the form T\$t% " a # b sin \$! t&6 \$ c% where T is the temperature and t is the time in months, with t " 1 corresponding to January. (b) Use a graphing utility to graph the model. How well does the model fit the data? (c) Find T& and use a graphing utility to graph the derivative. (d) Based on the graph of the derivative, during what times does the temperature change most rapidly? Most slowly? temperature changes? Explain. 140 CHAPTER 2 Differentiation 107. Modeling Data The cost of producing x units of a product is C " 60x # 1350. For one week management determined the number of units produced at the end of t hours during an eight-hour shift. The average values of x for the week are shown in the table. t 0 1 2 3 4 5 6 7 8 x 0 16 60 130 205 271 336 384 392 (a) Use a graphing utility to fit a cubic model to the data. (b) Use the Chain Rule to find dC&dt. (c) Explain why the cost function is not increasing at a constant rate during the 8-hour shift. 108. Finding a Pattern Consider the function f \$x% " sin ,x, where , is a constant. (a) Find the first-, second-, third-, and fourth-order derivatives of the function. (b) Verify that the function and its second derivative satisfy the equation f + \$x% # , 2 f \$x% " 0. (c) Use the results in part (a) to write general rules for the even- and odd-order derivatives f \$2k%\$x% and f \$2k\$1%\$x%. [Hint: \$\$1%k is positive if k is even and negative if k is odd.] 109. Conjecture Let f be a differentiable function of period p. (b) Consider the function g\$x% " f \$2x%. Is the function g& \$x% 110. Think About It Let r\$x% " f \$g\$x%% and s\$x% " g\$ f \$x%% where f and g are shown in the figure. Find (a) r&\$1% and (b) s&\$4%. y 7 6 5 4 3 2 1 ++ d u ' u ( " u& , dx u ++ ++ u * 0. In Exercises 114–117, use the result of Exercise 113 to find the derivative of the function. + + + + h\$x% " +x+ cos x f \$x% " +sin x+ 114. g\$x% " 2x \$ 3 115. f \$x% " x 2 \$ 4 116. 117. Linear and Quadratic Approximations The linear and quadratic approximations of a function f at x # a are P1)x* # f")a)x % a & f )a* and 1 P2)x* # 2 f\$ )a)x % a 2 & f")a)x % a & f )a). In Exercises 118 and 119, (a) find the specified linear and quadratic approximations of f, (b) use a graphing utility to graph f and the approximations, (c) determine whether P1 or P2 is the better approximation, and (d) state how the accuracy changes as you move farther from x # a. 118. f \$x% " tan !x 4 119. f \$x% " sec 2x a"1 a" ! 6 True or False? In Exercises 120–122, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 1 120. If y " \$1 \$ x%1)2, then y& " 2\$1 \$ x%\$1)2. (6, 6) 121. If f \$x% " sin 2\$2x%, then f&\$x% " 2\$sin 2x%\$cos 2x%. g (2, 4) 113. Let u be a differentiable function of x. Use the fact that u " #u 2 to prove that 122. If y is a differentiable function of u, u is a differentiable function of v, and v is a differentiable function of x, then (6, 5) f x dy du dv dy " . dx du dv dx 1 2 3 4 5 6 7 111. (a) Find the derivative of the function g\$x% " sin 2 x # cos 2 x in two ways. (b) For f \$x% " sec2 x and g\$x% " tan 2 x, show that f&\$x% "g&\$x%. 112. (a) Show that the derivative of an odd function is even. That is, if f \$\$x% " \$f \$x%, then f&\$\$x% " f&\$x%. (b) Show that the derivative of an even function is odd. That is, if f \$\$x% " f \$x%, then f&\$\$x% " \$f&\$x%. Putnam Exam Challenge 123. Let f \$x% " a1 sin x # a2 sin 2x # . . . # an sin nx, where a1, a2, . . ., an are real numbers and where n is a positive integer. Given that f \$x% f sin x for all real x, prove that a1 # 2a2 # . . . # nan f 1. + + + + + + 124. Let k be a fixed positive integer. The nth derivative of has the form 1 xk \$ 1 Pn\$x% \$x k \$ 1%n#1 where Pn\$x% is a polynomial. Find Pn\$1%. These problems were composed by the Committee on the Putnam Prize Competition. SECTION 2.5 Section 2.5 Implicit Differentiation 141 Implicit Differentiation • Distinguish between functions written in implicit form and explicit form. • Use implicit differentiation to find the derivative of a function. E X P L O R AT I O N Graphing an Implicit Equation How could you use a graphing utility to sketch the graph of the equation 2 x \$ 2y 3 " 4y ! 2? Here are two possible approaches. a. Solve the equation for x. Switch the roles of x and y and graph the two resulting equations. The combined graphs will show a 90% rotation of the graph of the original equation. b. Set the graphing utility to parametric mode and graph the equations x ! \$ (2t 3 \$ 4t " 2 y!t and x ! (2t 3 \$ 4t " 2 y ! t. . From either of these two approaches, can you decide whether the graph has a tangent line at the point #0, 1\$? Implicit and Explicit Functions Up to this point in the text, most functions have been expressed in explicit form. For example, in the equation y ! 3x 2 \$ 5 Explicit form the variable y is explicitly written as a function of x. Some functions, however, are only implied by an equation. For instance, the function y ! 1'x is defined implicitly by the equation xy ! 1. Suppose you were asked to find dy'dx for this equation. You could begin by writing y explicitly as a function of x and then differentiating. Implicit Form Explicit Form xy ! 1 y! 1 ! x\$1 x Derivative dy 1 ! \$x\$2 ! \$ 2 dx x This strategy works whenever you can solve for the function explicitly. You cannot, however, use this procedure when you are unable to solve for y as a function of x. For instance, how would you find dy'dx for the equation x 2 \$ 2y 3 " 4y ! 2 where it is very difficult to express y as a function of x explicitly? To do this, you can use implicit differentiation. To understand how to find dy'dx implicitly, you must realize that the differentiation is taking place with respect to x. This means that when you differentiate terms involving x alone, you can differentiate as usual. However, when you differentiate terms involving y, you must apply the Chain Rule, because you are assuming that y is defined implicitly as a differentiable function of x. Video EXAMPLE 1 a. Differentiating with Respect to x d 3 %x & ! 3x 2 dx Variables agree: use Simple Power Rule. Variables agree un b. nu n\$1 u# d 3 dy % y & ! 3y 2 dx dx Variables disagree: use Chain Rule. Variables disagree d dy %x " 3y& ! 1 " 3 dx dx d d d d. %xy 2& ! x % y 2& " y 2 %x& dx dx dx dy ! x 2y " y 2#1\$ dx dy ! 2xy " y2 dx c. ! . Try It " Exploration A Chain Rule: Product Rule Chain Rule Simplify. d %3y& ! 3y# dx 142 CHAPTER 2 Differentiation Implicit Differentiation Guidelines for Implicit Differentiation 1. Differentiate both sides of the equation with respect to x. 2. Collect all terms involving dy'dx on the left side of the equation and move all other terms to the right side of the equation. 3. Factor dy'dx out of the left side of the equation. 4. Solve for dy'dx. EXAMPLE 2 Implicit Differentiation Find dy'dx given that y 3 " y 2 \$ 5y \$ x 2 ! \$4. Solution NOTE In Example 2, note that implicit differentiation can produce an expression for dy'dx that contains both x and y. 1. Differentiate both sides of the equation with respect to x. d 3 d % y " y 2 \$ 5y \$ x 2& ! %\$4& dx dx d 3 d 2 d d 2 d % y & " % y & \$ %5y& \$ %x & ! %\$4& dx dx dx dx dx dy dy dy 3y 2 " 2y \$ 5 \$ 2x ! 0 dx dx dx 2. Collect the dy'dx terms on the left side of the equation and move all other terms to the right side of the equation. y 3y 2 2 (1, 1) 1 −3 −2 −1 −1 −2 . −4 1 (2, 0) 2 3. Factor dy'dx out of the left side of the equation. x 3 (1, −3) y 3 + y 2 − 5y − x 2 = − 4 Point on Graph Slope of Graph #2, 0\$ #1, \$ 3\$ \$ 45 x!0 0 #1, 1\$ Undefined 1 8 The implicit equation y3 " y 2 \$ 5y \$ x 2 ! \$ 4 has the derivative dy 2x ! 2 . dx 3y " 2y \$ 5 Figure 2.27 dy dy dy " 2y \$ 5 ! 2x dx dx dx dy #3y 2 " 2y \$ 5\$ ! 2x dx 4. Solve for dy'dx by dividing by #3y 2 " 2y \$ 5\$. dy 2x ! dx 3y 2 " 2y \$ 5 Try It Exploration A Video Video To see how you can use an implicit derivative, consider the graph shown in Figure 2.27. From the graph, you can see that y is not a function of x. Even so, the derivative found in Example 2 gives a formula for the slope of the tangent line at a point on this graph. The slopes at several points on the graph are shown below the graph. With most graphing utilities, it is easy to graph an equation that explicitly represents y as a function of x. Graphing other equations, however, can require some ingenuity. For instance, to graph the equation given in Example 2, use a graphing utility, set in parametric mode, to graph the parametric representations x ! (t 3 " t 2 \$ 5t " 4, y ! t, and x ! \$ (t 3 " t 2 \$ 5t " 4, y ! t, for \$5 ≤ t ≤ 5. How does the result compare with the graph shown in Figure 2.27? TECHNOLOGY SECTION 2.5 y + y2 =0 (0, 0) x −1 143 It is meaningless to solve for dy'dx in an equation that has no solution points. (For example, x 2 " y 2 ! \$4 has no solution points.) If, however, a segment of a graph can be represented by a differentiable function, dy'dx will have meaning as the slope at each point on the segment. Recall that a function is not differentiable at (a) points with vertical tangents and (b) points at which the function is not continuous. 1 x2 Implicit Differentiation 1 EXAMPLE 3 −1 Representing a Graph by Differentiable Functions . If possible, represent y as a differentiable function of x. (a) a. x 2 " y 2 ! 0 Editable Graph a. The graph of this equation is a single point. So, it does not define y as a differentiable function of x. See Figure 2.28(a). b. The graph of this equation is the unit circle, centered at #0, 0\$. The upper semicircle is given by the differentiable function 1 − x2 y= (−1, 0) (1, 0) −1 x 1 −1 . y ! (1 \$ x 2, \$1 < x < 1 and the lower semicircle is given by the differentiable function 1 − x2 y=− y ! \$ (1 \$ x 2, \$1 < x < 1. (b) At the points #\$1, 0\$ and #1, 0\$, the slope of the graph is undefined. See Figure 2.28(b). c. The upper half of this parabola is given by the differentiable function Editable Graph y y= y ! (1 \$ x, 1−x 1 x < 1 and the lower half of this parabola is given by the differentiable function (1, 0) . c. x " y 2 ! 1 Solution y 1 b. x 2 " y 2 ! 1 −1 y ! \$ (1 \$ x, x < 1. x 1 −1 At the point #1, 0\$, the slope of the graph is undefined. See Figure 2.28(c). y=− . 1−x Editable Graph EXAMPLE 4 Some graph segments can be represented by differentiable functions. Figure 2.28 Exploration B Exploration A Try It (c) Finding the Slope of a Graph Implicitly Determine the slope of the tangent line to the graph of x 2 " 4y 2 ! 4 at the point #(2, \$1'(2 \$. See Figure 2.29. Solution y 2 x 2 + 4y 2 = 4 x −1 . . Figure 2.29 1 −2 ( 2, − 1 2 ) x 2 " 4y 2 ! 4 dy 2x " 8y ! 0 dx dy \$2x \$x ! ! dx 8y 4y Write original equation. Differentiate with respect to x. Solve for dy . dx So, at #(2, \$1'(2 \$, the slope is dy \$ (2 1 ! ! . dx \$4'(2 2 Evaluate dy 1 when x ! (2 and y ! \$ . dx (2 Editable Graph Try It Exploration A Exploration B Open Exploration NOTE To see the benefit of implicit differentiation, try doing Example 4 using the explicit function y ! \$ 12(4 \$ x 2. 144 CHAPTER 2 Differentiation EXAMPLE 5 Finding the Slope of a Graph Implicitly Determine the slope of the graph of 3#x 2 " y 2\$ 2 ! 100xy at the point #3, 1\$. Solution d d %3#x 2 " y 2\$ 2& ! %100xy& dx dx dy dy 3#2\$#x 2 " y 2\$ 2x " 2y ! 100 x " y#1\$ dx dx dy dy 12y #x 2 " y 2\$ \$ 100x ! 100y \$ 12x#x 2 " y 2\$ dx dx dy %12y #x 2 " y 2\$ \$ 100x& ! 100y \$ 12x#x 2 " y 2\$ dx dy 100y \$ 12x#x 2 " y 2\$ ! dx \$100x " 12y#x 2 " y 2\$ 25y \$ 3x#x 2 " y 2\$ ! \$25x " 3y#x 2 " y 2\$ ! y 4 3 2 1 (3, 1) x −4 −2 −1 1 3 4 " ) * At the point #3, 1\$, the slope of the graph is −4 dy 25#1\$ \$ 3#3\$#32 " 12\$ 25 \$ 90 \$65 13 ! ! ! ! dx \$25#3\$ " 3#1\$#32 " 12\$ \$75 " 30 \$45 9 3(x 2 + y 2) 2 = 100xy . Lemniscate as shown in Figure 2.30. This graph is called a lemniscate. Figure 2.30 Try It y EXAMPLE 6 sin y = x Exploration B Determining a Differentiable Function Find dy'dx implicitly for the equation sin y ! x. Then find the largest interval of the form \$a < y < a on which y is a differentiable function of x (see Figure 2.31). (1, π2 ) π 2 Exploration A x −1 (−1, − π2 ) −π 2 1 − 3π 2 The . derivative is Figure 2.31 dy 1 ! . dx (1 \$ x 2 Editable Graph Solution d d %sin y& ! %x& dx dx dy cos y ! 