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# Selina Solutions Concise Mathematics Class 6 Chapter 22 Simple (Linear) Equations Exercise 22 (D) Selina Solutions Concise Mathematics Class 6 Chapter 22 Simple (Linear) Equations Exercise 22(D) are created by highly experienced faculty in the education industry, as per the ICSE syllabus. Various solved examples present before each exercise help students analyse the questions that would appear in the annual exam. Those who aspire to obtain high marks are suggested to practice Selina Solutions repeatedly. In order to improve problem solving skills, students can access Selina Solutions Concise Mathematics Class 6 Chapter 22 Simple (Linear) Equations Exercise 22(D) PDF, free from the available links below. ## Selina Solutions Concise Mathematics Class 6 Chapter 22 Simple (Linear) Equations Exercise 22(D) Download PDF ### Access Selina Solutions Concise Mathematics Class 6 Chapter 22 Simple (Linear) Equations Exercise 22(D) Exercise 22(D) 1. A number increased by 17 becomes 54. Find the number. Solution: Let us assume the number is x Hence, x + 17 = 54 x = 54 â€“ 17 We get, x = 37 Therefore, the required number is 37 2. A number decreased by 8 equals 26, find the number. Solution: Let us assume the number is x Hence, x â€“ 8 = 26 x = 26 + 8 We get, x = 34 Therefore, the required number is 34 3. One-fourth of a number add to two-seventh of it gives 135; find the number. Solution: Let us assume the number is x Hence, (1 / 4) Ã— x + (2 / 7) Ã— x = 135 (x / 4) + (2x / 7) = 135 Taking L.C.M. we get, (7x + 8x) / 28 = 135 15x = 135 Ã— 28 x = (135 Ã— 28) / 15 We get, x = 9 Ã— 28 x = 252 Therefore, the required number is 252 4. Two-fifths of a number subtracted from three-fourths of it gives 56, find the number. Solution: Let us assume the number is x (3 / 4) x â€“ (2 / 5) x = 56 (3x / 4) â€“ (2x / 5) = 56 Taking L.C.M. we get, (15x â€“ 8x) / 20 = 56 7x = 56 Ã— 20 x = (56 Ã— 20) / 7 We get, x = 8 Ã— 20 x = 160 Therefore, the required number is 160 5. A number is increased by 12 and the new number obtained is multiplied by 5. If the resulting number is 95, find the original number. Solution: Let us assume the number is x Hence, (x + 12) 5 = 95 5x + 60 = 95 5x = 95 â€“ 60 5x = 35 x = 35 / 5 We get, x = 7 Therefore, the original number is 7 6.A number is increased by 26 and the new number obtained is divided by 3. If the resulting number is 18; find the original number. Solution: Let us assume the number is x Hence, (x + 26) Ã· 3 = 18 This can be written as, (x + 26) / 3 = 18 (x + 26) = 18 Ã— 3 x + 26 = 54 x = 54 â€“ 26 We get, x = 28 Therefore, the original number is 28 7. The age of a man is 27 years more than the age of his son. If the sum of their ages is 47 years, find the age of the son and his father. Solution: Given Age of a man is 27 years more than the age of his son Sum of their ages = 47 years Let us assume the age of son is x Hence, Age of son = x years Age of father = x + 27 To find the value of x, x + x + 27 = 47 2x + 27 = 47 2x = 47 â€“ 27 2x = 20 x = 20 / 2 We get, x = 10 Thus age of son = 10 years Age of father = x + 27 Substituting the value of x, we get = 10 + 27 = 37 years Hence, the age of father is 37 years. Therefore, the age of son is 10 years and the age of father is 37 years 8. The difference between the ages of Gopal and his father is 26 years. If the sum of their ages is 56 years, find the ages of Gopal and his father. Solution: Given Difference between the ages of Gopal and his father is 26 years Sum of their ages = 56 years Let us assume the age of Gopal is x Hence, Age of Gopal = x Age of father = x + 26 To find the value of x x + x + 26 = 56 2x = 56 â€“ 26 2x = 30 x = 30 / 2 We get, x = 15 Thus the age of Gopal is 15 years Age of father = x + 26 Substituting the value of x, we get = 15 + 26 = 41 years Hence, the age of father is 41 years Therefore, the age of Gopal is 15years and the age of his father is 41 years 9. When two consecutive natural numbers are added, the sum is 31; find the numbers. Solution: Given Two consecutive natural numbers sum is 31 Let us assume the first natural number is x Hence, First number = x Second number = x + 1 To find the value of x, x + x + 1 = 31 2x = 31 â€“ 1 2x = 30 x = 30 / 2 We get, x = 15 Thus the first number is 15 Second number = x + 1 Substituting the value of x, we get = 15 + 1 = 16 Hence, the second number is 16 Therefore, the first natural number is 15 and the second natural number is 16 10. When three consecutive natural numbers are added, the sum is 66, find the numbers. Solution: Given Three consecutive natural numbers sum = 66 Let us assume the first natural number is x Hence, First number = x Second number = x + 1 Third number = x + 2 Now, calculation to find the value of x, x + x + 1 + x + 2 = 66 3x + 3 = 66 3x = 66 â€“ 3 3x = 63 x = 63 / 3 We get, x = 21 Hence, First number = x = 21 Second number = x + 1 = 21 + 1 = 22 Third number = x + 2 = 21 + 2 = 23 Therefore, three consecutive natural numbers are 21, 22 and 23 11. A natural number decreased by 7 is 12. Find the number. Solution: Let us assume the natural number is x Hence, x â€“ 7 = 12 x = 12 + 7 We get, x = 19 Therefore, the required number is 19 12. One-fourth of a number added to one-sixth of itself is 15. Find the number. Solution: Let us assume the number is x Hence, (1 / 4) Ã— x + (1 / 6) Ã— x = 15 x / 4 + x / 6 = 15 Taking L.C.M. we get, (3x + 2x) / 12 = 15 5x = 15 Ã— 12 x = (15 Ã— 12) / 5 x = 3 Ã— 12 We get, x = 36 Therefore, the required number is 36 13. A whole number is increased by 7 and the new number so obtained is multiplied by 5; the result is 45. Find the number. Solution: Let us assume the whole number is x Hence, (x + 7) 5 = 45 5x + 35 = 45 5x = 45 â€“ 35 We get, 5x = 10 x = 10 / 5 We get, x = 2 Therefore, the required whole number is 2 14. The age of a man and the age of his daughter differ by 23 years and the sum of their ages is 41 years. Find the age of the man. Solution: Given Difference between the ages of a man and his daughter = 23 years Sum of their ages = 41 years Let us assume the age of a daughter is x Hence, Age of a man = x + 23 Now, calculating to find the value of x, x + x + 23 = 41 2x + 23 = 41 2x = 41 â€“ 23 2x = 18 x = 18 / 2 We get, x = 9 Hence, Age of a man = x + 23 = 9 + 23 = 32 Therefore, the age of a man is 32 years 15. The difference between the ages of a woman and her son is 19 years and the sum of their ages is 37 years; find the age of the son. Solution: Given Difference between the ages of a woman and her son = 19 years Sum of their ages = 37 years Let us assume the age of a son is x Hence, Age of woman = x + 19 Now, calculating to find the value of x, x + x + 19 = 37 2x + 19 = 37 2x = 37 â€“ 19 2x = 18 x = 18 / 2 We get, x = 9 Therefore, the age of her son is 9 years
## Basic Probabilities Probability is a measure of the likelihood of an event. It is frequently calculated as a rational number, but often converted to a percentage. Rational numbers are numbers which can be expressed as a ratio of two whole numbers. So if we say the probability, P, of something happening is 1 chance in 2, we can express this in several ways: $P=\frac{1}{2}=0.5=50\%$ The above example happens to relate well to a simple real world example, the coin flip. In a single coin flip, we will either get a heads or a tails. So, if we ask the question "What is the probability of getting a heads in a coin flip?" we can answer "One chance in two, or 50%. An interesting feature of probabilities of cumulative events is that they are not additive. To illustrate this, we need the cumulative probability equation. The cumulative probability, $$P_C$$, is a function of multiple individual probabilities, $$P_1, P_2...P_n$$, with the relationship, $P_C=1-(1-P_1) \times (1-P_2) \times ...(1-P_n).$ So if we ask the question, "What are the chances of getting heads at least once during two coin flips?" we cannot say 100% for a very important reason. If we get a tails on the first try and try again, the coin does not know that it was tails last time, so the chances of getting a heads on the second coin flip are still 50%. To get the answer to this question, we must use the cumulative probability equation. The cumulative probability, {tex inline}P_C{/tex}, of getting a heads on either of two coin flips, $$P_1$$ and $$P_2$$, is $P_C = 1-[(1-P_1)\times(1-P_2)] = 1-[(1-0.5)\times(1-0.5)] = 0.75 = 75\%.$ # Notation in Probability As you can see from the examples above, probability is often noted with a P. More formally, the probability of event A occurring is written as $P(A).$ The probability of events A and B occurring is written as $P(A \cap B).$ The probability of events A or B occurring is written as $P(A \cup B).$ Finally, we need some notation for a relation of an event to a condition. Conditions are also a form of events, so this can be a bit confusing. The probability of event A occurring, given that event B has occurred, is noted as $P(A\|B).$ As an example of how we would write a conditional probability, lets make a statement based on the data in Koester's book. In a Search and Rescue operation, the probability, P, of locating a subject in an injured condition, I, given that the subject is a solo male, M, is 9%. We would state this as follows: The probability of an injury, given that the subject of a search is male, is 9%, or $P(I\|M) = 9\%.$
1986 AHSME Problems/Problem 19 Problem A park is in the shape of a regular hexagon $2$ km on a side. Starting at a corner, Alice walks along the perimeter of the park for a distance of $5$ km. How many kilometers is she from her starting point? $\textbf{(A)}\ \sqrt{13}\qquad \textbf{(B)}\ \sqrt{14}\qquad \textbf{(C)}\ \sqrt{15}\qquad \textbf{(D)}\ \sqrt{16}\qquad \textbf{(E)}\ \sqrt{17}$ Solution We imagine this problem on a coordinate plane and let Alice's starting position be the origin. We see that she will travel along two edges and then go halfway along a third. Therefore, her new $x$-coordinate will be $1 + 2 + \frac{1}{2} = \frac{7}{2}$ because she travels along a distance of $2 \cdot \frac{1}{2} = 1$ km because of the side relationships of an equilateral triangle, then $2$ km because the line is parallel to the $x$-axis, and the remaining distance is $\frac{1}{2}$ km because she went halfway along and because of the logic for the first part of her route. For her $y$-coordinate, we can use similar logic to find that the coordinate is $\sqrt{3} + 0 - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$. Therefore, her distance is $$\sqrt{\left(\frac{7}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{49}{4} + \frac{3}{4}} = \sqrt{\frac{52}{4}} = \sqrt{13},$$ giving an answer of $\boxed{A}$.
# 2.6 Function concepts -- composite functions Page 1 / 1 This module describes composite functions in Algebra. You are working in the school cafeteria, making peanut butter sandwiches for today’s lunch. • The more classes the school has, the more children there are. • The more children there are, the more sandwiches you have to make. • The more sandwiches you have to make, the more pounds (lbs) of peanut butter you will use. • The more peanut butter you use, the more money you need to budget for peanut butter. ...and so on. Each sentence in this little story is a function. Mathematically, if $c$ is the number of classes and $h$ is the number of children, then the first sentence asserts the existence of a function $h\left(c\right)$ . The principal walks up to you at the beginning of the year and says “We’re considering expanding the school. If we expand to 70 classes, how much money do we need to budget? What if we expand to 75? How about 80?” For each of these numbers, you have to calculate each number from the previous one, until you find the final budget number. $\underset{\to }{\text{# Classes}}$ $\underset{\to }{\text{# Children}}$ $\underset{\to }{\text{# Sandwiches}}$ $\underset{\to }{\text{lb.}}$ $\underset{\to }{\text{}}$ But going through this process each time is tedious. What you want is one function that puts the entire chain together: “You tell me the number of classes, and I will tell you the budget.” $\underset{\to }{\text{# Classes}}$ $\underset{\to }{\text{}}$ This is a composite function —a function that represents in one function, the results of an entire chain of dependent functions . Since such chains are very common in real life, finding composite functions is a very important skill. ## How do you make a composite function? We can consider how to build composite functions into the function game that we played on the first day. Suppose Susan takes any number you give her, quadruples it, and adds 6. Al takes any number you give him and divides it by 2. Mathematically, we can represent the two functions like this: $S\left(x\right)=4x+6$ $A\left(x\right)=\frac{x}{2}$ To create a chain like the one above, we give a number to Susan; she acts on it, and gives the resulting number to Al; and he then acts on it and hands back a third number. $3\to \text{Susan}\to S\left(3\right)=18\to \text{Al}\to A\left(18\right)=9$ In this example, we are plugging $S\left(3\right)$ —in other words, 18— into Al’s function. In general, for any $x$ that comes in, we are plugging $S\left(x\right)$ into $A\left(x\right)$ . So we could represent the entire process as $A\left(S\left(x\right)\right)$ . This notation for composite functions is really nothing new: it means that you are plugging $S\left(x\right)$ into the $A$ function. But in this case, recall that $S\left(x\right)=4x+6$ . So we can write: $A\left(S\left(x\right)\right)=\frac{S\left(x\right)}{2}=\frac{4x+6}{2}=2x+3$ What happened? We’ve just discovered a shortcut for the entire process. When you perform the operation $A\left(S\left(x\right)\right)$ —that is, when you perform the Al function on the result of the Susan function—you are, in effect, doubling and adding 3. For instance, we saw earlier that when we started with a 3, we ended with a 9. Our composite function does this in one step: $3\to 2x+3\to 9$ Understanding the meaning of composite functions requires real thought. It requires understanding the idea that this variable depends on that variable, which in turn depends on the other variable; and how that idea is translated into mathematics. Finding composite functions, on the other hand, is a purely mechanical process—it requires practice, but no creativity. Whenever you are asked for $f\left(g\left(x\right)\right)$ , just plug the $g\left(x\right)$ function into the $f\left(x\right)$ function and then simplify. ## Building and testing a composite function $f\left(x\right)={x}^{2}-4x$ $g\left(x\right)=x+2$ What is $f\left(g\left(x\right)\right)$ ? • To find the composite, plug $g\left(x\right)$ into $f\left(x\right)$ , just as you would with any number. $f\left(g\left(x\right)\right)=\left(x+2{\right)}^{2}-4\left(x+2\right)$ • Then simplify. $f\left(g\left(x\right)\right)=\left(x{}^{2}\text{}+4x+4\right)-\left(4x+8\right)$ $f\left(g\left(x\right)\right)={x}^{2}-4$ • Let’s test it. $f\left(g\left(x\right)\right)$ means do $g$ , then $f$ . What happens if we start with $x=9$ ? $7\to g\left(x\right)\to 7+2=9\to f\left(x\right)\to \left(9{\right)}^{2}-4\left(9\right)=45$ • So, if it worked, our composite function should do all of that in one step. $7\to {x}^{2}-4=\left(7{\right)}^{2}-4=45$ $✓$ It worked! There is a different notation that is sometimes used for composite functions. This book will consistently use $f\left(g\left(x\right)\right)$ which very naturally conveys the idea of “plugging $g\left(x\right)$ into $f\left(x\right)$ .” However, you will sometimes see the same thing written as $f°g\left(x\right)$ , which more naturally conveys the idea of “doing one function, and then the other, in sequence.” The two notations mean the same thing. show that the set of all natural number form semi group under the composition of addition explain and give four Example hyperbolic function _3_2_1 felecia ⅗ ⅔½ felecia _½+⅔-¾ felecia The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu 1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3 Pawel 2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31 Pawel ok Ifeanyi on number 2 question How did you got 2x +2 Ifeanyi combine like terms. x + x + 2 is same as 2x + 2 Pawel x*x=2 felecia 2+2x= felecia Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113? Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113 Pawel how do I set up the problem? what is a solution set? Harshika find the subring of gaussian integers? Rofiqul hello, I am happy to help! Abdullahi hi mam Mark find the value of 2x=32 divide by 2 on each side of the equal sign to solve for x corri X=16 Michael Want to review on complex number 1.What are complex number 2.How to solve complex number problems. Beyan yes i wantt to review Mark use the y -intercept and slope to sketch the graph of the equation y=6x how do we prove the quadratic formular Darius hello, if you have a question about Algebra 2. I may be able to help. I am an Algebra 2 Teacher thank you help me with how to prove the quadratic equation Seidu may God blessed u for that. Please I want u to help me in sets. Opoku what is math number 4 Trista x-2y+3z=-3 2x-y+z=7 -x+3y-z=6 can you teacch how to solve that🙏 Mark Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411 Brenna (61/11,41/11,−4/11) Brenna x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11 Brenna Need help solving this problem (2/7)^-2 x+2y-z=7 Sidiki what is the coefficient of -4× -1 Shedrak A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place. Jeannette has $5 and$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives. What is the expressiin for seven less than four times the number of nickels How do i figure this problem out. how do you translate this in Algebraic Expressions why surface tension is zero at critical temperature Shanjida I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason s. Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)= . After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight? Got questions? Join the online conversation and get instant answers!
The components of 252 can one of two people be element or composite. Subsequently, 252 is an even composite number itself which comprises of components that are either prime or composite. How about we number out exactly how to compute the components of 252, prime determinants of 252, and also factors of 252 in pairs alongside solved instances for a better understanding. You are watching: What are the factors of 252 Factors of 252: 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 28, 36, 42, 63, 84, 126, 252Prime factorization of 252: 22 × 32 × 7 1 What are the determinants of 252? 2 How to Calculate components of 252? 3 Factors the 252 by element Factorization 4 Factors that 252 in Pairs 5 FAQs on components of 252 What room the components of 252? A variable is a number that divides the offered number there is no leaving a remainder, so the number which offer the remainder together 0 when separated by 252 will it is in the components of 252. Because that Example: 252 ÷ 2 = 126. Right here we acquire the quotient 126 and the remainder is 0. For this reason 2 and 126 are the factors of 252. How to calculate the components of 252? To discover the factors of 252 we have actually to uncover the numbers which when separated by 252 give the remainder together 0. We will start separating 252 with natural numbers i.e., 1, 2, 3, …. As much as 126 (Half of 252). So, every the number which will provide the remainder as 0 will be the determinants of 252. This method is well-known as Division Method. The table listed below provides the representation of the over method. Division Factor 252 ÷ 1 Remainder = 0Hence, element = 1 252 ÷ 2 Remainder = 0Hence, aspect = 2 252 ÷ 3 Remainder = 0Hence, aspect = 3 252 ÷ 4 Remainder = 0Hence, factor = 4 252 ÷ 6 Remainder = 0Hence, aspect = 6 252 ÷ 7 Remainder = 0Hence, factor = 7 252 ÷ 9 Remainder = 0Hence, element = 9 252 ÷ 12 Remainder = 0Hence, aspect = 12 252 ÷ 14 Remainder = 0Hence, factor = 14 252 ÷ 18 Remainder = 0Hence, element = 18 252 ÷ 21 Remainder = 0Hence, variable = 21 252 ÷ 28 Remainder = 0Hence, factor = 4 252 ÷ 36 Remainder = 0Hence, aspect = 36 252 ÷ 42 Remainder = 0Hence, aspect = 42 252 ÷ 63 Remainder = 0Hence, variable = 63 252 ÷ 84 Remainder = 0Hence, variable = 84 252 ÷ 126 Remainder = 0Hence, variable = 126 252 ÷ 252 Remainder = 0Hence, element = 252 Therefore, the factors that 252 are 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 28, 36, 42, 63, 84, 126, 252. Explore determinants using illustrations and interactive examples Factors the 252 by prime Factorization Prime administrate is a an approach of converting any kind of composite number into the product the its element factors. We will divide the number with succeeding prime numbers until we acquire the quotient together 1. Division Method Step 1. Divide the number 252 with the the smallest prime number which gives the remainder as 0.Step 3. Repeat step 1 with the obtained quotient. The the smallest prime number as a variable of 252 is 2. So, splitting 252 by 2. 252 ÷ 2 = 126 Now the quotient is 126 with the smallest prime factor as 2. Now 126 ÷ 2 = 63 Similarly, 63 ÷ 3 = 2121 ÷ 3 = 77 ÷ 7 = 1 So, the prime administrate of 252 is. 252 = 2 × 2 × 3 × 3 × 7 = 22 × 32 × 7. Factor Tree Method The factor tree method can be practiced as displayed below So, the element factorization that 252 is. 252 = 2 × 2 × 3 × 3 × 7 = 22 × 32 × 7. See more: How Many Quarter Notes Equal A Whole Note Values, Answers For Lesson #4 Quiz (Note Durations) Factors the 252 in Pairs Pair determinants of a number indicate two numbers who product will provide the provided number. The pair determinants of 252 will be the pair the numbers whose product will provide 252 as result. The table shown below represents the calculation of determinants of 252 in pairs: Factor pair Pair factorization 1 and also 252 1 × 252 = 252 2 and also 126 2 × 126 = 252 3 and also 84 3 × 84 = 252 4 and also 63 4 × 63 = 252 6 and also 42 6 × 42 = 252 7 and 36 7 × 36 = 252 9 and also 28 9 × 28 = 252 12 and also 21 12 × 21 = 252 14 and also 18 14 × 18 = 252 Also, the product of two an adverse numbers results in a positive number. So, the product of negative values of the over factors bag will an outcome in 252. They are known as an adverse pair factors. Hence, the an adverse factor pairs of 252 would it is in (-1, -252), (-2, -126), (-3, -84), (-4, -63), (-6, -42), (-7, -36), (-9, -28), (-12, -21) and also (-14, -18). Important Notes As the sum of number of 252 is 9, it is divisible by 3 and also 9. Hence, 3 and 9 are factors of 252. Challenging Questions: What is the average of every the even factors of 252?Using element factorization, discover the number whereby 252 should be separated to make it a perfect square?
6 9 equivalent fractions • 2 3 is equivalent to 6 9 because 2 x 9 = 3 x 6 = 18 • 4 6 is equivalent to 6 9 because 4 x 9 = 6 x 6 = 36 • 8 12 is equivalent to 6 9 because 8 x 9 = 12 x 6 = 72 2/3 How do you write an equivalent fraction? Summary: • You can make equivalent fractions by multiplying or dividing both top and bottom by the same amount. • You only multiply or divide, never add or subtract, to get an equivalent fraction. • Only divide when the top and bottom stay as whole numbers. How to make an equivalent fraction? It is possible by these methods: • Method 1: Make the Denominators the same • Method 2: Cross Multiply • Method 3: Convert to decimals What are the rules of equivalent fractions? Equivalent Fractions Rule. A rule stating that if the numerator and denominator of a fraction are multiplied by the same nonzero number, the result is a fraction that is equivalent to the original fraction. This rule can be represented as: a//b = (n * a)//(n * b). Which number is equivalent to the fraction? Two or more fractions are said to be equivalent if they are equal to the same fraction when simplified. For example, the equivalent fractions of 1/5 are 5/25, 6/30, and 4/20, which on simplification, result in the same fraction, that is, 1/5. Are the fractions 6 9 and 4 6 equivalent? They are the same. What is the simplified fraction of 6 9? 23So 69 in simplest form is 23 . What is 6’4 equivalent to as a fraction? Thus, 12/8 is equivalent to 6:4. What is the fraction 6/8 equivalent to? 3/4Thus, 6/8 is an equivalent fraction of 3/4. We can find some other equivalent fractions by multiplying the numerator and the denominator of the given fraction by the same number. Thus, the equivalent fractions of 3/4 are 6/8, 9/12, 12/16, and 15/20. What is 6 9 in a mixed number? Since 69 is a proper fraction, it cannot be written as a mixed number. What is the whole number of 6 9? 6 / 9 = 0.66666667 If you insist on converting 6/9 to a whole number, the best we can do is round the decimal number up or down to the nearest whole number. What is 1/3 the same as? Answer: The fractions equivalent to 1/3 are 2/6, 3/9, 4/12, etc. Equivalent fractions have the same value in the reduced form. Explanation: Equivalent fractions can be written by multiplying or dividing both the numerator and the denominator by the same number. How do u do equivalent fractions? How to Find Equivalent Fractions. Multiply both the numerator and denominator of a fraction by the same whole number. As long as you multiply both top and bottom of the fraction by the same number, you won’t change the value of the fraction, and you’ll create an equivalent fraction. What is equivalent calculator? Equivalent Expression Calculator is a free online tool that displays the equivalent expressions for the given algebraic expression. BYJU’S online equivalent expression calculator tool makes the calculations and simplification faster and it displays the equivalent expression in a fraction of seconds. 1 2 and 6 12 . What is 3/5 equivalent to as a fraction? So, 3/5 = 6/10 = 9/15 = 12/20. What fraction is 2/3 equivalent to? An equivalent fraction of two-thirds (2/3) is sixteen twenty-fourths (16/24). How do you simplify fractions? So, reducing or simplifying fractions means we make the fraction as simple as possible. We do this by dividing the numerator and the denominator by the largest number that can divide into both numbers exactly. In other words, we divide the top and bottom by the biggest number they have in common. What is in simplified form? 1:184:10What is the Simplest Form of a Fraction? \| Don’t Memorise – YouTubeYouTubeStart of suggested clipEnd of suggested clip12 by 6 is 2 and 18 by 6 is 3 12 by 18 and 2 by 3 are the same and this 2 by 3 here is called theMore12 by 6 is 2 and 18 by 6 is 3 12 by 18 and 2 by 3 are the same and this 2 by 3 here is called the simplest form it means that this fraction cannot be reduced further. How do you simplify 6 8? Explanation: The simplest form of 68 can’t be divided further with N numbers other than 1 . What’s 6’9 as a percentage? 66.666667%Fraction to percent conversion tableFractionPercent3/933.333333%4/944.444444%5/955.555556%6/966.666667%41 more rows How to find equivalent fractions? To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 2 3) by the same natural number, ie, multiply by 2, 3, 4, 5, 6 Can you convert fractions to decimals? This Equivalent Fractions Table/Chart contains common practical fractions. You can easily convert from fraction to decimal, as well as, from fractions of inches to millimeters. What are Equivalent Fractions? Equivalent fractions are fractions with different numbers representing the same part of a whole. They have different numerators and denominators, but their fractional values are the same. How to make a fraction equivalent? Multiply both the numerator and denominator of a fraction by the same whole number. As long as you multiply both top and bottom of the fraction by the same number, you won’t change the value of the fraction , and you’ll create an equivalent fraction. What is half of a fraction? For example, think about the fraction 1/2. It means half of something. You can also say that 6/12 is half, and that 50/100 is half. They represent the same part of the whole. These equivalent fractions contain different numbers but they mean the same thing: 1/2 = 6/12 = 50/100 What are Equivalent Fractions? In math, equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole. How many equivalent fractions can a child show? Instead of handing out regular fraction worksheets to your children, ask your child to divide the pizza or pancakes such that he can show three equivalent fractions. What is equivalent fraction? Equivalent fractions are fractions that, although they have different digits in the numerator and denominator, represent the same value. For example, the fractions , , and all represent the same part of a whole: 1/2. This can be depicted using fraction bars: What happens when unlike fractions are converted into equivalent fractions? If they are unlike fractions, then the unlike fractions must be converted into equivalent fractions that share the same denominator in order to be added or subtracted. What happens when you multiply a fraction by 1? Since multiplying any number by 1 results in the same number, multiplying any fraction by a fractional representation of the number 1 results in an equivalent fraction. What is the least common multiple of 2 and 5? 1. : The least common multiple of 2 and 5 is 10 , so we will convert the above fractions to equivalent fractions that have a denominator of 10 in order to add them. 2. : The least common multiple of 7 and 21 is 21, so we only need to convert the first fraction to an equivalent fraction. Is each fraction bar the same? Even though each of the fraction bars is broken up into a different number of parts, each of the fraction bars represents the same fraction, albeit with different numerators and denominators. Any given fraction has an infinite number of equivalent fractions. Can you divide to find equivalent fractions? Depending on the starting fraction, it is also possible to divide to find equivalent fractions . For example, dividing the numerator and denominator of by 4 would give us , another equivalent fraction of .
Factors of 6 Introduction Do you know what these numbers have in common? 2, 4, 6, 8, 10, 12, 14…That’s right, they are even numbers. One important thing to remember that even numbers can do that other sets of numbers can’t do is that they are equally divisible by 2. This is important to remember when trying to find the factors of a number. If it is an even number, then you will automatically have the number 2 as a factor. Another good tip is that multiples of 3 have something in common too that’s important for finding factors. You guessed it. Those numbers can be divisible by 3. So, 3 would be a factor for those numbers. A Small Number with Factors The good thing to know is that the smaller numbers usually have fewer factors. So, if you get stuck trying to find all of the factors for a large number, it will probably take you longer. Let’s compare the number 6 with the number 1200. Obviously, six will have 6 or fewer factors. However, 1200 could have as many as 1200. Of course, it won’t because there are numbers that won’t evenly divide into 1200, but you will have to spend time seeing if several numbers can divide into 1200 evenly. Back to the number 6, let’s find the factors for this small number. Look at the tips below that you can use for all numbers and not just small numbers. Tip 1: Start off with 1 and itself. (1,6) Tip 2: Even numbers will always have 2 as a factor (1, 2, 6) Also, remember its partner. $$2\;\times\;\_\_\_\_\;=\;6$$ Two and three are factors because 2 times 3 equals 6. Tip 3: Multiples of 3 are evenly divisible by 3. (1, 2, 3, 6) Also, remember its partner. $$3\;\times\;\_\_\_\_\;=\;6$$ After starting with these tips, you will need to find the other factors, but this will be a good start and knock some of the factors out of the way. For a small number like 6, this shouldn’t take you much time.
# Difference between revisions of "2009 AMC 10A Problems/Problem 8" ## Problem 8 Three Generations of the Wen family are going to the movies, two from each generation. The two members of the youngest generation receive a $50$% discount as children. The two members of the oldest generation receive a $25\%$ discount as senior citizens. The two members of the middle generation receive no discount. Grandfather Wen, whose senior ticket costs $6.00$, is paying for everyone. How many dollars must he pay? $\mathrm{(A)}\ 34 \qquad \mathrm{(B)}\ 36 \qquad \mathrm{(C)}\ 42 \qquad \mathrm{(D)}\ 46 \qquad \mathrm{(E)}\ 48$ ## Solution A senior ticket costs $6.00$, so a regular ticket costs $6 \cdot \frac{1}{\frac{3}{4}}\:=\:6\cdot\frac{4}{3}\:=\:8$ dollars. Therefore children's tickets cost half that, or $4.00$, so we have: $2(6+8+4)\:=\:36$ So Grandfather Wen pays $36$, or $\fbox{B}$. ## Solution 2 Using average, we know that assuming the middle-aged people's ticket cost x dollars, 2(100%x+75%x+50%x). We average this into 2(75%x+75%x+75%x). We know that 75%x=6.00, which means 6*6 is the answer, or 36. -RealityWrites
# How to find the probability of a compound event ## How do you calculate the probability of a compound event? Compound probability is equal to the probability of the first event multiplied by the probability of the second event. ## How do you find the probability of simple and compound events? The probability of simple events is finding the probability of a single event occurring. When finding the probability of an event occurring, we will use the formula: number of favorable outcomes over the number of total outcomes. Compound events involve the probability of more than one event happening together. ## What is the formula of probability? Formula for the probability of A and B (independent events): p(A and B) = p(A) * p(B). If the probability of one event doesn’t affect the other, you have an independent event. All you do is multiply the probability of one by the probability of another. ## What are the 5 rules of probability? Basic Probability Rules • Probability Rule One (For any event A, 0 ≤ P(A) ≤ 1) • Probability Rule Two (The sum of the probabilities of all possible outcomes is 1) • Probability Rule Three (The Complement Rule) • Probabilities Involving Multiple Events. • Probability Rule Four (Addition Rule for Disjoint Events) • Finding P(A and B) using Logic. ## What is the probability in math? Probability = the number of ways of achieving success. the total number of possible outcomes. For example, the probability of flipping a coin and it being heads is ½, because there is 1 way of getting a head and the total number of possible outcomes is 2 (a head or tail). ## What is simple probability in math? The ratio of the number of outcomes favourable for the event to the total number of possible outcomes is termed as probability. In other words, a measure of the likelihood of an event (or measure of chance) is called probability. Sample space is the possible outcomes of the experiment. … You might be interested: Google calendar delete recurring event ## What is simple events in probability? A simple event is an event where all possible outcomes are equally likely to occur. So the probability of simple events will have all possible outcomes equally likely to happen or occur. For example, when you toss a coin, there are two possible outcomes – heads or tails, and the probability of heads or tails is equal. ## How do you find the probability of an event not happening? If you know the probability of an event occurring, it is easy to compute the probability that the event does not occur. If P(A) is the probability of Event A, then 1 – P(A) is the probability that the event does not occur. ## What are the 3 types of probability? Three Types of Probability • Classical: (equally probable outcomes) Let S=sample space (set of all possible distinct outcomes). … • Relative Frequency Definition. … • Subjective Probability. ## What is the formula of probability class 9? Formula for Probability The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
# How to solve system of linear equations College algebra students learn How to solve system of linear equations, and manipulate different types of functions. We can solve math problems for you. ## How can we solve system of linear equations In addition, there are also many books that can help you How to solve system of linear equations. Algebra is the branch of mathematics that deals with the equations and rules governing the manipulation of algebraic expressions. Algebra is used in solving mathematical problems and in discovering new mathematical truths. Algebra is based on the concept of variables, which are symbols that represent unknown numbers or quantities. Algebra is used to solve equations, which are mathematical statements that state that two expressions are equal. The process of solving an equation for a variable is called solving for x. To solve for x, one must first identify the equation's variables and then use algebraic methods to solve for the variable. Algebraic methods include using addition, subtraction, multiplication, and division to solve for a variable. In some cases, algebraic equations can be solve by using exponential or logarithmic functions. Algebra is a powerful tool that can be used to solve mathematical problems and discover new mathematical truths. How to solve perfect square trinomial. First, identify a, b, and c. Second, determine if a is positive or negative. Third, find two factors of ac that add to b. Fourth, write as the square of a binomial. Fifth, expand the binomial. Sixth, simplify the perfect square trinomial 7 eighth, graph the function to check for extraneous solutions. How to solve perfect square trinomial is an algebraic way to set up and solve equations that end in a squared term. The steps are simple and easy to follow so that you will be able to confidently solve equations on your own! The distance formula is generally represented as follows: d=√((x_2-x_1)^2+(y_2-y_1)^2) In this equation, d represents the distance between the points, x_1 and x_2 are the x-coordinates of the points, and y_1 and y_2 are the y-coordinates of the points. This equation can be used to solve for the distance between any two points in two dimensions. To solve for the distance between two points in three dimensions, a similar equation can be used with an additional term for the z-coordinate: d=√((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2) This equation can be used to solve for the distance between any two points in three dimensions. Online math graph As a math student, there are times when a picture is worth a thousand words. When it comes to graphing functions, this is especially true. Being able to visualize a function can help you understand its behavior and uncover patterns that may not be immediately apparent from looking at the equation alone. There are a number of online tools that allow you to enter an equation and see the corresponding graph. These tools can be a valuable resource for studying mathematics and exploring new concepts. Best of all, they're free and easy to use. So next time you're stuck on a problem, give one of these online math graphs a try. You may just find that the solution is right in front of you. ## We will support you with math difficulties OMG this is helping so much with grade 9 math it solves anything a swear AND IT’S FREE! I recommend this app 100%. if I could rate it higher I would. This is by far the coolest apo ever This app is very helpful NO MORE DIFFICULT MATH PROBLEMS. Thanks to this app ### Taliah Hall It's wonderful. It helps me with my homework and it shows me how to do it and it does not just give me the answers. It doesn't take long. Download it and it's great. helps me understand math questions or topics in a better and understandable way. Really great app!! ### Qiana Carter Fill in the blank math problems Math equation solver Math homework calculator Enter math problem show me solve Solve my algebra problem
# Negative Numbers Search my site: Level: Primary 6 / Grade 6 / PSLE Here is a simple way that I use to explain Negative Numbers. Before I introduce negative numbers to my students, I remind them of the number line. Numbers are usually written this way on a number line. Things to note: • The line starts at 0 • The numbers increase as you move to the right of the line Now I show them the number line with negative numbers added to it. Things to note: • The number line has to beginning and no ending, indicated by the arrow on both ends of the line • The negative numbers also increases digitally as you move to the left of the line, but the negative sign means the value of the number actually decreases To help my students understand better, I tell them to think of negative numbers this way: • The negative sign tells you how far away the number is from the zero. • So -3 means you are 3 steps away from 0 and -5 means you are 5 steps away from zero. • Therefore, -5 is smaller than -3 because you are further away from zero. ### Adding and Subtracting with Negative Numbers When it comes to using negative numbers in addition and subtraction, using a number line is crucial for understanding how it all works. The first thing to understand is the difference in the meaning between the subtraction symbol (-) and the negative number symbol (-). They both look the same but have different meanings. The subtraction symbol tells us which direction we want to go along the number line. The negative symbol applies only to the number it is paired with. The other thing to note is that the addition symbol (+) tells us to move to the right of the number line while the subtraction symbol (-) tells us to move to the left of the number line. Let's look at some examples. Here is the number line again. Let's find the answer to: 1 subtract 3. We write it as: 1 - 3. The subtraction tells us to move to the left on the number line. Starting at number 1, move 3 steps to the left on the number line. You will end up at negative two. So, 1 - 3 = -2 Here is another way to look at it. If I have 1 cookie (1) and I want to give you 3 cookies (-3), that means I owe you 2 cookies (-2). The negative sign means I owe you. 1 - 3 = -2 Let's try another example. Find the answer to -1 plus 3. We write it as: -1 + 3. The plus or addition sign tells us to move to the right on the number line. Starting at -1, move 3 steps to the right on the number line. You will end up at 2. So, -1 + 3 = 2 Here is the alternative explanation. I owe you 1 cookie (-1). Someone else gives me 3 cookies (+3). I have to give you what I owe so I end up with only 2 cookies. -1 + 3 = 2 Try a few more yourself. Remember: To add - move to the right. To subtract - move to the left. Or think in terms of giving or receiving cookies. 1. -2 + 5 2. 3 - 5 3. 0 - 3 4. -1 - 2 ### Exceptions to the Rule When a negative number comes after the addition or subtraction sign, it means you have to reverse the normal direction. That is, if a negative number comes after the addition sign, it means you move to the left instead of the right on the number line. Vice versa, if a negative number comes after the subtraction sign, it means you move to the right instead of to the left on the number line. Let's look at the example below. 2 + (-2) You start at the number 2. Then you move 2 steps to the left instead of the right and end up at 0. This means that 2 + (-2) is the same as 2 - 2 which gives the answer 0. 2 + (-2) = 0 Here is another example. Let's find the answer to (-3) - (-1). Start at -3. Then move 1 step to the right instead of the left. You will end up at -2. So, -3 - (-1) is the same as -3 + 1 which gives the answer -2 -3 - (-1) = -2
Linearity of the determinant I'd like to prove the following properties of the determinant map. 1. $\det I = 1$ 2. $\det$ is linear in the rows of the input matrix The determinant map is defined on $n\times n$ matrices $A$ by: $$\det \begin{bmatrix}a\end{bmatrix} = a$$$$\det A = a_{11}\det A_{11}-a_{21}\det A_{21}\pm \dots\pm a_{n1}\det A_{n1}$$ Where $A_{xy}$ is the matrix obtained from $A$ by removing the $xth$ row and the $yth$ column and $a_{xy}$ are the entries of the matrix $A$. Proof. 1. $\det I_n = 1\times \det I_{n-1}+0=\det I_{n-2}=\dots=det I_1 = 1$ 2. The proof is by induction on the size of the matrix. For $2\times 2$ matrices: $\det \begin{bmatrix}a & b\\ c+x & d+y\end{bmatrix}=a(d+y)-b(c+x)=(ad-bc)+(ay-bx)=\det \begin{bmatrix}a & b\\ c & d\end{bmatrix}+\det \begin{bmatrix}a & b\\ x & y\end{bmatrix}$ Let $C=\begin{bmatrix} \dots\\ R+S\\ \dots \end{bmatrix}$ be a $n\times n$ matrix for $n\gt 2$, we compute $\det C$ in terms of $\det R$ and $\det S$ where $R = \begin{bmatrix}\dots\\R\\\dots\end{bmatrix}$ and $S = \begin{bmatrix}\dots\\S\\\dots\end{bmatrix}$. Let $i$ denote the index of the rows $R, S, R+S$. $$\det C = \sum_{j\neq i}\pm c_{j1} \det C_{j1} \pm c_{i1} \det C_{i1}$$. By the induction hypothesis, we can write: $$\det C = \sum_{j\neq i}\pm c_{j1}(\det R_{j1}+\det S_{j1}) \pm c_{i1}\det C_{i1}$$ We know that $\det C_{i1}=\det R_{i1}=\det S_{i1}$ and $c_{i1}=r_{i1}+s_{i1}$ and $c_{j1}=r_{j1}=s_{j1}$ for $j\neq i$. So $$\det C = \sum_{j\neq i}\pm r_{j1}\det R_{j1} + \sum_{j\neq i}\pm s_{j1}\det S_{j1} \pm (r_{i1} + s_{i1})\det C_{i1}= \sum_{j\neq i}\pm r_{j1}\det R_{j1} + \sum_{j\neq i}\pm s_{j1}\det S_{j1} \pm ( r_{i1}\det C_{i1} + s_{i1}\det C_{i1})= \sum_{j\neq i}\pm r_{j1}\det R_{j1} + \sum_{j\neq i}\pm s_{j1}\det S_{j1} \pm (r_{i1}\det R_{i1} + s_{i1}\det S_{i1})= \sum_{j}\pm r_{j1}\det R_{j1} + \sum_{j}\pm s_{j1}\det S_{j1}=\det R + \det S$$. A simular induction argument shows that $$\det \begin{bmatrix}\dots \\ c\times R \\ \dots\end{bmatrix}= c\times \det \begin{bmatrix}\dots \\ R \\ \dots\end{bmatrix}$$ Is my proof correct? • The determinant is multi-linear in its rows, not linear. I believe you also left out some $\rm{det}$s in your definition of the determinant - this looks like it should be recursive to me. – Anthony Carapetis Aug 6 '13 at 15:56 • The usual $\det$ is defined in a different way (by Laplace): $$\det A = a_{11} \det A_{11} - a_{21} \det A_{21} + a_{31} \det A_{31} - \ldots$$ See en.wikipedia.org/wiki/… – AlexR Aug 6 '13 at 15:56 • It might go down easier if you first show that the determinant can be evaluated along any line/row, and then use the 'special' line for calculations. – Jonathan Y. Aug 6 '13 at 16:01 • @AnthonyCarapetis Artin, in his book, Algebra says it's linear. – saadtaame Aug 6 '13 at 16:06 • For 2., it is easier to start the induction at $n=1$. – Tony Huynh Aug 6 '13 at 16:47
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are viewing an older version of this Concept. Go to the latest version. # Gas Pressure and Force ## Pressure is the force exerted over a given area and the behavior of most gases can be described using the ideal gas law. 0% Progress Practice Gas Pressure and Force Progress 0% Gas Pressure and Force Students will learn the basics of pressure and force, how pressure varies with altitude in our atmosphere, and the ideal gas law, which connects the concepts of volume, pressure, temperature, and the number of particles that compose a gas. ### Key Equations P=F/AP=P0ehapressure equals the force divided by the area of applicationexponential model for atmospheric pressure\begin{align}P = F / A && \text{pressure equals the force divided by the area of application} \\ P = P_0 e^\frac{-h} {a} && \text{exponential model for atmospheric pressure}\end{align} PV=NkTIdeal Gas Law\begin{align} PV = NkT && \text{Ideal Gas Law}\end{align} An ideal gas is a gas where the atoms are treated as point-particles and assumed to never collide or interact with each other. If you have N\begin{align}N\end{align} molecules of such a gas at temperature T\begin{align}T\end{align} and volume V\begin{align}V\end{align}, the pressure can be calculated from this formula. Note that k=1.38×1023J/K\begin{align}k=1.38\times10^{-23}\;\mathrm{J/K}\end{align} Guidance • The pressure of a gas is the force the gas exerts on a certain area. For a gas in a container, the amount of pressure is directly related to the number and intensity of atomic collisions on a container wall. • An ideal gas is a gas for which interactions between molecules are negligible, and for which the gas atoms or molecules themselves store no potential energy. For an “ideal” gas, the pressure, temperature, and volume are simply related by the ideal gas law. • Atmospheric pressure (1atm=101,000\begin{align}1 \;\mathrm{atm} = 101,000\end{align} Pascals) is the pressure we feel at sea level due to the weight of the atmosphere above us. As we rise in elevation, there is less of an atmosphere to push down on us and thus less pressure. #### Example 1 How many molecules of gas does it take to equalize the pressure inside a 1 liter box, that originally starts with no gas inside, with the atmospheric pressure at room temperature (300 K)? ##### Solution To solve this problem, we just plug in the given values into the ideal gas law. PVNNN=NkT=PVkT=101000Pa1L1.381023J/K300K=2.441025molecules\begin{align} PV&=NkT\\ N&=\frac{PV}{kT}\\ N&=\frac{101000\;\text{Pa} 1\;\text{L}}{1.3810^{-23}\;\text{J/K} 300\;\text{K}}\\ N&=2.4410^{25}\;\text{molecules}\\ \end{align} ### Time for Practice 1. If the number of molecules is increased, how is the pressure on a particular area of the box affected? Explain conceptually, in words rather than with equations. 2. Use the formula P=F/A\begin{align}P = F / A\end{align} to argue why it is easier to pop a balloon with a needle than with a finger (pretend you don’t have long fingernails). 3. Take an empty plastic water bottle and suck all the air out of it with your mouth. The bottle crumples. Why, exactly, does it do this? 4. You will notice that if you buy a large drink in a plastic cup, there will often be a small hole in the top of the cup, in addition to the hole that your straw fits through. Why is this small hole necessary for drinking? 5. Why is it a good idea for Noreen to open her bag of chips before she drives to the top of a high mountain? 6. How much pressure are you exerting on the floor when you stand on one foot? (You will need to estimate the area of your foot in square meters.) 7. Calculate the amount of force exerted on a 2cm×2cm\begin{align}2 \;\mathrm{cm} \times 2 \;\mathrm{cm}\end{align} patch of your skin due to atmospheric pressure (P0=101,000Pa)\begin{align}(P_0 = 101,000 \;\mathrm{Pa})\end{align}. Why doesn’t your skin burst under this force? 8. Assuming that the pressure of the atmosphere decreases exponentially as you rise in elevation according to the formula P=P0eha\begin{align}P = P_0 e^\frac{-h} {a}\end{align}, where P0\begin{align}P_0\end{align} is the atmospheric pressure at sea level (101,000Pa)\begin{align}(101,000 \;\mathrm{Pa})\end{align}, h\begin{align}h\end{align} is the altitude in km and a is the scale height of the atmosphere (a8.4km)\begin{align}(a \approx 8.4 \;\mathrm{km})\end{align}. 1. Use this formula to determine the change in pressure as you go from San Francisco to Lake Tahoe, which is at an elevation approximately 2km\begin{align}2 \;\mathrm{km}\end{align} above sea level. 2. If you rise to half the scale height of Earth’s atmosphere, by how much does the pressure decrease? 3. If the pressure is half as much as on sea level, what is your elevation? 9. An instructor has an ideal monatomic helium gas sample in a closed container with a volume of 0.01m3\begin{align}0.01\;\mathrm{m}^3\end{align}, a temperature of 412K\begin{align}412 \;\mathrm{K}\end{align}, and a pressure of 474kPa\begin{align}474 \;\mathrm{kPa}\end{align}. 1. Approximately how many gas atoms are there in the container? 2. Calculate the mass of the individual gas atoms. 3. Calculate the speed of a typical gas atom in the container. 4. The container is heated to 647K\begin{align}647 \;\mathrm{K}\end{align}. What is the new gas pressure? 5. While keeping the sample at constant temperature, enough gas is allowed to escape to decrease the pressure by half. How many gas atoms are there now? 6. Is this number half the number from part (a)? Why or why not? 7. The closed container is now compressed isothermally so that the pressure rises to its original pressure. What is the new volume of the container? 8. Sketch this process on a P-V diagram. 9. Sketch cubes with volumes corresponding to the old and new volumes. 1. . 2. . 3. . 4. . 5. . 6. . 7. 40N\begin{align}40 \;\mathrm{N}\end{align} 8. a. 21,000Pa\begin{align}21,000 \;\mathrm{Pa}\end{align} b. Decreases to 61,000Pa\begin{align}61,000 \;\mathrm{Pa}\end{align} c. 5.8km\begin{align}5.8 \;\mathrm{km}\end{align} 9. a.8.34×1023\begin{align} 8.34 \times 10^{23}\end{align} b. 6.64×1027kg\begin{align}6.64 \times 10^{-27}\;\mathrm{kg}\end{align} c. 1600m/s\begin{align}1600 \;\mathrm{m/s}\end{align} d. 744kPa\begin{align}744 \;\mathrm{kPa}\end{align} e. 4.2×1020\begin{align}4.2 \times 10^{20}\end{align} or 0.0007moles\begin{align}0.0007 \;\mathrm{moles}\end{align} g. 0.00785m3\begin{align}0.00785 \;\mathrm{m}^3\end{align}
17 Q: # (x-2) men can do a piece of work in x days and (x+7) men can do 75% of the same work in (x-10)days. Then in how many days can (x+10) men finish the work? A) 27 days B) 12 days C) 25 days D) 18 days Answer: B) 12 days Explanation: $\frac{3}{4}×\left(x-2\right)x=\left(x+7\right)\left(x-10\right)$ => x= 20 and x=-14 so, the acceptable values is x=20 Therefore, Total work =(x-2)x = 18 x 20 =360 unit Now 360 = 30 x k => k=12 days Q: A, B and C can do a piece of work in 72, 48 and 36 days respectively. For first p/2 days, A & B work together and for next ((p+6))/3days all three worked together. Remaining 125/3% of work is completed by D in 10 days. If C & D worked together for p day then, what portion of work will be remained? A) 1/5 B) 1/6 C) 1/7 D) 1/8 Answer & Explanation Answer: B) 1/6 Explanation: Total work is given by L.C.M of 72, 48, 36 Total work = 144 units Efficieny of A = 144/72 = 2 units/day Efficieny of B = 144/48 = 3 units/day Efficieny of C = 144/36 = 4 units/day According to the given data, 2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100 3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54 p = 198/16.5 p = 12 days. Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day (C+D) in p days = (4 + 6) x 12 = 120 unit Remained part of work = (144-120)/144 = 1/6. Report Error 2 370 Q: 10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work? A) 215 days B) 225 days C) 235 days D) 240 days Answer & Explanation Answer: B) 225 days Explanation: Given that (10M + 15W) x 6 days = 1M x 100 days => 60M + 90W = 100M => 40M = 90W => 4M = 9W. From the given data, 1M can do the work in 100 days => 4M can do the same work in 100/4= 25 days. => 9W can do the same work in 25 days. => 1W can do the same work in 25 x 9 = 225 days. Hence, 1 woman can do the same work in 225 days. Report Error 6 555 Q: A,B,C can complete a work in 15,20 and 30 respectively.They all work together for two days then A leave the work,B and C work some days and B leaves 2 days before completion of that work.how many days required to complete the whole work? Answer Given A,B,C can complete a work in 15,20 and 30 respectively. The total work is given by the LCM of 15, 20, 30 i.e, 60. A's 1 day work = 60/15 = 4 units B's 1 day work = 60/20 = 3 units C's 1 day work = 60/30 = 2 units (A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units Let B + C worked for x days = (3 + 2) x = 5x units C worked for 2 days = 2 x 2 = 4 units Then, 18 + 5x + 4 = 60 22 + 5x = 60 5x = 38 x = 7.6 Therefore, total number of days taken to complete the work = 2 + 7.6 + 2 = 11.6 = 11 3/5 days. Report Error 351 Q: M, N and O can complete the work in 18, 36 and 54 days respectively. M started the work and worked for 8 days, then N and O joined him and they all worked together for some days. M left the job one day before completion of work. For how many days they all worked together? A) 4 B) 5 C) 3 D) 6 Answer & Explanation Answer: B) 5 Explanation: Let M, N and O worked together for x days. From the given data, M alone worked for 8 days M,N,O worked for x days N, O worked for 1 day But given that M alone can complete the work in 18 days N alone can complete the work in 36 days O alone can complete the work in 54 days The total work can be the LCM of 18, 6, 54 = 108 units M's 1 day work = 108/18 = 6 units N's 1 day work = 108/36 = 3 units O's 1 day work = 108/54 = 2 units Now, the equation is 8 x 6 + 11x + 5 x 1 = 108 48 + 11x + 5 = 108 11x = 103 - 48 11x = 55 x = 5 days. Hence, all M,N and O together worked for 5 days. Report Error 2 353 Q: P, Q, and R can do a job in 12 days together. If their efficiency of working be in the ratio 3 : 8 : 5, Find in what time Q can complete the same work alone? A) 36 days B) 30 days C) 24 days D) 22 days Answer & Explanation Answer: A) 36 days Explanation: Given the ratio of efficiencies of P, Q & R are 3 : 8 : 5 Let the efficiencies of P, Q & R be 3x, 8x and 5x respectively They can do work for 12 days. => Total work = 12 x 16x = 192x Now, the required time taken by Q to complete the job alone = days. Report Error 2 438 Q: 5 men and 3 boys can together cultivate a 23 acre field in 4 days and 3 men and 2 boys together can cultivate a 7 acre field in 2 days. How many boys will be needed together with 7 men, if they cultivate 45 acre of field in 6 days. A) 6 B) 4 C) 2 D) 3 Answer & Explanation Answer: C) 2 Explanation: Let work done by 1 man in i day be m and Let work done by 1 boy in 1 day be b From the given data, 4(5m + 3b) = 23 20m + 12b = 23....(1) 2(3m + 2b) = 7 6m + 4b = 7 ....(2) By solving (1) & (2), we get m = 1, b = 1/4 Let the number of required boys = n 6(7 1 + n x 1/4) = 45 => n = 2. Report Error 4 478 Q: The ratio of efficiencies of P, Q and R is 2 : 3 : 4. While P and R work on alternate days and Q work for all days. Now the work completed in total 10 days and the total amount they get is Rs. 1200. Find the amount of each person(respectively). A) 200, 600, 400 B) 400, 600, 200 C) 600, 200, 400 D) 400, 200, 600 Answer & Explanation Answer: A) 200, 600, 400 Explanation: Ratio of efficiencies of P, Q and R = 2 : 3 : 4 From the given data, Number of working days of P, Q, R = 5 : 10 : 5 Hence, ratio of amount of p, Q, R = 2x5 : 3x10 : 4x5 = 10 : 30 : 20 Amounts of P, Q, R = 200, 600 and 400. Report Error 1 415 Q: Two persons Shyam and Rahim can do a job in 32 days together. Rahim can do the same job in 48 days alone. They started working together and after working 8 days Rahim is replaced by a third person Ram whose efficiency is half that of Rahim. Find in how many days the remaining work will be completed by both Shyam and Ram together? A) 16 days B) 72/5 days C) 15 days D) 96/5 days Answer & Explanation Answer: B) 72/5 days Explanation: Work done by Shyam and Rahim in 8 days = 8/32 = 1/4 Remaining work to be done by Shyam and Ram = 1 - 1/4 = 3/4 Given efficieny of Ram is half of Rahim i.e, as Rahim can do the work in 48 days, Ram can do the work in 24 days. One day work of Ram and Shyam = (1/32 - 1/48) + 1/24 = 5/96 Hence, the total work can be done by Shyam and Ram together in 96/5 days. Therefore, remaining work 3/4 can be done by them in 3/4 x 96/5 = 72/5 = 14.4 days. Report Error 5 554
## Elementary Math (K-6th) Tutorial #### Intro In this tutorial, we will look at a simple mixed addition and multiplication to illustrate the reason for order of operations from unit perspective. #### Sample Problem On Sat morning, I jogged from my home to the field. After I got to the field, I jogged along the field 2 times and then jogged back to my home. If the field is 600 m from my home and the perimeter of the field is 1 km, find the total distance I jogged on Sat morning? #### Solution First, consider this problem as a multi-steps word problem. We can solve this problem step-by-step. Jogging from home to the field = 600 m Jogging along the field = 2 x 1 km = 2 km = 2 x 1000 m = 2000 m Jogging from the field back to home = 600 m Total distance = 600 m + 2000 m + 600 m = 3200 m As we can see, a lot of word problems are naturally solved by following the order of operations. Even if we do not know the rule, we can still solve this problem by analyzing it step-by-step. If we want to write the calculation in ONE mathematical sentence, we will need to follow the order of operations. The mathematical sentence is 600 + 2 x 1 x 1000 + 600 We need to perform the calculation 2 x 1 x 1000 first before addition. If we take a closer look of the unit, the perimeter of the field is in km but the distance between the home and the field is in m. In order to perform addition (and subtraction), we need to make sure that the units for different numbers are the same. In this example, performing multiplication first is to convert the total distance that I jogged along the field from km to m. Therefore, we have 600 m + 2000 m + 600 m Now, all numbers are in the same unit. We can perform addition to get the final answer of 3200 m. In real-life word problems, multiplication and division can change the unit of a number. Performing multiplication and division first ensures that units for different numbers are the same before we can add (or subtract) these numbers.
Tips for Creating 8th Grade Level Word Problems Question Using hockey sticks and pucks helps the student relate to more abstract math concept of 00:00 00:00 Tips for Creating 8th Grade Level Word Problems Hi, I'm Scott for About.com. Today I have a few tips for you on how to create 8th grade level word problems from Math.About.com.Word problems are a great way to teach students how the math they are learning in school applies to everyday practical situations. Word problems or story problems help students develop critical thinking skills.Start by becoming familiar with the general concepts the student is learning in math. Students in the eighth grade should be studying more complex algebra and patterning, including being able to write algebraic equations. Additionally, the eighth grade curriculum includes more advanced levels of probability and explaining patterns in data, more complex geometry skills, and finally measurement and number concepts including factors and square roots. The word problems you develop for your student should focus on testing these skills. For an overview of fourth grade math skills, go to: Math.About.com.Once you've decided the type of word problem you're going to create, relate the problem to a real-life situation. Students in the eighth grade are advancing their understanding of algebra, geometry, probability, measurement and more complex number concepts. Eighth grade word problems should test these specific skills. When creating word problems for eighth graders, use people, objects, places, or concepts that they are familiar with. In the following word problem, familiar sports references help the student both better relate to abstract eighth grade number concepts and to take more interest in the math problem.Here is a sample problem: 5 hockey pucks and three hockey sticks cost \$23. 5 hockey pucks and 1 hockey stick cost \$20. How much does 1 hockey puck cost? To solve the equation, the student should set two variables: one for hockey pucks (I used p) and the other for sticks (I used s). Using these variables, create two linear algebraic equations. The first would be 5p + 3s = 23 and the second 5p + s = 20. Then, it's useful to first solves the bottom equation because you can easily isolate the variable S on one side of the equation. Do this, and find that s = 20 - 5p and plug this into the first equation, 5p + 3s = 23, to solve for p. Then you get p = 3.7 or 3 dollars and 70 cents per puck. Here, using hockey sticks and pucks helps the student relate to more abstract math concept of linear algebraic expressions.Math is all about problem solving. One of the best ways to help students learn math is to present them with a problem in which they have to devise their own strategies to find the solution(s). There is usually more than one way to solve many math problems, so try to devise a problem that can be solved in two or more different ways using math concepts familiar to the student.Here is the sample problem: What is the best buy for the following advertised sale on pop? a.) 8 cans for \$4.88 b.) 10 cans for \$5.80 c.) 3 cans for \$1.68 or d.) 12 cans for \$7.20? Here, one way to solve the question is to write out simple linear algebraic equations for each option and solve. The answer could also be arrived at by simply using basic division. You can find word problem worksheets according to each grade on Math.About.com along with more practical problem solving tips.