1 dx dy 1 ! dx cos y The largest interval about the origin for which y is a differentiable function of x is \$ &'2 < y < &'2. To see this, note that cos y is positive for all y in this interval and is 0 at the endpoints. If you restrict y to the interval \$ &'2 < y < &'2, you should be able to write dy'dx explicitly as a function of x. To do this, you can use cos y ! (1 \$ sin2 y ! (1 \$ x 2, \$ & & < y < 2 2 and conclude that . 1 dy ! . dx (1 \$ x 2 Try It Exploration A SECTION 2.5 ISAAC BARROW (1630–1677) . The graph in Figure 2.32 is called the kappa curve because it resembles the Greek letter kappa, ' . The general solution for the tangent line to this curve was discovered by the English mathematician Isaac Barrow. Newton was Barrow’s student, and they corresponded frequently regarding their work in the early development of calculus. Implicit Differentiation 145 With implicit differentiation, the form of the derivative often can be simplified (as in Example 6) by an appropriate use of the original equation. A similar technique can be used to find and simplify higher-order derivatives obtained implicitly. Finding the Second Derivative Implicitly EXAMPLE 7 Given x 2 " y 2 ! 25, find d 2y . dx 2 Solution Differentiating each term with respect to x produces dy !0 dx dy 2y ! \$2x dx dy \$2x x ! !\$ . dx 2y y MathBio 2x " 2y Differentiating a second time with respect to x yields d 2y # y\$#1\$ \$ #x\$#dy'dx\$ !\$ 2 dx y2 y \$ #x\$#\$x'y\$ !\$ y2 y 2 " x2 !\$ y3 25 !\$ 3. y . Try It Quotient Rule Substitute \$x'y for dy . dx Simplify. Substitute 25 for x 2 " y 2. Exploration A Finding a Tangent Line to a Graph EXAMPLE 8 Find the tangent line to the graph given by x 2#x 2 " y 2\$ ! y 2 at the point #(2'2, (2'2\$, as shown in Figure 2.32. Solution By rewriting and differentiating implicitly, you obtain x 4 " x 2y 2 \$ y 2 ! 0 dy dy 4x 3 " x 2 2y " 2xy 2 \$ 2y !0 dx dx dy 2y#x 2 \$ 1\$ ! \$2x#2x 2 " y 2\$ dx dy x #2x 2 " y 2\$ . ! dx y #1 \$ x 2\$ ! y 1 ( 22 , 22) At the point #(2'2, (2'2\$, the slope is x −1 1 −1 . The kappa curve Figure 2.32 " x 2(x 2 + y 2) = y 2 dy #(2'2\$%2#1'2\$ " #1'2\$& 3'2 ! ! !3 dx 1'2 #(2'2\$%1 \$ #1'2\$& and the equation of the tangent line at this point is y\$ (2 2 ! !3 x\$ (2 2 y ! 3x \$ (2. Try It " Exploration A 146 CHAPTER 2 Differentiation Exercises for Section 2.5 The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. Click on to view the complete solution of the exercise. Click on to print an enlarged copy of the graph. In Exercises 1–16, find dy /dx by implicit differentiation. 31. Bifolium: 32. Folium of Descartes: ! ! " # Point: !1, 1" x2 y2 2 1. x 2 ! y 2 # 36 2. x 2 " y 2 # 16 3. x1%2 ! y1%2 # 9 4. x3 ! y 3 # 8 5. x3 " xy ! y 2 # 4 6. x 2 y ! y 2x # "2 7. x3y 3 " y # x 8. &xy # x " 2y 2 10. 2 sin x cos y # 1 1 9. x3 " 3x 2 y ! 2xy 2 # 12 11. sin x ! 2 cos 2y # 1 12. !sin \$ x ! cos \$ y" 2 # 2 13. sin x # x!1 ! tan y" 14. cot y # x " y 15. y # sin!xy" 1 16. x # sec y 17. 19. 9x 2 ! y2 ! # 16 16y 2 # 144 18. x2 20. 9y 2 ! y2 2 " 28. 2 x 1 1 2 x 2 1 x2 " 4 , !2, 0" x2 ! 4 (y 2)2 = 4(x 3) (x + 1)2 + (y 2)2 = 20 y 8 6 4 2 8 x 2 4 2 10 12 14 8 6 36. Rotated ellipse 7x 2 6 xy = 1 3xy + 13y 2 16 = 0 y 2 3 (1, 1) 1 2 x 3 1 2 y 37. Cruciform x 2y 2 2 1 2 2 38. Astroid 4y 2 x 2/3 + y 2/3 = 5 =0 y 12 4 (4, 2 2 1 1 9x 2 6 x 1 3 x y 1 x 2 3 2 1 3, 1) 2 y 3 ( 3 Famous Curves In Exercises 29–32, find the slope of the tangent line to the graph at the given point. !4 " x"y 2 # x3 Point: !2, 2" 4 4 3 !x 2 ! 4"y # 8 Point: !2, 1" x 2 3 30. Cissoid: (3, 4) 4 (4, 0) y # \$ 4 34. Circle 35. Rotated hyperbola !"1, 1" 2%3 2%3 x ! y # 5, !8, 1" x 3 ! y 3 # 4xy ! 1, !2, 1" tan!x ! y" # x, !0, 0" \$ x cos y # 1, 2, 3 3 2 33. Parabola 2 4 6 8 29. Witch of Agnesi: 2 Famous Curves In Exercises 33–40, find an equation of the tangent line to the graph at the given point. To print an enlarged copy of the graph, select the MathGraph button. #9 24. !x ! y"3 # x3 ! y 3, 27. 3 y 22. x 2 " y 3 # 0, !1, 1" 26. 4 2 21. xy # 4, !"4, "1" 25. y 1 In Exercises 21–28, find dy/ dx by implicit differentiation and evaluate the derivative at the given point. 23. y 2 # Point: ! 3, 3 " 4 8 1 " 4x ! 6y ! 9 # 0 x2 x3 ! y 3 " 6xy # 0 y In Exercises 17–20, (a) find two explicit functions by solving the equation for y in terms of x, (b) sketch the graph of the equation and label the parts given by the corresponding explicit functions, (c) differentiate the explicit functions, and (d) find dy/ dx and show that the result is equivalent to that of part (c). x2 4x 2 y 3 3) 6 4 2 (8, 1) x 2 4 x 6 12 4 12 SECTION 2.5 39. Lemniscate 40. Kappa curve 3(x 2 + y 2)2 = 100(x 2 y 2) y 6 3 4 2 2 (4, 2) x 6 In Exercises 57 and 58, find the points at which the graph of the equation has a vertical or horizontal tangent line. y 2(x 2 + y 2) = 2x2 y 6 57. 25x 2 ! 16y 2 ! 200x " 160y ! 400 # 0 58. 4x 2 ! y 2 " 8x ! 4y ! 4 # 0 (1, 1) x 3 2 2 4 2 6 3 3 41. (a) Use implicit differentiation to find an equation of the x2 y2 tangent line to the ellipse ! # 1 at !1, 2". 2 8 (b) Show that the equation of the tangent line to the ellipse x x y y x2 y2 ! 2 # 1 at !x0, y0" is 02 ! 02 # 1. 2 a b a b 42. (a) Use implicit differentiation to find an equation of the x2 y2 tangent line to the hyperbola " # 1 at !3, "2". 6 8 (b) Show that the equation of the tangent line to the hyperbola x x y y x2 y2 " 2 # 1 at !x0, y0" is 02 " 02 # 1. 2 a b a b In Exercises 43 and 44, find dy/dx implicitly and find the largest interval of the form !a < y < a or 0 < y < a such that y is a differentiable function of x. Write dy/dx as a function of x. 43. tan y # x 45. x 2 ! y2 # 36 46. x 2 y 2 " 2x # 3 47. x 2 " y 2 # 16 48. 1 " xy # x " y 49. y 2 # x 3 50. y 2 # 4x y2 60. y 2 # x 3 2x 2 ! 3y 2 # 5 # 4x 62. x3 # 3! y " 1" 61. x ! y # 0 x!3y " 29" # 3 x # sin y Orthogonal Trajectories In Exercises 63 and 64, verify that the two families of curves are orthogonal where C and K are real numbers. Use a graphing utility to graph the two families for two values of C and two values of K. 63. xy # C, x2 " y 2 # K 64. x 2 ! y 2 # C 2, y # Kx In Exercises 65–68, differentiate (a) with respect to x ( y is a function of x) and (b) with respect to t ( x and y are functions of t). 65. 2y 2 " 3x 4 # 0 66. x 2 " 3xy 2 ! y 3 # 10 67. cos \$ y " 3 sin \$ x # 1 68. 4 sin x cos y # 1 69. Describe the difference between the explicit form of a function and an implicit equation. Give an example of each. 70. In your own words, state the guidelines for implicit differentiation. 52. y 2 # x"1 , x2 ! 1 #2, 55\$ & In Exercises 53 and 54, find equations for the tangent line and normal line to the circle at the given points. (The normal line at a point is perpendicular to the tangent line at the point.) Use a graphing utility to graph the equation, tangent line, and normal line. 71. Orthogonal Trajectories The figure below shows the topographic map carried by a group of hikers. The hikers are in a wooded area on top of the hill shown on the map and they decide to follow a path of steepest descent (orthogonal trajectories to the contours on the map). Draw their routes if they start from point A and if they start from point B. If their goal is to reach the road along the top of the map, which starting point should they use? To print an enlarged copy of the graph, select the MathGraph button. 54. x 2 ! y 2 # 9 55. Show that the normal line at any point on the circle x 2 ! y 2 # r 2 passes through the origin. 56. Two circles of radius 4 are tangent to the graph of y 2 # 4x at the point !1, 2". Find equations of these two circles. 1671 00 !0, 3", !2, &5 " 18 !4, 3", !"3, 4" 59. 2x 2 ! y 2 # 6 In Exercises 51 and 52, use a graphing utility to graph the equation. Find an equation of the tangent line to the graph at the given point and graph the tangent line in the same viewing window. 53. x 2 ! y 2 # 25 Orthogonal Trajectories In Exercises 59–62, use a graphing utility to sketch the intersecting graphs of the equations and show that they are orthogonal. [Two graphs are orthogonal if at their point(s) of intersection their tangent lines are perpendicular to each other.] 44. cos y # x In Exercises 45–50, find d 2 y /dx 2 in terms of x and y. 51. &x ! &y # 4, !9, 1" 147 Implicit Differentiation B 1994 A 00 18 148 CHAPTER 2 Differentiation 72. Weather Map The weather map shows several isobars— curves that represent areas of constant air pressure. Three high pressures H and one low pressure L are shown on the map. Given that wind speed is greatest along the orthogonal trajectories of the isobars, use the map to determine the areas having high wind speed. 76. Slope Find all points on the circle x2 ! y2 # 25 where the 3 slope is 4. 77. Horizontal Tangent Determine the point(s) at which the graph of y 4 # y2 " x2 has a horizontal tangent. 78. Tangent Lines Find equations of both tangent lines to the x2 y2 ellipse ! # 1 that passes through the point !4, 0". 4 9 79. Normals to a Parabola The graph shows the normal lines from the point !2, 0" to the graph of the parabola x # y2. How many normal lines are there from the point !x0, 0" to the graph 1 1 of the parabola if (a) x0 # 4, (b) x0 # 2, and (c) x0 # 1? For what value of x0 are two of the normal lines perpendicular to each other? H H L H y 73. Consider the equation x 4 # 4!4x 2 " y 2". (2, 0) (a) Use a graphing utility to graph the equation. x (b) Find and graph the four tangent lines to the curve for y # 3. x = y2 (c) Find the exact coordinates of the point of intersection of the two tangent lines in the first quadrant. 74. Let L be any tangent line to the curve &x ! &y # &c. Show that the sum of the x- and y-intercepts of L is c. 75. Prove (Theorem 2.3) that d n 'x ( # nx n"1 dx for the case in which n is a rational number. (Hint: Write y # x p%q in the form y q # x p and differentiate implicitly. Assume that p and q are integers, where q > 0.) 80. Normal Lines (a) Find an equation of the normal line to the ellipse x2 y2 ! #1 32 8 at the point !4, 2". (b) Use a graphing utility to graph the ellipse and the normal line. (c) At what other point does the normal line intersect the ellipse? SECTION 2.6 Section 2.6 Related Rates 149 Related Rates • Find a related rate. • Use related rates to solve real-life problems. r Finding Related Rates h You have seen how the Chain Rule can be used to find dy#dx implicitly. Another important use of the Chain Rule is to find the rates of change of two or more related variables that are changing with respect to time. For example, when water is drained out of a conical tank (see Figure 2.33), the volume V, the radius r, and the height h of the water level are all functions of time t. Knowing that these variables are related by the equation V! r # 2 r h 3 Original equation you can differentiate implicitly with respect to t to obtain the related-rate equation h d !V " ! dt dV ! dt d dt # 3 # ! 3 &#3 r h' (r dhdt " h &2r drdt') &r dhdt " 2rh drdt'. 2 2 Differentiate with respect to t. 2 From this equation you can see that the rate of change of V is related to the rates of change of both h and r. r E X P L O R AT I O N h Finding a Related Rate In the conical tank shown in Figure 2.33, suppose that the height is changing at a rate of \$0.2 foot per minute and the radius is changing at a rate of \$0.1 foot per minute. What is the rate of change in the volume when the radius is r ! 1 foot and the height is h ! 2 feet? Does the rate of change in the volume depend on the values of r and h? Explain. EXAMPLE 1 . Volume is related to radius and height. Figure 2.33 Animation FOR FURTHER INFORMATION To Related Rates” by Bill Austin, Don .. Barry, and David Berman in Mathematics Magazine. MathArticle Two Rates That Are Related Suppose x and y are both differentiable functions of t and are related by the equation y ! x 2 " 3. Find dy#dt when x ! 1, given that dx#dt ! 2 when x ! 1. Solution Using the Chain Rule, you can differentiate both sides of the equation with respect to t. y ! x2 " 3 d d \$ y% ! \$x 2 " 3% dt dt dy dx ! 2x dt dt When x ! 1 and dx#dt ! 2, you have dy ! 2!1"!2" ! 4. dt Try It Exploration A Write original equation. Differentiate with respect to t. Chain Rule 150 CHAPTER 2 Differentiation Problem Solving with Related Rates In Example 1, you were given an equation that related the variables x and y and were asked to find the rate of change of y when x ! 1. Equation: Given rate: Find: y ! x2 " 3 dx ! 2 when x ! 1 dt dy when x ! 