# Lesson 3 ## Warm-up: How Many Do You See: Groups of 10 (10 minutes) ### Narrative The purpose of this How Many Do You See is for students to subitize or use grouping strategies to describe the images they see. This leads into the next activity, in which students add or subtract 10 from multiples of 10. The third image introduces a base-ten drawing, which students will see throughout the rest of the unit. ### Launch • Groups of 2 • “How many do you see? How do you see them?” • Flash the image. • 30 seconds: quiet think time ### Activity • Display the image. • 1 minute: partner discussion • Record responses. • Repeat for each image. ### Student Facing How many do you see? How do you see them? ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis • “How are the cube towers of 10 the same as the drawing in the last image? How are they different?” (They both show 4 tens in different ways. One looks like cubes and one looks like a drawing.) ## Activity 1: How Many Tens Now? (10 minutes) ### Narrative The purpose of this activity is for students to use towers of 10 to physically add or subtract a ten from a multiple of 10. The structure of this task encourages students to notice patterns in the count of tens and the numbers used to represent the count. Students connect adding and subtracting a ten to skip-counting forward or backward by ten and what they've learned about counting groups of tens from previous lessons (MP7, MP8). ### Launch • Groups of 3 • Give each group 9 towers of 10 connecting cubes. • “Show 1 ten.” • “Add a ten. How many do you have now?” • 30 seconds: partner discussion • Share responses. • “Now that we have answered the first one together, you and your partner will keep adding 1 ten and record how many you have each time. As you work, talk to your partner about what you notice about the numbers.” ### Activity • 5 minutes: partner work time • Monitor for students who represent the value with: • base-ten drawings • __ tens • two-digit number ### Student Facing 1. Show 1 ten. How many do you have now? How many do you have now? Show your thinking using drawings, numbers, or words. How many do you have now? Show your thinking using drawings, numbers, or words. How many do you have now? Show your thinking using drawings, numbers, or words. How many do you have now? Show your thinking using drawings, numbers, or words. How many do you have now? Show your thinking using drawings, numbers, or words. How many do you have now? Show your thinking using drawings, numbers, or words. How many do you have now? Show your thinking using drawings, numbers, or words. 1. Show 9 tens Take away a ten. How many do you have now? Show your thinking using drawings, numbers, or words. 2. Take away another ten. How many do you have now? Show your thinking using drawings, numbers, or words. 3. Take away another ten. How many do you have now? Show your thinking using drawings, numbers, or words. 4. Take away another ten. How many do you have now? Show your thinking using drawings, numbers, or words. 5. Take away another ten. How many do you have now? Show your thinking using drawings, numbers, or words. 6. Take away another ten. How many do you have now? Show your thinking using drawings, numbers, or words. 7. Take away another ten. How many do you have now? Show your thinking using drawings, numbers, or words. 8. Take away another ten. How many do you have now? Show your thinking using drawings, numbers, or words. ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis • Display 7 towers of 10. • “I have 7 tens, which is 70 cubes. How many will I have if I add one more ten? How can I represent it?” • Invite selected students to share their representations. • “How are these representations related?” (They all show the same value.) • “What did you notice each time you added a ten?” (The number was the next number you say when you count by ten. The number of tens was 1 more. One of the digits changed. It went up by 1.) • “What did you notice each time you subtracted a ten?” (The number was the number you say when you count back by 10. The number of tens was 1 less. One of the digits changed. It was 1 less.) ## Activity 2: Introduce Five in a Row, Add or Subtract 10 (15 minutes) ### Narrative The purpose of this activity is for students to learn stage 4 of the Five in a Row center. Students choose a card that shows a multiple of 10. They choose whether to add or subtract 10 from the number on the card to cover a number on their gameboard. MLR8 Discussion Supports. Display sentence frames to encourage partner discussion during the game: “I will add 10. The sum is _____.” and “I will subtract 10. The difference is _____.” Action and Expression: Internalize Executive Functions. To support working memory, provide students with access to sticky notes or mini whiteboards to keep track of solutions for adding 10 or subtracting 10 before making a choice of where to place the counter on the gameboard. Supports accessibility for: Memory, Organization ### Required Materials Materials to Gather Materials to Copy • Five in a Row Addition and Subtraction Stage 4 Gameboard • Number Cards, Multiples of 10 (0-90) ### Required Preparation • Create a set of cards from the blackline master for each group of 2. ### Launch • Groups of 2 • Give each group a set of cards and a gameboard. Give students access to connecting cubes in towers of 10 and two-color counters. • “We are going to learn a new way to play Five in a Row. It is called Five in a Row, Add or Subtract 10.” • Display the gameboard and pile of cards. • “I am going to flip one card over. Now I will decide if I want to add 10 to the number or subtract 10 from the number. I am going to choose to add 10. What is the sum? How do you know?” • 30 seconds: quiet think time • Share responses. • “Now, I put a counter on the sum on my gameboard. Now it’s my partner’s turn.” • “Start by deciding who will use the yellow side and who will use the red side of the counters. Then, take turns flipping one card over, choosing to add 10 to or subtract 10 from the number, and placing your counter on the gameboard to cover the sum or difference. The first person to cover five numbers in a row is the winner.” ### Activity • 8 minutes: partner work time • “Now choose your favorite round. Record how you added or subtracted 10.” • 2 minutes: independent work time • Monitor for students who: • draw towers of 10 • count on or back by 10 • say or write “2 tens and 1 ten is 3 tens” • record with expressions or equations ### Student Facing Show your thinking using drawings, numbers, or words. ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis • Invite 2-3 previously identified students to share how they added or subtracted 10. • Display gameboard with some numbers covered by counters. • Display a card. • “Would you add or subtract 10 from this number? Why?” (I would add 10 because then I could cover _____. I would subtract 10 because then I could cover _____.) • Repeat as time allows. ## Activity 3: Centers: Choice Time (15 minutes) ### Narrative The purpose of this activity is for students to choose from activities that offer practice adding and subtracting. Students choose from any stage of previously introduced centers. • Five in a Row • How Close? • Number Puzzles ### Required Materials Materials to Gather ### Required Preparation • Gather materials from previous centers: • Five in a Row, Stages 1–4 • How Close, Stages 1 and 2 • Number Puzzles, Stages 1 and 2 ### Launch • Groups of 2 • “Now you are going to choose from centers we have already learned.” • Display the center choices in the student book. • “Think about what you would like to do.” • 30 seconds: quiet think time ### Activity • Invite students to work at the center of their choice. • 10 minutes: center work time Choose a center. Five in a Row How Close? Number Puzzles ### Activity Synthesis • “Han is playing Five in a Row, Add or Subtract 10. He picks a card with the number 60. What are the two numbers he could cover on his gameboard? How do you know?” ## Lesson Synthesis ### Lesson Synthesis “Today we added and subtracted 10. Tell your partner how you add 10 to a number.” (I count by ten until I get to the number and then I count one more number. I look at the number to see how many tens are in it and I add one more.) ## Cool-down: Unit 4, Section A Checkpoint (0 minutes) ### Cool-Down For access, consult one of our IM Certified Partners.
Education.com # Independent and Dependent Events Practice Questions (based on 1 rating) To review these concepts, go to Independent and Dependent Events Study Guide. ## Independent and Dependent Events Practice Questions ### Practice 1 #### Problems 1. One card is selected at random from a standard deck of 52 cards. The card is replaced, and then another card is selected. What is the probability that the first card chosen is a red card and the second card chosen is a queen? 2. One card is selected at random from a standard deck of 52 cards. The card is replaced, and then another card is selected. What is the probability that the first card chosen is a heart and the second card chosen is a diamond? 3. A red die and a white die are each rolled while playing a board game. What is the probability of getting a 1 on the red die and a 4 on the white die? 4. A game uses 26 letter tiles. Each tile has one letter on it and each letter of the alphabet appears once on the tiles. If one tile is selected at random, replaced, and then another selected, what is the probability that the first tile is an A and the second another vowel? #### Solutions 1. Read and understand the question. This question is looking for the probability of selecting a red card first, and then a queen second. The events are independent events because the first card is replaced after it is selected. 2. Make a plan. Use the facts that there are 26 red cards and 4 queens in the deck, and there are a total of 52 cards. Because the first card is replaced after it is chosen, the number of total possible outcomes is always the same (52). Find the probability of each event and multiply them. Carry out the plan. The probability of first selecting a red card at random is . The probability of selecting a queen second is . Multiply these probabilities to find the probability that both events will happen: . Check your answer. To check this solution, multiply the probabilities again. Because , this answer is checking. 3. Read and understand the question. This question is looking for the probability of selecting a heart first, and a diamond second. The events are independent events because the first card is replaced after it is selected. 4. Make a plan. Use the facts that there are 13 hearts and 13 diamonds in the deck, and there are a total of 52 cards. Because the first card is replaced after it is chosen, the number of total possible outcomes is always the same (52). Find the probability of each event and multiply them. Carry out the plan. The probability of first selecting a heart at random is. The probability of selecting a diamond second is . Multiply these probabilities to find the probability that both events will happen: . Check your answer. To check this solution, multiply the probabilities again. Because , this solution is checking. 5. Read and understand the question. This question is looking for the probability of rolling a 1 on the red die and a 4 on the white die. The events are independent events because the outcome of one die does not depend on the outcome of the other. 6. Make a plan. Use the facts that there are 6 sides to a die, and one side has 1 dot and one side has 4 dots. Find the probabilities for each event and multiply them. Carry out the plan. The probability of rolling a 1 on the red die is , and the probability of rolling a 4 on the white die is also . Multiply these to find the total probability: . Check your answer. To check this solution, multiply the probabilities again. Because , this solution is checking. 7. Read and understand the question. This question is looking for the probability of selecting an A and another vowel when choosing two letter tiles at random. Because the first tile is replaced, the events are independent events. 8. Make a plan. Use the fact that there are 26 different tiles, each with a different letter. There is one tile with an A and 5 tiles in total with vowels. Find the probability of each situation and multiply them together. Carry out the plan. The probability of selecting an A is and the probability of selecting a vowel is . The probability of both events happening is . Check your answer. To check this solution, multiply the probabilities together again. Because , this solution is checking. 150 Characters allowed ### Related Questions #### Q: See More Questions ### Today on Education.com #### WORKBOOKS May Workbooks are Here! #### WE'VE GOT A GREAT ROUND-UP OF ACTIVITIES PERFECT FOR LONG WEEKENDS, STAYCATIONS, VACATIONS ... OR JUST SOME GOOD OLD-FASHIONED FUN! Get Outside! 10 Playful Activities #### PARENTING 7 Parenting Tips to Take the Pressure Off Welcome!
# How do you solve v+ 303= - 307? Apr 12, 2018 $v = - 610$ #### Explanation: $v + 303 = - 307$ Subtract 303 from both sides $v + 303 - \left(303\right) = - 307 - \left(303\right)$ Simplify $v = - 610$ Apr 12, 2018 -610 #### Explanation: $v + 303 = - 307$ $v = - 307 - 303$ $v = - 610$ In order to solve, you want to move all variables ($v$, in this case) to one side. To do this, you will need to add, subtract, multiply, or divide numbers to one side or the other. Here, we moved $303$ over by subtracting, since it was being added to $v$. Were $303$ being subtracted from $v$ , you would add it over to the other side. Same with multiplication and division being opposites of one another. Apr 13, 2018 $v = - 610$. #### Explanation: You can get this answer by subtracting the $303$ from your left side and from your right side. The reason for this is because it makes both sides of the equation equal and whatever you do to one side you must do to another. When you subtract $303$ from $- 307$ you will get the number $- 610$. This gives you $v = - 610$. You will divide by $1$ because when v is alone there is an invisible $1$ in front of it because you only have $1$ $v$. You will get your answer $v = - 610$. To check your answer you can plug in $- 610$ in place of v which would look like this : $- 610 + 303 = - 307$ which is correct. Hope this helps!!!!!
Upcoming SlideShare × # March 13, 2014 545 views Published on 0 Likes Statistics Notes • Full Name Comment goes here. Are you sure you want to Yes No • Be the first to comment • Be the first to like this Views Total views 545 On SlideShare 0 From Embeds 0 Number of Embeds 387 Actions Shares 0 1 0 Likes 0 Embeds 0 No embeds No notes for slide ### March 13, 2014 1. 1. 13 Today: Warm-Up Review for Test Special Products Complete Class Work from Yesterday Chaka Khan Academy Due Sunday/Monday 2. 2. Warm-Up/Test Prep:5 5. (2x - 3)(x + 2)(x - 1) 3. (y - y2 ) - (2y - 2y2 ) 1. The sum of 2 binomials is 5x2 - 6x. if one of the binomials is 3x2 - 2x, what is the other binomial? All Questions are very similar to tomorrow's. 2. 9x6 - 18x4 + 5x2 2x4 4. (2y2 + 3xy - 4) - (- 5x2 + 5xy + 2) = 3. 3. Binomial Sign Summary (x + 4)(x + 3) Middle Term Last Term positive positive (x - 4)(x + 3) negative negative (x + 4)(x - 3) positive negative (x - 4)(x - 3) negative positive Which term is bigger doesn't matter when both signs are the same, but it does when the signs are different. In this case, because the negative coefficient is larger than the positive. See above 4. 4. Special Products are three types of binomials that occur frequently in algebra. Because of this, you will benefit by memorizing their patterns. It will help a great deal when we start factoring (soon). Here they are: P.S. You'll want to write this down. You should be able to recognize the products both ways. (x2 + 16x + 64) = ( )2x + 8 (4p2 + 16pr + 16r2 ) = ( )22p + 4r 5. 5. 2. (9a2 – 12ab + 4b2 ) = ( )23a – 2b 3.
# Proportional Relationships ## Objective Find the unit rate of ratios involving fractions. ## Common Core Standards ### Core Standards ? • 7.RP.A.1 — Compute unit rates associated with ratios of fractions, including ratios of lengths, areas and other quantities measured in like or different units. For example, if a person walks 1/2 mile in each 1/4 hour, compute the unit rate as the complex fraction 1/2/1/4 miles per hour, equivalently 2 miles per hour. ? • 6.RP.A.2 • 6.RP.A.3.B • 6.NS.A.1 ## Criteria for Success ? 1. Recall a rate, associated with a ratio $a:b$, as $a/b$ units of the first quantity per 1 unit of the second quantity, especially in the case where $a$ and $b$ are both fractions. For example, if a person walks $6\frac{3}{4}$ miles in $2\frac{1}{4}$ hours, the person is traveling at a rate of 3 miles per hour. 2. Recall a unit rate as the numerical part of a rate. For example, in the rate of 3 miles per hour, the 3 represents the unit rate. 3. Identify two unit rates for a ratio $a:b$, as $a/b$ and $b/a$. 4. Compute the value of a complex fraction. 5. Use precision in communicating with units (MP.6). ## Tips for Teachers ? • In the last few lessons of this unit, students use their understanding of proportional relationships to solve problems. They learn strategies, including using a unit rate and using proportions, to efficiently solve problems, and they make connections back to the representations of proportional relationships. • In order to fully access this lesson, students may need to review or recall concepts and skills from 6.NS.1 and 6.RP.2, particularly dividing two fractions and finding the unit rate of whole numbers. #### Remote Learning Guidance If you need to adapt or shorten this lesson for remote learning, we suggest prioritizing Anchor Problem 1 (benefits from worked example). Find more guidance on adapting our math curriculum for remote learning here. #### Fishtank Plus • Problem Set • Student Handout Editor • Vocabulary Package ## Anchor Problems ? ### Problem 1 Find the unit rate in each situation below in miles per hour. 1. Andy drives 180 miles in 3 hours. 2. Brianna bikes ${36{1 \over 2}}$ miles in 2 hours. 3. Chris walks ${4{1 \over 2}}$ miles in ${1{1 \over 4}}$ hours. 4. Deandre runs ${6{1 \over 3}}$miles in 50 minutes. ### Problem 2 Angel and Jayden were at track practice. The track is ${{2 \over 5}}$ kilometers around. • Angel ran 1 lap in 2 minutes. • Jayden ran 3 laps in 5 minutes. 1. How many minutes does it take Angel to run one kilometer? What about Jayden? 2. How far does Angel run in one minute? What about Jayden? 3. Who is running faster? Explain your reasoning. #### References Illustrative Mathematics Track Practice Track Practice, accessed on Aug. 2, 2017, 11:06 a.m., is licensed by Illustrative Mathematics under either the CC BY 4.0 or CC BY-NC-SA 4.0. For further information, contact Illustrative Mathematics. ### Problem 3 Which is the better deal? Justify your response. 3${1 \over3}$ lb. of turkey for $10.50 OR 2${1 \over2}$ lb. of turkey for$6.25 #### References EngageNY Mathematics Grade 7 Mathematics > Module 1 > Topic C > Lesson 11Exit Ticket Grade 7 Mathematics > Module 1 > Topic C > Lesson 11 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m.. ## Problem Set ? The following resources include problems and activities aligned to the objective of the lesson that can be used to create your own problem set. • Include problems where students are given two ratios and asked to find the difference between the rates. For example, in Anchor Problem #1, how much faster, in mph, is Brianna biking than Chris is walking? • Include problems where students are given a list of items with amounts and cost, and are asked to determine which item has the lowest or highest cost per unit. On rollerskates, Ian can travel $\frac{3}{4}$ of a mile in 5 minutes. On his bike, he can travel $3 \frac{1}{2}$ miles in 12 minutes. How fast, in miles per hour, can Ian travel on rollerskates and on his bike?
# What is the equation of the line between (-9,16) and (4,2)? Apr 15, 2018 $14 x + 13 y = 82$ #### Explanation: Equation of the line involves: 2)using the point gradient formula to find your equation (in this case, this the second step) $G r a \mathrm{di} e n t \left(m\right) = \frac{16 - 2}{- 9 - 4}$ = $\frac{14}{-} 13$ Equation of line: We also using the point $\left(4 , 2\right)$ $\left(y - 2\right) = - \frac{14}{13} \left(x - 4\right)$ $13 y - 26 = - 14 x + 56$ $14 x + 13 y = 82$
# two-digit number • Oct 10th 2009, 07:09 AM ronaldj two-digit number The digit 3 is written at the right of a certain two-digit number thus forming a three-digit number. The new number is 372 more than the original two-digit number. What was the original two-digit number? my answer is 94? • Oct 10th 2009, 07:16 AM Defunkt Quote: Originally Posted by ronaldj The digit 3 is written at the right of a certain two-digit number thus forming a three-digit number. The new number is 372 more than the original two-digit number. What was the original two-digit number? my answer is 94? Let x be the original two-digit number. Then, the new number is: $\displaystyle 10x + 3$. Now, we know that $\displaystyle 10x + 3 = x + 372$ Solving this gives us: $\displaystyle 9x = 369 \Rightarrow x = 41$ How did you arrive at 94? You should have easily noted that if you modify the number as stated you get 943, which is obviously not 372 more than 94. • Oct 10th 2009, 07:19 AM earboth Quote: Originally Posted by ronaldj The digit 3 is written at the right of a certain two-digit number thus forming a three-digit number. The new number is 372 more than the original two-digit number. What was the original two-digit number? my answer is 94? Let the original number be 10x +y with $\displaystyle x\in\{1, 2, ..., 9\}$ and $\displaystyle y\in\{0, 1, 2, ..., 9\}$ Then the 3-digit-number is 100x+10y+3. According to the wording of the question you get: $\displaystyle 100x+10y+3-372=10x+y$ $\displaystyle 10x+y=41$ The only possible value for x = 4 and y = 1. Thus the original number is 41. • Oct 10th 2009, 07:28 AM ronaldj Quote: Originally Posted by Defunkt Let x be the original two-digit number. Then, the new number is: $\displaystyle 10x + 3$. Now, we know that $\displaystyle 10x + 3 = x + 372$ Solving this gives us: $\displaystyle 9x = 369 \Rightarrow x = 41$ How did you arrive at 94? You should have easily noted that if you modify the number as stated you get 943, which is obviously not 372 more than 94. Mea culpa. You are right - thanks! • Oct 10th 2009, 07:35 AM ronaldj Quote: Originally Posted by earboth Let the original number be 10x +y with $\displaystyle x\in\{1, 2, ..., 9\}$ and $\displaystyle y\in\{0, 1, 2, ..., 9\}$ Then the 3-digit-number is 100x+10y+3. According to the wording of the question you get: $\displaystyle 100x+10y+3-372=10x+y$ $\displaystyle 10x+y=41$ The only possible value for x = 4 and y = 1. Thus the original number is 41. QUESTION - my kid wants to know if it could also be 21 since 21 + 372 = 393/ • Oct 10th 2009, 07:53 AM Defunkt Quote: Originally Posted by ronaldj QUESTION - my kid wants to know if it could also be 21 since 21 + 372 = 393/ If you add 3 to the right of 21, do you get 393? • Oct 10th 2009, 07:59 AM ronaldj you are sooh right! My head hurts but you are right. Tx for pointing out our mistake!
Courses Courses for Kids Free study material Offline Centres More # How do you simplify $\sqrt{0.36}$? Last updated date: 22nd Feb 2024 Total views: 339.3k Views today: 8.39k Verified 339.3k+ views Hint:We try to form the indices formula for the value 2. This is a square root of $0.36$. We convert the decimal to fractions. The root of the whole fraction happens by finding the root of the denominator and numerator separately as $\sqrt{\dfrac{p}{q}}=\dfrac{\sqrt{p}}{\sqrt{q}}$. We find the root values and place to find the solution. We find the prime factorisation of 36 and 100. Complete step by step solution: We need to find the value of the algebraic form of $\sqrt{0.36}$. This is a square root form. The given value is the form of indices. We are trying to find the root value of $0.36$. We know the theorem of indices ${{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}$. Putting value 2 we get ${{a}^{\dfrac{1}{2}}}=\sqrt{a}$. Converting decimal to fraction we get $0.36=\dfrac{36}{100}$. We have to solve the answer of $\sqrt{\dfrac{36}{100}}$. We know $\sqrt{\dfrac{p}{q}}=\dfrac{\sqrt{p}}{\sqrt{q}}$. This means to find the answer of $\sqrt{\dfrac{36}{100}}$, we need to find $\sqrt{36}$ and $\sqrt{100}$. For $\sqrt{36}=\sqrt{{{\left( 6 \right)}^{2}}}=6$ and for $\sqrt{100}=\sqrt{{{\left( 10 \right)}^{2}}}=10$ We now place the values and get $\sqrt{\dfrac{36}{100}}=\dfrac{\sqrt{36}}{\sqrt{100}}=\dfrac{6}{10}=\dfrac{3}{5}$. Now we convert the fraction into fraction and get $\dfrac{3}{5}=0.6$ The simplified value of the square root $\sqrt{0.36}$ is $0.6$. Note: We can also use the variable from where we can take $x={{\left( 0.36 \right)}^{\dfrac{1}{2}}}$. But we need to remember that we can’t use the square on both sides of the equation $x={{\left( 0.36 \right)}^{\dfrac{1}{2}}}$ as in that case we are taking an extra value of negative as in $0.36$ as a root value. Then this linear equation becomes a quadratic equation.
# 2014 AMC 8 Problems/Problem 18 ## Problem Four children were born at City Hospital yesterday. Assume each child is equally likely to be a boy or a girl. Which of the following outcomes is most likely? $\textbf{(A) }\text{all 4 are boys}\\ \textbf{(B) }\text{all 4 are girls}\\ \textbf{(C) }\text{2 are girls and 2 are boys}\\ \textbf{(D) }\text{3 are of one gender and 1 is of the other gender}\\ \textbf{(E) }\text{all of these outcomes are equally likely}$ ## Solution 1 We'll just start by breaking cases down. The probability of A occurring is $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$. The probability of B occurring is $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$. The probability of C occurring is $\dbinom{4}{2}\cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}$, because we need to choose 2 of the 4 children to be girls. For D, there are two possible cases, 3 girls and 1 boy or 3 boys and 1 girl. The probability of the first case is $\dbinom{4}{1}\cdot\left(\frac{1}{2}\right)^4 = \frac{1}{4}$ because we need to choose 1 of the 4 children to be a boy. However, the second case has the same probability because we are choosing 1 of the 4 children to be a girl, so the total probability is $\frac{1}{4} \cdot 2 = \frac{1}{2}$. So out of the four fractions, D is the largest. So our answer is $\boxed{\text{(D) 3 of one gender and 1 of the other}}.$
# AP Statistics Curriculum 2007 Distrib Dists (Difference between revisions) Revision as of 02:03, 7 November 2009 (view source)IvoDinov (Talk \| contribs) (→Parameter estimation)← Older edit Revision as of 02:05, 7 November 2009 (view source)IvoDinov (Talk \| contribs) (→Parameter estimation)Newer edit → Line 198: Line 198: * Thus, we can solve the equation above $\Chi^2(k_0) = 5.261948$ for the single variable of interest -- the unknown parameter $k_0$. Suppose we are using the same example as before, $x=\{x_1=5,x_2=1,x_3=5\}$. Then the solution is an asymptotic chi-squared distribution driven estimate of the parameter $k_0$. * Thus, we can solve the equation above $\Chi^2(k_0) = 5.261948$ for the single variable of interest -- the unknown parameter $k_0$. Suppose we are using the same example as before, $x=\{x_1=5,x_2=1,x_3=5\}$. Then the solution is an asymptotic chi-squared distribution driven estimate of the parameter $k_0$. - : $\Chi^2(k_0) = \sum_{i=1}^3{\frac{(x_i-\hat{\mu_i})^2}{\hat{\mu_i} \left (1+ \frac{\hat{\mu_i}}{k_0} \right )}}$. + $\Chi^2(k_0) = \sum_{i=1}^3{\frac{(x_i-\hat{\mu_i})^2}{\hat{\mu_i} \left (1+ \frac{\hat{\mu_i}}{k_0} \right )}}$. - :: $\Chi^2(k_0) = \frac{(5-61.67)^2}{61.67(1+61.67/k_0)}+\frac{(1-41.67)^2}{41.67(1+41.67/k_0)}+\frac{(5-18.67)^2}{18.67(1+18.67/k_0)}=5.261948.$ Solving this equation for $k_0$ provides the desired estimate for the last parameter. + $\Chi^2(k_0) = \frac{(5-61.67)^2}{61.67(1+61.67/k_0)}+\frac{(1-41.67)^2}{41.67(1+41.67/k_0)}+\frac{(5-18.67)^2}{18.67(1+18.67/k_0)}=5.261948.$ Solving this equation for $k_0$ provides the desired estimate for the last parameter. :: [http://www.mathematica.com/ Mathematica] provides 3 distinct ($k_0$) solutions to this equation: {'''50.5466''', -21.5204, '''2.40461'''}. Since $k_0>0$ there are 2 candidate solutions. :: [http://www.mathematica.com/ Mathematica] provides 3 distinct ($k_0$) solutions to this equation: {'''50.5466''', -21.5204, '''2.40461'''}. Since $k_0>0$ there are 2 candidate solutions. - * '''Estimates of Probabilities''': As $\frac{\mu_i}{k_0}p_0=p_i$, we have: + * '''Estimates of Probabilities''': Assume $k_0=2$ and $\frac{\mu_i}{k_0}p_0=p_i$, we have: : $\frac{61.67}{k_0}p_0=31p_0=p_1$ : $\frac{61.67}{k_0}p_0=31p_0=p_1$ : $20p_0=p_2$ : $20p_0=p_2$ : $9p_0=p_3$ : $9p_0=p_3$ : Hence, $1-p_0=p_1+p_2+p_3=60p_0$. Therefore, $p_0=\frac{1}{61}$, $p_1=\frac{31}{61}$, $p_2=\frac{20}{61}$ and $p_3=\frac{9}{61}$. : Hence, $1-p_0=p_1+p_2+p_3=60p_0$. Therefore, $p_0=\frac{1}{61}$, $p_1=\frac{31}{61}$, $p_2=\frac{20}{61}$ and $p_3=\frac{9}{61}$. - : Therefore, the best model distribution for the observed sample $x=\{x_1=5,x_2=1,x_3=5\}$ is $X \sim NMD(2, \{\frac{31}{61}, \frac{20}{61},\frac{9}{61}\}).$ + : Therefore, the best model distribution for the observed sample $x=\{x_1=5,x_2=1,x_3=5\}$ is $X \sim NMD\left (2, \left \{\frac{31}{61}, \frac{20}{61},\frac{9}{61}\right\} \right ).$ : Notice that in this calculation, we explicitely used the complete cancer data table, not only the sample $x=\{x_1=5,x_2=1,x_3=5\}$, as we need multiple samples (multiple sites or columns) to estimate the $k_0$ parameter. : Notice that in this calculation, we explicitely used the complete cancer data table, not only the sample $x=\{x_1=5,x_2=1,x_3=5\}$, as we need multiple samples (multiple sites or columns) to estimate the $k_0$ parameter. ## General Advance-Placement (AP) Statistics Curriculum - Geometric, HyperGeometric, Negative Binomial Random Variables and Experiments ### Geometric • Definition: The Geometric Distribution is the probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set {1, 2, 3, ...}. The name geometric is a direct derivative from the mathematical notion of geometric series. • Mass Function: If the probability of successes on each trial is P(success)=p, then the probability that x trials are needed to get one success is $P(X = x) = (1 - p)^{x-1} \times p$, for x = 1, 2, 3, 4,.... • Expectation: The Expected Value of a geometrically distributed random variable X is ${1\over p}.$ • Variance: The Variance is ${1-p\over p^2}.$ ### HyperGeometric The hypergeometric distribution is a discrete probability distribution that describes the number of successes in a sequence of n draws from a finite population without replacement. An experimental design for using Hypergeometric distribution is illustrated in this table: Type Drawn Not-Drawn Total Defective k m-k m Non-Defective n-k N+k-n-m N-m Total n N-n N • Explanation: Suppose there is a shipment of N objects in which m are defective. The Hypergeometric Distribution describes the probability that in a sample of n distinctive objects drawn from the shipment exactly k objects are defective. • Mass function: The random variable X follows the Hypergeometric Distribution with parameters N, m and n, then the probability of getting exactly k successes is given by $P(X=k) = {{{m \choose k} {{N-m} \choose {n-k}}}\over {N \choose n}}.$ This formula for the Hypergeometric Mass Function may be interpreted as follows: There are ${{N}\choose{n}}$ possible samples (without replacement). There are ${{m}\choose{k}}$ ways to obtain k defective objects and there are ${{N-m}\choose{n-k}}$ ways to fill out the rest of the sample with non-defective objects. The mean and variance of the hypergeometric distribution have the following closed forms: Mean: $n \times m\over N$ Variance: ${ {nm\over N} ( 1-{m\over N} ) (N-n)\over N-1}$ #### Examples • SOCR Activity: The SOCR Ball and Urn Experiment provides a hands-on demonstration of the utilization of Hypergeometric distribution in practice. This activity consists of selecting n balls at random from an urn with N balls, R of which are red and the other N - R green. The number of red balls Y in the sample is recorded on each update. The distribution and moments of Y are shown in blue in the distribution graph and are recorded in the distribution table. On each update, the empirical density and moments of Y are shown in red in the distribution graph and are recorded in the distribution table. Either of two sampling models can be selected with the list box: with replacement and without replacement. The parameters N, R, and n can be varied with scroll bars. • A lake contains 1,000 fish; 100 are randomly caught and tagged. Suppose that later we catch 20 fish. Use SOCR Hypergeometric Distribution to: • Compute the probability mass function of the number of tagged fish in the sample of 20. • Compute the expected value and the variance of the number of tagged fish in this sample. • Compute the probability that this random sample contains more than 3 tagged fish. • Hypergeometric distribution may also be used to estimate the population size: Suppose we are interested in determining the population size. Let N = number of fish in a particular isolated region. Suppose we catch, tag and release back M=200 fish. Several days later, when the fish are randomly mixed with the untagged fish, we take a sample of n=100 and observe m=5 tagged fish. Suppose p=200/N is the population proportion of tagged fish. Notice that when sampling fish we sample without replacement. Thus, hypergeometric is the exact model for this process. Assuming the sample-size (n) is < 5% of the population size(N), we can use binomial approximation to hypergeometric. Thus if the sample of n=100 fish had 5 tagged, the sample-proportion (estimate of the population proportion) will be $\hat{p}={5\over 100}=0.05$. Thus, we can estimate that $0.05=\hat{p}={200\over N}$, and $N\approx 4,000$, as shown on the figure below. ### Negative Binomial The family of Negative Binomial Distributions is a two-parameter family; p and r with 0 < p < 1 and r > 0. There are two (identical) combinatorial interpretations of Negative Binomial processes (X or Y). #### X=Trial index (n) of the rth success, or Total # of experiments (n) to get r successes • Probability Mass Function: $P(X=n) = {n-1 \choose r-1}\cdot p^r \cdot (1-p)^{n-r} \!$, for n = r,r+1,r+2,.... (n=trial number of the rth success) • Mean: $E(X)= {r \over p}$ • Variance: $Var(X)= {r(1-p) \over p^2}$ #### Y = Number of failures (k) to get r successes • Probability Mass Function: $P(Y=k) = {k+r-1 \choose k}\cdot p^r \cdot (1-p)^k \!$, for k = 0,1,2,.... (k=number of failures before the rth successes) • $Y \sim NegBin(r, p)$, the probability of k failures and r successes in n=k+r Bernoulli(p) trials with success on the last trial. • Mean: $E(Y)= {r(1-p) \over p}$. • Variance: $Var(Y)= {r(1-p) \over p^2}$. • Note that X = Y + r, and E(X) = E(Y) + r, whereas VAR(X)=VAR(Y). #### Application Suppose Jane is promoting and fund-raising for a presidential candidate. She wants to visit all 50 states and she's pledged to get all electoral votes of 6 states before she and the candidate she represents are satisfied. In every state, there is a 30% chance that Jane will be able to secure all electoral votes and 70% chance that she'll fail. • What's the probability mass function of the number of failures (k=n-r) to get r=6 successes? In other words, What's the probability mass function that the last 6th state she succeeds to secure all electoral votes happens to be the at the nth state she campaigns in? NegBin(r, p) distribution describes the probability of k failures and r successes in n=k+r Bernoulli(p) trials with success on the last trial. Looking to secure the electoral votes for 6 states means Jane needs to get 6 successes before she (and her candidate) is happy. The number of trials (i.e., states visited) needed is n=k+6. The random variable we are interested in is X={number of states visited to achieve 6 successes (secure all electoral votes within these states)}. So, n = k+6, and $X\sim NegBin(r=6, p=0.3)$. Thus, for $n \geq 6$, the mass function (giving the probabilities that Jane will visit n states before her ultimate success is: $P(X=n) = {n-1 \choose r-1}\cdot p^r \cdot (1-p)^{n-r} = {n - 1 \choose r-1} \cdot 0.3^6 \cdot 0.7^{n-r}$ • What's the probability that Jane finishes her campaign in the 10th state? Let $X\sim NegBin(r=6, p=0.3)$, then $P(X=10) = {10-1 \choose 6-1}\cdot 0.3^6 \cdot 0.7^{10-6} = 0.022054.$ • What's the probability that Jane finishes campaigning on or before reaching the 8th state? $P(X\leq 8) = 0.011292$ • Suppose the success of getting all electoral votes within a state is reduced to only 10%, then X~NegBin(r=6, p=0.1). Notice that the shape and domain the Negative-Binomial distribution significantly chance now (see image below)! What's the probability that Jane covers all 50 states but fails to get all electoral votes in any 6 states (as she had hoped for)? $P(X\geq 50) = 0.632391$ ### Negative Multinomial Distribution (NMD) The Negative Multinomial Distribution is a generalization of the two-parameter Negative Binomial distribution (NB(r,p)) to $m\ge 1$ outcomes. Suppose we have an experiment that generates $m\ge 1$ possible outcomes, $\{X_0,\cdots,X_m\}$, each occurring with probability $\{p_0,\cdots,p_m\}$, respectively, where with 0 < pi < 1 and $\sum_{i=0}^m{p_i}=1$. That is, $p_0 = 1-\sum_{i=1}^m{p_i}$. If the experiment proceeds to generate independent outcomes until $\{X_0, X_1, \cdots, X_m\}$ occur exactly $\{k_0, k_1, \cdots, k_m\}$ times, then the distribution of the m-tuple $\{X_1, \cdots, X_m\}$ is Negative Multinomial with parameter vector $(k_0,\{p_1,\cdots,p_m\})$. Notice that the degree-of-freedom here is actually m, not (m+1). That is why we only have a probability parameter vector of size m, not (m+1), as all probabilities add up to 1 (so this introduces one relation). Contrast this with the combinatorial interpretation of Negative Binomial (special case with m=1): X˜NegativeBinomial(NumberOfSuccesses = r,ProbOfSuccess = p), X=Total # of experiments (n) to get r successes (and therefore n-r failures); X˜NegativeMultinomial(k0,{p0,p1}), X=Total # of experiments (n) to get k0 (dafault variable) and nk0 outcomes of 1 other possible outcome (X1). #### Negative Multinomial Summary • Probability Mass Function: $P(k_0, \cdots, k_m) = \left (\sum_{i=0}^m{k_i}-1\right)!\frac{p_0^{k_0}}{(k_0-1)!} \prod_{i=1}^m{\frac{p_i^{k_i}}{k_i!}}$, or equivalently: $P(k_0, \cdots, k_m) = \Gamma\left(\sum_{i=1}^m{k_i}\right)\frac{p_0^{k_0}}{\Gamma(k_0)} \prod_{i=1}^m{\frac{p_i^{k_i}}{k_i!}}$, where Γ(x) is the Gamma function. • Mean (vector): $\mu=E(X_1,\cdots,X_m)= (\mu_1=E(X_1), \cdots, \mu_m=E(X_m)) = \left ( \frac{k_0p_1}{p_0}, \cdots, \frac{k_0p_m}{p_0} \right)$. • Variance-Covariance (matrix): Cov(Xi,Xj) = {cov[i,j]}, where $cov[i,j] = \begin{cases} \frac{k_0 p_i p_j}{p_0^2},& i\not= j,\\ \frac{k_0 p_i (p_i + p_0)}{p_0^2},& i=j.\end{cases}$. #### Cancer Example The Probability Theory Chapter of the EBook shows the following example using 400 Melanoma (skin cancer) Patients where the Type and Site of the cancer are recorded for each subject, as in the Table below. Type Site Totals Head and Neck Trunk Extremities Hutchinson's melanomic freckle 22 2 10 34 Superficial 16 54 115 185 Nodular 19 33 73 125 Indeterminant 11 17 28 56 Column Totals 68 106 226 400 The sites (locations) of the cancer may be independent, but there may be positive dependencies of the type of cancer for a given location (site). For example, localized exposure to radiation implies that elevated level of one type of cancer (at a given location) may indicate higher level of another cancer type at the same location. We want to use the Negative Multinomial distribution to model the sites cancer rates and try to measure some of the cancer type dependencies within each location. Let's denote by xi,j the cancer rates for each site ($0\leq i \leq 2$) and each type of cancer ($0\leq j \leq 3$). For each (fixed) site ($0\leq i \leq 2$), the cancer rates are independent Negative Multinomial distributed random variables. That is, for each column index (site) the column-vector X has the following distribution: X = {X1,X2,X3NMD(k0,{p1,p2,p3}). Different columns (sites) are considered to be different instances of the random multinomially distributed X vector. Then we have the following estimates: $\hat{\mu}_{i,j} = \frac{x_{i,.}\times x_{.,j}}{x_{.,.}}$ $x_{i,.} = \sum_{j=0}^{3}{x_{i,j}}$ $x_{.,j} = \sum_{i=0}^{2}{x_{i,j}}$ $x_{.,.} = \sum_{i=0}^{2}\sum_{j=0}^{3}{{x_{i,j}}}$ Example: $\hat{\mu}_{1,1} = \frac{x_{1,.}\times x_{.,1}}{x_{.,.}}=\frac{34\times 68}{400}=5.78$ • Variance-Covariance: For a single column vector, X = {X1,X2,X3NMD(k0,{p1,p2,p3}), covariance between any pair of Negative Multinomial counts (Xi and Xj) is: $cov[X_i,X_j] = \begin{cases} \frac{k_0 p_i p_j}{p_0^2},& i\not= j,\\ \frac{k_0 p_i (p_i + p_0)}{p_0^2},& i=j.\end{cases}$. Example: For the first site (Head and Neck, j=0), suppose that $X=\left \{X_1=5, X_2=1, X_3=5\right \}$ and X˜NMD(k0 = 10,{p1 = 0.2,p2 = 0.1,p3 = 0.2}). Then: $p_0 = 1 - \sum_{i=1}^3{p_i}=0.5$ NMD(X \| k0,{p1,p2,p3}) = 0.00465585119998784 $cov[X_1,X_3] = \frac{10 \times 0.2 \times 0.2}{0.5^2}=1.6$ $\mu_2=\frac{k_0 p_2}{p_0} = \frac{10\times 0.1}{0.5}=2.0$ $\mu_3=\frac{k_0 p_3}{p_0} = \frac{10\times 0.2}{0.5}=4.0$ $corr[X_2,X_3] = \left (\frac{\mu_2 \times \mu_3}{(k_0+\mu_2)(k_0+\mu_3)} \right )^{\frac{1}{2}}$ and therefore, $corr[X_2,X_3] = \left (\frac{2 \times 4}{(10+2)(10+4)} \right )^{\frac{1}{2}} = 0.21821789023599242.$ You can also use the interactive SOCR negative multinomial distribution calculator to compute these quantities, as shown on the figure below. • There is no MLE estimate for the NMD k0 parameter (see this reference). However, there are approximate protocols for estimating the k0 parameter, see the example below. • Correlation: correlation between any pair of Negative Multinomial counts (Xi and Xj) is: $Corr[X_i,X_j] = \begin{cases} \left (\frac{\mu_i \times \mu_j}{(k_0+\mu_i)(k_0+\mu_j)} \right )^{\frac{1}{2}} = \left (\frac{p_i p_j}{(p_0+p_i)(p_0+p_j)} \right )^{\frac{1}{2}}, & i\not= j, \\ 1, & i=j.\end{cases}$. • The marginal distribution of each of the Xi variables is negative binomial, as the Xi count (considered as success) is measured against all the other outcomes (failure). But jointly, the distribution of $X=\{X_1,\cdots,X_m\}$ is negative multinomial, i.e., $X \sim NMD(k_0,\{p_1,\cdots,p_m\})$ . Notice that the pair-wise NMD correlations are always positive, where as the correlations between multinomail counts are always negative. Also note that as the parameter k0 increases, the paired correlations go to zero! Thus, for large k0, the Negative Multinomial counts Xi behave as independent Poisson random variables with respect to their means $\left ( \mu_i= k_0\frac{p_i}{p_0}\right )$. #### Parameter estimation • Estimation of the mean (expected) frequency counts (μj) of each outcome (Xj): The MLE estimates of the NMD mean parameters μj are easy to compute. If we have a single observation vector $\{x_1, \cdots,x_m\}$, then $\hat{\mu}_i=x_i.$ If we have several observation vectors, like in this case we have the cancer type frequencies for 3 different sites, then the MLE estimates of the mean counts are $\hat{\mu}_j=\frac{x_{j,.}}{I}$, where $0\leq j \leq J$ is the cancer-type index and the summation is over the number of observed (sampled) vectors (I). For the cancer data above, we have the following MLE estimates for the expectations for the frequency counts: Hutchinson's melanomic freckle type of cancer (X0) is $\hat{\mu}_0 = 34/3=11.33$. Superficial type of cancer (X1) is $\hat{\mu}_1 = 185/3=61.67$. Nodular type of cancer (X2) is $\hat{\mu}_2 = 125/3=41.67$. Indeterminant type of cancer (X3) is $\hat{\mu}_3 = 56/3=18.67$. • Estimation of the k0 (gamma) parameter: There is no MLE for the k0 parameter; however, there is a protocol for estimating k0 using the chi-squared goodness of fit statistic. In the usual chi-squared statistic: $\Chi^2 = \sum_i{\frac{(x_i-\mu_i)^2}{\mu_i}}$, we can replace the expected-means (μi) by their estimates, $\hat{\mu_i}$, and replace denominators by the corresponding negative multinomial variances. Then we get the following test statistic for negative multinomial distributed data: $\Chi^2(k_0) = \sum_{i}{\frac{(x_i-\hat{\mu_i})^2}{\hat{\mu_i} \left (1+ \frac{\hat{\mu_i}}{k_0} \right )}}$. Now we can derive a simple method for estimating the k0 parameter by varying the values of k0 in the expression Χ2(k0) and matching the values of this statistic with the corresponding asymptotic chi-squared distribution. The following protocol summarizes these steps using the cancer data above: df = (# rows – 1)(# columns – 1) = (3-1)*(4-1) = 6 • Mean Counts Estimates: The mean counts estimates (μj) for the 4 different cancer types are: $\hat{\mu}_1 = 185/3=61.67$; $\hat{\mu}_2 = 125/3=41.67$; and $\hat{\mu}_3 = 56/3=18.67$. • Thus, we can solve the equation above Χ2(k0) = 5.261948 for the single variable of interest -- the unknown parameter k0. Suppose we are using the same example as before, x = {x1 = 5,x2 = 1,x3 = 5}. Then the solution is an asymptotic chi-squared distribution driven estimate of the parameter k0. $\Chi^2(k_0) = \sum_{i=1}^3{\frac{(x_i-\hat{\mu_i})^2}{\hat{\mu_i} \left (1+ \frac{\hat{\mu_i}}{k_0} \right )}}$. $\Chi^2(k_0) = \frac{(5-61.67)^2}{61.67(1+61.67/k_0)}+\frac{(1-41.67)^2}{41.67(1+41.67/k_0)}+\frac{(5-18.67)^2}{18.67(1+18.67/k_0)}=5.261948.$ Solving this equation for k0 provides the desired estimate for the last parameter. Mathematica provides 3 distinct (k0) solutions to this equation: {50.5466, -21.5204, 2.40461}. Since k0 > 0 there are 2 candidate solutions. • Estimates of Probabilities: Assume k0 = 2 and $\frac{\mu_i}{k_0}p_0=p_i$, we have: $\frac{61.67}{k_0}p_0=31p_0=p_1$ 20p0 = p2 9p0 = p3 Hence, 1 − p0 = p1 + p2 + p3 = 60p0. Therefore, $p_0=\frac{1}{61}$, $p_1=\frac{31}{61}$, $p_2=\frac{20}{61}$ and $p_3=\frac{9}{61}$. Therefore, the best model distribution for the observed sample x = {x1 = 5,x2 = 1,x3 = 5} is $X \sim NMD\left (2, \left \{\frac{31}{61}, \frac{20}{61},\frac{9}{61}\right\} \right ).$ Notice that in this calculation, we explicitely used the complete cancer data table, not only the sample x = {x1 = 5,x2 = 1,x3 = 5}, as we need multiple samples (multiple sites or columns) to estimate the k0 parameter.
# How to find all Pythagorean triples containing a given number? I am solving one question related to right triangle in which one side is $12$ and I have to find the greatest possible perimeter of such a triangle. Is there any way to find all Pythagorean triples if only one side is given? Let a Pythagorean triple be of the form $(x,y,z)$ where one of $x, y$ or $z$ is 12. From Elementary Number Theory we know that $z$ is odd, and $x \not\equiv y \pmod{2}$ (i.e. one of $x$ and $y$ is odd and one is even). (I will omit the proof of this but it is fairly simple). We also know that for primitive Pythagorean triples: $$x=2pq, \\ y=p^2-q^2 \\ z=p^2+q^2\\$$ for some $p>q>0$. We therefore know that 12 must be $x$ as $x$ is even, so we look at all the factors of $6$ to find the possible values for $p,q$. $$6 = 6 \cdot 1=2 \cdot 3$$ So our possible values for $p$ are: $p=6$ or $p=3$, and our possible values for $q$ are: $q=1$ and $q=2$. With $p=6,q=1$ we get: $x=12,y=35,z=37$ and $P=x+y+z=84$. With $p=3,q=2$ we get: $x=12,y=5,z=13$ and $P=30$. So the maximal perimeter corresponds to the triangle with sides $12, 35, 37$ and perimeter $84$. ---EDIT--- The above only considered primitive Pythagorean triples. To consider all P.ts we must look at those where all numbers are less than 12 and find the non primitive triples which arise from them. These are: $$(4,3,5) \rightarrow (12,9,15),(16,12,20)$$ Upon multiplying by $3$ and $4$ respectively. Neither of these triples give rise to a perimeter greater than $84$, so the triangle with one side of length $12$ with greatest perimeter is the triangle corresponding to $(12,35,37)$ • You reversed the roles of $y$ and $z$. You meant $y = p^2 \color{red}{-} q^2$ and $z = p^2 \color{red}{+} q^2$. The formula you are citing is for primitive Pythagorean triples, that is, those that are relatively prime. Commented Dec 27, 2015 at 18:00 • Sorry, yes you are correct in both cases. I will edit the part relative to your first point now. WRT your second point, could one not look at all primitive Pythagorean triples where all sides are less than 12, and find the non primitive triples with one side 12 that arise from these? Commented Dec 27, 2015 at 18:05 Hint: There are no integer right triangles with a hypotenuse of 12 (why?), so the side of length $12$ must be a leg, and then $144 = c^2-b^2 = (c+b)(c-b)$. • why? (for first line) Commented Dec 27, 2015 at 15:37 • @Ayushakj If it were, you could express $144$ as a sum of two nonzero squares. Can you? Commented Dec 27, 2015 at 15:38 • I think the OP is asking for a right triangle with one side 12, You're answering about a hypotenuse of 12. Commented Dec 27, 2015 at 15:40 • @EthanBolker I clarified the answer; I was talking about a leg of 12. Commented Dec 27, 2015 at 15:42 • So now I have to substitute value of c and b. or any other way. Commented Dec 27, 2015 at 15:43 If we solve any of Euclid's formula functions for $$n$$, we can find triples for any given side, if they exist, with a finite search of $$m$$ values. For $$A=m^2-n^2$$, we let $$n=\sqrt{m^2-A}$$ where $$\lceil\sqrt{A+1}\space\rceil\le m\le \bigl\lceil\frac{A}{2}\bigr\rceil$$. If any $$m$$ yields a positive integer $$n$$, we have $$(m,n)$$ for a Pythagorean triple. For example, if $$A=27$$, then $$5\le m \le 14$$ and we find $$(m,n)=(6,3)$$ and $$(14,13)$$; we find $$f(6,3)=(27,36,45)$$ and $$f(14,13)=(27,364,365)$$. If $$A=12$$, we find only $$f(4,2)=(12,16,20)$$ which is $$4*(3,4,5)$$. For $$B=2mn$$, $$n=\frac{B}{2m}$$ where $$\lceil\sqrt{2B}\space\space\rceil\le m \le \frac{B}{2}$$; for $$B=12,\space\space3\le m\le 6$$ and we find only $$f(6,1)=(35,12,37).$$ For $$C= m^2+n^2,\space n=\sqrt{C-m^2}$$ where $$\biggl\lceil\sqrt{\frac{C}{2}}\space\space\biggr\rceil \le m\le\bigl\lfloor\sqrt{C}\bigr\rfloor$$. For example, if $$C=1105,\space 24\le m \le 33$$ and we find four triples that match. $$f(24,23)=(47,1104,1105)\quad f(31,12)=(817,744,1105)\quad f(32,9)=(943,576,1105)\quad f(33,4)=(1073,264,1105)$$
# Video: KS2-M19 • Paper 1 • Question 26 1/5 + 2 1/10 = ＿. 04:25 ### Video Transcript One and one-fifth plus two and one-tenth equals what. A common mistake with problems like this is to think that when we add the fractions part, all we have to do is add the numerators, one add one equals two, and then add the denominators, five plus 10 equals 15. But we can’t just expect that one of these plus one of these equals two of these. All three fractions are different sizes. They have different denominators. So we can’t just add fifths and tenths and expect to get an answer in fifteenths. So when the time comes to add the fraction part of these mixed numbers together, we’re going to have to convert them so they have the same denominator. Let’s partition both of our mixed numbers into a whole number and a fraction. One and one-fifth is the same as one plus one-fifth. And two and one-tenth is the same as two plus one-tenth. So really, we could write the whole addition as one plus one-fifth plus two plus one-tenth. First let’s start by adding the whole number part of our mixed numbers. One plus two equals three. Now, we can find the total of the fraction part of each mixed number. And as we spotted already, we need to convert these fractions so they have a common denominator before we add them together. What number is a multiple of both five and 10? We could convert them both into twentieths because 20 is a multiple of both five and 10. But it’s much quicker to convert fifths into tenths. And that way, both fractions will have a common denominator of 10. We only need to convert the first one. But how do you know what the answer is going to be without writing it out? To turn fifths into tenths, we double the denominator. And to keep the fraction the same value, we do the same to the numerator. One-fifth is the same as two-tenths. And so our calculation becomes two-tenths plus one-tenth, which equals three-tenths. So when we added the whole number part of our mixed numbers, we got an answer of three. And after adding our fraction part of our mixed numbers, we got an answer of three-tenths. So to find the overall total, all we have to do is to add the two answers. Three plus three-tenths equals three and three-tenths. Can you think of any other ways that we could write three and three-tenths? But remember, if we use a place value grid, three and three-tenths is the same as three ones and three-tenths. And so if we write the answer 3.3, that would be just as good as three and three-tenths. There’s another way we could write the answer and that’s as a fraction. We know that the tenths column shows us that we’ve already got three-tenths. But if we have three wholes as well, three ones, how many tenths is that all together? Well, we know that one whole or one one is the same as ten-tenths. So three wholes or three ones must be the same as three lots of ten-tenths or thirty-tenths. And so if we also add the three-tenths part of our number, we can write the answer as an improper fraction, thirty-three tenths. All three answers are exactly the same. And you could write any one of them and still get the correct answer. First, we split up the mixed numbers so we could see the whole numbers and the fractions separately. Then, all we needed to do was to add the whole numbers then add the fraction parts. And we had to remember to convert the first fraction so that they both had common denominator and then, finally, find the total. One and one-fifth plus two and one-tenth equals three and three-tenths or 3.3 or thirty-three tenths. Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.