1 dt In each of the remaining examples in this section, you must create a mathematical model from a verbal description. EXAMPLE 2 Ripples in a Pond A pebble is dropped into a calm pond, causing ripples in the form of concentric circles, as shown in Figure 2.34. The radius r of the outer ripple is increasing at a constant rate of 1 foot per second. When the radius is 4 feet, at what rate is the total area A of the disturbed water changing? Solution The variables r and A are related by A ! # r 2. The rate of change of the radius r is dr#dt ! 1. Equation: Given rate: Find: A ! #r2 dr !1 dt dA when dt r!4 With this information, you can proceed as in Example 1. Total area increases as the outer radius increases. . Figure 2.34 d d \$A% ! \$# r 2% dt dt dA dr ! 2# r dt dt dA ! 2# !4"!1" ! 8# dt Differentiate with respect to t. Chain Rule Substitute 4 for r and 1 for dr#dt. When the radius is 4 feet, the area is changing at a rate of 8# square feet per second. Try It Exploration A Video Video Guidelines For Solving Related-Rate Problems NOTE When using these guidelines, be sure you perform Step 3 before Step 4. Substituting the known values of the variables before differentiating will produce an inappropriate derivative. 1. Identify all given quantities and quantities to be determined. Make a sketch and label the quantities. 2. Write an equation involving the variables whose rates of change either are given or are to be determined. 3. Using the Chain Rule, implicitly differentiate both sides of the equation with respect to time t. 4. After completing Step 3, substitute into the resulting equation all known values for the variables and their rates of change. Then solve for the required rate of change. SECTION 2.6 Related Rates 151 The table below lists examples of mathematical models involving rates of change. For instance, the rate of change in the first example is the velocity of a car. Verbal Statement Mathematical Model The velocity of a car after traveling for 1 hour is 50 miles per hour. x ! distance traveled dx ! 50 when t ! 1 dt Water is being pumped into a swimming pool at a rate of 10 cubic meters per hour. V ! volume of water in pool dV ! 10 m3#hr dt A gear is revolving at a rate of 25 revolutions per minute (1 revolution ! 2# rad). % ! angle of revolution d% dt An Inflating Balloon EXAMPLE 3 Air is being pumped into a spherical balloon (see Figure 2.35) at a rate of 4.5 cubic feet per minute. Find the rate of change of the radius when the radius is 2 feet. Solution Let V be the volume of the balloon and let r be its radius. Because the volume is increasing at a rate of 4.5 cubic feet per minute, you know that at time t the rate of change of the volume is dV#dt ! 92. So, the problem can be stated as shown. Given rate: Find: dV 9 (constant rate) ! dt 2 dr when r ! 2 dt To find the rate of change of the radius, you must find an equation that relates the radius r to the volume V. Equation: V! 4 #r3 3 Volume of a sphere Differentiating both sides of the equation with respect to t produces dV dr ! 4# r 2 dt dt dr 1 dV . ! dt 4# r 2 dt & ' Differentiate with respect to t. Solve for dr#dt. Finally, when r ! 2, the rate of change of the radius is . Inflating a balloon Figure 2.35 Animation &' dr 1 9 ! * 0.09 foot per minute. dt 16# 2 Try It Exploration A Video In Example 3, note that the volume is increasing at a constant rate but the radius is increasing at a variable rate. Just because two rates are related does not mean that they are proportional. In this particular case, the radius is growing more and more slowly as t increases. Do you see why? 152 CHAPTER 2 Differentiation EXAMPLE 4 The Speed of an Airplane Tracked by Radar An airplane is flying on a flight path that will take it directly over a radar tracking station, as shown in Figure 2.36. If s is decreasing at a rate of 400 miles per hour when s ! 10 miles, what is the speed of the plane? s Solution Let x be the horizontal distance from the station, as shown in Figure 2.36. Notice that when s ! 10, x ! +10 2 \$ 36 ! 8. 6 mi Given rate: x Find: Not drawn to scale An airplane is flying at an altitude of 6 miles, s miles from the station. ds#dt ! \$400 when s ! 10 dx#dt when s ! 10 and x ! 8 You can find the velocity of the plane as shown. Equation: Figure 2.36 x2 " 62 ! s2 dx ds 2x ! 2s dt dt dx s ds ! dt x dt dx 10 ! !\$400" dt 8 ! \$500 miles per hour Pythagorean Theorem Differentiate with respect to t. & ' . Solve for dx#dt. Substitute for s, x, and ds#dt. Simplify. Because the velocity is \$500 miles per hour, the speed is 500 miles per hour. Try It EXAMPLE 5 Exploration A Open Exploration A Changing Angle of Elevation Find the rate of change in the angle of elevation of the camera shown in Figure 2.37 at 10 seconds after lift-off. Solution Let % be the angle of elevation, as shown in Figure 2.37. When t ! 10, the height s of the rocket is s ! 50t 2 ! 50!10" 2 ! 5000 feet. Given rate: Find: ds#dt ! 100t ! velocity of rocket d%#dt when t ! 10 and s ! 5000 Using Figure 2.37, you can relate s and % by the equation tan % ! s#2000. Equation: tan θ = s 2000 s θ 2000 ft Not drawn to scale A television camera at ground level is filming the lift-off of a space shuttle that is rising vertically according to the position equation s ! 50t 2, where s is measured in feet and t is measured in seconds. The camera is 2000 feet from Figure 2.37 Animation s 2000 d% 1 ds !sec 2%" ! dt 2000 dt d% 100t ! cos 2 % dt 2000 2000 ! +s 2 " 2000 2 tan % ! See Figure 2.37. & ' & Differentiate with respect to t. Substitute 100t for ds#dt. ' 2 100t 2000 cos % ! 2000#+s 2 " 2000 2 When t ! 10 and s ! 5000, you have 2 d% 2000!100"!10" ! ! dt 50002 " 20002 29 2 So, when t ! 10, % is changing at a rate of 29 Try It Exploration A SECTION 2.6 EXAMPLE 6 Related Rates 153 The Velocity of a Piston In the engine shown in Figure 2.38, a 7-inch connecting rod is fastened to a crank of radius 3 inches. The crankshaft rotates counterclockwise at a constant rate of 200 revolutions per minute. Find the velocity of the piston when % ! ##3. 7 3 θ θ . x The velocity of a piston is related to the angle of the crankshaft. Figure 2.38 Animation Solution Label the distances as shown in Figure 2.38. Because a complete revolution corresponds to 2# radians, it follows that d%#dt ! 200!2#" ! 400# b a θ c Law of Cosines: b 2 ! a 2 " c 2 \$ 2ac cos % Figure 2.39 Given rate: Find: d% ! 400# (constant rate) dt dx # when % ! dt 3 You can use the Law of Cosines (Figure 2.39) to find an equation that relates x and %. Equation: 7 2 ! 3 2 " x 2 \$ 2!3"!x" cos % dx d% dx 0 ! 2x \$ 6 \$x sin % " cos % dt dt dt dx d% !6 cos % \$ 2x" ! 6x sin % dt dt dx 6x sin % d% ! dt 6 cos % \$ 2x dt & ' & ' When % ! ##3, you can solve for x as shown. 7 2 ! 3 2 " x 2 \$ 2!3"!x" cos 49 ! 9 " x 2 \$ 6x # 3 &12' 0 ! x 2 \$ 3x \$ 40 0 ! !x \$ 8"!x " 5" x!8 Choose positive solution. So, when x ! 8 and % ! ##3, the velocity of the piston is . dx 6!8"!+3#2" ! !400#" dt 6!1#2" \$ 16 9600#+3 ! \$13 * \$4018 inches per minute. Try It Exploration A NOTE Note that the velocity in Example 6 is negative because x represents a distance that is decreasing. 154 CHAPTER 2 Differentiation Exercises for Section 2.6 The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. Click on to view the complete solution of the exercise. Click on to print an enlarged copy of the graph. In Exercises 1– 4, assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt. Equation Find 1. y ! \$x 2. y ! 2"x 2 % 3x# 3. xy ! 4 4. x 2 \$ y 2 ! 25 Given (a) dy when x ! 4 dt dx !3 dt (b) dx when x ! 25 dt dy !2 dt (a) dy when x ! 3 dt dx !2 dt (b) dx when x ! 1 dt dy !5 dt (a) dy when x ! 8 dt dx ! 10 dt (b) dx when x ! 1 dt dy ! %6 dt (a) dy when x ! 3, y ! 4 dt dx !8 dt dy ! %2 dt dx (b) when x ! 4, y ! 3 dt 12. In your own words, state the guidelines for solving relatedrate problems. 13. Find the rate of change of the distance between the origin and a moving point on the graph of y ! x2 \$ 1 if dx!dt ! 2 centimeters per second. 14. Find the rate of change of the distance between the origin and a moving point on the graph of y ! sin x if dx!dt ! 2 centimeters per second. 15. Area The radius r of a circle is increasing at a rate of 3 centimeters per minute. Find the rates of change of the area when (a) r ! 6 centimeters and (b) r ! 24 centimeters. 16. Area Let A be the area of a circle of radius r that is changing with respect to time. If dr!dt is constant, is dA!dt constant? Explain. 17. Area The included angle of the two sides of constant equal length s of an isosceles triangle is #. 1 (a) Show that the area of the triangle is given by A ! 2s 2 sin #. 1 In Exercises 5–8, a point is moving along the graph of the given function such that dx/dt is 2 centimeters per second. Find dy/dt for the given values of x. 5. y ! x 2 \$ 1 (a) x ! %1 (b) x ! 0 (c) x ! 1 1 6. y ! 1 \$ x2 (a) x ! %2 (b) x ! 0 (c) x ! 2 7. y ! tan x " (a) x ! % 3 " (b) x ! % 4 (c) x ! 0 8. y ! sin x " (a) x ! 6 " (b) x ! 4 " (c) x ! 3 In Exercises 9 and 10, using the graph of f, (a) determine whether dy/dt is positive or negative given that dx/dt is negative, and (b) determine whether dx/dt is positive or negative given that dy/dt is positive. 9. 10. y 6 5 4 3 2 4 2 f 1 x 1 2 y 3 4 3 2 1 (b) If # is increasing at the rate of 2 radian per minute, find the rates of change of the area when # ! "!6 and # ! "!3. (c) Explain why the rate of change of the area of the triangle is not constant even though d#!dt is constant. 18. Volume The radius r of a sphere is increasing at a rate of 2 inches per minute. (a) Find the rate of change of the volume when r ! 6 inches and r ! 24 inches. (b) Explain why the rate of change of the volume of the sphere is not constant even though dr!dt is constant. 19. Volume A spherical balloon is inflated with gas at the rate of 800 cubic centimeters per minute. How fast is the radius of the balloon increasing at the instant the radius is (a) 30 centimeters and (b) 60 centimeters? 20. Volume All edges of a cube are expanding at a rate of 3 centimeters per second. How fast is the volume changing when each edge is (a) 1 centimeter and (b) 10 centimeters? 21. Surface Area The conditions are the same as in Exercise 20. Determine how fast the surface area is changing when each edge is (a) 1 centimeter and (b) 10 centimeters. 1 22. Volume The formula for the volume of a cone is V ! 3" r 2 h. Find the rate of change of the volume if dr!dt is 2 inches per minute and h ! 3r when (a) r ! 6 inches and (b) r ! 24 inches. f x 1 2 3 11. Consider the linear function y ! ax \$ b. If x changes at a constant rate, does y change at a constant rate? If so, does it change at the same rate as x? Explain. 23. Volume At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 10 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when the pile is 15 feet high? SECTION 2.6 24. Depth A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep. 25. Depth A swimming pool is 12 meters long, 6 meters wide, 1 meter deep at the shallow end, and 3 meters deep at the deep end (see figure). Water is being pumped into the pool at 14 cubic meter per minute, and there is 1 meter of water at the deep end. (a) What percent of the pool is filled? (b) At what rate is the water level rising? 1 m3 4 min 3 2 ft min 1m 3m 29. Construction A winch at the top of a 12-meter building pulls a pipe of the same length to a vertical position, as shown in the figure. The winch pulls in rope at a rate of %0.2 meter per second. Find the rate of vertical change and the rate of horizontal change at the end of the pipe when y ! 6. y 12 3 ft ds = 0.2 m sec dt 12 m (x, y) s 9 h ft 3 ft 13 ft 12 ft 6 Figure for 25 12 m 3 Figure for 26 26. Depth A trough is 12 feet long and 3 feet across the top (see figure). Its ends are isosceles triangles with altitudes of 3 feet. (a) If water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising when h is 1 foot deep? 3 Figure for 30 (a) The winch pulls in rope at a rate of 4 feet per second. Determine the speed of the boat when there is 13 feet of rope out. What happens to the speed of the boat as it gets closer to the dock? (b) Suppose the boat is moving at a constant rate of 4 feet per second. Determine the speed at which the winch pulls in rope when there is a total of 13 feet of rope out. What happens to the speed at which the winch pulls in rope as the boat gets closer to the dock? (a) How fast is the top of the ladder moving down the wall when its base is 7 feet, 15 feet, and 24 feet from the wall? (c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 7 feet from the wall. Not drawn to scale x 30. Boating A boat is pulled into a dock by means of a winch 12 feet above the deck of the boat (see figure). 27. Moving Ladder A ladder 25 feet long is leaning against the wall of a house (see figure). The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (b) Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall. 6 Figure for 29 (b) If the water is rising at a rate of 38 inch per minute when h ! 2, determine the rate at which water is being pumped into the trough. 31. Air Traffic Control An air traffic controller spots two planes at the same altitude converging on a point as they fly at right angles to each other (see figure). One plane is 150 miles from the point moving at 450 miles per hour. The other plane is 200 miles from the point moving at 600 miles per hour. (a) At what rate is the distance between the planes decreasing? (b) How much time does the air traffic controller have to get one of the planes on a different flight path? y m y 25 ft 5m ft 2 sec Figure for 27 Figure for 28 mathematics of moving ladders, see the article “The Falling The College Mathematics Journal. MathArticle Distance (in miles) 0.15 sec r 155 28. Construction A construction worker pulls a five-meter plank up the side of a building under construction by means of a rope tied to one end of the plank (see figure). Assume the opposite end of the plank follows a path perpendicular to the wall of the building and the worker pulls the rope at a rate of 0.15 meter per second. How fast is the end of the plank sliding along the ground when it is 2.5 meters from the wall of the building? 12 ft 6m Related Rates x 200 5 mi s s 100 x Not drawn to scale 100 200 Distance (in miles) Figure for 31 x Figure for 32 156 CHAPTER 2 Differentiation 32. Air Traffic Control An airplane is flying at an altitude of 5 miles and passes directly over a radar antenna (see figure on previous page). When the plane is 10 miles away "s ! 10#, the radar detects that the distance s is changing at a rate of 240 miles per hour. What is the speed of the plane? 33. Sports A baseball diamond has the shape of a square with sides 90 feet long (see figure). A player running from second base to third base at a speed of 28 feet per second is 30 feet from third base. At what rate is the player’s distance s from home plate changing? y 2nd 16 1st 8 4 90 ft 4 Home Figure for 33 and 34 39. Evaporation As a spherical raindrop falls, it reaches a layer of dry air and begins to evaporate at a rate that is proportional to its surface area "S ! 4" r 2#. Show that the radius of the raindrop decreases at a constant rate. 40. Electricity The combined electrical resistance R of R1 and R2, connected in parallel, is given by 1 1 1 ! \$ R R1 R2 where R, R1, and R2 are measured in ohms. R1 and R2 are increasing at rates of 1 and 1.5 ohms per second, respectively. At what rate is R changing when R1 ! 50 ohms and R2 ! 75 ohms? 12 3rd 38. Machine Design Repeat Exercise 37 for a position function 3 of x"t# ! 35 sin " t. Use the point "10 , 0# for part (c). 8 12 16 20 x Figure for 35 34. Sports For the baseball diamond in Exercise 33, suppose the player is running from first to second at a speed of 28 feet per second. Find the rate at which the distance from home plate is changing when the player is 30 feet from second base. 35. Shadow Length A man 6 feet tall walks at a rate of 5 feet per second away from a light that is 15 feet above the ground (see figure). When he is 10 feet from the base of the light, 41. Adiabatic Expansion When a certain polyatomic gas undergoes adiabatic expansion, its pressure p and volume V satisfy the equation pV 1.3 ! k, where k is a constant. Find the relationship between the related rates dp!dt and dV!dt. circular arc of radius r. In order not to rely on friction alone to overcome the centrifugal force, the road is banked at an angle of magnitude # from the horizontal (see figure). The banking angle must satisfy the equation rg tan # ! v 2, where v is the velocity of the cars and g ! 32 feet per second per second is the acceleration due to gravity. Find the relationship between the related rates dv!dt and d#!dt. (a) at what rate is the tip of his shadow moving? (b) at what rate is the length of his shadow changing? 36. Shadow Length Repeat Exercise 35 for a man 6 feet tall walking at a rate of 5 feet per second toward a light that is 20 feet above the ground (see figure). y V y 20 r 16 (0, y) 12 1m 8 4 (x, 0) 4 8 12 16 20 x x Figure for 36 Figure for 37 37. Machine Design The endpoints of a movable rod of length 1 meter have coordinates "x, 0# and "0, y# (see figure). The position of the end on the x-axis is x"t# ! 1 "t sin 2 6 where t is the time in seconds. (a) Find the time of one complete cycle of the rod. (b) What is the lowest point reached by the end of the rod on the y-axis? (c) Find the speed of the y-axis endpoint when the x-axis endpoint is "14, 0#. 43. Angle of Elevation A balloon rises at a rate of 3 meters per second from a point on the ground 30 meters from an observer. Find the rate of change of the angle of elevation of the balloon from the observer when the balloon is 30 meters above the ground. 44. Angle of Elevation A fish is reeled in at a rate of 1 foot per second from a point 10 feet above the water (see figure). At what rate is the angle between the line and the water changing when there is a total of 25 feet of line out? 10 ft x V SECTION 2.6 45. Angle of Elevation An airplane flies at an altitude of 5 miles toward a point directly over an observer (see figure). The speed of the plane is 600 miles per hour. Find the rates at which the angle of elevation # is changing when the angle is (a) # ! 30&, (b) # ! 60&, and (c) # ! 75&. Related Rates 157 50. Think About It Describe the relationship between the rate of change of y and the rate of change of x in each expression. Assume all variables and derivatives are positive. (a) dy dx !3 dt dt (b) dy dx ! x"L % x# , 0 f x f L dt dt Acceleration In Exercises 51 and 52, find the acceleration of the specified object. (Hint: Recall that if a variable is changing at a constant rate, its acceleration is zero.) 5 mi 51. Find the acceleration of the top of the ladder described in Exercise 27 when the base of the ladder is 7 feet from the wall. V Not drawn to scale 46. Linear vs. Angular Speed A patrol car is parked 50 feet from a long warehouse (see figure). The revolving light on top of the car turns at a rate of 30 revolutions per minute. How fast is the light beam moving along the wall when the beam makes angles of (a) # ! 30&, (b) # ! 60&, and (c) # ! 70& with the line perpendicular from the light to the wall? 52. Find the acceleration of the boat in Exercise 30(a) when there is a total of 13 feet of rope out. 53. Modeling Data The table shows the numbers (in millions) of single women (never married) s and married women m in the civilian work force in the United States for the years 1993 through 2001. (Source: U.S. Bureau of Labor Statistics) Year 1993 1994 1995 1996 1997 1998 1999 2000 2001 P V V 50 ft 30 cm x x Figure for 47 47. Linear vs. Angular Speed A wheel of radius 30 centimeters revolves at a rate of 10 revolutions per second. A dot is painted at a point P on the rim of the wheel (see figure). (a) Find dx!dt as a function of #. (b) Use a graphing utility to graph the function in part (a). (c) When is the absolute value of the rate of change of x greatest? When is it least? (d) Find dx!dt when # ! 30& and # ! 60&. 48. Flight Control An airplane is flying in still air with an airspeed of 240 miles per hour. If it is climbing at an angle of 22&, find the rate at which it is gaining altitude. 49. Security Camera A security camera is centered 50 feet above a 100-foot hallway (see figure). It is easiest to design the camera with a constant angular rate of rotation, but this results in a variable rate at which the images of the surveillance area are recorded. So, it is desirable to design a system with a variable rate of rotation and a constant rate of movement of the scanning beam along the hallway. Find a model for the variable rate of rotation if dx!dt ! 2 feet per second. % % 15.0 15.3 15.5 15.8 16.5 17.1 17.6 17.8 18.0 m 32.0 32.9 33.4 33.6 33.8 33.9 34.4 34.6 34.7 (a) Use the regression capabilities of a graphing utility to find a model of the form m"s# ! as3 \$ bs2 \$ cs \$ d for the data, where t is the time in years, with t ! 3 corresponding to 1993. x Figure for 46 s (b) Find dm!dt. Then use the model to estimate dm!dt for t ! 10 if it is predicted that the number of single women in the work force will increase at the rate of 0.75 million per year. 54. Moving Shadow A ball is dropped from a height of 20 meters, 12 meters away from the top of a 20-meter lamppost (see figure). The ball’s shadow, caused by the light at the top of the lamppost, is moving along the level ground. How fast is the shadow moving 1 second after the ball is released? (Submitted by Dennis Gittinger, St. Philips College, San Antonio, TX) 20 m y (0, 50) V 12 m x 100 ft 158 CHAPTER 2 Differentiation Review Exercises for Chapter 2 The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. Click on to view the complete solution of the exercise. Click on to print an enlarged copy of the graph. In Exercises 1–4, find the derivative of the function by using the definition of the derivative. 1. f "x# ! x 2 " 2x % 3 3. f "x# ! x%1 x"1 4. f "x# ! 5. f "x# ! "x % 1#2&3 6. f "x# ! 19. h"t# ! 3t 4 20. f "t# ! "8t 5 21. f "x# ! x " 3x 3 2 x 25. g"t# ! 4x x%3 2 3t 2 2 "3x# 2 28. g"## ! 4 cos # % 6 5 sin # " 2# 30. g"## ! 3 26. h"x# ! 27. f "\$# ! 2\$ " 3 sin \$ 29. f "\$# ! 3 cos \$ " sin \$ 4 3 8 2 4 31. y 12 4 1 1 8 x 1 4 1 8 1 2 ( (a) Is f continuous at x ! 2? ( 1 (b) Is f differentiable at x ! 2? Explain. 8. Sketch the graph of f "x# ! ' x2 % 4x % 2, 1 " 4x " x2, x < "2 x v "2. (a) Is f continuous at x ! "2? (b) Is f differentiable at x ! "2? Explain. In Exercises 9 and 10, find the slope of the tangent line to the graph of the function at the given point. 2 x 9. g"x# ! x 2 " , 3 6 3x " 2x 2, 8 %"1, 65& %"2, " 354& In Exercises 11 and 12, (a) find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of the graphing utility to confirm your results. 2 , "0, 2# 11. f "x# ! x 3 " 1, ""1, "2# 12. f "x# ! x%1 In Exercises 13 and 14, use the alternative form of the derivative to find the derivative at x " c (if it exists). 1 , c!2 13. g"x# ! x 2"x " 1#, c ! 2 14. f "x# ! x%1 In Exercises 15–30, find the derivative of the function. 15. y ! 25 16. y ! "12 x8 18. g"x# ! x12 y 2 x 4 32. y U 2 7. Sketch the graph of f "x# ! 4 " x " 2 . 17. f "x# ! 24. f "x# ! x1\$2 " x"1\$2 Writing In Exercises 31 and 32, the figure shows the graphs of a function and its derivative. Label the graphs as f or f! and write a short paragraph stating the criteria used in making the selection. To print an enlarged copy of the graph, select the MathGraph button. y 10. h"x# ! 22. g"s# ! 4s 4 " 5s 2 2 3 x 23. h"x# ! 6!x % 3! 2. f "x# ! !x % 1 In Exercises 5 and 6, describe the x-values at which f is differentiable. 2 . U 2 x x 1 33. Vibrating String When a guitar string is plucked, it vibrates with a frequency of F ! 200!T, where F is measured in vibrations per second and the tension T is measured in pounds. Find the rates of change of F when (a) T ! 4 and (b) T ! 9. 34. Vertical Motion A ball is dropped from a height of 100 feet. One second later, another ball is dropped from a height of 75 feet. Which ball hits the ground first? 35. Vertical Motion To estimate the height of a building, a weight is dropped from the top of the building into a pool at ground level. How high is the building if the splash is seen 9.2 seconds after the weight is dropped? 36. Vertical Motion A bomb is dropped from an airplane at an altitude of 14,400 feet. How long will it take for the bomb to reach the ground? (Because of the motion of the plane, the fall will not be vertical, but the time will be the same as that for a vertical fall.) The plane is moving at 600 miles per hour. How far will the bomb move horizontally after it is released from the plane? 