or Find what you need to study Light # 6.4 The Fundamental Theorem of Calculus and Accumulation Functions Now that we have introduced integrals, you may be wondering how they connect with derivatives. Today, we’ll introduce the Fundamental Theorem of Calculus which connects these two topics. ## ∫ Fundamental Theorem of Calculus The Fundamental Theorem of Calculus connects differentiation and integration. It states that the derivative of an integral is just the function inside the integral. 😊 We can define a new function, $F$, that represents the antiderivative of $f$. $F(x)=\int_{a}^{x}f(t)dt$ The Fundamental Theorem of Calculus states that if $f$ is continuous on the interval $(a,b)$ then for every $x$ in the interval: $\frac{d}{dx}\left[\int_{a}^{x}f(t)dt\right]=F'(x)=f(x)$ This means that we can use the Fundamental Theorem of Calculus to find derivatives. Let’s walk through an example to see what this means. ### ✏️ Fundamental Theorem of Calculus Example Find $g'(16)$ if $g(x)$ is the function below. $g(x)=\int_{5}^{x}\sqrt[4]{t}dt$ How should we proceed? If we ignore what $g(x)$ is specifically defined as for a minute and think about how we would proceed for a function that is not defined with integrals, we would do so by first finding the derivative of $g(x)$ and then plugging $16$ in. Well for functions defined by definite integrals, also called accumulation functions, we do the same! How do we differentiate $\int_{5}^{x}\sqrt[4]{t}dt$? This is where the Fundamental Theorem of Calculus comes in. Since it states that the derivative of an integral is just the function inside the integral… $g'(x)=\sqrt[4]{x}$ Now all we have to do is substitute 16. $g'(16)=\sqrt[4]{16}=2$ ### ✏️ Fundamental Theorem of Calculus Example 2 In the above example, the upper bound function was just $x$, making it super easy! All we had to do was substitute $x$ into the integrated and multiply by the derivative of $x$. Questions become slightly more complicated when the functional bound is something other than x, such as the following function. Let’s find $F'(x)$. $F(x)=\int_{3}^{x^2}(t+4)dt$ The upper bound in this question is $x^2$, so we have to multiply our answer by the derivative of the functional bound. $F'(x)=\frac{d}{dx}\int_{3}^{x^2}(t+4)dt$ Let’s use the chain rule. $F'(x)=(x^2+4)\frac{d}{dx}x^2$ $F'(x)=(x^2+4)(2x)$ So, when the upper bound of the definite integral is a function of x, you need to multiply the integral's integrand by the derivative of the upper bound. It’s not too difficult, but it does mean you have to keep an eye on the upper bound! 👀 ## 📝 Practice Problems Time to try some practice on our own! ### ❓ Problems #### Question 1 Let $g(x)=\int_{0}^{x}\sqrt{8+\cos(t)}dt$. Find $g'(0)$. #### Question 2 Let $g(x)=\int_{1}^{x}(5t^2+2t)dt$. Find $g'(3)$. #### Question 3 Let $g(x)=\int_{0}^{x}\sqrt{\sin\left(t\right)+15}dt$. Find $g'(\frac{\pi}{2})$. #### Question 4 Let $F(x)=\int_{3x}^{1}sec^2(t)dt$. Find $F'(x)$. #### Question 1 To find $g'(0)$, we simply have to find $g'(x)$ and evaluate it at $x=0$ since the upper bound is already $x$! By the Fundamental Theorem of Calculus, $g'(x)=\sqrt{8+\cos(x)}$. Therefore, $g'(0)=\sqrt{8+\cos(0)}=\sqrt{8+1}=\sqrt{9}=3$. #### Question 2 To find $g'(3)$, we simply have to find $g'(x)$ and evaluate it at $x=3$. By the Fundamental Theorem of Calculus, $g'(x)=5x^2+2x$. Therefore, $g'(3)=5\cdot3^2+2\cdot 3=45+6=51$. #### Question 3 To find $g'(\frac{\pi}{2})$, we simply have to find $g'(x)$ and evaluate it at $x=\frac{\pi}{2}$. By the Fundamental Theorem of Calculus, $g'(x)=\sqrt{\sin(x)+15}$. Therefore, $g'(\frac{\pi}{2})=\sqrt{15+\sin(\frac{\pi}{2})}=\sqrt{15+1}=\sqrt{16}=4$. #### Question 4 To find $F'(x)$, we first have to notice that the upper bound is not x. The upper bound needs to have the variable, so we have to use integral rules to switch the bounds. If you’re going through our guides in order, you haven’t come across integral rules yet. We’re covering that in two key topics, 6.6! Here’s the integral rule you’d be using in this question: $\int_{a}^{b}f(x)\, dx = -\int_{b}^{a}f(x)\, dx$ Here’s how we apply it… $F'(x)=\frac{d}{dx}(-\int_{1}^{3x}sec^2(t)dt)$ And here’s the rest of the problem, using what we already know! $F'(x)=-sec^2(3x)\frac{d}{dx}3x$ $=-3sec^2(3x)$ ## 💫 Closing Great work! You've learned about the Fundamental Theorem of Calculus, which, remember, bridges the gap between differentiation and integration. You've also seen how to apply this theorem to find derivatives of functions defined by definite integrals, including cases where the upper bound is a function of $x$. When the upper bound is not a simple $x$, you multiply the integrand by the derivative of the upper bound. Keep practicing and working through the practice problems to solidify your understanding. Calculus can be challenging, but with dedication and practice, you'll master it! Good luck. 📚 # 6.4 The Fundamental Theorem of Calculus and Accumulation Functions Now that we have introduced integrals, you may be wondering how they connect with derivatives. Today, we’ll introduce the Fundamental Theorem of Calculus which connects these two topics. ## ∫ Fundamental Theorem of Calculus The Fundamental Theorem of Calculus connects differentiation and integration. It states that the derivative of an integral is just the function inside the integral. 😊 We can define a new function, $F$, that represents the antiderivative of $f$. $F(x)=\int_{a}^{x}f(t)dt$ The Fundamental Theorem of Calculus states that if $f$ is continuous on the interval $(a,b)$ then for every $x$ in the interval: $\frac{d}{dx}\left[\int_{a}^{x}f(t)dt\right]=F'(x)=f(x)$ This means that we can use the Fundamental Theorem of Calculus to find derivatives. Let’s walk through an example to see what this means. ### ✏️ Fundamental Theorem of Calculus Example Find $g'(16)$ if $g(x)$ is the function below. $g(x)=\int_{5}^{x}\sqrt[4]{t}dt$ How should we proceed? If we ignore what $g(x)$ is specifically defined as for a minute and think about how we would proceed for a function that is not defined with integrals, we would do so by first finding the derivative of $g(x)$ and then plugging $16$ in. Well for functions defined by definite integrals, also called accumulation functions, we do the same! How do we differentiate $\int_{5}^{x}\sqrt[4]{t}dt$? This is where the Fundamental Theorem of Calculus comes in. Since it states that the derivative of an integral is just the function inside the integral… $g'(x)=\sqrt[4]{x}$ Now all we have to do is substitute 16. $g'(16)=\sqrt[4]{16}=2$ ### ✏️ Fundamental Theorem of Calculus Example 2 In the above example, the upper bound function was just $x$, making it super easy! All we had to do was substitute $x$ into the integrated and multiply by the derivative of $x$. Questions become slightly more complicated when the functional bound is something other than x, such as the following function. Let’s find $F'(x)$. $F(x)=\int_{3}^{x^2}(t+4)dt$ The upper bound in this question is $x^2$, so we have to multiply our answer by the derivative of the functional bound. $F'(x)=\frac{d}{dx}\int_{3}^{x^2}(t+4)dt$ Let’s use the chain rule. $F'(x)=(x^2+4)\frac{d}{dx}x^2$ $F'(x)=(x^2+4)(2x)$ So, when the upper bound of the definite integral is a function of x, you need to multiply the integral's integrand by the derivative of the upper bound. It’s not too difficult, but it does mean you have to keep an eye on the upper bound! 👀 ## 📝 Practice Problems Time to try some practice on our own! ### ❓ Problems #### Question 1 Let $g(x)=\int_{0}^{x}\sqrt{8+\cos(t)}dt$. Find $g'(0)$. #### Question 2 Let $g(x)=\int_{1}^{x}(5t^2+2t)dt$. Find $g'(3)$. #### Question 3 Let $g(x)=\int_{0}^{x}\sqrt{\sin\left(t\right)+15}dt$. Find $g'(\frac{\pi}{2})$. #### Question 4 Let $F(x)=\int_{3x}^{1}sec^2(t)dt$. Find $F'(x)$. #### Question 1 To find $g'(0)$, we simply have to find $g'(x)$ and evaluate it at $x=0$ since the upper bound is already $x$! By the Fundamental Theorem of Calculus, $g'(x)=\sqrt{8+\cos(x)}$. Therefore, $g'(0)=\sqrt{8+\cos(0)}=\sqrt{8+1}=\sqrt{9}=3$. #### Question 2 To find $g'(3)$, we simply have to find $g'(x)$ and evaluate it at $x=3$. By the Fundamental Theorem of Calculus, $g'(x)=5x^2+2x$. Therefore, $g'(3)=5\cdot3^2+2\cdot 3=45+6=51$. #### Question 3 To find $g'(\frac{\pi}{2})$, we simply have to find $g'(x)$ and evaluate it at $x=\frac{\pi}{2}$. By the Fundamental Theorem of Calculus, $g'(x)=\sqrt{\sin(x)+15}$. Therefore, $g'(\frac{\pi}{2})=\sqrt{15+\sin(\frac{\pi}{2})}=\sqrt{15+1}=\sqrt{16}=4$. #### Question 4 To find $F'(x)$, we first have to notice that the upper bound is not x. The upper bound needs to have the variable, so we have to use integral rules to switch the bounds. If you’re going through our guides in order, you haven’t come across integral rules yet. We’re covering that in two key topics, 6.6! Here’s the integral rule you’d be using in this question: $\int_{a}^{b}f(x)\, dx = -\int_{b}^{a}f(x)\, dx$ Here’s how we apply it… $F'(x)=\frac{d}{dx}(-\int_{1}^{3x}sec^2(t)dt)$ And here’s the rest of the problem, using what we already know! $F'(x)=-sec^2(3x)\frac{d}{dx}3x$ $=-3sec^2(3x)$ ## 💫 Closing Great work! You've learned about the Fundamental Theorem of Calculus, which, remember, bridges the gap between differentiation and integration. You've also seen how to apply this theorem to find derivatives of functions defined by definite integrals, including cases where the upper bound is a function of $x$. When the upper bound is not a simple $x$, you multiply the integrand by the derivative of the upper bound. Keep practicing and working through the practice problems to solidify your understanding. Calculus can be challenging, but with dedication and practice, you'll master it! Good luck. 📚
# Prelab 7 You are not logged in. Note that this link will take you to an external site (https://shimmer.csail.mit.edu) to authenticate, and then you will be redirected back to this page. ## 1) Measuring Frequency For lab this week we're going to be working on the pieces in our system which do the math to detect the resulting frequency shift in our ultrasonic wave from the Doppler Effect. The high-level block diagram below shows the pieces of our system that we've built so far and what we'll specifically be building in Lab 7. From our work in Lab 4, we generate a 40kHz square wave, amplify it with an operational amplifier circuit before transmitting it out via an ultrasonic transducer; this was our transmitter stage. The sent waves get reflected off of objects and a small fraction of that energy will be recovered by the ultrasonic receiver we built up and tested in Lab 6, which recovers the signal and amplifies it back up so we can do some useful work and processing on it. What we'll be doing in Lab 7 is the first part of this processing. In particular, we want to compare the original frequency we sent out to the frequency we're getting back, since the difference in those frequencies contains information about movement (thanks to the Doppler Effect). We can express this shift using the following equation where \delta f is our frequency shift, v is the velocity of an object, and c is the speed of sound. \delta f(t) = 2\frac{v}{c}\, f_0(t) ## 2) Frequency Difference If we want to figure out the difference between two frequencies, how can we do it? Trig identities are the key. Let's just say we have two signals. The first is v_i: v_i(t) = V_0 \cos \left(2 \pi f_0 t\right) The second is v_r which is shifted from the frequency of v_i by \delta f: v_r(t) = V_1 \cos \left(2 \pi \left(f_0 +\delta f\right) t \right) If we multiply these two signals, we can use the following trig identity: \cos\left(\alpha\right) \cos\left(\beta\right) = \frac{1}{2} \left( \cos\left(\alpha-\beta\right) + \cos \left(\alpha + \beta\right)\right) This indicates that the result of multiplying v_i(t) and v_r(t), would come out to be: v_i(t)\cdot v_r(t) = \frac{V_0V_1}{2}\left[\cos\left(2\pi\left(f_0-\left(f_0+\delta ft\right)\right)\right) +\cos\left(2\pi\left(f_0+\left(f_0+\delta f\right)\right)t\right)\right] which simplifies to (using the fact that \cos\left(-x\right)=\cos\left(x\right): v_i(t)\cdot v_r(t) = \frac{V_0V_1}{2}\left(\cos\left(2\pi\delta ft\right) +\cos\left(2\pi\left(2f_0+\delta f\right)t\right)\right] And this is really useful because it is telling us that one of the resulting signals will have a frequency based on the difference of our two input frequencies, and the other one will be based on the sum of our two input frequencies. One of those signals (the difference) has all key information we want, namely \delta f. In case you have any doubts that this trig identity actually works, below is an image of the result of multiplying two sinusoids together, with f_0=30Hz and \delta f=5Hz, resulting in a second frequency of 35Hz: (in lab we'll be working at a much higher frequency. The 30 Hz signal here is just for an example) Pretty neat! You can see solid evidence of two distinct sinusoidal behaviors. One is oscillating at approximately 65 Hz (the sum of 35 Hz and 30 Hz) and one at roughly 5 Hz (the difference between 35 Hz and 30 Hz). We're still not out of the woods. We have two signals in our result but only want one. While we'll talk about this more in lab, it turns out that we'll be able to remove the high-frequency component (our sum) with some circuitry and end up only getting the difference signal out. ## 3) Multiplying We're almost done. We've been using multiplication in our discussion above, but how do we multiply two signals in circuits? We can multiply a signal by a constant (many of our previous op amp circuits), and we can do things like add signals and take their difference (summing amplifier and difference amplifier), but how do we multiply two signals? We haven't covered that yet. There are actual analog multipliers in existence. These are circuits that can take in two analog signals and do exactly what we say: multiply them. For our system we're going to take advantage of a few things to keep it simple, however: • While we care a lot about the difference in frequency between our transmitted and received signals, since that's what contains information about the velocity of the moving object, we don't care about the shape of our signal. In fact if you think about what our initial reference signal is in our circuit, it is not a sine wave, rather a 40 kHz square wave! • Some traditional digital logic gates (as in AND gates, NOR gates, and a particular pair which we'll discuss below) can effectively perform the multiplication action on incoming signals. The caveat is that these signals need to be digital in nature. To get started, if we don't care about our signal shape, let's just convert our sine waves so that they are one of two values (one form of digital). We will discretize the signals in the following way: • +1 if the value of the sine wave is above 0 • -1 if the value of the sine wave is below 0 When we do that to our two frequencies above, and we then take this result and just multiply it as usual (like we did before with our full analog sine wave from before) we get the following: This looks really interesting. There are two signals above, one at 30 Hz (f_0), and one at 35 Hz (f_0+\delta f) just like before, and the bottom is the result of their multiplication. If we zoom in as shown below, you can see there's a digital signal at roughly twice the frequency of our two original signals (their sum f_0+f_0+\delta f), albeit at varying duty cycle. In looking at the image above, you can also see that there's a much slower pattern going on as well at around 5Hz, and that should make sense since that is the difference signal (our f_0 - (f_0+\delta f)=-\delta f) So it seems that even with digital signals the aforementioned trig identity holds up! (we can thank Fourier for that). Moving on, this digital multiplication that we have above looks like the following if we do a mapping of (inputs)\to output: • (-1, -1) \to +1 (when both signals are -1, generate a +1) • (-1, +1) \to -1 (when signals are -1 and +1, generate a -1) • (+1, -1) \to -1 (when signals are +1 and -1, generate a -1) • (+1, +1) \to +1 (when both signals are +1, generate a +1) If we relabel our "-1" signal as "0" and "+1+ as just "1", we could rewrite these four combinations as follows (all -1 become 0 and all +1 become 1, but we keep the input/output results above): We're not saying 0 times 0 is 1...we're using binary "0" to represent the value -1 and binary 1 to represent the value +1. This is really important to understand! • (0, 0) \to 1 • (0, 1) \to 0 • (1, 0) \to 0 • (1, 1) \to 1 This relationship of 1's and 0's is exactly the same as a type of logic gate known as an Exclusive NOR gate, which is shown below (along with it's Truth table): What this means is that an XNOR gate, which we can purchase for buy off the street for very little money (<$0.50, versus ~$10 for a true analog multiplier), can effectively multiply two binary signals for us in this representation. So what we need to do to implement this (and what we'll be doing in lab) is the following: • Take the amplified recevier signal and convert it to digital using a comparator. The reference signal we generate to go into the transmitter is already a square wave so we don't need to put that signal through a comparator. • Multiply the resultant signal with the original square wave signal using an XNOR gate to produce a signal containing the signals that express the sum and (more importantly to us) the difference of the two input signals' frequencies!
Solve 3x^3+6x^2+4x+2=0 \| Uniteasy.com # Solve the cubic equation: ## $$3x^3+6x^2+4x+2=0$$ Since the discriminant $$\Delta >0$$, the cubic equation has one real root and two conjugate complex roots. $$\Delta=0.034293552812071$$ $$\begin{cases} x_1=-\dfrac{\sqrt[3]{10}}{3}-\dfrac{2}{3} \\ x_2=\dfrac{\sqrt[3]{10}}{6}+\dfrac{\sqrt[3]{10}}{6}\sqrt{3}i-\dfrac{2}{3} \\ x_3=\dfrac{\sqrt[3]{10}}{6}-\dfrac{\sqrt[3]{10}}{6}\sqrt{3}i-\dfrac{2}{3} \end{cases}$$ In decimals, $$\begin{cases} x_1=-1.384811563344 \\ x_2=-0.5+0.28867513459481i \\ x_3=-0.5-0.28867513459481i \end{cases}$$ Detailed Steps on Solution ## 1. Convert to depressed cubic equation The idea is to convert general form of cubic equation $$ax^3+bx^2+cx+d = 0$$ to the form without quadratic term. $$t^3+pt+q = 0$$ By substituting $$x$$ with $$t - \dfrac{b}{3a}$$, the general cubic equation could be transformed to $$t^3+\dfrac{3ac-b^2}{3a^2}t+\dfrac{2b^3-9abc+27a^2d}{27a^3} = 0$$ Compare with the depressed cubic equation. Then, $$p = \dfrac{3ac-b^2}{3a^2}$$ $$q = \dfrac{2b^3-9abc+27a^2d}{27a^3}$$ Substitute the values of coefficients, $$p, q$$ is obtained as $$p = \dfrac{3\cdot 3\cdot 4-6^2}{3\cdot 3^2}=0$$ $$q = \dfrac{2\cdot 6^3-9\cdot3\cdot 6\cdot 4+27\cdot 3^2\cdot2}{27\cdot 3^3}=\dfrac{10}{27}$$ ### Use the substitution to transform Let $$p$$ and $$q$$ being the coefficient of the linean and constant terms, the depressed cubic equation is expressed as. $$t^3 +pt+q=0$$ Let $$x=t-\dfrac{2}{3}$$ The cubic equation $$3x³ + 6x² + 4x + 2=0$$ is transformed to $$t^3 +\dfrac{10}{27}=0$$ For the equation $$t^3 +\dfrac{10}{27}$$, we have $$p=0$$ and $$q = \dfrac{10}{27}$$ ### Calculate the discriminant The nature of the roots are determined by the sign of the discriminant. \begin{aligned} \\\Delta&=\dfrac{q^2}{4}+\dfrac{p^3}{27}\\ & =\dfrac{\Big(\dfrac{10}{27}\Big)^2}{4}+0\\ & =\dfrac{25}{729}\\ & =0.034293552812071\\ \end{aligned} ### 1.1 Use the root formula directly If the discriminant is greater than zero, we can use the root formula to determine the roots of the cubic equation. $$t_{1,2,3} =\begin{cases} \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } +\sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}& \\ ω\cdotp \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + \overline{ω} \sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}&\ \\ \overline{ω}\cdotp \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + ω\cdotp \sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }} \end{cases}$$ in which, $$ω = \dfrac{-1+i\sqrt{3}}{2}$$ and $$\overline{ω} =\dfrac{-1-i\sqrt{3}}{2}$$ Substitute the values of $$p, q$$ and $$\Delta$$ which we have calculated. Then, \begin{aligned} \\t_1&=\sqrt[3]{-\dfrac{5}{27}+\sqrt{\dfrac{25}{729}}}+\sqrt[3]{-\dfrac{5}{27}-\sqrt{\dfrac{25}{729}}}\\ & =\sqrt[3]{-\dfrac{5}{27}+\sqrt{\dfrac{5^2}{27^2}}}+\sqrt[3]{-\dfrac{5}{27}-\sqrt{\dfrac{5^2}{27^2}}}\\ & =\sqrt[3]{-\dfrac{5}{27}+\dfrac{5}{27}}+\sqrt[3]{-\dfrac{5}{27}-\dfrac{5}{27}}\\ & =\sqrt[3]{\dfrac{-5}{1\cdot 3^3}+\dfrac{5}{1\cdot 3^3}}+\sqrt[3]{\dfrac{-5}{1\cdot 3^3}-\dfrac{5}{1\cdot 3^3}}\\ & =\dfrac{1}{3}\Big(\sqrt[3]{\dfrac{-5}{1}+\dfrac{5}{1}}+\sqrt[3]{\dfrac{-5}{1}-\dfrac{5}{1}}\Big)\\ & =\dfrac{1}{3}\sqrt[3]{-10}\\ & =-\dfrac{\sqrt[3]{10}}{3}\\ \end{aligned} If we denote $$R = -\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }$$ $$\overline{R} = -\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }$$ then, $$\sqrt[3]{R} = 0$$, $$\sqrt[3]{\overline{R}} =-\dfrac{\sqrt[3]{10}}{3}$$ \begin{aligned} \\t_2&= ω\cdotp \sqrt[3]{R}+ \overline{ω} \sqrt[3]{\overline{R} }\\ & =\dfrac{-\sqrt[3]{R}-\sqrt[3]{\overline{R} }}{2} +\dfrac{\sqrt{3}( \sqrt[3]{R} - \sqrt[3]{\overline{R} }) }{2} i\\ & =\dfrac{1}{2}\Big[0-\Big(-\dfrac{\sqrt[3]{10}}{3}\Big)\Big]+\dfrac{\sqrt{3}}{2}\Big[0-\Big(-\dfrac{\sqrt[3]{10}}{3}\Big)\Big]i\\ & =\dfrac{\sqrt[3]{10}}{6}+\dfrac{\sqrt[3]{10}}{6}\sqrt{3}i\\ \end{aligned} \begin{aligned} \\t_3&= \overline{ω}\cdotp \sqrt[3]{R}+ ω\cdotp \sqrt[3]{\overline{R}}\\ & =\dfrac{-\sqrt[3]{R}-\sqrt[3]{\overline{R} }}{2} +\dfrac{\sqrt{3}(- \sqrt[3]{R} + \sqrt[3]{\overline{R} }) }{2}i \\ & =\dfrac{1}{2}\Big[0-\Big(-\dfrac{\sqrt[3]{10}}{3}\Big)\Big]-\dfrac{\sqrt{3}}{2}\Big[0-\Big(-\dfrac{\sqrt[3]{10}}{3}\Big)\Big]i\\ & =\dfrac{\sqrt[3]{10}}{6}-\dfrac{\sqrt[3]{10}}{6}\sqrt{3}i\\ \end{aligned} ## Roots of the general cubic equation Since $$x = t - \dfrac{b}{3a}$$, substituting the values of $$t$$, $$a$$ and $$b$$ gives $$x_1 = t_1-\dfrac{2}{3}$$ $$x_2 = t_2-\dfrac{2}{3}$$ $$x_3 = t_3-\dfrac{2}{3}$$ ## 2. Summary In summary, we have tried the method of cubic root formula to explore the solutions of the equation. The cubic equation $$3x³ + 6x² + 4x + 2=0$$ is found to have one real root and two complex roots. Exact values and approximations are given below. $$\begin{cases} x_1=-\dfrac{\sqrt[3]{10}}{3}-\dfrac{2}{3} \\ x_2=\dfrac{\sqrt[3]{10}}{6}+\dfrac{\sqrt[3]{10}}{6}\sqrt{3}i-\dfrac{2}{3} \\ x_3=\dfrac{\sqrt[3]{10}}{6}-\dfrac{\sqrt[3]{10}}{6}\sqrt{3}i-\dfrac{2}{3} \end{cases}$$ in decimal notation, $$\begin{cases} x_1=-1.384811563344 \\ x_2=-0.5+0.28867513459481i \\ x_3=-0.5-0.28867513459481i \end{cases}$$ ## 3. Graph for the function $$f(x) = 3x³ + 6x² + 4x + 2$$ Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = 3x³ + 6x² + 4x + 2$$ has one intersection point with the x-axis. Scroll to Top
# How To Calculate Variance? Variance is used to calculate how spread out a dataset is. Calculation of variance is very helpful while creating statistical models. The low variance can alert the statistician about the over-fitting of the data. Often while calculating the variance for statistical measurement, it can be tricky. But once you got to the final formula, it can be straightforward. In this section let’s find out how to find the variance of a sample data set. Most of the times while working with different data sets statisticians only have access to a sample or the subset of the main data set. For instance, instead of calculating the “cost of every car in Germany”, a statistician could find the cost of a random sample of a few thousand cars. They can use this sample to get an estimated value for the price of the cars in Germany but most likely not the actual numbers. So in the first step, write down the sample set on which you want to work on. After you got the sample set, start working with the sample variance formula. Write down the sample variance formula. As we discussed earlier, the variance lets you know about the spread out of your dataset. The closer the variance to zero, the clustered are they together. The sample variance formula is, s2= ∑[(xi – x̅)2]/(n – 1) Here, • s2 is the variance. Remember that the variance is always measured in squared units. • xi represents any term from the sample data set. • ∑ Meaning “summation,” refers to calculate the following terms for each value of xi, then add them together. • x̅ is the mean of the sample. • n is the number of data points Now as we have the sample variance formula, let’s explore more about the methods. In this step, you have to calculate the mean of the sample. The term x̅ or “x-bar” refers to the mean of the sample data set. To calculate the mean of the dataset you can use the regular method. First, add all the data points together and then divide the result by the no. of data points present in the sample set. Let’s say we have a sample data set of 6 data points. The data points are 17, 15, 23, 7, 9, & 13. Now to calculate the mean of this sample data set we will add all the data points together. 17 + 15 + 23 + 7 + 9 + 13 = 84 After adding the data points now, we will divide the result with the no of data points present in the sample set to find the mean of the sample data set. 84/6 = 14 Hence, the mean of the data set is 14. Now as per the formula, we have x̅, the mean of the dataset. In next step subtract mean from each data point of the sample set to find xi – x̅. After subtracting, we have x1 – x̅ = 17 – 14 = 3 x2 – x̅ = 15 – 14 = 1 x3 – x̅ = 23 – 14 = 9 x4 – x̅ = 7 – 14 = -7 x5 – x̅ = 9 – 14 = -5 x6 – x̅ = 13 – 14 = -1 Following the formula, now take the square of each result. After taking the squares, we have 9, 1, 81, 49, 25, 1. As we have the term (xi – x̅)2, now find the summation of (xi – x̅)2 for all the values of “i”. i.e. 9 + 1 + 81 + 49 + 25 + 1 = 166. Now we have solved the critical part of the formula. In the next step, we have to divide the result with n-1. As we know, the value of n is 6. Then n-1 = 5. Hence after dividing the result will be 166/5 = 33.2 That means, s2 = 33.2 => s = square root of 33.2 = 5.76. Hence the variance of the sample set is 5.76
• (a) Find the length of the curve between two points. (b) Find the area bounded above by the catenary and below by the x-axis. (c) Find the equation of the tangent line to the curve at the point (c;f(c)). • If the tangent at a point P on the curve y = x 3 intersects the curve again at Q, let A be the area of the region bounded by the curve and the line segment PQ. Let B be the area of the region defined in the same way starting with Q instead of P. • The bèzier equations are very simple. You basically take the line segment created by the points, get the lines between them, then you linearly interpolate a point between the starting and ending point of each line. This way you get one less point than before. You do the same with the lines defined by these, and so on until you get a single point. This will draw your bezier curve. You can also just use specialized equations, like for instance for the 4 point version: The tangent line to a curve at a given point is a straight line that just "touches" the curve at that point. So if the function is f(x) and if the tangent "touches" its curve at x=c, then the tangent will pass through the point (c,f(c)). The slope of this tangent line is f'(c) ( the derivative of the function f(x) at x=c). The equation $$x^2+y^2=1$$ defines a curve that is the unit circle. All points $$(x, y)$$ satisfying the equation lie on the curve. The curve is implicitly defined by the equation. GeoGebra categorizes a curve defined by an equation as a Line, a Conic, or as an Implicit curve. "The Line Test" One thing you may note about a convex shape is that, no matter where you draw a line that passes through the shape, it will always pass through only two of the lines or polygons making up the shape: If you try the same thing with a concave shape it can pass through more than two of the lines: + Dr rahman mohammed • # Svg draw curved line between two points Feb 17, 2013 · Hi, I recently learned that to find the tangent at a point on any curve, you can simply place a mirror on that point and reflect the part of the curve on one side of that point such that the reflection flows smoothly into the other part of the curve on the other side. Once this is done, draw a... For each 'i', an arrow is drawn between the point '(x0[i], y0[i])' and the point '(x1[i],y1[i])'. If code=1 an arrowhead is drawn at '(x0[i],y0[i])' if code=2 an arrowhead is drawn at '(x1[i],y1[i])'. If code=3 an arrowhead is drawn at both ends of the arrow. unless arr.length = 0, when no head is drawn. The Add Anchor Point tool allows you to insert additional anchor points into your path, effectively splitting a line segment into two. This is useful if, say, you’ve created a curve already but you want to turn the curve into a more complex shape. Get the free "Intersection points of two curves/lines" widget for your website, blog, Wordpress, Blogger, or iGoogle. A neat widget that will work out where two curves/lines will intersect.Mar 01, 2011 · draw a curved line between two points esri arcgis silverlight api. Discussion created by spatialideaswebsolutions on Mar 1, 2011 Latest reply on Mar 1, 2011 by charanp. Apr 04, 2016 · Curved Lines. Drawing curves follows a similar theory to drawing diagonal lines. This will let you draw circles that look perfectly round, and not bumpy. The easiest place to start is cirlcles. Like in straight lines, you’ll have to pay attention to the length of each line segment. The points determined in this way are then joined with straight lines. fn(x) or expr (with x inside) must return a numeric of the same length as x . This used to be a quick hack which seems to serve a useful purpose, but can give bad results for functions which are not smooth. A plot item, that represents a series of points. A curve is the representation of a series of points in the x-y plane. It supports different display styles, interpolation ( f.e. spline ) and symbols. Usage a) Assign curve properties When a curve is created, it is configured to draw black solid lines with in QwtPlotCurve::Lines style A bezier curve is specified by four points; it passes through two of those points exactly, and the tangent vectors at those two endpoints are determined by the two "control points". The interface in terms of "control points" makes no sense to me - I wrote code to convert endpoints and tangent vectors at endpoints to endpoints and control points. The first point you click on is the start of the curve and the point you release on is the end of the curve. Now you must click and drag two more points. These are known as the "control points". You can probably find a rigorous mathematical definition of how the control points effect the appearance of the curve on Google. Feb 16, 2019 · With these two options, you can lift off for the faintest marks or press hard to get a nice, strong line. When working with charcoal or a chisel-point pencil, varying the angle of the tip can create a great variation in line width. We offer two of the most popular choices: normalize.css and a reset. Or, choose Neither and nothing will be applied. We offer two popular choices: Autoprefixer (which processes your CSS server-side) and -prefix-free (which applies prefixes via a script, client-side).• Curve: A list of four Points de ning a B ezier curve. The rst and last points are the anchor points, the second and third points are the control points. • Layer: A list of up to ten Curves, which can be manipulated and drawn as a unit. 2.2 Atom Types The only legal atom type allowed in Curve is an int: An int object may be used to ... Located in the right-click menu, the “add pulse” and “add step” options provide operations for editing the curve in these two manners. A “pulse” is a quick and tight drop composed of four points. A “step” consists of leveling the Y value between two points with a sharp transition between the two sections. Nov 28, 2016 · In those programs, any line can be converted into a sketch by clicking any point on the line and dragging it. I would eventually like to import a picture into Fusion, sketch out the outline of the object with basic lines, further refine it by clicking the dragging the points of the lines to move them and make curves, and then extrude it. between documents copying -- draw objects between documents pasting -- draw A symmetrical anchor point has the same line curvature on either side, and two control lines that move together as a straight line. On the Drawing toolbar, open the Curves toolbar and select the Freeform Line tool.Free slope calculator - find the slope of a curved line, step-by-step This website uses cookies to ensure you get the best experience. By using this website, you agree to our Cookie Policy. Well organized and easy to understand Web building tutorials with lots of examples of how to use HTML, CSS, JavaScript, SQL, PHP, Python, Bootstrap, Java and XML. When the two points coincide, the line is tangent to the curve. This limit process is represented in the animation to the right. On a position-time graph… average velocity is the slope of the straight line connecting the endpoints of a curve; instantaneous velocity is the slope of the line tangent to a curve at any point A bezier curve is specified by four points; it passes through two of those points exactly, and the tangent vectors at those two endpoints are determined by the two "control points". The interface in terms of "control points" makes no sense to me - I wrote code to convert endpoints and tangent vectors at endpoints to endpoints and control points. If TRUE, then curves are marked at each censoring time. If mark.time is a numeric vector, then curves are marked at the specified time points. fun: an arbitrary function defining a transformation of the survival curve. For example fun=log is an alternative Another nice property of these curves is that the tangent vector at a point P is parallel to the line joining P's two surrounding points. To do more than two points just step through the array of points using the previous point, the current point and the next two points as the four points for the spline. For each of these segments draw a curve for 0<t<1. This curve will be between the current point and the next point. This method is not particularly fast though. A line has only one dimension and it is called length. A straight line is the shortest distance between two points; A curve line is a line no part of which is straight; Parallel line are those which are equal distance apart through their entire length. Perpendicular lines are those which intersect with an angle of 90 degrees. Graphviz has many useful features for concrete diagrams, such as options for colors, fonts, tabular node layouts, line styles, hyperlinks, and custom shapes. Roadmap. dot - “hierarchical” or layered drawings of directed graphs. This is the default tool to use if edges have directionality. neato - “spring model” layouts. This is the ... May 30, 2008 · The basic concept is to bisect each of the straight lines, then draw a curve between the bisections, using the original points as the curve’s control point. Nothing new I’m sure, but it works really well for me, because I can continue to think in straight lines, but draw in curves. Click to add points. Click and drag points to move them. A design is a closed curve when you begin and end a drawing at the same point, without lifting your pencil, and you cannot get into or out of the shape. A simple closed curve is when the lines of ... See full list on vanseodesign.com When these points are connected, they form a curve. Within the curved lines, we can find open and closed curves. The open curves are the ones that are impossible for us to reach the starting point again if we were to follow the succession of points with a pencil, without lifting it from the paper. This is an example of an open curved line: • ## Baltimore city police academy dates 2020 Mega neon shadow dragon adopt me ebay German 800 silver Evenly distributes object copies along a path or a portion of a path. Find The path can be a line, polyline, 3D polyline, spline, helix, arc, circle, or ellipse. 2D path array 3D path array This command is equivalent to the Path option in ARRAY. Note: The 3D objects used in the example cannot be created in AutoCAD LT. The following prompts are displayed. Select objects Select the objects to ... ## Draco malfoy x reader imagines • Nov 20, 2011 · A line is formed by connecting two data points. 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By using this website, you agree to our Cookie Policy. • This assistant allows you to position and adjust four points to create a cubic bezier curve. You can then draw along the curve, snapping your brush stroke directly to the curve line. Perfect curves every time! If you press the Shift key while holding the first two handles, they will snap to perfectly horizontal or vertical lines. Naira to dollar Sep 03, 2019 · "Bell curve" refers to the bell shape that is created when a line is plotted using the data points for an item that meets the criteria of normal distribution. In a bell curve, the center contains the greatest number of a value and, therefore, it is the highest point on the arc of the line. ## Coordinate plane sandbox Phosphoric acid and lithium hydroxide net ionic equation The dialog box has two groups. Object Type: You can select a line or arc to draw a profile. Selected line or arc box color changes to dark blue. 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Lesson hub ## Can't find the answer? Try online tutoring We have the UK’s best selection of online tutors, when and for how long you need them. Getting 1-on-1 support is cheaper than you might think. ### Participating users Welcome to our free-to-use Q&A hub, where students post questions and get help from other students and tutors. You can ask your own question or look at similar Mathematics questions. Let's solve your equation step-by-step.3x(x+4)+3(x+4)=0Step 1: Simplify both sides of the equation3x2+15x+12=0Step 2: Factor left side of equation.3(x+1)(x+4)=0Step 3: Set factors equal to 0.x+1=0 or x+4=0ans:x=−1 or x=−4 3x(x+4)+3(x+4)=0 (3x+3)(x+4)=0 3(x+1)(x+4)=0 x+1=0 and x+4=0 x=-1 or x=-4 3x(x+4) = 3x^2 +12x 3(x+4) = 3x + 12 So: 3x(x+4) + 3(x+4) = 3x^2 +12x + 3x + 12 ----> just expanded the brackets Therefore: 3x^2 +12x + 3x + 12 = 0 Simplify it to: 3x^2 + 15x + 12 = 0 Factorise to get: (3x + 3) (x+4) = 0 Therefore: (3x + 3) = 0 or (x+4) = 0 If: 3x + 3 = 0 --> 3x = -3 so x= -3/3 = -1 If: x+4 = 0 ---> x = -4 x = -1 or -4 Hope that helped! Firstly take the bracket of (x+4), if x+4=0 then x must be equal to -4. The other numbers make another bracket of (3x+3). If 3x+3=0, then 3x=-3 and so x is equal to -1 3( x+4)(x+1)=0 X= -4, -1 3( x+4)(x+1)=0 X= -4, -1 Browndaisha ! simply remember in math there is rule of DMAS. it says 1. Division,Multlipication,3.Addition and 4.Subtraction. in any equation follow these steps. you will get result. given the equation, we can say that we have (x+4) in common. So, we take that out, (x+4), we will have another bracket, (3x+3). We can say that (x+4)(3x+3)=0 now we can equate, we can say either (x+4)=0 or (3x+3)=0 x=0 -4 or 3x=0-3 x= -4 or x=-3/3 x=-4 or x=-1 expand the brackets 3x^2 + 12x + 3x + 12 = 0 Gather like terms 3x^2 + 15x + 12=0 simplify by dividing the whole equation by 3 x^2 + 5x + 4 = 0 Factorise (two numbers that added give you 5 and multiplied give you 4) (x+1)(x+4) = 0 Thus, the two solutions are x=-1 and x=-4 If you need further help, feel free to contact me for tuition services First, expand the brackets So 3x^2+12x + 3x+12 =0 Then simplify it all out - ax^2+bx+c=0 is the format you want the question in to solve 3x^2 + 15x +12 =0 This is the same as 3(x^2+5x+4)=0 Now you need to find 2 numbers to multiply to make 4 (c in the equation) that also add to make 5 - 1 and 4. Therefore x=-1 and x=-4 Check it in the equation and see if it equals 0, if it does it's right
# 2.2. Symmetric monoidal preorders# ## 2.2.1. Definition and first examples# $\newcommand{\Cat}[1]{\mathbf{#1}}$ Definition 2.2. See Monoidal category - Monoidal preorders and Symmetric monoidal category for an alternative definition. Exercise 2.5. This won’t work, because the monoidal product doesn’t satisfy requirement (a). Consider the example: \begin{split} \begin{align} \\ x_1 & = -2 \\ x_2 & = -2 \\ y_1 & = 1 \\ y_2 & = 1 \\ x_1 \leq y_1 & = true \\ x_2 \leq y_2 & = true \\ x_1 \otimes x_2 & = 4 \not\leq 1 = y_1 \otimes y_2 \\ \end{align} \end{split} Exercise 2.8. The element $$e$$ in the monoid $$(M, \ast, e)$$ serves as $$I$$ in the discrete preorder $$(\Cat{Disc}_M, =, \ast, e)_{}$$. The monoid multiplication $$\ast$$ serves as the monoidal product $$\otimes$$ and satisfies properties (b), (c), and (d) by definition of being a monoid. By virtue of being a discrete preorder, it satisfies (a) because $$x_1$$ will always equal $$y_1$$ and therefore with “=” as the order operator this condition comes down to $$x_1 \otimes x_2 = x_1 \otimes x_2$$. ## 2.2.2. Introducing wiring diagrams# For more background, see String diagram. Exercise 2.20. Using reflexivity, we have that $$u \leq u$$. Add this to the first of the equations in (2.16), using (a) in the definition of a symmetrical monoidal preorder: \begin{split} \begin{align} \\ t & \leq v + w \\ u & \leq u \\ t + u & \leq u + v + w \\ \end{align} \end{split} Similarly for the second equation: \begin{split} \begin{align} \\ w + u & \leq x + z \\ v & \leq v \\ v + w + u & \leq v + x + z \\ \end{align} \end{split} And the third equation: \begin{split} \begin{align} \\ v + x & \leq y \\ z & \leq z \\ v + x + z & \leq y + z \\ \end{align} \end{split} Using transitivity, we can combine all three of these equations to get Equation (2.18) in the text. To answer 3., the symmetry axiom does not need to be invoked because wires do not cross. ## 2.2.3. Applied examples# Exercise 2.21. The condition (a) holds by the conservation of matter. The condition (b) holds because adding nothing to a reactant leaves you with the reactant as the product. The condition (d) holds because it shouldn’t matter what order reactants are added together in a reaction. It’s not clear that (c) would hold in all circumstances; adding two reactants may produce a product that no longer reacts with another reactant that the original reactants may have reacted with. ## 2.2.4. Abstract examples# Exercise 2.29. The monoidal unit must be $$false$$ (satisfying the unitality condition (a)). It’s not clear that the author is presenting these tables to show they are symmetric (as in “symmetric” monoidal preorder), but they show that symmetric condition (d) is satisfied. The associative condition (c) is satisfied because booleans over $$\vee$$ are associative. The monotonicity requirement (a) is satisfied because the booleans over $$\vee$$ are also already monotone, and this is apparent in the tables as well. See also Boolean algebra - Monotone laws and Monotonic function - In Boolean functions. Exercise 2.31. Notice this question only asks for a monoidal structure, not a symmetric monoidal preorder structure. Since it seems to have both, we’ll show both here. The monoidal unit should be 1 (satisfies unitality). The natural numbers under addition satisfy the associativity condition, so this symmetric monoidal preorder will as well. See Natural number - Algebraic properties satisfied by the natural numbers. The natural numbers also satisfy the commutativity condition (symmetry condition in the text). The following is based (roughly) on Propositional calculus - Natural deduction system and Propositional calculus - Proofs in propositional calculus. To see the monotonicity condition is satisfied we’ll use the normal rules of algebra on the natural (positive) numbers, with these two premises: \begin{split} \begin{align} \\ x_1 & \leq y_1 \\ x_2 & \leq y_2 \\ \end{align} \end{split} We have, multiplying both sides of the first premise by $$y_2$$: $\begin{split} $$\\ x_1 \ast y_2 \leq y_1 \ast y_2 \\ \label{eq:31a} \tag{a}$$ \end{split}$ Multiplying both sides of the second premise by $$x_1$$: $\begin{split} $$\\ x_1 \ast x_2 \leq x_1 \ast y_2 \\ \label{eq:31b} \tag{b}$$ \end{split}$ And combining $$\eqref{eq:31a}$$ and $$\eqref{eq:31b}$$ with transitivity: $x_1 \ast x_2 \leq x_1 \ast y_2 \leq y_1 \ast y_2$ Exercise 2.33. No, because $$0 \| n = \infty \neq 0$$. Exercise 2.34. In the following table we use 0 for no, 1 for maybe, and 2 for yes: import pandas as pd preorder = [0, 1, 2] min = pd.DataFrame(index=preorder, columns=preorder, data=[[0,0,0], [0,1,1], [0,1,2]]) display(min) 0 1 2 0 0 0 0 1 0 1 1 2 0 1 2 The “min” function is associative; see Associative property - Examples. The “yes” element can be used as a monoidal unit. It’s easy to see the symmetry/commutativity property is satisfied in the truth table. To see the monotonicity condition is satisfied we’ll use as an inference rule that a partially-applied “min” function is order-preserving. That is, if we have that $$a_1 \leq a_2$$, it follows that $$min(a_1, b) \leq min(a_2, b)$$. Then, starting from these two premises: \begin{split} \begin{align} \\ x_1 & \leq y_1 \\ x_2 & \leq y_2 \\ \end{align} \end{split} Applying $$min(\_, y_2)$$ to the first premise: $\begin{split} $$\\ min(x_1, y_2) \leq min(y_1, y_2) \\ \label{eq:34a} \tag{a}$$ \end{split}$ Applying $$min(x_1, \_)$$ to the second premise: $\begin{split} $$\\ min(x_1, x_2) \leq min(x_1, y_2) \\ \label{eq:34b} \tag{b}$$ \end{split}$ And combining $$\eqref{eq:34a}$$ and $$\eqref{eq:34b}$$ with transitivity: $min(x_1, x_2) \leq min(x_1, y_2) \leq min(y_1, y_2)$ Exercise 2.35. It should satisfy the identity/unitality condition because $$S \cap X$$ for any $$X$$ that is a subset of $$S$$ should be $$X$$. See also Intersection (set theory); the intersection operation is associative, and commutative. To see the monotonicity condition is satisfied we’ll use as an inference rule that a partially-applied $$\cap$$ is order-preserving. That is, if we have that $$a_1 \leq a_2$$ (that is, $$a_1 \subset a_2$$), it follows that $$a_1 \cap b \leq a_2 \cap b$$. Then, starting from these two premises: \begin{split} \begin{align} \\ x_1 & \leq y_1 \\ x_2 & \leq y_2 \\ \end{align} \end{split} Applying $$\_ \cap y_2$$ to the first premise: $\begin{split} $$\\ x_1 \cap y_2 \leq y_1 \cap y_2 \\ \label{eq:35a} \tag{a}$$ \end{split}$ Applying $$x_1 \cap \_$$ to the second premise: $\begin{split} $$\\ x_1 \cap x_2 \leq x_1 \cap y_2 \\ \label{eq:35b} \tag{b}$$ \end{split}$ And combining $$\eqref{eq:35a}$$ and $$\eqref{eq:35b}$$ with transitivity: $x_1 \cap x_2 \leq x_1 \cap y_2 \leq y_1 \cap y_2$ Exercise 2.36. $$\def\NN{{\bf N}}\def\bold#1{{\bf #1}}$$A list of example statements (including those that were already made) that may be made in this language: • $$R$$ = “n is prime” • $$R$$ = “n is even” • $$S$$ = $$0 \leq n$$ • $$T$$ = $$n = 15$$ Clearly $$S \leq P$$ for all $$P$$ in $$n \in \NN$$, but $$T$$ is only true for a single $$P$$. If we use “and” as our monoidal product, we must pick something that is always true (like $$S$$) as our monoidal unit. If we had chosen “or” then we’d have to come up with a statement that is always false. An “and” operation is clearly already associative and commutative. It’s also order-preserving, which we can use in the following proof (in a style suggested by Example 1.123): Exercise 2.39. The $$I$$ remains the identity because in the original symmetric monoidal preorder we had $$I \otimes x = x$$ and $$x \otimes I = x$$ and nothing about $$\otimes$$ has changed. Similarly, if $$\otimes$$ was associative and commutative it should remain so. Exercise 2.40. As a preorder $$\bold{Cost}^{op}$$ is $$([0, ∞], \leq)$$. In this preorder smaller numbers are “better” (greater) as in golf. The monoidal unit remains 0, and the monoidal product remains +. ## 2.2.5. Monoidal monotone maps# Exercise 2.43. To show this is indeed a monotone map, we must show $$x ≤_B y$$ implies $$f(x) ≥_{Cost} f(y)$$ for all $$x,y \in B$$. There are only four cases to check: \begin{split} \begin{align} \\ false \leq false & \to \infty \geq \infty \\ false \leq true & \to \infty \geq 0 \\ true \leq false & \to 0 \geq \infty \\ true \leq true & \to 0 \geq 0 \\ \end{align} \end{split} Checking condition (a) of Definition 2.41: $0 \leq g(true) = 0$ Checking condition (b) of Definition 2.41: \begin{split} \begin{align} \\ \infty + \infty \leq g(false) = \infty \\ \infty + 0 \leq g(false) = \infty \\ 0 + \infty \leq g(false) = \infty \\ 0 + 0 \leq g(true) = 0 \\ \end{align} \end{split} Yes, $$g$$ is a strict monoidal monotone. Exercise 2.44. To show $$d$$ is a monotone map, we must show $$x ≥_{Cost} y$$ implies $$d(x) ≤_B d(y)$$ for all $$x,y \in B$$. There are only nine cases to check, if we treat all numbers between 0 and $$\infty$$ as one of three options for each variable: \begin{split} \begin{align} \\ 0 \geq 0 & \to T \leq T \\ 0 \geq 7 & \to T \leq F \\ 0 \geq ∞ & \to T \leq F \\ 7 \geq 0 & \to F \leq T \\ 7 \geq 7 & \to F \leq F \\ 7 \geq ∞ & \to F \leq F \\ ∞ \geq 0 & \to F \leq T \\ ∞ \geq 7 & \to F \leq F \\ ∞ \geq ∞ & \to F \leq F \\ \end{align} \end{split} Checking condition (a) of Definition 2.41: $T \leq d(0) = T$ Checking condition (b) of Definition 2.41: \begin{split} \begin{align} \\ d(0) \wedge d(0) = T & \to d(0 + 0) = T \\ d(0) \wedge d(7) = F & \to d(0 + 7) = F \\ d(0) \wedge d(∞) = F & \to d(0 + ∞) = F \\ d(7) \wedge d(0) = F & \to d(7 + 0) = F \\ d(7) \wedge d(7) = F & \to d(7 + 7) = F \\ d(7) \wedge d(∞) = F & \to d(7 + ∞) = F \\ d(∞) \wedge d(0) = F & \to d(∞ + 0) = F \\ d(∞) \wedge d(7) = F & \to d(∞ + 7) = F \\ d(∞) \wedge d(∞) = F & \to d(∞ + ∞) = F \\ \end{align} \end{split} Yes, $$d$$ is a strict monoidal monotone. To show $$u$$ is a monotone map, we must show $$x ≥_{Cost} y$$ implies $$u(x) ≤_B u(y)$$ for all $$x,y \in B$$. There are only nine cases to check, if we treat all numbers between 0 and $$\infty$$ as one of three options for each variable: \begin{split} \begin{align} \\ 0 \geq 0 & \to T \leq T \\ 0 \geq 7 & \to T \leq T \\ 0 \geq ∞ & \to T \leq F \\ 7 \geq 0 & \to T \leq T \\ 7 \geq 7 & \to T \leq T \\ 7 \geq ∞ & \to T \leq T \\ ∞ \geq 0 & \to F \leq T \\ ∞ \geq 7 & \to F \leq T \\ ∞ \geq ∞ & \to F \leq F \\ \end{align} \end{split} Checking condition (a) of Definition 2.41: $T \leq u(0) = T$ Checking condition (b) of Definition 2.41: \begin{split} \begin{align} \\ u(0) \wedge u(0) = T & \to u(0 + 0) = T \\ u(0) \wedge u(7) = T & \to u(0 + 7) = T \\ u(0) \wedge u(∞) = F & \to u(0 + ∞) = F \\ u(7) \wedge u(0) = T & \to u(7 + 0) = T \\ u(7) \wedge u(7) = T & \to u(7 + 7) = T \\ u(7) \wedge u(∞) = F & \to u(7 + ∞) = F \\ u(∞) \wedge u(0) = F & \to u(∞ + 0) = F \\ u(∞) \wedge u(7) = F & \to u(∞ + 7) = F \\ u(∞) \wedge u(∞) = F & \to u(∞ + ∞) = F \\ \end{align} \end{split} Yes, $$u$$ is a strict monoidal monotone. Exercise 2.45. To answer 1., see the same question in Exercise 2.31. To answer 2., we simply need to guess and check different kinds of functions for the monoidal monotone. An answer that works is $$f(p) := n^p$$ where $$n$$ is any integer (e.g. 2). To answer 3., see nearly the same question in Exercise 2.5.
B Hajja, Mowaffaq, Extremal properties of the incentre and the excenters of a triangle", Book IV, Proposition 4: To inscribe a circle in a given triangle, "The distance from the incenter to the Euler line", http://forumgeom.fau.edu/FG2012volume12/FG201217index.html, http://forumgeom.fau.edu/FG2014volume14/FG201405index.html, http://forumgeom.fau.edu/FG2011volume11/FG201102index.html, https://en.wikipedia.org/w/index.php?title=Incenter&oldid=989898020, All Wikipedia articles written in American English, Creative Commons Attribution-ShareAlike License, This page was last edited on 21 November 2020, at 17:29. The distance from the vertex to the incenter is equal to the length of the angle bisector multiplied by the sum of the lengths of the sides forming this vertex divided by the sum of the lengths of all three sides: In an isosceles triangle, all of centroid, orthocentre, incentre and circumcentre lie on the same line. From the given figure, three medians of a triangle meet at a centroid “G”. meet at , and {\displaystyle A} A Incenter of a triangle - formula A point where the internal angle bisectors of a triangle intersect is called the incenter of the triangle. a = BC = √ [ (0+3)2 + (1-1)2] = √9 = 3. b = AC = √ [ (3+3)2 + (1-1)2] = √36 = 6. c = AB = √ [ (3-0)2 + (1-1)2] = √9 = 3. The incenter is one of the triangle's points of concurrency formed by the intersection of the triangle's 3 angle bisectors.. {\displaystyle (x_{B},y_{B})} a This provides a way of finding the incenter of a triangle using a ruler with a square end: First find two of these tangent points based on the length of the sides of the triangle, then draw lines perpendicular to the sides of the triangle. Coordinates of the three vertices: $$A(x_1, y_1)$$, $$B(x_2, y_2)$$, and $$C(x_3, y_3)$$ Method C Let the side AB = a, BC = b, AC = c then the coordinates of the in-center is given by the formula: {\displaystyle {\overline {AC}}:{\overline {BC}}={\overline {AF}}:{\overline {BF}}} D Well, there is no specific circumcenter formula to find it. ¯ F {\displaystyle \angle {ACB}} Incenter Incenter is the center of the inscribed circle (incircle) of the triangle, it is the point of intersection of the angle bisectors of the triangle. In geometry, the incenter of a triangle is a triangle center, a point defined for any triangle in a way that is independent of the triangle's placement or scale. The three medians of a triangle meet in the centroid. B △ b In a right angled triangle, orthocentre is the point where right angle is formed. ¯ {\displaystyle {\overrightarrow {CI}}} B . C {\displaystyle \angle {BAC}} The incentre of a triangle is the point of concurrency of the angle bisectors of angles of the triangle. The radii of the incircles and excircles are closely related to the area of the triangle. y where {\displaystyle {\overline {BC}}:{\overline {BF}}={\overline {CI}}:{\overline {IF}}} The incenter is the center of the circle inscribed in the triangle. F This geometry video tutorial explains how to identify the location of the incenter, circumcenter, orthocenter and centroid of a triangle. In ( Every triangle has three distinct excircles, each tangent to one of the triangle's sides. The incenter is the center of the Adams' circle, Conway circle, and incircle. c The orthocenter is the intersecting point for all the altitudes of the triangle. {\displaystyle b} : B A , Geometrically, a triangle’s incenter can be located by drawing any two of its three angle bisectors and finding where they intersect, which is called the point of concurrency . well you need coordinates for the points. I-- we'll see in about five seconds-- is the center of a circle that can be put inside the triangle that's tangent to the three sides. : The incenter may be equivalently defined as the point where the internal angle bisectors of the triangle cross, as the point equidistant from the triangle's sides, as the junction point of the medial axis and innermost point of the grassfire transform of the triangle, and as the center point of the inscribed circle of the triangle. Together with the centroid, circumcenter, and orthocenter, it is one of the four triangle centers known to the ancient Greeks, and the only one that does not in general lie on the Euler line. → B Adjust the triangle above by dragging any vertex and see that it will never go outside the triangle , In Euclid's Elements, Proposition 4 of Book IV proves that this point is also the center of the inscribed circle of the triangle. Find the measure of the third angle of triangle CEN and then cut the angle in half: 4 B This math recipe will help you find the incenter of a triangle, coordinates of whose vertices are known. , C Let the side AB = a, BC = b, AC = c then the coordinates of the in-center is given by the formula: The radius (or inradius ) of the inscribed circle can be found by using the formula: Summary You can't make a circle hitting all five points. {\displaystyle \angle {ABC}} Adjust the triangle above by dragging any vertex and see that it will never go outside the triangle Wondering how to calculate circumcenter without using circumcenter formula calculator? D {\displaystyle c} The incenter (I) of a triangle is the center of its inscribed circle (also called, incircle). This location gives the incenter an interesting property: The incenter is equally far away from the triangle’s three sides. So When the vertices of a triangle are combined with its orthocenter, any one of the points is the orthocenter of the other three, as … In the case of quadrilaterals, an incircle exists if and only if the sum of the lengths of opposite sides are equal: Both pairs of opposite sides sum to. Let’s observe the same in the applet below. Let a be the length of BC, b the length of AC, and c the length of AB. When one exists, the polygon is called tangential. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. And let A circle is inscribed in the triangle if the triangle's three sides are all tangents to a circle. • B A centroid is also known as the centre of gravity. The point where the altitudes of a triangle meet is known as the Orthocenter. : {\displaystyle {F}} {\displaystyle {\overline {AD}}} ∠ for the incenter are given by[2], The collection of triangle centers may be given the structure of a group under coordinatewise multiplication of trilinear coordinates; in this group, the incenter forms the identity element. . This geometry video tutorial explains how to identify the location of the incenter, circumcenter, orthocenter and centroid of a triangle. What is Incenter formula? {\displaystyle {I}} You find a triangle’s incenter at the intersection of the triangle’s three angle bisectors. Distance between the Incenter and the Centroid of a Triangle. Therefore $\triangle IAB$ has base length c and height r, and so has ar… Dragutin Svrtan and Darko Veljan, "Non-Euclidean versions of some classical triangle inequalities", Marie-Nicole Gras, "Distances between the circumcenter of the extouch triangle and the classical centers". Then X = I (the incenter) maximizes or minimizes the ratio Incenters, like centroids, are always inside their triangles.The above figure shows two triangles with their incenters and inscribed circles, or incircles (circles drawn inside the triangles so the circles barely touc… = I Orthocenter Formula - Learn how to calculate the orthocenter of a triangle by using orthocenter formula prepared by expert teachers at Vedantu.com. The method to find circumcenter of triangle is given below. The point that is equidistant to all sides of a triangle is called the incenter: A median is a line segment that has one of its endpoints in the vertex of a triangle and the other endpoint in the midpoint of the side opposite the vertex. = The distance between the incenter and circumcenter is, where is the circumradius and is the inradius, a result known as the Euler triangle formula. A Performance & security by Cloudflare, Please complete the security check to access. The incenter is the center of the incircle. You may need to download version 2.0 now from the Chrome Web Store. Line of Euler B C In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. ¯ y This page will define the following: incenter, circumcenter, orthocenter, centroid, and Euler line. and The center of the incircle is a triangle center called the triangle's incenter. , and A I'm trying to figure out how to find the incenter of a triangle with (x, y, z) coordinates for the verteces. ¯ And you're going to see in a second why it's called the incenter. {\displaystyle E} B y Proof of Existence. It is a theorem in Euclidean geometry that the three interior angle bisectors of a triangle meet in a single point. Circumcenter Formula. The intersection point will be the incenter. . The radius of the incircle is the length of DH, FH, and EH. A I B ‹ Derivation of Formula for Radius of Circumcircle up Derivation of Heron's / Hero's Formula for Area of Triangle › Log in or register to post comments 54292 reads B Any other point within the orthocentroidal disk is the incenter of a unique triangle.[15]. ¯ c ¯ C C {\displaystyle a} = The trilinear coordinates for a point in the triangle give the ratio of distances to the triangle sides. Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The incenter lies on the Nagel line and Soddy line, and lies on the Euler line only for an isosceles triangle. The point of concurrency is known as the centroid of a triangle. For polygons with more than three sides, the incenter only exists for tangential polygons—those that have an incircle that is tangent to each side of the polygon. C A ∠ When the vertices of a triangle are combined with its orthocenter, any one of the points is the orthocenter of the other three, as … A quadrilateral that does have an incircle is called a Tangential Quadrilateral. https://www.khanacademy.org/.../v/incenter-and-incircles-of-a-triangle The center of the incircle is called the triangle's incenter. C Therefore, Definition. For a triangle, the center of the incircle is … B In this situation, the circle is called an inscribed circle, and its center is called the inner center, or incenter. In the case above, where I and H are along BO, that means I, B, H, and O are on the same line segment, with C off elsewhere. of the Incenter of a Triangle. The formula for the radius. Steps to construct the circumcenter of a triangle: Step 1: Draw the perpendicular bisectors of all the sides of the triangle using a compass.. ¯ , and A Definition: For a two-dimensional shape “triangle,” the centroid is obtained by the intersection of its medians. are the angles at the three vertices. Another way to prevent getting this page in the future is to use Privacy Pass. {\displaystyle b} The formula of the distance from the vertex to the incenter in terms of the sides and the angle bisector The incenter is the point where the angle bisectors intersect.. : C {\displaystyle {\overline {CF}}} meet at B The radius of incircle is given by the formula r=At/s where At = area of the triangle and s = ½ (a + b + c). Suppose the vertices of the triangle are A(x1, y1), B(x2, y2) and C(x3, y3). A C x A The incenter is one of the triangle's points of concurrency formed by the intersection of the triangle's 3 angle bisectors.. Incenter I, of the triangle is given by. ( • Since the triangle's three sides are all tangents to the inscribed circle, the distances from the circle's center to the three sides are all equal to the circle's radius. Approx. and The area of any triangle is where is the Semiperimeter of the triangle. {\displaystyle \triangle {ACF}} {\displaystyle {\overline {BE}}} , and Always inside the triangle: The triangle's incenter is always inside the triangle. C Time. E The incenter is the center of the triangle's incircle, the largest circle that will fit inside the triangle and touch all three sides. Arie Bialostocki and Dora Bialostocki, "The incenter and an excenter as solutions to an extremal problem". F {\displaystyle {\overline {AC}}} An excircle or escribed circle of the triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two. {\displaystyle (x_{A},y_{A})} . A bisector divides an angle into two congruent angles.. Find the measure of the third angle of triangle CEN and then cut the angle in half:. {\displaystyle a} Use the calculator above to calculate coordinates of the incenter of the triangle ABC.Enter the x,y coordinates of each vertex, in any order. There is no direct formula to calculate the orthocenter of the triangle… I {\displaystyle D} If the coordinates of all the vertices of a triangle are given, then the coordinates of incircle are given by, (a + b + c a x 1 + b x 2 + c x 3 , a + b + c a y 1 + b y 2 + c y 3 ) where ) : △ As in a triangle, the incenter (if it exists) is the intersection of the polygon's angle bisectors. and The radius of incircle is given by the formula r=At/s where At = area of the triangle and s = ½ (a + b + c). Please enable Cookies and reload the page. ¯ The formula of the distance from the vertex to the incenter in terms of the sides and the angle bisector The incenter is the point where the angle bisectors intersect.. A circle is inscribed in the triangle if the triangle's three sides are all tangents to a circle. The angle bisectors of a triangle are each one of the lines that divide an angle into two equal angles. Every triangle and regular polygon has a unique incircle, but in general polygons with 4 or more sides (such as non- square rectangles) do not have an incircle. C If the triangle is acute, the orthocenter is in the interior of the triangle.In a right triangle, the orthocenter is the polygon vertex of the right angle.. Every triangle has an incenter and an incircle. Cloudflare Ray ID: 617201378e7fdff3 There are either one, two, or three of these lines for any given triangle. {\displaystyle \angle {ACB}} . A Incenter - The incenter of a triangle is located where all three angle bisectors intersect. The incenter of a triangle is the point of intersection of all the three interior angle bisectors of the triangle. One method for computing medial axes is using the grassfire transform, in which one forms a continuous sequence of offset curves, each at some fixed distance from the polygon; the medial axis is traced out by the vertices of these curves. F ¯ The center of the incircle is called the triangle's incenter.. An excircle or escribed circle of the triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two. Then we have to prove that [20][21], Relative distances from an angle bisector. , and , The incenter is the center of the triangle's incircle, the largest circle that will fit inside the triangle and touch all three sides. This will occur inside acute triangles, outside obtuse triangles, and for right triangles, it will occur at the midpoint of the hypotenuse. All three medians meet at a single point (concurrent). X The incenter of a triangle is the point where the bisectors of each angle of the triangle intersect. if you keep repeating that with the mid points being turned into corners of the progressively smaller triangles you have in effect the center of a triangle. [4], The medial axis of a polygon is the set of points whose nearest neighbor on the polygon is not unique: these points are equidistant from two or more sides of the polygon. B Circumcenter Geometry. In the below mentioned diagram orthocenter is denoted by the letter ‘O’. F It is the only point equally distant from the line segments, but there are three more points equally distant from the lines, the excenters, which form the centers of the excircles of the given triangle. ¯ I The incenter of a triangle can also be explained as the center of the circle which is inscribed in a triangle $$\text{ABC}$$. Step 2: Extend all the perpendicular bisectors to meet at a point.Mark the intersection point as $$\text O$$, this is the circumcenter. The distance between the incenter and circumcenter is , where is the circumradius and is the inradius, a result known as the Euler triangle formula. To download free study materials like NCERT Solutions, Revision Notes, Sample Papers and Board … Definition. x {\displaystyle c} A {\displaystyle C} , so that An angle bisector is the ray that divides any angle into two congruent smaller angles. The incenter is the point of intersection of the three angle bisectors. ¯ A {\displaystyle {\overline {AC}}:{\overline {AF}}={\overline {CI}}:{\overline {IF}}} meet at When a circle is inscribed in a triangle such that the circle touches each side of the triangle, the center of the circle is called the incenter of the triangle. The centre of the circle that touches the sides of a triangle is called its incenter. [9], By Euler's theorem in geometry, the squared distance from the incenter I to the circumcenter O is given by[10][11], where R and r are the circumradius and the inradius respectively; thus the circumradius is at least twice the inradius, with equality only in the equilateral case.