37. Projectile Motion y ! x " 0.02x 2. A ball thrown follows a path described by (a) Sketch a graph of the path. (b) Find the total horizontal distance the ball is thrown. (c) At what x-value does the ball reach its maximum height? (Use the symmetry of the path.) (d) Find an equation that gives the instantaneous rate of change of the height of the ball with respect to the horizontal change. Evaluate the equation at x ! 0, 10, 25, 30, and 50. (e) What is the instantaneous rate of change of the height when the ball reaches its maximum height? 159 REVIEW EXERCISES 38. Projectile Motion The path of a projectile thrown at an angle of 45+ with level ground is y!x" 32 2 "x # v02 where the initial velocity is v0 feet per second. 39. Horizontal Motion The position function of a particle moving along the x-axis is for (a) Find the velocity of the particle. (b) Find the open t-interval(s) in which the particle is moving to the left. (c) Find the position of the particle when the velocity is 0. (d) Find the speed of the particle when the position is 0. 40. Modeling Data The speed of a car in miles per hour and the stopping distance in feet are recorded in the table. Speed, x 20 30 40 Stopping Distance, y 25 55 105 188 300 50 53. y ! x cos x " sin x 54. g"x# ! 3x sin x % x2 cos x 60 (a) Use the regression capabilities of a graphing utility to find a quadratic model for the data. (b) Use a graphing utility to plot the data and graph the model. 56. f "x# ! x%1 , x"1 57. f "x# ! "x tan x, "0, 0# 58. f "x# ! 1 % sin x , "', 1# 1 " sin x 59. Acceleration The velocity of an object in meters per second is v"t# ! 36 " t 2, 0 f t f 6. Find the velocity and acceleration of the object when t ! 4. 60. Acceleration An automobile’s velocity starting from rest is v"t# ! 90t 4t % 10 where v is measured in feet per second. Find the vehicle’s velocity and acceleration at each of the following times. (b) 5 seconds (c) 10 seconds In Exercises 61–64, find the second derivative of the function. 61. g"t# ! t 3 " 3t % 2 4 x 62. f "x# ! 12! 63. f "\$# ! 3 tan \$ 64. h"t# ! 4 sin t " 5 cos t In Exercises 65 and 66, show that the function satisfies the equation. Equation Function 65. y ! 2 sin x % 3 cos x y) % y ! 0 10 " cos x 66. y ! x xy( % y ! sin x In Exercises 67–78, find the derivative of the function. %xx "% 31 & % 2 (c) Use a graphing utility to graph dy\$dx. 67. h"x# ! (d) Use the model to approximate the stopping distance at a speed of 65 miles per hour. 69. f "s# ! "s 2 " 1#5\$2"s 3 % 5# 70. h"\$# ! (e) Use the graphs in parts (b) and (c) to explain the change in stopping distance as the speed increases. 71. y ! 3 cos"3x % 1# 72. In Exercises 41–54, find the derivative of the function. x sin 2x 73. y ! " 2 4 74. 41. f "x# ! "3x 2 % 7#"x 2 " 2x % 3# 75. y ! 2 3\$2 2 sin x " sin7\$2 x 3 7 76. 77. y ! sin 'x x%2 78. 42. g"x# ! "x " 3x#"x % 2# 3 43. h"x# ! !x sin x 45. f "x# ! x2 % x " 1 x2 " 1 44. f "t# ! t 3 cos t 46. f "x# ! 6x " 5 x2 % 1 1 47. f "x# ! 4 " 3x 2 9 48. f "x# ! 2 3x " 2x x2 49. y ! cos x sin x 50. y ! 2 x %12, "3& 2x3 " 1 , "1, 1# x2 55. f "x# ! (a) 1 second . " < t < 52. y ! 2x " x 2 tan x In Exercises 55–58, find an equation of the tangent line to the graph of f at the given point. (a) Find the x-coordinate of the point where the projectile strikes the ground. Use the symmetry of the path of the projectile to locate the x-coordinate of the point where the projectile reaches its maximum height. (b) What is the instantaneous rate of change of the height when the projectile is at its maximum height? (c) Show that doubling the initial velocity of the projectile multiplies both the maximum height and the range by a factor of 4. (d) Find the maximum height and range of a projectile thrown with an initial velocity of 70 feet per second. Use a graphing utility to graph the path of the projectile. x"t# ! t 2 " 3t % 2 51. y ! 3x 2 sec x 68. f "x# ! x 2 % 2 1 x & 5 \$ "1 " \$#3 y ! 1 " cos 2x % 2 cos 2 x sec7 x sec5 x y! " 7 5 3x f "x# ! !x 2 % 1 cos"x " 1# y! x"1 In Exercises 79–82, find the derivative of the function at the given point. 79. f "x# ! !1 " x3, ""2, 3# 3 x2 " 1, "3, 2# 80. f "x# ! ! 160 CHAPTER 2 81. y ! 1 csc 2x, 2 Differentiation In Exercises 101–106, use implicit differentiation to find dy/dx. %'4 , 12& 82. y ! csc 3x % cot 3x, % & ' ,1 6 In Exercises 83–86, use a computer algebra system to find the derivative of the function. Use the utility to graph the function and its derivative on the same set of coordinate axes. Describe the behavior of the function that corresponds to any zeros of the graph of the derivative. 2x 83. g"x# ! !x % 1 85. f "t# ! !t % 1 84. f "x# ! "x " 2#"x % 4#) 3 t ! %1 2 86. y ! !3x "x % 2# 3 In Exercises 87–90, (a) use a computer algebra system to find the derivative of the function at the given point, (b) find an equation of the tangent line to the graph of the function at the point, and (c) graph the function and its tangent line on the same set of coordinate axes. 87. f "t# ! t2"t " 1#5, "2, 4# 88. g"x# ! x!x2 % 1, 89. y ! tan!1 " x, 90. y ! 2 csc3"!x #, "3, 3!10 # ""2, tan !3 # "1, 2 csc3 1# In Exercises 91–94, find the second derivative of the function. 1 % tan x x 91. y ! 2x 2 % sin 2x 92. y ! 93. f "x# ! cot x 94. y ! sin 2 x In Exercises 95 –98, use a computer algebra system to find the second derivative of the function. 95. f "t# ! t (1 " t#2 96. g"x# ! 101. x 2 % 3xy % y 3 ! 10 102. x 2 % 9y 2 " 4x % 3y ! 0 103. y!x " x!y ! 16 104. y 2 ! "x " y#"x 2 % y# 105. x sin y ! y cos x 106. cos"x % y# ! x In Exercises 107 and 108, find the equations of the tangent line and the normal line to the graph of the equation at the given point. Use a graphing utility to graph the equation, the tangent line, and the normal line. 107. x 2 % y 2 ! 20, "2, 4# 109. A point moves along the curve y ! !x in such a way that the y-value is increasing at a rate of 2 units per second. At what rate is x changing for each of the following values? 1 (a) x ! 2 700 t 2 % 4t % 10 where t is the time in hours. Find the rate of change of T with respect to t at each of the following times. (a) t ! 1 (b) t ! 3 (c) t ! 5 (c) x ! 4 111. Depth The cross section of a five-meter trough is an isosceles trapezoid with a two-meter lower base, a three-meter upper base, and an altitude of 2 meters. Water is running into the trough at a rate of 1 cubic meter per minute. How fast is the water level rising when the water is 1 meter deep? 112. Linear and Angular Velocity A rotating beacon is located 1 kilometer off a straight shoreline (see figure). If the beacon rotates at a rate of 3 revolutions per minute, how fast (in kilometers per hour) does the beam of light appear to be 1 moving to a viewer who is 2 kilometer down the shoreline? V 1 km rev 3 min 1 km 2 97. g"\$# ! tan 3\$ " sin"\$ " 1# 98. h"x# ! x!x 2 " 1 T! (b) x ! 1 110. Surface Area The edges of a cube are expanding at a rate of 5 centimeters per second. How fast is the surface area changing when each edge is 4.5 centimeters? 6x " 5 x2 % 1 99. Refrigeration The temperature T of food put in a freezer is 108. x 2 " y 2 ! 16, "5, 3# Not drawn to scale 113. Moving Shadow A sandbag is dropped from a balloon at a height of 60 meters when the angle of elevation to the sun is 30+ (see figure). Find the rate at which the shadow of the sandbag is traveling along the ground when the sandbag is at a height of 35 meters. Hint: The position of the sandbag is given by s"t# ! 60 " 4.9t 2.) (d) t ! 10 100. Fluid Flow The emergent velocity v of a liquid flowing from a hole in the bottom of a tank is given by v ! !2gh, where g is the acceleration due to gravity (32 feet per second per second) and h is the depth of the liquid in the tank. Find the rate of change of v with respect to h when (a) h ! 9 and (b) h ! 4. (Note that g ! %32 feet per second per second. The sign of g depends on how a problem is modeled. In this case, letting g be negative would produce an imaginary value for v.) Rays Position: s(t) = 60 4.9t 2 60 m 30° P.S. P.S. 161 Problem Solving The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system. Click on to view the complete solution of the exercise. Click on to print an enlarged copy of the graph. 1. Consider the graph of the parabola y ! x 2. (a) Find the radius r of the largest possible circle centered on the y-axis that is tangent to the parabola at the origin, as show in the figure. This circle is called the circle of curvature (see Section 12.5). Find the equation of this circle. Use a graphing utility to graph the circle and parabola in the same (b) Find the center !0, b" of the circle of radius 1 centered on the y-axis that is tangent to the parabola at two points, as shown in the figure. Find the equation of this circle. Use a graphing utility to graph the circle and parabola in the same viewing y y 2 1 1 x 1 x Figure for 1(b) 2. Graph the two parabolas y ! and y ! "x 2 % 2x " 5 in the same coordinate plane. Find equations of the two lines simultaneously tangent to both parabolas. x2 3. (a) Find the polynomial P1!x" ! a0 % a1x whose value and slope agree with the value and slope of f !x" ! cos x at the point x ! 0. (b) Find the polynomial P2!x" ! a0 % a1x % a2 whose value and first two derivatives agree with the value and first two derivatives of f !x" ! cos x at the point x ! 0. This polynomial is called the second-degree Taylor polynomial of f !x" ! cos x at x ! 0. x2 (c) Complete the table comparing the values of f and P2. What do you observe? "1.0 "0.1 "0.001 0 0.001 y!x% 3 \$ " 2 4 at the point #\$4 , 32\$. 7. The graph of the eight curve, x 4 ! a2!x 2 " y 2", a # 0, 1 1 Figure for 1(a) 6. Find a function of the form f !x" ! a % b cos cx that is tangent to the line y ! 1 at the point !0, 1", and tangent to the line y 1 r 5. Find a third-degree polynomial p!x" that is tangent to the line y ! 14x " 13 at the point !1, 1", and tangent to the line y ! "2x " 5 at the point !"1, "3". is shown below. 2 (0, b) x Problem Solving 0.1 a a x (a) Explain how you could use a graphing utility to graph this curve. (b) Use a graphing utility to graph the curve for various values of the constant a. Describe how a affects the shape of the curve. (c) Determine the points on the curve where the tangent line is horizontal. 8. The graph of the pear-shaped quartic, b2y 2 ! x3!a " x", a, b > 0, is shown below. y 1.0 cos x a x P2%x& (d) Find the third-degree Taylor polynomial of f !x" ! sin x at x ! 0. 4. (a) Find an equation of the tangent line to the parabola y ! x 2 at the point !2, 4". (b) Find an equation of the normal line to y ! x 2 at the point !2, 4". (The normal line is perpendicular to the tangent line.) Where does this line intersect the parabola a second time? (c) Find equations of the tangent line and normal line to y ! x 2 at the point !0, 0". (d) Prove that for any point !a, b" # !0, 0" on the parabola y ! x 2, the normal line intersects the graph a second time. (a) Explain how you could use a graphing utility to graph this curve. (b) Use a graphing utility to graph the curve for various values of the constants a and b. Describe how a and b affect the shape of the curve. (c) Determine the points on the curve where the tangent line is horizontal. 162 CHAPTER 2 Differentiation 9. A man 6 feet tall walks at a rate of 5 feet per second toward a streetlight that is 30 feet high (see figure). The man’s 3-foot-tall child follows at the same speed, but 10 feet behind the man. At times, the shadow behind the child is caused by the man, and at other times, by the child. (a) Suppose the man is 90 feet from the streetlight. Show that (b) Suppose the man is 60 feet from the streetlight. Show that (c) Determine the distance d from the man to the streetlight at which the tips of the two shadows are exactly the same distance from the streetlight. (d) Determine how fast the tip of the shadow is moving as a function of x, the distance between the man and the street light. Discuss the continuity of this shadow speed function. y 3 30 ft 1 6 ft Not drawn to scale 3 ft 10 ft V 2 Figure for 9 x 4 6 8 lim zq0 sin z z for z in degrees. What is the exact value of this limit? (Hint: (c) Use the limit definition of the derivative to find d sin z dz for z in degrees. (d) Define the new functions S!z" ! sin!cz" and C!z" ! cos!cz", where c ! \$'180. Find S!90" and C!180". Use the Chain Rule to calculate d S!z". dz (8, 2) 2 (b) Use the table to estimate 10 14. An astronaut standing on the moon throws a rock into the air. The height of the rock is 27 2 t % 27t % 6 10 1 s!" Figure for 10 where s is measured in feet and t is measured in seconds. 3 x (see figure). 10. A particle is moving along the graph of y ! ( When x ! 8, the y-component of its position is increasing at the rate of 1 centimeter per second. (a) How fast is the x-component changing at this moment? (b) How fast is the distance from the origin changing at this moment? (c) How fast is the angle of inclination ( changing at this moment? 11. Let L be a differentiable function for all x. Prove that if L!a % b" ! L!a" % L!b" for all a and b, then L& !x" ! L& !0" for all x. What does the graph of L look like? 12. Let E be a function satisfying E!0" ! E& !0" ! 1. Prove that if E!a % b" ! E!a"E!b" for all a and b, then E is differentiable and E& !x" ! E!x" for all x. Find an example of a function satisfying E!a % b" ! E!a"E!b". sin x ! 1 assumes that x is measured x in radians. What happens if you assume that x is measured in (a) Find expressions for the velocity and acceleration of the rock. (b) Find the time when the rock is at its highest point by finding the time when the velocity is zero. What is the height of the rock at this time? (c) How does the acceleration of the rock compare with the acceleration due to gravity on Earth? 15. If a is the acceleration of an object, the jerk j is defined by j ! a&!t". (a) Use this definition to give a physical interpretation of j. (b) Find j for the slowing vehicle in Exercise 117 in Section 2.3 and interpret the result. (c) The figure shows the graph of the position, velocity, acceleration, and jerk functions of a vehicle. Identify each y 13. The fundamental limit lim a xq0 (a) Set your calculator to degree mode and complete the table. b x c z (in degrees) sin z z 0.1 0.01 0.0001 d ```