[12]:p. I Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. Circumcenter - The circumcenter is located at the intersection of the perpendicular bisectors of all sides. . B C How to Find the Incenter of a Triangle on the XY Plane. Geometry Problem 1492. , ¯ ¯ Suppose the vertices of the triangle are A(x1, y1), B(x2, y2) and C(x3, y3). A The incenter lies on the Nagel line and Soddy line, and lies on the Euler line only for an isosceles triangle. ) {\displaystyle {\overline {CI}}} F The incenter of a triangle is the intersection of its (interior) angle bisectors. ¯ F A triangle (black) with incircle (blue), incenter (I), excircles (orange), excenters (J A,J B,J C), internal angle bisectors (red) and external angle bisectors (green) In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. b {\displaystyle {\tfrac {BX}{CX}}} ) is the bisection of One can derive the formula as below. Thus the radius C'Iis an altitude of $\triangle IAB$. Inradius respectively Solutions to an extremal problem '' and is equally distant from all sides right,! Longest median of the circle is inscribed in the future is to use Privacy Pass in this situation, circle! As in a triangle center called the inner center, or three these... 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mathleaks.com mathleaks.com Start chapters home Start History history History expand_more Community Community expand_more {{ filterOption.label }} {{ item.displayTitle }} {{ item.subject.displayTitle }} arrow_forward {{ searchError }} search {{ courseTrack.displayTitle }} {{ printedBook.courseTrack.name }} {{ printedBook.name }} It may not be understood at first glance, but all squares, rectangles, and rhombi are parallelograms. Therefore, all the properties of parallelograms apply to these quadrilaterals as well! In this lesson, theorems about parallelograms will be discussed to understand this concept better. ### Catch-Up and Review Here are a few recommended readings before getting started with this lesson. ## Investigating Properties of Parallelograms In the applet below, a parallelogram, a rectangle, a rhombus, and a square can be selected and rotated clockwise about the point of intersection of its diagonals. What can be noted when these polygons are rotated Note that all the quadrilaterals in the applet are parallelograms. Additionally, when rotated each quadrilateral is mapped onto itself. Considering this information, what conclusions can be made about the opposite sides of a parallelogram? ## Properties of a Parallelogram's Opposite Sides A conclusion that can be made from the previous exploration is that the opposite sides of a parallelogram are congruent. This is explained in detail in the following theorem. ## Parallelogram Opposite Sides Theorem The opposite sides of a parallelogram are congruent. In respects to the characteristics of the diagram, the following statement holds true. ### Proof Using Congruent Triangles This theorem can also be proven by using congruent triangles. Consider the parallelogram PQRS and its diagonal PR. It can be noted that two triangles are formed with PR as a common side. By the definition of a parallelogram, PQ and SR are parallel. Therefore, by the Alternate Interior Angles Theorem, it can be stated that QPRSRP and that QRPSPR. Furthermore, by the Reflexive Property of Congruence, PR is congruent to itself. Consequently, PQR and RSP have two pairs of congruent angles and an included congruent side. Therefore, by the Angle-Side-Angle Congruence Theorem, PQR and RSP are congruent triangles. Since corresponding parts of congruent figures are congruent, PS is congruent to QR and PQ is congruent to RS. Furthermore, it can be stated whether a quadrilateral is a parallelogram just by checking if its opposite sides are congruent. ## Converse Parallelogram Opposite Sides Theorem If the opposite sides of a quadrilateral are congruent, then the polygon is a parallelogram. Following the above diagram, the statement below holds true. If PQSR and QRPS, then PQRS is a parallelogram. ### Proof This theorem can be proven by using congruent triangles. Consider the quadrilateral PQRS, whose opposite sides are congruent, and its diagonal PR. By the Reflexive Property of Congruence, this diagonal is congruent to itself. Therefore, by the Side-Side-Side Congruence Theorem, PQR and RSP are congruent triangles. Since corresponding parts of congruent figures are congruent, corresponding angles of PQR and RSP are congruent. Finally, by the Converse of the Alternate Interior Angles Theorem, PQ is parallel to RS and QR is parallel to SP. Therefore, by the definition of a parallelogram, PQRS is a parallelogram. This proves the theorem. If PQSR and QRPS, then PQRS is a parallelogram. Another conclusion that can be made from the exploration is that the opposite angles of a parallelogram are congruent. This is explained in detail in the following theorem. ## Parallelogram Opposite Angles Theorem In a parallelogram, the opposite angles are congruent. For the parallelogram PQRS, the following statement holds true. ### Proof This theorem can be proved by using congruent triangles. Consider the parallelogram PQRS and its diagonal PR. Opposite sides of a parallelogram are parallel. Therefore, by the Alternate Interior Angles Theorem it can be stated that QPRSRP and QRPSPR. Furthermore, by the Reflexive Property of Congruence, PR is congruent to itself. Two angles of PQR and their included side are congruent to two angles of RSP and their included side. By the Angle-Side-Angle Congruence Theorem, PQR and RSP are congruent triangles. Since corresponding parts of congruent figures are congruent, Q and S are congruent angles. By drawing the diagonal QS and using a similar procedure, it can be shown that P and R are also congruent angles. Furthermore, it can be determined whether a quadrilateral is a parallelogram just by looking at its opposite angles. ## Converse Parallelogram Opposite Angles Theorem If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram. Based on the above diagram, the following statement holds true. If AC and BD, then ABCD is a parallelogram. ### Proof Assume that ABCD is a quadrilateral with opposite congruent angles. It should be noted that congruent angles have the same measure. Then, let be the measure of A and C, and be the measure of B and D. By the Polygon Interior Angles Theorem, the sum of the interior angles of a quadrilateral is With this information, a relation can be found between the consecutive interior angles of ABCD. x+y+x+y=360 Simplify 2x+2y=360 2(x+y)=360 x+y=180 Since x+y=180, the consecutive interior angles of ABCD are supplementary. Therefore, by the Converse Consecutive Interior Angles Theorem, it can be concluded that opposite sides of ABCD are parallel. Consequently, by the definition of a parallelogram, ABCD is a parallelogram. Proof: ## Solving Problems Using Properties of a Parallelogram To be able to be carefree and enjoy a soccer match over the weekend, Vincenzo wants to complete his Geometry homework immediately after school. He is given a diagram showing a parallelogram, and asked to find the values of a, b, and x. Find the values of a, b, and x to help Vincenzo be carefree for the match! ### Hint In a parallelogram, opposite sides are congruent and opposite angles are congruent. ### Solution First, for simplicity, the value of x will be found. After that, the values of a and b will be calculated. ### Value of x According to the Parallelogram Opposite Sides Theorem, the opposite sides of a parallelogram are congruent. Therefore, the opposite sides have the same length. With this information, an equation in terms of x can be expressed. This equation can now be solved for x. 2x+5=5x10 Solve for x -3x+5=-10 -3x=-15 x=5 ### Values of a and b According to the Parallelogram Opposite Angles Theorem, the opposite angles of a parallelogram are congruent. That means the opposite angles have the same measure. Knowing this, a system of equations can be expressed. This system will be solved by using the Substitution Method. For simplicity, the degree symbol will be removed. Solve by substitution ## Investigating the Diagonals of a Parallelogram In the following applet, a parallelogram is shown. By dragging one of its vertices, different types of parallelograms such as squares, rectangles, and rhombi can be formed. By using the measuring tool provided, investigate what relationships exist between the diagonals of each parallelogram. After investigating each parallelogram type, what relationships between the diagonals of the parallelograms were discovered? ## Properties of a Parallelogram's Diagonals A conclusion that can be made from the previous exploration is that the diagonals of a parallelogram intersect at their midpoint. This is explained in detail in the following theorem. ### Parallelogram Diagonals Theorem In a parallelogram, the diagonals bisect each other. If PQRS is a parallelogram, then the following statement holds true. ### Proof Using Congruent Triangles This theorem can be proven by using congruent triangles. Consider the parallelogram PQRS and its diagonals PR and QS. Let M be the point intersection of the diagonals. Since PQ and SR are parallel, by the Alternate Interior Angles Theorem it can be stated that QPRSRP and that PQSRSQ. Furthermore, by the Parallelogram Opposite Sides Theorem it can be said that PQSR. Here, two angles of PMQ and their included side are congruent to two angles of RMS and their included side. Therefore, by the Angle-Side-Angle Congruence Theorem PMQ and RMS are congruent triangles. Since corresponding parts of congruent triangles are congruent, PM is congruent to RM and QM is congruent to SM. By the definition of a segment bisector, both segments PR and QS are bisected at point M. Therefore, it has been proven that the diagonals of a parallelogram bisect each other. Also, a quadrilateral can be identified as a parallelogram just by looking at its diagonals. ## Converse Parallelogram Diagonals Theorem If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. Based on the diagram above, the following relation holds true. If AC and BD bisect each other, then ABCD is a parallelogram. ### Proof Let E be point of intersection of the diagonals of a quadrilateral. Since the diagonals bisect each other, E is the midpoint of each diagonal. Because AEB and CED are vertical angles, they are congruent by the Vertical Angles Theorem. Therefore, by the Side-Angle-Side Congruence Theorem, AEB and CED are congruent triangles. Since corresponding parts of congruent figures are congruent, AB and CD are congruent. Applying a similar reasoning, it can be concluded that AED and CEB are congruent triangles. Consequently, AD and BC are also congruent. Finally, since both pairs of opposite sides of quadrilateral ABCD are congruent, the Converse Parallelogram Opposite Sides Theorem states that ABCD is a parallelogram. ## Solving Problems Using Properties of a Parallelogram's Diagonals Vincenzo has one last exercise to finish before going to a soccer match. He has been given a diagram showing a parallelogram. He is asked to find the value of x and y. Find the values of x and y and help Vincenzo finish his homework! ### Hint The diagonals of a parallelogram bisect each other. ### Solution According to the Parallelogram Diagonals Theorem, the diagonals of a parallelogram bisect each other. With this information, two equations can be written. These can be solved one at a time. Equation (I) will be solved first. 9=2x+1 Solve for x 8=2x 4=x x=4 The value of x is 4. Finally, Equation (II) will be solved. y+4=3y Solve for y 4=2y 2=y y=2 The value of y is 2. ## Investigating Rotations of Parallelograms Consider a rigid motion that rotates a parallelogram about the point of intersection of its diagonals. Since a rotation is a rigid motion, the preimage and the image are congruent figures. Furthermore, because corresponding parts of congruent figures are congruent, the three statements below hold true. • Opposite sides of a parallelogram are congruent. • Opposite angles of a parallelogram are congruent. • The diagonals of a parallelogram bisect each other. Note that the Parallelogram Opposite Sides Theorem, the Parallelogram Opposite Angles Theorem, and the Parallelogram Diagonals Theorem have been proved using a rotation about the point of intersection of the diagonals. ## Diagonals of a Rectangle It can be determined whether a parallelogram is a rectangle just by looking at its diagonals. Furthermore, if a parallelogram is a rectangle, a statement about its diagonals can be made. ## Rectangle Diagonals Theorem A parallelogram is a rectangle if and only if its diagonals are congruent. Based on the diagram, the following relation holds true. PQRS is a rectangle PRQS Two proofs will be provided for this theorem. Each proof will consist of two parts. • Part I: If PQRS is a rectangle, then PRQS. • Part II: If PRQS, then PQRS is a rectangle. ### Proof Using Similar Triangles This proof will use similar triangles to prove the theorem. ### Part I: PQRS Is a Rectangle ⇒PR≅QS Suppose PQRS is a rectangle and PR and QS are its diagonals. By the Parallelogram Opposite Sides Theorem, the opposite sides of a parallelogram are congruent. Therefore, RS and QP are congruent. Additionally, by the Reflexive Property of Congruence, SP, or PS, is congruent to itself. Since the angles of a rectangle are right angles, by the definition of congruent angles, RSPQPS. Consequently, RSP and QPS have two pairs of congruent sides and congruent included angles. Therefore, by the Side-Angle-Side Congruence Theorem, the triangles are congruent. Because corresponding parts of congruent triangles are congruent, PR and QS, which are the diagonals of PQRS, are congruent. ### Part II: PR≅QS⇒PQRS Is a Rectangle Consider the parallelogram PQRS and its diagonals PR and QS such that PRQS. By the Parallelogram Opposite Sides Theorem, PQSR. Additionally, by the Reflexive Property of Congruence, PS is congruent to itself. The sides of QPS are congruent to the sides of RSP. Therefore, by the Side-Side-Side Congruence Theorem, QPSRSP. Moreover, since corresponding parts of congruent triangles are congruent, QPS is congruent to RSP. Note that QPS and RSP are consecutive angles. By the Parallelogram Consecutive Angles Theorem, these angles are supplementary. With this information, it can be concluded that both QPS and RSP are right angles. Additionally, by the Parallelogram Opposite Angles Theorem, QPSSRQ and RSPPQR. Because all of the angles are right angles, PQRS is a rectangle. ### Proof Using Transformations This proof will use transformations to prove the theorem. ### Part I: PQRS Is a Rectangle ⇒PR≅QS Consider the rectangle PQRS and its diagonals PR and QS. Let M be the point of intersection of the diagonals. Let A and B be the midpoints of PS and RQ. Then, a line through M and the midpoints A and B can be drawn. Note that QA, AR, PB, and BS are congruent segments. Because congruent segments have the same length, the distance between Q and A equals the distance between R and A. Therefore, Q is the image of R after a reflection across Similarly, P is the image of S after the same reflection. Since M lies on a reflection across maps M onto itself. Reflection Across Preimage Image R Q S P M M The table shows that the images of the vertices of RSM are the vertices of QPM. Therefore, QPM is the image of RSM after a reflection across Since a reflection is a rigid motion, this proves that the triangles are congruent. Because corresponding parts of congruent figures are congruent, QMRM and PMSM. Additionally, by the Parallelogram Diagonals Theorem, the diagonals of the rectangle bisect each other. Therefore, all four segments are congruent. Each diagonal of the parallelogram consists of the same two congruent segments. By the Segment Addition Postulate, the diagonals are congruent. ### Part II: PR≅QS⇒PQRS Is a Rectangle Consider the parallelogram PQRS and its diagonals PR and QS such that PRQS. By the Parallelogram Diagonals Theorem, the diagonals of a rectangle bisect each other at M. By the Parallelogram Opposite Sides Theorem, PQSR and QRPS. Let A and B be the midpoints of PS and RQ. Then, a line through M and the midpoints A and B can be drawn. As shown before, Q, P, and M are the respective images of R, S, and M after a reflection across Therefore, since QPM is the image of RSM after a reflection across the triangles are congruent. Let C and D be the midpoints of of PQ and RS. By following the same reasoning, PMS is the image of QMR after a reflection across Therefore, the triangles are congruent. The parallelogram consists of four triangles in which the opposite triangles are congruent. Therefore, the corresponding angles of these triangles are congruent. Additionally, because all triangles are isosceles, the angles opposite congruent sides are congruent as well. Each angle of the parallelogram is the sum of the same two congruent angles. Therefore, all angles of the parallelogram are congruent. Moreover, by the Parallelogram Consecutive Angles Theorem, P and S are supplementary. With this information, it can be concluded that both angles are right triangles. Because all of the angles are congruent, the angles of the parallelogram are right triangles. Therefore, PQRS is a rectangle. ## Solving Problems Using the Diagonals of a Rectangle Zosia arrives early to a Harry Styles concert! She notices something about the stage, so she uses a napkin as paper and draws a diagram. The stage is a rectangle that she labels as ABCD. Find the lengths of its diagonals to help Zosia understand the stage that she will see her favorite artist sing on! ### Hint In a rectangle, the diagonals are congruent. ### Solution By the Rectangle Diagonals Theorem, the diagonals of a rectangle are congruent. This means that AC and BD have the same length. With this information, an equation in terms of x can be written. The above equation will be solved for x. 4x+3=452x Solve for x 6x+3=45 6x=42 x=7 The value of x was found. Next, this value can be substituted in any of the expressions for the length of a diagonal. In this case, x=7 will be arbitrarily substituted in the expression for AC. AC=4x+3 AC=4(7)+3 Evaluate right-hand side AC=28+3 AC=31 It was found that the length of AC is 31. Since the diagonals of a rectangle are congruent, the length of BD is also 31. ## Diagonals of a Rhombus As with rectangles, it can also be determined whether a parallelogram is a rhombus just by looking at its diagonals. ## Rhombus Diagonals Theorem A parallelogram is a rhombus if and only if its diagonals are perpendicular. Based on the diagram, the following relation holds true. Parallelogram ABCD is a rhombus ### Proof This proof will be written in two parts. ### If a Parallelogram Is a Rhombus, Then Its Diagonals Are Perpendicular A rhombus is a parallelogram with four congruent sides. By the Parallelogram Diagonals Theorem, it can be said that its diagonals bisect each other. Let Let ABCD be a rhombus with P at the midpoint of both diagonals. Note that AB is congruent to AD and DP is congruent to BP. Additionally, by the Reflexive Property of Congruence, AP is congruent to itself. Therefore, by the Side-Side-Side Congruence Theorem, APB is congruent to APD. Because corresponding parts of congruent triangles are congruent, APB and APD are congruent angles. Furthermore, these angles form a linear pair, which means they are supplementary. With this information, it can be concluded that both APB and APD are right angles. This implies that AC is perpendicular to BD. Therefore, the diagonals of a rhombus are perpendicular. Parallelogram ABCD is a rhombus ### If Its Diagonals Are Perpendicular, Then a Parallelogram is a Rhombus Conversely, let ABCD be a parallelogram whose diagonals are perpendicular. By the Parallelogram Diagonals Theorem, the diagonals of the parallelogram bisect each other. If P is the midpoint of both diagonals, then AP and CP are congruent. Since AC and BD are perpendicular, APB and CPB measure and thus are congruent angles. By the Reflexive Property of Congruence, BP is congruent to itself. This means that two sides and their included angle are congruent. By the Side-Angle-Side Congruence Theorem, APB and CPB are congruent triangles. Because corresponding parts of congruent figures are congruent, it can be said that AB is congruent to CB. Furthermore, by the Parallelogram Opposite Sides Theorem, AB is congruent to DC and AD is congruent to BC. By the Transitive Property of Congruence, it follows that all sides of the parallelogram are congruent. This means that if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. parallelogram ABCD is a rhombus ## Solving Problems Using the Diagonals of a Rhombus Zosia is now listening to Dua Lipa at home. Staring at some of her album covers, Zosia decides to design a parallelogram as the background art for Dua's next cover! She has made a parallelogram ABCD in which the diagonals are perpendicular. To make a unique design, she wants to be sure of the length of AB. Help Zosia draw the perfect design by finding the length of AB! ### Hint If the diagonals of a parallelogram are perpendicular, then the quadrilateral is a rhombus. ### Solution By the Rhombus Diagonal Theorem, the diagonals of a parallelogram are perpendicular if and only if the quadrilateral is a rhombus. Therefore, since BD and AC are perpendicular, ABCD is a rhombus. This means that BC and CD have the same length. With this information, an equation in terms of x can be written. The above equation will be now solved for x. 6x+3=8x19 -2x+3=-19 -2x=-22 x=11 The value of x was found. Since the given quadrilateral is a rhombus, all sides are congruent and therefore have the same length. This means that the length of AB is the same as the length of BC. To find it, x=11 will be substituted into the expression for BC. BC=6x+3 BC=6(11)+3 Evaluate right-hand side BC=66+3 BC=69 The length of BC is 69. As it has been previously said, all sides have the same length. Therefore, AB is also 69. ## Additional Applications of a Parallelogram's Properties By using the theorems seen in this lesson, other properties can be derived. One of them is the Parallelogram Consecutive Angles Theorem. Parallelogram Consecutive Angles Theorem The consecutive angles of a parallelogram are supplementary. Furthermore, the theorems seen in this lesson can be applied to different parallelograms in different contexts. Consider a square. By definition, all its angles are right angles, and all its sides are congruent. Therefore, a square is both a rectangle and a rhombus. Therefore, by the Rectangle Diagonals Theorem and the Rhombus Diagonals Theorem, the diagonals of a square are congruent and perpendicular.
# Stars and bars (combinatorics) In the context of combinatorial mathematics, stars and bars (also called "sticks and stones",[1] "balls and bars",[2] and "dots and dividers"[3]) is a graphical aid for deriving certain combinatorial theorems. It can be used to solve many simple counting problems, such as how many ways there are to put n indistinguishable balls into k distinguishable bins.[4] ## Statements of theorems The stars and bars method is often introduced specifically to prove the following two theorems of elementary combinatorics concerning the number of solutions to an equation. ### Theorem one For any pair of positive integers n and k, the number of k-tuples of positive integers whose sum is n is equal to the number of (k − 1)-element subsets of a set with n − 1 elements. For example, if n = 10 and k = 4, the theorem gives the number of solutions to x1 + x2 + x3 + x4 = 10 (with x1, x2, x3, x4 > 0) as the binomial coefficient ${\displaystyle {\binom {n-1}{k-1}}={\binom {10-1}{4-1}}={\binom {9}{3}}=84.}$ This corresponds to compositions of an integer. ### Theorem two For any pair of positive integers n and k, the number of k-tuples of non-negative integers whose sum is n is equal to the number of multisets of cardinality n taken from a set of size k, or equivalently, the number of multisets of cardinality k − 1 taken from a set of size n + 1. For example, if n = 10 and k = 4, the theorem gives the number of solutions to x1 + x2 + x3 + x4 = 10 (with x1, x2, x3, x4 ${\displaystyle \geq 0}$ ) as: ${\displaystyle \left(\!\!{k \choose n}\!\!\right)={k+n-1 \choose n}={\binom {13}{10}}=286}$ ${\displaystyle \left(\!\!{n+1 \choose k-1}\!\!\right)={n+1+k-1-1 \choose k-1}={\binom {13}{3}}=286}$ ${\displaystyle {\binom {n+k-1}{k-1}}={\binom {10+4-1}{4-1}}={\binom {13}{3}}=286}$ This corresponds to weak compositions of an integer. ## Proofs via the method of stars and bars ### Theorem one proof Suppose there are n objects (represented here by stars) to be placed into k bins, such that all bins contain at least one object. The bins are distinguishable (say they are numbered 1 to k) but the n stars are not (so configurations are only distinguished by the number of stars present in each bin). A configuration is thus represented by a k-tuple of positive integers, as in the statement of the theorem. For example, with n = 7 and k = 3, start by placing the stars in a line: ★ ★ ★ ★ ★ ★ ★ Fig. 1: Seven objects, represented by stars The configuration will be determined once it is known which is the first star going to the second bin, and the first star going to the third bin, etc.. This is indicated by placing k − 1 bars between the stars. Because no bin is allowed to be empty (all the variables are positive), there is at most one bar between any pair of stars. For example: ★ ★ ★ ★ \| ★ \| ★ ★ Fig. 2: These two bars give rise to three bins containing 4, 1, and 2 objects There are n − 1 gaps between stars. A configuration is obtained by choosing k − 1 of these gaps to contain a bar; therefore there are ${\displaystyle {\tbinom {n-1}{k-1}}}$ possible combinations. ### Theorem two proof In this case, the weakened restriction of non-negativity instead of positivity means that we can place multiple bars between stars, before the first star and after the last star. For example, when n = 7 and k = 5, the tuple (4, 0, 1, 2, 0) may be represented by the following diagram: ★ ★ ★ ★ \| \| ★ \| ★ ★ \| Fig. 3: These four bars give rise to five bins containing 4, 0, 1, 2, and 0 objects To see that there are ${\displaystyle {\tbinom {n+k-1}{k-1}}}$ possible arrangements, observe that any arrangement of stars and bars consists of a total of n + k − 1 objects, n of which are stars and k − 1 of which are bars. Thus, we only need to choose k − 1 of the n + k − 1 positions to be bars (or, equivalently, choose n of the positions to be stars). Theorem 1 can now be restated in terms of Theorem 2, because the requirement that all the variables are positive is equivalent to pre-assigning each variable a 1, and asking for the number of solutions when each variable is non-negative. For example: ${\displaystyle x_{1}+x_{2}+x_{3}+x_{4}=10}$ with ${\displaystyle x_{1},x_{2},x_{3},x_{4}>0}$ is equivalent to: ${\displaystyle x_{1}+x_{2}+x_{3}+x_{4}=6}$ with ${\displaystyle x_{1},x_{2},x_{3},x_{4}\geq 0}$ ## Proofs by generating functions Both cases are very similar, we will look at the case when ${\displaystyle x_{i}\geq 0}$ first. The 'bucket' becomes ${\displaystyle {\frac {1}{1-x}}}$ This can also be written as ${\displaystyle 1+x+x^{2}+\dots }$ and the exponent of x tells us how many balls are placed in the bucket. Each additional bucket is represented by another ${\displaystyle {\frac {1}{1-x}}}$, and so the final generating function is ${\displaystyle {\frac {1}{1-x}}{\frac {1}{1-x}}\dots {\frac {1}{1-x}}={\frac {1}{(1-x)^{k}}}}$ As we only have n balls, we want the coefficient of ${\displaystyle x^{n}}$ (written ${\displaystyle [x^{n}]:}$) from this ${\displaystyle [x^{n}]:{\frac {1}{(1-x)^{k}}}}$ This is a well-known generating function - it generates the diagonals in Pascal's Triangle, and the coefficient of ${\displaystyle x^{n}}$ is ${\displaystyle {\binom {n+k-1}{k-1}}}$ For the case when ${\displaystyle x_{i}>0}$, we need to add x into the numerator to indicate that at least one ball is in the bucket. ${\displaystyle {\frac {x}{1-x}}{\frac {x}{1-x}}\dots {\frac {x}{1-x}}={\frac {x^{k}}{(1-x)^{k}}}}$ and the coefficient of ${\displaystyle x^{n}}$ is ${\displaystyle {\binom {n-1}{k-1}}}$ ## Examples Many elementary word problems in combinatorics are resolved by the theorems above. ### Example 1 If one wishes to count the number of ways to distribute seven indistinguishable one dollar coins among Amber, Ben, and Curtis so that each of them receives at least one dollar, one may observe that distributions are essentially equivalent to tuples of three positive integers whose sum is 7. (Here the first entry in the tuple is the number of coins given to Amber, and so on.) Thus stars and bars theorem 1 applies, with n = 7 and k = 3, and there are ${\displaystyle {\tbinom {7-1}{3-1}}=15}$ ways to distribute the coins. ### Example 2 If n = 5, k = 4, and a set of size k is {a, b, c, d}, then ★\|★★★\|\|★ could represent either the multiset {a, b, b, b, d} or the 4-tuple (1, 3, 0, 1). The representation of any multiset for this example should use SAB2 with n = 5, k – 1 = 3 bars to give ${\displaystyle {\tbinom {5+4-1}{4-1}}={\tbinom {8}{3}}=56}$. ### Example 3 SAB2 allows for more bars than stars, which isn't permitted in SAB1. So, for example, 10 balls into 7 bins is ${\displaystyle {\tbinom {16}{6}}}$, while 7 balls into 10 bins is ${\displaystyle {\tbinom {16}{9}}}$, with 6 balls into 11 bins as ${\displaystyle {\tbinom {16}{10}}={\tbinom {16}{6}}.}$ ### Example 4 If we have the infinite power series ${\displaystyle \left[\sum _{k=1}^{\infty }x^{k}\right],}$ we can use this method to compute the Cauchy product of m copies of the series. For the nth term of the expansion, we are picking n powers of x from m separate locations. Hence there are ${\displaystyle {\tbinom {n-1}{m-1}}}$ ways to form our nth power: ${\displaystyle \left[\sum _{k=1}^{\infty }x^{k}\right]^{m}=\sum _{n=m}^{\infty }{{n-1} \choose {m-1}}x^{n}}$ ### Example 5 The graphical method was used by Paul Ehrenfest and Heike Kamerlingh Onnes—with symbol ε (quantum energy element) in place of a star and the symbol 0 in place of a bar—as a simple derivation of Max Planck's expression for the number of "complexions" for a system of "resonators" of a single frequency.[5][6] By complexions (microstates) Planck meant distributions of P energy elements ε over N resonators.[7][8] The number R of complexions is ${\displaystyle R={\frac {(N+P-1)!}{P!(N-1)!}}.\ }$ The graphical representation of each possible distribution would contain P copies of the symbol ε and N – 1 copies of the symbol 0. In their demonstration, Ehrenfest and Kamerlingh Onnes took N = 4 and P = 7 (i.e., R = 120 combinations). They chose the 4-tuple (4, 2, 0, 1) as the illustrative example for this symbolic representation: εεεε0εε00ε. ## References 1. ^ Batterson, J. Competition Math for Middle School. Art of Problem Solving. 2. ^ Flajolet, Philippe; Sedgewick, Robert (June 26, 2009). Analytic Combinatorics. Cambridge University Press. ISBN 978-0-521-89806-5. 3. ^ "Art of Problem Solving". artofproblemsolving.com. Retrieved 2021-10-26. 4. ^ Feller, William (1968). An Introduction to Probability Theory and Its Applications. Vol. 1 (3rd ed.). Wiley. p. 38. 5. ^ Ehrenfest, Paul; Kamerlingh Onnes, Heike (1914). "Simplified deduction of the formula from the theory of combinations which Planck uses as the basis of his radiation theory". Proceedings of the KNAW. 17: 870–873. Retrieved 16 May 2024. 6. ^ Ehrenfest, Paul; Kamerlingh Onnes, Heike (1915). "Simplified deduction of the formula from the theory of combinations which Planck uses as the basis of his radiation theory". The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science. Series 6. 29 (170): 297–301. doi:10.1080/14786440208635308. Retrieved 5 December 2020. 7. ^ Planck, Max (1901). "Ueber das Gesetz der Energieverteilung im Normalspectrum". Annalen der Physik. 309 (3): 553–563. Bibcode:1901AnP...309..553P. doi:10.1002/andp.19013090310. 8. ^ Gearhart, C. (2002). "Planck, the Quantum, and the Historians" (PDF). Phys. Perspect. 4 (2): 170–215. Bibcode:2002PhP.....4..170G. doi:10.1007/s00016-002-8363-7. Retrieved 16 May 2024.