# How do you simplify \frac { - 3m ^ { 5} n ^ { 4} } { 2m ^ { - 6} n ^ { 0} }? Jan 20, 2018 See a solution process below: #### Explanation: First, rewrite the expression as: $\left(- \frac{3}{2}\right) \left({m}^{5} / {m}^{-} 6\right) \left({n}^{4} / {n}^{0}\right) \implies - \frac{3}{2} \left({m}^{5} / {m}^{-} 6\right) \left({n}^{4} / {n}^{0}\right)$ Next, use this rule of exponents to simplify the $m$ term: ${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$ $- \frac{3}{2} \left({m}^{\textcolor{red}{5}} / {m}^{\textcolor{b l u e}{- 6}}\right) \left({n}^{4} / {n}^{0}\right) \implies$ $- \frac{3}{2} \left({m}^{\textcolor{red}{5} - \textcolor{b l u e}{- 6}}\right) \left({n}^{4} / {n}^{0}\right) \implies$ $- \frac{3}{2} \left({m}^{\textcolor{red}{5} + \textcolor{b l u e}{6}}\right) \left({n}^{4} / {n}^{0}\right) \implies$ $- \frac{3}{2} \left({m}^{11}\right) \left({n}^{4} / {n}^{0}\right) \implies$ $\frac{- 3 {m}^{11}}{2} \left({n}^{4} / {n}^{0}\right)$ Now, use this rule of exponents to simplify the $n$ term: ${a}^{\textcolor{red}{0}} = 1$ $\frac{- 3 {m}^{11}}{2} \left({n}^{4} / {n}^{\textcolor{red}{0}}\right) \implies$ $\frac{- 3 {m}^{11}}{2} \left({n}^{4} / 1\right) \implies$ $\frac{- 3 {m}^{11}}{2} \left({n}^{4}\right) \implies$ $\frac{- 3 {m}^{11} {n}^{4}}{2}$
# A Matrix Representation of a Linear Transformation and Related Subspaces ## Problem 164 Let $T:\R^4 \to \R^3$ be a linear transformation defined by $T\left (\, \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} \,\right) = \begin{bmatrix} x_1+2x_2+3x_3-x_4 \\ 3x_1+5x_2+8x_3-2x_4 \\ x_1+x_2+2x_3 \end{bmatrix}.$ (a) Find a matrix $A$ such that $T(\mathbf{x})=A\mathbf{x}$. (b) Find a basis for the null space of $T$. (c) Find the rank of the linear transformation $T$. (The Ohio State University Linear Algebra Exam Problem) ## Solution. ### (a) A matrix representation of a linear transformation Let $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$, and $\mathbf{e}_4$ be the standard 4-dimensional unit basis vectors for $\R^4$. Then the matrix representation for the linear transformation is given by the formula $A:=\begin{bmatrix} T(\mathbf{e}_1) \quad T(\mathbf{e}_2) \quad T(\mathbf{e}_3)\quad T(\mathbf{e}_4) \end{bmatrix}.$ Then we calculate \begin{align} T(\mathbf{e}_1)=T \left(\,\begin{bmatrix} 1\\0\\0\\0 \end{bmatrix} \, \right) =\begin{bmatrix} 1\\3\\1 \end{bmatrix}. \end{align} Similarly, we compute and obtain $T(\mathbf{e}_2) =\begin{bmatrix} 2\\5\\1 \end{bmatrix}, T(\mathbf{e}_3)=\begin{bmatrix} 3\\8\\2 \end{bmatrix}, T(\mathbf{e}_4)=\begin{bmatrix} -1\\-2\\0 \end{bmatrix}.$ Therefore the matrix $A$ we are looking for is $A=\begin{bmatrix} 1 & 2 & 3 & -1\\ 3 & 5 & 8 & -2\\ 1 & 1& 2 & 0 \end{bmatrix}.$ ### (b) The null space of the linear transformation First, note that the null space of a linear transformation $T$ is the same as the null space of the matrix $A$ that represents $T$. Thus, we find a basis for the null space of the matrix $A$ we obtained in (a). The null space of $A$ consists of the solutions of $A\mathbf{x}=\mathbf{0}$ and so we apply Gauss-Jordan elimination process to solve the linear equation as follows. We apply the elementary row operations to the augmented matrix $[A\|\mathbf{0}]$ of $A$. \begin{align} [A\|\mathbf{0}]&= \left[\begin{array}{rrrr\|r} 1 & 2 & 3 & -1 &0\\ 3 & 5 & 8 & -2&0\\ 1 & 1& 2 & 0 &0 \end{array} \right] \xrightarrow[R_3-R_1]{R_2-3R_1} \left[\begin{array}{rrrr\|r} 1 & 2 & 3 & -1 &0\\ 0 & -1 & -1 & 1&0\\ 0 & -1& -1 & 1 &0 \end{array} \right]\\ & \xrightarrow[\text{then } -R_2]{R_3-R_2} \left[\begin{array}{rrrr\|r} 1 & 2 & 3 & -1 &0\\ 0 & 1 & 1 & -1&0\\ 0 & 0 & 0 & 0 &0 \end{array} \right] \xrightarrow{R_1-2R_2} \left[\begin{array}{rrrr\|r} 1 & 0 & 1 & 1 &0\\ 0 & 1 & 1 & -1&0\\ 0 & 0 & 0 & 0 &0 \end{array} \right]. \end{align} Therefore the solutions are \begin{align} x_1&=-x_3-x_4\\ x_2&=-x_3+x_4 \end{align} and thus the solution $\mathbf{x}$ of $A\mathbf{x}=\mathbf{0}$ is of the form $\mathbf{x}=\begin{bmatrix} -x_3-x_4\\ -x_3+x_4\\ x_3\\ x_4 \end{bmatrix} =x_3 \begin{bmatrix} -1\\-1\\1\\0 \end{bmatrix}+x_4\begin{bmatrix} -1\\1\\0\\1 \end{bmatrix}.$ Hence we have $\calN(T)=\calN(A)=\Span\left\{\, \begin{bmatrix} -1\\-1\\1\\0 \end{bmatrix}, \begin{bmatrix} -1\\1\\0\\1 \end{bmatrix} \, \right\}$ and the set $B:=\left\{\, \begin{bmatrix} -1\\-1\\1\\0 \end{bmatrix}, \begin{bmatrix} -1\\1\\0\\1 \end{bmatrix}\, \right\}$ is a spanning set for the null space $\calN(T)$. We claim that the set $B$ is linearly independent. Indeed, if we have a linear combination $a_1\begin{bmatrix} -1\\-1\\1\\0 \end{bmatrix}+a_2\begin{bmatrix} -1\\1\\0\\1 \end{bmatrix}=\mathbf{0},$ then we have $\begin{bmatrix} -a_1-a_2\\ -a_1+a_2\\ a_1\\ a_2 \end{bmatrix}=\mathbf{0}$ and thus $a_1=a_2=0$ and the set $B$ is linearly independent. Since $B$ is a linearly independent spanning set for the null space $\calN(T)$, the set $B$ is a basis for $\calN(T)$. In summary, we found a basis $B:=\left\{\, \begin{bmatrix} -1\\-1\\1\\0 \end{bmatrix}, \begin{bmatrix} -1\\1\\0\\1 \end{bmatrix}\, \right\}.$ ### (c) The rank of the linear transformation We apply the rank-nullity theorem to find the rank of $T$. From part (b), we see that nullity, the dimension of the null space $\calN(T)$, is $2$. The linear transformation is a map from $\R^4$ to $\R^3$. Thus the rank-nullity theorem states that we have $\text{rank of } T + \text{nullity of } T=4$ $\text{rank of } T + 2=4.$ Therefore the rank of the linear transformation $T$ is 2. Let $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ are vectors in $\R^n$. Suppose that vectors $\mathbf{u}_1$, $\mathbf{u}_2$ are orthogonal and the norm of $\mathbf{u}_2$...
# How do you solve the following system: 10y=42+2x , y= 3/4x+3 ? Apr 6, 2016 $x = \frac{24}{11}$ $y = \frac{51}{11}$ #### Explanation: $10 y = 42 + 2 x \text{ (1)}$ $\text{expand the second equation by 10}$ $y = \frac{3}{4} x + 3 \text{ (2)}$ $\textcolor{red}{10} \cdot y = \textcolor{red}{10} \cdot \left(\frac{3}{4} x + 3\right)$ $10 y = \frac{30}{4} x + 30 \text{ (3)}$ $\text{so left side of the equations (1) and (3)are equal}$ $\text{we can write:}$ $42 + 2 x = \frac{30}{4} x + 30$ $2 x - 30 \frac{x}{4} = 30 - 42$ $\frac{8 x - 30 x}{4} = - 12$ $- 22 x = - 4 \cdot 12$ $22 x = 48 \text{ } x = \frac{48}{22}$ $x = \frac{24}{11}$ $\text{ now use the equation (1)}$ $10 y = 42 + 2 \left(\frac{24}{11}\right)$ $10 y = 42 + \frac{48}{11}$ $10 y = \frac{462 + 48}{11}$ $y = \frac{51 \cancel{0}}{11 \cancel{0}}$ $y = \frac{51}{11}$
# Identity and Inverse Matrices Suppose that we have a matrix A, and we wish to find a matrix I, with the property that AI = A. We refer to I as a special type of matrix called the identity matrix. When we multiply any matrix by I, we get the matrix we started with back. Let’s derive the identity matrix for a 2 by 2 matrix. First, let’s define an arbitrary matrix . Now, let’s multiply it by our identity matrix, and use the result to determine which values in the identity matrix will give us the matrix A as a result. This gives us a system of equations to solve, which look like this: Now, you could use some of the system solving techniques we learned in previous sections to solve this, however the solution can be found rather easily just through inspection. For the first equation to be true, and . For the last equation to be true, and . From this, we can conclude that the identity matrix is: . We can derive a higher dimension identity matrices in a similar way. We simply define a general matrix, multiply by the identity, and solve for the correct values. Identity matrices are nice to know as they allow us to find matrices that can be equivalent to multiplying by 1. In addition to this, we can also use them to define an inverse operation for matrix multiplication. Suppose we have a matrix A, and a matrix B. If AB = I, then B is the inverse of the matrix A. We also call A an invertible matrix, since it has an inverse. Inverse matrices have a lot of power in linear algebra, and we will be able to conclude a lot of information using the concept of inverses, and information from previous sections. Theorem: Let A be a square matrix. Suppose that BA = AB = I and CA = AC = I. If this is true, then B = C. We know that B = BI, since the identity gives us back the matrix we started with. From here, we can substitute for I, giving us B(AC), since AC = I. If we regroup the brackets, we get (BA)C = IC = C, therefore B = C. Theorem: Suppose that A and B are n x n matrices, such that AB = I and BA = I. If this is true, B = , and B and A have a rank n. If we assume that matrix B does not have a rank of n, there is no way that it could be the inverse matrix. This is because the resulting system of equations we discussed when deriving the identity matrix would not have a trivial solution. Since this is the case, both B and A must have a rank of n for B to be the inverse matrix. Theorem: Suppose that A and B are invertible matrices and that t is some non-zero real number. The following statements are true. These theories give us a great idea of how we can manipulate inverses, but we still haven’t looked at how we find an inverse to start with. To find the inverse of a matrix A, we need to solve the equation AB = I. This is going to look pretty like our actual derivation of the identity matrix earlier, however, the right-hand side is the identity matrix instead of the original matrix A. One effective way to solve such a system would be using the reduced row echelon format. We begin with the left-hand side of the matrix being A, and the right-hand side of the matrix being I. We then reduce A until it is equal to I, and the resulting right-hand side will be the inverse matrix. Let’s look at an example to see how this can be done. Example: Find the inverse of First, we are going to create an augmented matrix, with our matrix on the left-hand side, and the identity on the right-hand side. From here, we are going to manipulate the right-hand side until it is equal to the identity on the left-hand side. Once this is done, we will have the inverse matrix on the right-hand side, and the identity on the left-hand side. We start by subtracting the first row by the second Next, we will multiply the first row by 2 and subtract it from row 3. We multiply the second row by 2 and subtract it from row 3. Multiply the second and third row by -1 Subtract the second row by the first row Add 2 times the third row to the first row We have now finished reducing the left-hand side, giving us the inverse matrix, , on the right-hand side.
# Trigonometric Functions Topics: Real number, Addition, Negative and non-negative numbers Pages: 6 (1830 words) Published: January 6, 2013 Section 5.2 Trigonometric Functions of Real Numbers The Trigonometric Functions EXAMPLE: Use the Table below to ﬁnd the six trigonometric functions of each given real number t. π π (a) t = (b) t = 3 2 1 EXAMPLE: Use the Table below to ﬁnd the six trigonometric functions of each given real number t. π π (a) t = (b) t = 3 2 Solution: (a) From the Table, we see that the terminal point determined by √ t = √ is P (1/2, 3/2). Since the coordinates are x = 1/2 and π/3 y = 3/2, we have √ √ π 3 3/2 √ π 1 π sin = cos = tan = = 3 3 2 3 2 3 1/2 √ √ π 3 2 3 π π 1/2 csc = = sec = 2 cot = √ 3 3 3 3 3 3/2 (b) The terminal point determined by π/2 is P (0, 1). So π π 1 π 0 π cos = 0 csc = = 1 cot = = 0 sin = 1 2 2 2 1 2 1 But tan π/2 and sec π/2 are undeﬁned because x = 0 appears in the denominator in each of their deﬁnitions. π . 4 Solution: √ From the Table above, we see that √ terminal point determined by t = π/4 is the √ √ P ( 2/2, 2/2). Since the coordinates are x = 2/2 and y = 2/2, we have √ √ √ π 2 2 2/2 π π sin = =1 cos = tan = √ 4 2 4 2 4 2/2 √ π √ π π √ 2/2 csc = 2 sec = 2 cot = √ =1 4 4 4 2/2 EXAMPLE: Find the six trigonometric functions of each given real number t = 2 Values of the Trigonometric Functions EXAMPLE: π π (a) cos > 0, because the terminal point of t = is in Quadrant I. 3 3 (b) tan 4 > 0, because the terminal point of t = 4 is in Quadrant III. (c) If cos t < 0 and sin t > 0, then the terminal point of t must be in Quadrant II. EXAMPLE: Determine the sign of each function. 7π (b) tan 1 (a) cos 4 Solution: (a) Positive (b) Positive EXAMPLE: Find each value. 2π π (a) cos (b) tan − 3 3 19π 4 (c) sin 3 EXAMPLE: Find each value. π 19π 2π (b) tan − (c) sin (a) cos 3 3 4 Solution: (a) Since 2π 3π − π 3π π π = = − =π− 3 3 3 3 3 the reference number for 2π/3 is π/3 (see Figure (a) below) and the terminal point of 2π/3 is in Quadrant II. Thus cos(2π/3) is negative and (b) The reference number for −π/3 is π/3 (see Figure (b) below). Since the terminal point of −π/3 is in Quadrant IV, tan(−π/3) is negative. Thus (c) Since 19π 20π − π 20π π π = = − = 5π − 4 4 4 4 4 the reference number for 19π/4 is π/4 (see Figure (c) below) and the terminal point of 19π/4 is in Quadrant II. Thus sin(19π/4) is positive and EXAMPLE: Find each value. 2π 4π (a) sin (b) tan − 3 3 (c) cos 14π 3 4 EXAMPLE: Find each value. 4π 2π (b) tan − (a) sin 3 3 Solution: (a) Since (c) cos 14π 3 2π 3π − π 3π π π = = − =π− 3 3 3 3 3 the reference number for 2π/3 is π/3 and the terminal point of 2π/3 is in Quadrant II. Thus sin(2π/3) is positive and √ 2π 3 π sin = sin = 3 3 2 (b) Since 4π 3π + π 3π π π =− =− − = −π − 3 3 3 3 3 the reference number for −4π/3 is π/3 and the terminal point of −4π/3 is in Quadrant II. Thus tan (−4π/3) is negative and − tan − (c) Since 4π 3 = − tan π 3 √ =− 3 14π 15π − π 15π π π = = − = 5π − 3 3 3 3 3 the reference number for 14π/3 is π/3 and the terminal point of 14π/3 is in Quadrant II. Thus cos(14π/4) is negative and 14π π 1 cos = − cos = − 3 3 2 EXAMPLE: Evaluate 7π π (b) cos (a) sin 3 6 11π 4 17π 3 17π 2 121π 6 (c) tan (d) sec (e) csc (f) cot 5 EXAMPLE: Evaluate 7π 11π 17π 17π 121π π (b) cos (c) tan (d) sec (e) csc (f) cot (a) sin 3 6 4 3 2 6 Solution: (a) The reference number for π/3 is π/3. Since the terminal point of π/3 is in Quadrant I, sin(π/3) is positive. Thus √ π 3 π sin = sin = 3 3 2 (b) Since 7π = 6π+π = 6π + π = π + π , the reference number for 7π/6 is π/6 and the terminal 6 6 6 6 6 point of 7π/6 is in Quadrant III. Thus cos(7π/6) is negative and √ π 7π 3 = − cos = − cos 6 6 2 (c) Since 11π = 12π−π = 12π − π = 3π − π , the reference number for 11π/4 is π/4 and the 4 4 4 4 4 terminal point of 11π/6 is in Quadrant II. Thus tan(11π/4) is negative and tan π 11π = − tan = −1 4 4 (d) Since 17π = 18π−π = 18π − π = 6π − π , the reference number for 17π/3 is π/3 and the 3 3 3 3 3 terminal point of 17π/3 is in Quadrant IV. Thus sec(17π/3) is...
# A set of data has a mean of 127 and a standard deviation of 15. What is the z-score of x = 118? Answer: The z-score is a measure of how statistically related a value is from the mean. It is a tool to standardize scores from different sets of data which is why it is also known as "Standard Score". It can be calculated using the formula: z-score = (any value in the data - mean) ÷ standard deviation Given: any value in the data = 118 standard deviation = 15 mean = 127 Solution: z-score = (118 - 127) ÷ 15 z-score = -0.6 ## Related Questions How do you round 38,288 to the nearest 100 Consider the number 38,288 Now, we have to round this number to the nearest hundreds place. The place value to the extreme right of the number is ones, then tens, hundreds, thousands, ten thousands and so on. So, the digit at hundreds place = 2 Consider the digit to the right of the hundred place which is '8', which is greater than 5. Therefore, we will add '1' to the digit at hundreds place. Therefore, the number 38,288 rounded to nearest hundreds is 38,300. Two angles are complementary. They have measures of (7x + 2)° and (3x – 2)°, respectively. What is the value of x? 10 180 9 90 Two angles are complementary angles if they have a sum of 90 degrees. We could write the expression as below. angle 1 + angle 2 = 90 7x + 2 + 3x - 2 = 90 solve for x by adding like terms 7x + 3x + 2 - 2 = 90 10x = 90 x = 90/10 x = 9 The value of x is 9 9 Step-by-step explanation: 25% of 176 is what number? The answer is 44 divide 176/100 then multiply by 25% The powerful survival impulse that leads infants to seek closeness to their caregivers is called:A)attachment.B)imprinting.C)habituation.D)assimilation.E)the rooting reflex A. Attachment Step-by-step explanation: The powerful survival impulse that leads infants to seek closeness to their caregivers is called Attachment. The infant can count on the caregiver possibly parent(s) for care which gives the infant a solid foundation for dependence and survival. Which is the inverse of the function f(x)=4x-9 = . Step-by-step explanation: Given : f(x)=4x-9. To find : Which is the inverse of the function. Solution : We have given f(x)=4x-9. Step 1 : replace y to x x = 4y -9. Step 2: solve for y On adding by 9 both side x + 9 = 4y On dividing by 4 both sides y = . y = . Then = . Therefore, = . To find the inverse of a function, switch the x and the y (the f(x) = y), then solve for y. x = 4y - 9 x + 9 = 4y y = (x+9)/4 So f^-1(x) = (x+9)/4 The scale factor for Maria’s dollhouse furniture is 1:8. If the sofa in Maria’s dollhouse is 7 1/2 inches long, how long is the actual sofa? The actual sofa is 60 inches or 5 feet long. Step-by-step explanation: 1. Let's review all the information given for solving this question: Scale factor for Maria's doll house furniture = 1:8 Size of the sofa in Maria's doll house = 7 1/2 inches 2. Let's find the size of the real sofa Size of the real sofa = 7 1/2 * 8 Size of the real sofa = 60 inches long. Size of the real sofa = 60/12 feet (1 feet = 12 inches) Size of the real sofa = 5 feet long. 60 inches Step-by-step explanation: How offen is simple interest used in the business and banking worlds? Very often. Step-by-step explanation: What fraction is less than one third with one as a numerator?
# Search by Topic #### Resources tagged with Generalising similar to Pros and Cons: Filter by: Content type: Stage: Challenge level: ### There are 122 results Broad Topics > Using, Applying and Reasoning about Mathematics > Generalising ### Mini-max ##### Stage: 3 Challenge Level: Consider all two digit numbers (10, 11, . . . ,99). In writing down all these numbers, which digits occur least often, and which occur most often ? What about three digit numbers, four digit numbers. . . . ### Repeaters ##### Stage: 3 Challenge Level: Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13. ### Cunning Card Trick ##### Stage: 3 Challenge Level: Delight your friends with this cunning trick! Can you explain how it works? ### Card Trick 2 ##### Stage: 3 Challenge Level: Can you explain how this card trick works? ### More Magic Potting Sheds ##### Stage: 3 Challenge Level: The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it? ### Special Sums and Products ##### Stage: 3 Challenge Level: Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48. ### Games Related to Nim ##### Stage: 1, 2, 3 and 4 This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning. ### Adding in Rows ##### Stage: 3 Challenge Level: List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it? ### One, Three, Five, Seven ##### Stage: 3 and 4 Challenge Level: A game for 2 players. Set out 16 counters in rows of 1,3,5 and 7. Players take turns to remove any number of counters from a row. The player left with the last counter looses. ### Picturing Triangular Numbers ##### Stage: 3 Challenge Level: Triangular numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers? ### Maths Trails ##### Stage: 2 and 3 The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind maths trails. ### Pentanim ##### Stage: 2, 3 and 4 Challenge Level: A game for 2 players with similaritlies to NIM. Place one counter on each spot on the games board. Players take it is turns to remove 1 or 2 adjacent counters. The winner picks up the last counter. ### Tourism ##### Stage: 3 Challenge Level: If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable. ### Consecutive Negative Numbers ##### Stage: 3 Challenge Level: Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers? ### Handshakes ##### Stage: 3 Challenge Level: Can you find an efficient method to work out how many handshakes there would be if hundreds of people met? ### Christmas Chocolates ##### Stage: 3 Challenge Level: How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes? ### Magic Letters ##### Stage: 3 Challenge Level: Charlie has made a Magic V. Can you use his example to make some more? And how about Magic Ls, Ns and Ws? ### Cubes Within Cubes Revisited ##### Stage: 3 Challenge Level: Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need? ### Nim-like Games ##### Stage: 2, 3 and 4 Challenge Level: A collection of games on the NIM theme ### Number Pyramids ##### Stage: 3 Challenge Level: Try entering different sets of numbers in the number pyramids. How does the total at the top change? ### Jam ##### Stage: 4 Challenge Level: A game for 2 players ### Nim ##### Stage: 4 Challenge Level: Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The loser is the player who takes the last counter. ### Nim-interactive ##### Stage: 3 and 4 Challenge Level: Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The winner is the player to take the last counter. ### Have You Got It? ##### Stage: 3 Challenge Level: Can you explain the strategy for winning this game with any target? ### Three Times Seven ##### Stage: 3 Challenge Level: A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why? ### One O Five ##### Stage: 3 Challenge Level: You can work out the number someone else is thinking of as follows. Ask a friend to think of any natural number less than 100. Then ask them to tell you the remainders when this number is divided by. . . . ### Chess ##### Stage: 3 Challenge Level: What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board? ### Chocolate Maths ##### Stage: 3 Challenge Level: Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . . ### Nim-7 ##### Stage: 1, 2 and 3 Challenge Level: Can you work out how to win this game of Nim? Does it matter if you go first or second? ### Magic Squares ##### Stage: 4 and 5 An account of some magic squares and their properties and and how to construct them for yourself. ### Winning Lines ##### Stage: 2, 3 and 4 An article for teachers and pupils that encourages you to look at the mathematical properties of similar games. ##### Stage: 3 Challenge Level: A little bit of algebra explains this 'magic'. Ask a friend to pick 3 consecutive numbers and to tell you a multiple of 3. Then ask them to add the four numbers and multiply by 67, and to tell you. . . . ### Intersecting Circles ##### Stage: 3 Challenge Level: Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have? ### Is There a Theorem? ##### Stage: 3 Challenge Level: Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel? ### Go Forth and Generalise ##### Stage: 3 Spotting patterns can be an important first step - explaining why it is appropriate to generalise is the next step, and often the most interesting and important. ### Konigsberg Plus ##### Stage: 3 Challenge Level: Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges. ### Shear Magic ##### Stage: 3 Challenge Level: What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles? ### Sum Equals Product ##### Stage: 3 Challenge Level: The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 × 1 [1/3]. What other numbers have the sum equal to the product and can this be so for. . . . ### Building Gnomons ##### Stage: 4 Challenge Level: Build gnomons that are related to the Fibonacci sequence and try to explain why this is possible. ### Lower Bound ##### Stage: 3 Challenge Level: What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 = ### 2001 Spatial Oddity ##### Stage: 3 Challenge Level: With one cut a piece of card 16 cm by 9 cm can be made into two pieces which can be rearranged to form a square 12 cm by 12 cm. Explain how this can be done. ### Jam ##### Stage: 4 Challenge Level: To avoid losing think of another very well known game where the patterns of play are similar. ### Seven Squares - Group-worthy Task ##### Stage: 3 Challenge Level: Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning? ### Pair Products ##### Stage: 4 Challenge Level: Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. What do you notice? ### Painted Cube ##### Stage: 3 Challenge Level: Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces? ### All Tangled Up ##### Stage: 3 Challenge Level: Can you tangle yourself up and reach any fraction? ### Multiplication Arithmagons ##### Stage: 4 Challenge Level: Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons? ### Plus Minus ##### Stage: 4 Challenge Level: Can you explain the surprising results Jo found when she calculated the difference between square numbers? ### Problem Solving, Using and Applying and Functional Mathematics ##### Stage: 1, 2, 3, 4 and 5 Challenge Level: Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information. ### Generating Triples ##### Stage: 4 Challenge Level: Sets of integers like 3, 4, 5 are called Pythagorean Triples, because they could be the lengths of the sides of a right-angled triangle. Can you find any more?
# Linear Equations – Definition and Forms Linear equations are equations that produce graphs that are linear, that is, their graphs are straight lines. The main characteristic of these equations is that all of the variables have a maximum degree of one. In this article, we will explore these equations in more detail. We will start by considering what linear equations are in general, and then we will look at the different forms in which they can be written. We will explore the slope-intercept form, the point-slope form, and the standard form. ##### ALGEBRA Relevant for Learning about linear equations and their graphs. See definition ##### ALGEBRA Relevant for Learning about linear equations and their graphs. See definition ## Definition of linear equations Linear equations are equations of the first order. These equations are defined by straight lines in the Cartesian plane. An equation for a straight line is called a linear equation. All the variables in linear equations have a maximum order of 1. Example: the equation $latex y = x + 2$ is linear: The following are examples of linear equations with 1 variable, with 2 variables, and with 3 variables: ### EXAMPLES Linear equations with one variable: • $latex 2x+4=17$ • $latex 3x=24$ • $latex \frac{1}{2}x+2=10$ Linear equations with two variables: • $latex x+3y=15$ • $latex y=3x-12$ • $latex \frac{1}{3}x+2y=5$ Linear equations with three variables: • $latex x+2y-2z=10$ • $latex 2a+4b-c=8$ • $latex \frac{1}{2}x+2y=3z-4$ ## Forms of linear equations There are varios different forms in which we can write linear equations, but generally, equations have constants (such as 2 or c) and must contain only simple variables (such as “x” or “y“). ### EXAMPLES These are linear equations: • $latex y=2x-4$ • $latex y-5=2(x-1)$ • $latex y+3x-5=0$ • $latex 3x=24$ • $latex \frac{y}{3}=12$ However, keep in mind that linear equations should not contain variables (such as “x” or “y“) with exponents, square roots, cube roots, etc. ### EXAMPLES These are not linear equations: • $latex {{y}^{2}}+3=9$ • $latex {{y}^{2}}+3=9$ • $latex \frac{{{{x}^{2}}}}{4}=20$ ## Slope-intercept form of a linear equation The slope-intercept form is the most common in linear equations. This form is represented as follows: The line we have above represents the equation y = 2x + 1. In this equation: • the slope is $latex m = 2$. • the y-intercept is 1. ### Slope The slope of the line is equal to the rate of change of y over the rate of change of x. The slope can be evaluated as follows: $latex m=\frac{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}{{\left( {{{x}_{2}}-{{x}_{1}}} \right)}}$ Basically, the slope indicates the inclination of the line in the plane. ## Point-intercept form of a linear equation Another common form for linear equations is the point-slope form. In this form, the equation of the linear equation is formed considering the points in the Cartesian plane: An example of this form is the equation $latex y-3=\frac{1}{2}\left( {x-2} \right)$, where: • $latex {{y}_{1}}=3$ • $latex {{x}_{1}}=2$ • $latex m=\frac{1}{2}$ ## Standard form of a linear equation The standard form of a linear equation with two variables is represented as $latex Ax + By + C = 0$ where A≠0, B≠0 and x and y are the variables. ### EXAMPLES The equation $latex 2x+4y+5=0$ is in the standard form and we have: • A=2 • B=4 • C=5 ### What are linear equations? Linear equations are the equations used to represent straight lines. The equations for lines in the Cartesian plane are linear equations. ### What are the three main forms of linear equations? The three main forms of linear equations are point-slope form, slope-intercept form, and standard form. ### How do we express the standard form of linear equations? The standard form of linear equations is given by $latex Ax + By + C = 0$. ### What is the slope-intercept form of linear equations? The slope-intercept form is given by $latex y = mx + b$. Where m is the slope and b is the y-intercept. ### What is the difference between linear equations and non-linear equations? A linear equation corresponds to straight lines. A non-linear equation does not form a straight line. It could be a curve, oscillations or others.
# The Mathematics of Shuffled Cards It is said that each time you shuffle a 52-card deck, each arrangement you make may have never existed in all history, or may never exist again. Why? Because of the enormous number of arrangements that can be made using 52 distinct objects (in this case, cards). To understand this, we can look at the number of arrangements that can be made with smaller number of objects. Lets start with 3 objects A, B, and C. The possible arrangements are ABC, ACB, BAC, BCA, CAB and CBA. Notice that for the first position, there are 3 possible choices (see figure below). Then, after you made the first choice, there are only 2 possible choices left. And after the second choice, you only have 1 possible choice. This means that the number of arrangements of 3 objects is $3 \times 2 \times 1 = 6$ Using the method above, we can calculate the number of arrangements of objects by multiplying the number of objects by all the numbers less than it down to 1. For example, the number of arrangements of 4 objects is $4 \times 3 \times 2 \times 1 = 24$ (read Introduction to Permutations for details). Therefore, the number possible arrangements for a 52-deck card is $52 \times 51 \times 50 \times \cdots \times 3 \times 2 \times 1$ or the product of 52 and all the integers less than it all the way down to 1. Note that the $\cdots$ symbol is used to denote that there are numbers that are not shown in the multiplication above. If we use a calculator to compute the expression above, we get 8.0658175e+67 or a number starting with 8 plus 67 more digits. Now, let’s compute how many possible arrangements of cards are made in a year. There are 86400 seconds in a day, so if we assume that on the average there are one thousand 52-deck cards shuffled every second in the world, then, the number of arrangements in a year is 1000 decks/second x 86400 second/day x 365 days = 31 536 000 000. If we divide 8.0658175e+67 by 31 536 000 000, we can exhaust all the arrangement in approximately $2.55 \times 10^{57}$ years, about 185 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 times the age of the universe. Scientists believe that the universe is only 13.8 billion years or $1.38 \times 10^{10}$ years. So, you see, it is possible that all the arrangements of well shuffled cards may not all exist until the end of the universe. ## 3 thoughts on “The Mathematics of Shuffled Cards” 1. Wow! This one is a fascinating math thought of… Thanks for sharing this… I am a math teacher.. But I did not get the 1000 cards part and hence even the stuff after that clearly…. Would be thankful if you can Plz elaborate a bit on this / make it bit easier so that I can share this with my students… Thanks again… • Thank you Rupesh for your comment. It was supposed to be one thousand 52-deck cards, not 1000 cards. My apologies for the confusion. I revised the post and made the calculations more detailed. Please tell me if you have more questions. 🙂 2. The likely hood of getting the kings queens jacks 10s 9s 8s 7s 6s 5s 4s 3s 2s aces together is 1.9577227e+65 and getting these in order is 1.219078e+75 which is a 73 figure number and an 83 figure number
# What Is 29/45 as a Decimal + Solution With Free Steps The fraction 29/45 as a decimal is equal to 0.644. The division of two numbers is one of the four fundamental arithmetic operations, along with addition, subtraction, and multiplication. The result of a division can either be an integer or decimal number depending on the values of the dividend and the divisor. Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers. Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division,Â which we will discuss in detail moving forward. So, letâ€™s go through the Solution of fraction 29/45. ## Solution First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively. This can be done as follows: Dividend = 29 Divisor = 45 Now, we introduce the most important quantity in our division process: theÂ Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents: Quotient = Dividend $\div$ Divisor = 29 $\div$ 45 This is when we go through the Long Division solution to our problem. Figure 1 ## 29/45 Long Division Method We start solving a problem using the Long Division Method by first taking apart the divisionâ€™s components and comparing them. As we have 29Â and 45, we can see how 29 is Smaller than 45, and to solve this division, we require that 29 be Bigger than 45. This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later. Now, we begin solving for our dividend 29, which after getting multiplied by 10 becomes 290. We take this 290 and divide it by 45; this can be done as follows: Â 290 $\div$ 45 $\approx$ 6 Where: 45 x 6 = 270 This will lead to the generation of a Remainder equal to 290 â€“ 270 = 20. Now this means we have to repeat the process by Converting the 20 into 200Â and solving for that: 200 $\div$ 45 $\approx$ 4Â Where: 45 x 4 = 180 This, therefore, produces another Remainder which is equal to 200 â€“ 180 = 20. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 200. 200 $\div$ 45 $\approx$ 4Â Where: 45 x 4 = 180 Finally, we have a Quotient generated after combining the three pieces of it as 0.644, with a Remainder equal to 20. Images/mathematical drawings are created with GeoGebra.
Courses Courses for Kids Free study material Offline Centres More Store # Solution of $\left( {1 + xy} \right)ydx + \left( {1 - xy} \right)xdy = 0$ is1.$\log \dfrac{x}{y} + \dfrac{1}{{xy}} = c$2.$\log \dfrac{x}{y} = c$3.$\log \dfrac{x}{y} - \dfrac{1}{{xy}} = c$4.$\log \dfrac{y}{x} - \dfrac{1}{{xy}} = c$ Last updated date: 14th Aug 2024 Total views: 402.6k Views today: 9.02k Verified 402.6k+ views Hint: In the given question, we have been given that there is a differentiation equation in two variables. We have to calculate the value of the differentiation equation. For doing that we are going to need to first of all, rearrange the terms with different differentiating factors. Then, we are going to have to apply the required formulae and solve for the given variables. Here, the terms are in the differentiating form, so we are going to have to apply integration on the two sides of the equality for solving this question. The given equation is $\left( {1 + xy} \right)ydx + \left( {1 - xy} \right)xdy = 0$. First, we are going to rearrange this equation, $ydx + xdy + x{y^2}dx - {x^2}ydy = 0$ Now, dividing both sides by ${x^2}{y^2}$, $\Rightarrow$ $\dfrac{1}{{{x^2}{y^2}}}\left( {ydx + xdy + x{y^2}dx - {x^2}ydy} \right) = \dfrac{0}{{{x^2}{y^2}}}$ or $\dfrac{{ydx + xdy}}{{{x^2}{y^2}}} + \dfrac{{dx}}{x} - \dfrac{{dy}}{y} = 0$ Now, we are going to apply the $d\left( { - \dfrac{1}{{xy}}} \right) = \dfrac{{ydx + xdy}}{{{x^2}{y^2}}}$ $d\left( {\log x} \right) = \dfrac{{dx}}{x}$ $d\left( {\log y} \right) = \dfrac{{dy}}{y}$ Now, substituting these in $\dfrac{{ydx + xdy}}{{{x^2}{y^2}}} + \dfrac{{dx}}{x} - \dfrac{{dy}}{y} = 0$, we get: $d\left( { - \dfrac{1}{{xy}}} \right) + d\left( {\log x} \right) - d\left( {\log y} \right) = 0$ Now, integrating both sides, we get: $\Rightarrow$ $\int {d\left( { - \dfrac{1}{{xy}}} \right) + d\left( {\log x} \right) - d\left( {\log y} \right)} = \int 0$ $\Rightarrow$ $- \dfrac{1}{{xy}} + \log x - \log y = c$ We know, $\log \dfrac{a}{b} = \log a - \log b$ Hence, $- \dfrac{1}{{xy}} + \log x - \log y = \log \dfrac{x}{y} - \dfrac{1}{{xy}}$ Hence, $\log \dfrac{x}{y} - \dfrac{1}{{xy}} = c$ Thus, the correct option is (3). Note: So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. Then we think about the formulae which contain the known and the unknown and pick the one which is the most suitable and the most effective for finding the answer of the given question. Then we put in the knowns into the formula, evaluate the answer and find the unknown. It is really important to follow all the steps of the formula to solve the given expression very carefully and in the correct order, because even a slightest error is going to make the whole expression awry and is going to give us an incorrect answer.

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