#### Need Help? Get in touch with us # Learn How to Measure The Area of Irregular Shapes Grade 3 Sep 25, 2022 ### Key concepts • How to break larger area into smaller areas • Divide the irregular shape into squares and rectangles • Find the area of each individual squares and rectangles • Find the area of any irregular shapes ## 1.1 Area of Square 1. All sides are of equal length in a square. 2. Area of a square = Side x Side 3. Side = 6 inches 4. Area = 6 in x 6 in. = 36 square inches ### 1.2 Area of Rectangle As the opposite sides are the same in a rectangle area Area of the rectangle = Length * Width Length = 8 inches, Width = 5 inches Area = 8 in x 5 in = 40 square inches Jack wants to lay artificial grass for playing golf in the ground. Let us help Jack find the area of the ground to be covered by grass. Method 1 Draw the figure on the grid paper and count the unit squares covered to find the area. Total number of squares = 56 Area to be covered by grass = 56 square feet Method 2 • Break the larger area into smaller parts. • Look for the possibilities in which the smaller shapes are part of a larger shape. • Divide the larger area into smaller rectangles and find the area. The golf area to be covered by grass is divided into rectangles A, B and C Find individual areas Area of rectangle A = 4 ft x 3 ft = 12 square feet Area of rectangle B = 4 ft x 3 ft = 12 square feet Area of rectangle C = 4 ft x 8 ft = 32 square feet Total area = 12 + 12 + 32 = 56 square feet ### Steps to find the area of an irregular shape 1. Find all the unknown sides. 2. Divide the irregular shape into squares and rectangles 3. Find the area of each individual squares and rectangles 4. Add all the individual areas to find the total area of the irregular shape. Total area = sum of all individual areas ### Steps to find the area of an irregular shape 1. Find all the unknown sides. In this example, find the values of side a and side b Side a = 10 – 3 = 7 cm Side b = 5 – 3 = 2 cm 2. Divide the irregular shape into squares and rectangles In the example, the figure is divided into one rectangle-A and one square-B Total area = Area of Rectangle A + Area of Square B ### Steps to find the area of an irregular shape 3. Find the area of each individual squares and rectangles Area of rectangle A = length x width = 10 x 2 = 20 square cm Area of square B = side x side = 3 x 3 = 9 square cm 4. Add all the individual areas to find the total area of the irregular shape Total area = Area of rectangle A + Area of rectangle B = 20 sq. cm + 9 sq. cm = 29 sq. cm ### Example: Find the area of the shaded part Total area of the shaded part = Area of the outer rectangle – Area of the inner rectangle Area of outer rectangle = 10 cm x 8 cm = 80 square cm Area of inner rectangle = 4 cm x 6 cm = 24 square cm Total area of shaded part = 80 sq. cm – 24 sq. cm = 56 square cm ### Assessment Find the area of the irregular shapes ## Exercise: Find the area of the shapes shown below. 1. 2. 3. 4. 5. ### What we have learned: • The area of a surface or a plane figure is the number of square units needed to cover the surface or the figure • Area of the square = S x S • Area of the rectangle = Length x Width • Area is measured using standard units • To find area of irregular shape • Break the larger area into smaller parts. • Look for the possibilities in which smaller shapes are part of the larger shape. • Divide the larger area into smaller rectangles and find the area. ### Let Summarize: To find the area of irregular shapes, the first thing to do is to divid the irregular shape into regular shapes that you can recognize such as triangles, rectangles, circles, squares and so forth. Then, find the area of these individual shapes and add them up. Comments: #### Addition and Multiplication Using Counters & Bar-Diagrams Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […] Read More >> #### Dilation: Definitions, Characteristics, and Similarities Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […] Read More >> #### How to Write and Interpret Numerical Expressions? Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […] Read More >> #### System of Linear Inequalities and Equations Introduction: Systems of Linear Inequalities: A system of linear inequalities is a set of two or more linear inequalities in the same variables. The following example illustrates this, y < x + 2…………..Inequality 1 y ≥ 2x − 1…………Inequality 2 Solution of a System of Linear Inequalities: A solution of a system of linear inequalities […] Read More >>
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths! Have you got a Super Brain? Adapting and exploring maths challenge problems is an excellent way of finding ideas for IB maths explorations and extended essays. This problem is taken from the book: The first 25 years of the Superbrain challenges. I’m going to see how many different ways I can solve it. The problem is to find all the integer solutions to the equation above. Finding only integer solutions is a fundamental part of number theory – a branch of mathematics that only deals with integers. Method number 1: Brute force This is a problem that computers can make short work of. Above I wrote a very simple Python program which checked all values of x and y between -99 and 99. This returned the only solution pairs as: Clearly we have not proved these are the only solutions – but even by modifying the code to check more numbers, no more pairs were found. Method number 2: Solving a linear equation We can notice that the equation is linear in terms of y, and so rearrange to make y the subject. We can then use either polynomial long division or the method of partial fractions to rewrite this. I’ll use partial fractions. The general form for this fraction can be written as follows: Next I multiply by the denominator and the compare coefficients of terms. This therefore gives: I can now see that there will only be an integer solution for y when the denominator of the fraction is a factor of 6. This then gives (ignoring non integer solutions): I can then substitute these back to find my y values, which give me the same 4 coordinate pairs as before: Method number 3: Solving a quadratic equation I start by making a quadratic in x: I can then use the quadratic formula to find solutions: Which I can simplify to give: Next I can note that x will only be an integer solution if the expression inside the square root is a square number. Therefore I have: Next I can solve a new quadratic as follows: As before I notice that the expression inside my square root must be a square number. Now I can see that I need to find m and n such that I have 2 square numbers with a difference of 24. I can look at the first 13 square numbers to see that from the 12th and 13th square numbers onwards there will also be a difference of more than 24. Checking this list I can find that m = 1 and m = 5 will satisfy this equation. This then gives: which when I solve for integer solutions and then sub back into find x gives the same four solutions: Method number 4: Graphical understanding Without rearranging I could imagine this as a 3D problem by plotting the 2 equations: This gives the following graph: We can see that the plane intersects the curve in infinite places. I’ve marked A, B on the graph to illustrate 2 of the coordinate pairs which we have found. This is a nice visualization but doesn’t help find our coordinates, so lets switch to 2D. In 2D we can use our rearranged equation: This gives the following graph: Here I have marked on the solution pairs that we found. The oblique asymptote (red) is y = 2x-1 because as x gets large the fraction gets very small and so the graph gets closer and closer to y = 2x -1. All points on this curve are solutions to the equation – but we can see that the only integer solution pairs will be when x is small. When x is a large integer then the curve will be close to the asymptote and hence will return a number slightly bigger than an integer. So, using this approach we would check all possible integer solutions when x is small, and again should be able to arrive at our coordinate pairs. So, 4 different approaches that would be able to solve this problem. Can you find any others? Essential resources for IB students: Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications. There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful! The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s. Essential Resources for IB Teachers If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes: 1. Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour. 2. Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination. 3. Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework. 4. A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course. There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring! Essential Resources for both IB teachers and IB students I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
# Percent Error Formula • Last Updated : 20 Jan, 2022 Percent Error calculates the percentage difference between actual value and measured value. It tells us how close the measured value is to be true. There may be slight manufacturing errors for the instruments that are used for measurements and can never be assumed to be exact. For this kind of problem percent error helps us to specify to what extent it will be true. The formula for this percent error is given by Percent Error = (\|Measured Value – Actual Value\|/Actual Value) × 100 To get more understanding from the formula, let’s look into a few examples Question 1: The actual length of a straight line is 8cm and when we measured with a scale it was 7cm (Wrong value) due to some marginal error. What is the percentage error? Solution: Given, Measured value = 7cm Actual Value = 8cm Percent error = (\|Measured Value-Actual Value\| / Actual Value) x 100 = (\|7-8\| / 8) x 100 = (\|-1\| / 8) x 100 = (1/8) x 100 = 12.5% So percentage error is 12.5% Question 2: What is the percentage error if the measured value is 40cm and the Actual value is 50cm? Solution: Given, Measured value=40cm Actual Value=50cm Percent error = (\|Measured Value-Actual Value\| / Actual Value) x 100 = (\|40-50\| / 50) x 100 = (\|-10\| / 50) x 100 = (10/50) x 100 = 20% So percentage error is 20% Question 3: A scale measures wrongly a value as 15 cm due to some marginal errors. Find the percentage error if the actual measurement of the value is 14 cm. Solution: Given, Measured value=15cm Actual Value=14cm Percent error= ( \|Measured Value-Actual Value\| / Actual Value ) x 100 = ( \|15-14\| / 14 ) x 100 = ( \|1\| / 14 ) x 100 = 100/14 = 7.14% So percentage error is 7.14% Question 4: What is the percentage error if the measured value is 11cm and the Actual value is 10cm? Solution: Given, Measured value=11cm (wrongly measured) Actual Value=10cm (Exact value) Percent error= ( \|Measured Value-Actual Value\| / Actual Value ) x 100 = ( \|11-10\| / 10 ) x 100 = ( \|1\| / 10 ) x 100 = 100/10 = 10% So percentage error is 10% Question 5: Find the percentage error if the wrongly measured value is 100cm and the Actual value is 105cm? Solution: Given, Wrongly Measured value=100cm Actual Value=105cm Percent error= (\|Measured Value-Actual Value\| / Actual Value) x 100 = (\|100-105\| / 105) x 100 = (\|-5\| / 105) x 100 = (5/105) x 100 = 100/21 = 4.76% So percentage error for the above data is 4.76% My Personal Notes arrow_drop_up
# How do you calculate specific gravity example? The specific gravity of a substance is characteristic; it is the same for different samples of a substance (if pure, the same in composition, and free from cavities or inclusions) and is used to help identify unknown substances. ## How do you solve specific gravity problems? At 25°C (77°F or 298.15K), the weight of water is1 gram per cubic centimeter, or 1,000 kilograms per cubic meter. ## What is the formula for corrected specific gravity? If the density of a substance is known, to calculate the specific gravity of a substance simply divide the density of the substance by the density of water or air. Because the density of water is 1000 kg/m3 or 1 g/cm3, it is simple to calculate. The density of air is 1.205 kg/m3. ## What is the specific gravity of water at 25 C? Use the equation “m / v = D” where m is mass in grams or kilograms, v is volume in milliliters or liters, and D is density. For example, if you had a sample that was 8 grams and 9 milliliters, your equation would be: “8.00 g / 9.00 mL = 0.89 g/mL.” ## Why do we calculate specific gravity? Specific gravity refers to the ratio of the density of an object and the reference material. Furthermore, the specific gravity can tell us if the object will sink or float in reference material. Besides, the reference material is water that always has a density of 1 gram per cubic centimeter or 1 gram per millimeter. ## How do you calculate specific gravity of water? Density is defined as mass per unit volume. It has the SI unit kg m-3 or kg/m3 and is an absolute quantity. Specific gravity is the ratio of a material’s density with that of water at 4 °C (where it is most dense and is taken to have the value 999.974 kg m-3). It is therefore a relative quantity with no units. ## How do you calculate specific gravity of a liquid? Specific gravity is defined as the ratio of the density of the solid part of a material to the density of water at 20°C. Typically, the specific gravity of soils is in the range 2.60 to about 2.80. ## What is the specific gravity of an object? For example, the specific gravity of 0.84 corresponds to the density of 840 (0.84 x 1000) kg/cubic meters. Multiply the density by the acceleration of gravity (9.81) to calculate the specific weight. In our example, the specific weight is 840 x 9.81 = 8,240.4. ## Is specific gravity same as density? The temperature of measurement should be specified and controlled in all specific gravity measurements because the specific gravity of a solution is affected by temperature. Standards of specific gravity are set by analysing standard mixes at 27 degrees Celsius (80.6 Fahrenheit). ## What is the value of specific gravity? The density and specific gravity of milk is usually given at 15.6°C (60° F). (The specific gravity of water is usually expressed at 4°C). The specific gravity of some major milk constituents are: water: 1.00; fat: 0.93, protein: 1.346, lactose: 1.666, salts: 4.23, and SNF: 1.616. ## What is the volume of a solution that has a specific gravity of 1.2 and a mass of 185g? Now we have the density and we have the mass of the solution, so we can find out the volume that will be mass divided by density here. The mass of the solution given 185 grams, divided by the density of 1.2 grams per milliliters and we get the volume of the solution- 154.17 milliliters. ## How do you convert specific gravity to weight? Substances with a lesser density than water would have an SG of less than 1, and a denser substance more. Since water has a density of 1 gram per cubic centimeter, the equation for establishing a liquid’s specific gravity would be: Specific Gravity = (density of liquid) / (1 gm/cm³). ## Is specific gravity affected by temperature? Specific gravity is defined as a ratio of a density of a fluid to the density of water at 4∘C. It has no unit. ## What is the specific gravity of milk? Specific gravity values for a few common substances are: Au-19.3; mercury-13.6; alcohol-0.7893; benzene- 0.8786. ## What is specific gravity of sand? Sand particles composed of quartz have a specific gravity ranging from 2.65 to 2.67. Inorganic clays generally range from 2.70 to 2.80. Soils with large amounts of organic matter or porous particles (such as diatomaceous earth) have specific gravities below 2.60. ## Why is the specific gravity of water 1? What is the specific gravity of 10 Kg of water occupied in 10 m3 with respect to 200 g/m3? Explanation: Specific gravity = (10/10)/0.2 = 5. ## What is specific gravity and its unit? The “Specific Gravity” of a substance is the ratio of its mass to that of an equal volume of water at the same temperature and pressure. This is also the ratio of the densities of the two substances: SG = (mass of a volume V of a material)/(mass of a volume V of water) = (rMaterial/rWater). ## What is the specific gravity of ice? Specific gravity of ice = 0.9. ## What is the specific gravity of alcohol? The specific gravity of a substance or liquid, including water, the reference liquid, is going to change depending on temperature and pressure. That is why a standard temperature and pressure are used in the calculation of specific gravity. If those outside influences are not regulated, specific gravity will change. ## What is the specific gravity of 10 kg of water? Factors Affecting Specific Gravity The density of the substance that is being studied or considered. Mass (and weight) of the substance. Since specific gravity is measured under control, the temperature also affects it. We also consider the pressure conditions while talking about the specific gravity. ## How do you find specific gravity from mass and volume? 1 Answer. The concentration of a sample in a sample can be determined by means of density. Specific gravity is the density of solute divided by the density of solvent. ## What is the specific gravity of a substance that weights 75 g and occupies a volume of 150 ml? Putting the values mass is 75. Gram, divided by a volume is 150 milliliters, so this will be equal to 0.5 gram per milliliter now specific gravity. ## Does specific gravity change with pressure? Density = Mass / Volume Specific gravity is the density of a substance divided by the density of water. Since (at standard temperature and pressure) water has a density of 1 gram/cm3, and since all of the units cancel, specific gravity is usually very close to the same value as density (but without any units).
Kirk Braunius Assignment 1 Statement of problem: Find 2 linear functions g(x) and f(x) such that their product h(x) = g(x) f(x) is tangent to both g(x) and h(x) at two distinct points. The product of two linear functions will be a quadratic equation of the form . My first step was to take a "guess" and graph the equations. I pictured the h(x) being a parabola with the g(x) and f(x) lines crossing (one with positive slope, one with negative slope). I used and : Obviously neither line is tangent. It appears that moving either line (or both) such that the y value of their intersection moves up will get us closer. So I tryed graphing these two lines, moving the y intercepts up (I simply changed the b values from 3 to 5): That made it worse! While the intersection of the lines moved up, the "y" value of the vertex of the parabola went from 9 to 25. What if I move just one line? Try our y=3x+3, and change the second equation to y=-3x - 2. The composite function is , or Our graph is as follows: That looks like it works! Was I that lucky? Setting our first linear equation equal to the composite, we have 3x + 3 = (3x + 3)(-3x -2). This has only one solution at x=-1, so that is tangent. Likewise -3x - 2 = (3x + 3)(-3x -2) has just one solution, at x = -2/3. Great! But why did that work? And can we find others? Let's try making the slopes of the lines 1 and -1 instead of 3 and -3. I have shown both graphs together below: Here is yet another added, with slopes of 7 and -7. Notice that the intersection of the lines appear to have the same value of y for each pair. I have shown a line at y = .5 to illustrate this. Likewise, the vertices of the parabolas are all at y = .25, and I have shown a line. So how can we choose 2 lines that will create this effect without just guessing? From our observation, it appears that if the lines have slopes that are "negative of each other" and they intersect at y = .5 they should work. Lets try that with an arbitrary line of y = 4x - 1. At y = .5 we have .5 = 4x - 1 or x = 3/8 So our "companion" line should be y = -4x + b, with y=.5 when x=3/8. So again at y = .5, with x = 3/8, .5 = -4 (3/8) + b, or b = 2. So our other line must be y = -4x + 2. Lets try it! So, without proof, we have observed that the product of two linear functions is tangent to each of the two functions when the linear functions are of the form y = mx + b and y = -mx +c with intersection at y=.5.
# ACT Math : Cosine ## Example Questions ← Previous 1 3 4 5 6 7 ### Example Question #1 : Cosine In the above triangle, and . Find . Explanation: With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. With this information, we can use the cosine function to find the angle. ### Example Question #2 : Cosine For the above triangle, and . Find . Explanation: With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. With this information, we can use the cosine function to find the angle. ### Example Question #1 : How To Find An Angle With Cosine For the above triangle, and . Find . This triangle cannot exist. This triangle cannot exist. Explanation: With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. However, if we plug the given values into the formula for cosine, we get: This problem does not have a solution. The sides of a right triangle must be shorter than the hypotenuse. A triangle with a side longer than the hypotenuse cannot exist. Similarly, the domain of the arccos function is . It is not defined at 1.3. ### Example Question #4 : Cosine A rope is thrown down from a building to the ground and tied up at a distance of from the base of the building. What is the angle measure between the rope and the ground? Round to the nearest hundredth of a degree. Explanation: You can draw your scenario using the following right triangle: Recall that the cosine of an angle is equal to the ratio of the adjacent side to the hypotenuse of the triangle. You can solve for the angle by using an inverse cosine function: or degrees. ### Example Question #5 : Cosine What is the value of in the right triangle above? Round to the nearest hundredth of a degree. Explanation: Recall that the cosine of an angle is equal to the ratio of the adjacent side to the hypotenuse of the triangle. You can solve for the angle by using an inverse cosine function: or . ### Example Question #6 : Cosine A support beam (buttress) lies against a building under construction. If the beam is feet long and strikes the building at a point feet up the wall, what angle does the beam strike the building at? Round to the nearest degree. Explanation: Our answer lies in inverse functions. If the buttress is feet long and is feet up the ladder at the desired angle, then: Thus, using inverse functions we can say that Thus, our buttress strikes the buliding at approximately a angle. ### Example Question #7 : Cosine A stone monument stands as a tourist attraction. A tourist wants to catch the sun at just the right angle to "sit" on top of the pillar. The tourist lies down on the ground meters away from the monument, points the camera at the top of the monument, and the camera's display reads "DISTANCE -- METERS". To the nearest degree, what angle is the sun at relative to the horizon? Explanation: Our answer lies in inverse functions. If the monument is meters away and the camera is meters from the monument's top at the desired angle, then: Thus, using inverse functions we can say that Thus, our buttress strikes the buliding at approximately a angle. ### Example Question #1 : How To Find A Missing Side With Cosine If angle A measures 30 degrees and the hypotenuse is 4, what is the length of AB in the given right triangle? 8√3 2√3 4 2 √3 2√3 Explanation: Cosine A = Adjacent / Hypotenuse = AB / AC = AB / 4 Cosine A = AB / 4 Cos (30º) = √3 / 2 = AB / 4 Solve for AB √3 / 2 = AB / 4 AB = 4 * (√3 / 2) = 2√3 ### Example Question #9 : Cosine Explanation: To solve this problem you need to make the triangle that the problem is talking about. Cosine is equal to the adjacent side over the hypotenuse of a right triangle So this is what our triangle looks like: Now use the pythagorean theorem to find the other side: Sine is equal to the opposite side over the hypotenuse, the opposite side is 12 ### Example Question #1 : Cosine The hypotenuse of right triangle HLM shown below is long. The cosine of angle is . How many inches long is ?
# Simplifying 0.67 As A Fraction: A Step-By-Step Guide Simplifying 0.67 As A Fraction: A Step-By-Step Guide. Fractions are a fundamental aspect of mathematics, allowing us to express numbers that are not whole or decimal. If you’ve ever wondered how to convert the decimal number 0.67 into a fraction, you’re in the right place. In this blog post, we will guide you through the process, step by step, to simplify 0.67 as a fraction. By the end, you will have a clear understanding of how to express 0.67 as a fraction. Let’s get started! ## Understanding Fractions Before delving into the conversion process, it’s important to grasp the concept of fractions. A fraction consists of two parts: a numerator and a denominator. The numerator represents the part or quantity being considered, while the denominator signifies the total number of equal parts the whole is divided into. For example, in the fraction 3/4, 3 is the numerator and 4 is the denominator. ## Converting Decimal To Fraction Converting a decimal to a fraction involves a systematic approach. Let’s go through the steps to convert 0.67 into a simplified fraction. Step 1: Write down the decimal as a fraction with the decimal value as the numerator. To convert 0.67 to a fraction, we can write it as 0.67/1. Step 2: Remove the decimal by multiplying both the numerator and denominator by 10, 100, or any power of 10 that eliminates the decimal places. In this case, we multiply both the numerator and denominator by 100 to remove the decimal places: (0.67 × 100) / (1 × 100) = 67/100. Step 3: Simplify the fraction, if possible. The fraction 67/100 can be simplified by dividing both the numerator and denominator by their greatest common divisor (GCD). In this case, the GCD of 67 and 100 is 1. Therefore, the fraction 67/100 is already in its simplest form. ## The Final Result Following the conversion steps, we find that 0.67 as a fraction is equal to 67/100. This means that if we divide a whole into 100 equal parts, 67 of those parts would represent the decimal value 0.67. You can search for more about similar topics like these on Tipsfeed. ## FAQ ### How Much Is .067 As A Fraction? 0.067 as a fraction is 67/1000. ### What Is .62 As A Fraction? 0.62 as a fraction is 31/50. ### What Is .65 As A Fraction? 65% as a fraction is 13/20. ### Is 0.67 A Decimal? 0.67 percent = 0.0067 decimal. ## Conclusion: Converting decimals to fractions is a valuable skill in mathematics, allowing us to express decimal values in a more precise and meaningful way. In this blog post, we have explored the step-by-step process of simplifying 0.67 as a fraction. By following these steps, we determined that 0.67 can be represented as the fraction 67/100. With practice, you can become more proficient in converting decimals to fractions, enhancing your mathematical abilities. Keep exploring and expanding your knowledge of fractions for continued success in your mathematical journey. I Have Covered All The Following Queries And Topics In The Above Article 0.67 Repeating As A Fraction What Is 0.67 As A Fraction 0.67 As A Fraction In Simplest Form Write 0.67 As A Fraction In Simplest Form -0.67 As A Fraction 0.67 As A Fraction In Simplest Form 0.67 As A Percentage 0.67 Repeating As A Fraction 67 As A Fraction Of An Inch 0.67 In Cups 0.6 As A Fraction 1.5 As A Fraction 0.25 As A Fraction 0.67 As A Fraction
# Componendo and Dividendo: Proof & Example Componendo and Dividendo is one basic property about proportions that provide a quick way to perform calculations and reduce the number of required extensions. This is especially useful when working with equations involving fractions or rational functions in math competitions, especially when you’re looking at fractions. i.e \frac{a+b}{a-b}=\frac{c+d}{c-d} Proof of Componendo and Dividendo \frac{a+b}{a-b}=\frac{c+d}{c-d}: Add one both sides of the left and right-hand side fractions. \begin{array}{l} \Rightarrow \frac{a}{b}+1=\frac{c}{d}+1 \\ \Rightarrow \frac{a+b}{b}=\frac{c+d}{d} \end{array} Let this equation be (A) Step: 2 Subtract one both sices of the left and right hand side fractions: \begin{array}{l} \Rightarrow \frac{a}{b}-1=\frac{c}{d}-1 \\ \Rightarrow \frac{a-b}{b}=\frac{c-d}{d} \end{array} Let this equation be (B) Now divide equation divide A by B: \begin{array}{l} \Rightarrow \frac{\left[\frac{a+b}{b}\right]}{\left[\frac{a-b}{b}\right]}=\frac{\left[\frac{c+d}{d}\right]}{\left[\frac{c-d}{d}\right]}\\ \Rightarrow\left[\frac{a+b}{b}\right] \times\left[\frac{b}{a-b}\right]=\left[\frac{c+d}{d}\right] \times\left[\frac{d}{c-d}\right]\\ \Rightarrow\left[\frac{a+b}{a-b}\right] \times\left[\frac{b}{b}\right]=\left[\frac{c+d}{c-d}\right] \times\left[\frac{d}{d}\right]\\ \Rightarrow\left[\frac{a+b}{a-b}\right] \times\left[\frac{\not b}{\not b}\right]=\left[\frac{c+d}{c-d}\right] \times\left[\frac{\not d}{\not d}\right]\\ \Rightarrow\left[\frac{a+b}{a-b}\right] \times 1=\left[\frac{c+d}{c-d}\right] \times 1\\ \therefore \quad \frac{a+b}{a-b}=\frac{c+d}{c-d} \end{array} #### Definitions Related To Componendo and Dividendo If a / b=c / d • Invertendo: b / a=d / c • Alternendo : a / c=b / d • Componendo: (a+b), b=(c+d) / d • Dividendo : (a-b) / b=(c-d) / d • Componendo – Dividendo: (a+b) /(a-b)=(c+d)/(c-d) Componendo and Dividendo Example: If \frac{a}{\delta}=\frac{c}{d}, prove that \frac{3 a-8 b}{2 a+8 b}=\frac{2 c-8 d}{3 c-8 d} Method 1. We have \begin{array}{l} \frac{a}{b}=\frac{c}{d} \\ \frac{3 a}{8 b}=\frac{3 c}{8 d} \\ \frac{3 a+9 b}{2 a-8 b}=\frac{3 c+8 d}{3 c-8 d} \\ \frac{3 a-9 b}{2 a+8 b}=\frac{3 c-8 d}{3 c+8 d} \end{array} Method 2. Alternatively, we can make use of the property that \frac{a+k h}{a-k b}=\frac{c+k d}{c-k d} If \frac{a}{b}=\frac{c}{d} Then, \frac{a+\left(\frac{8}{3}\right) b}{a-\left(\frac{8}{3}\right) b}=\frac{c+\left(\frac{8}{3}\right) d}{c-\left(\frac{8}{3}\right) d} Multiplying both the numerator and the denominator by 2 on both sides: \begin{array}{l} \frac{3 a+8 b}{3 a-8 b}=\frac{3 c+8 d}{2 c-8 d} \\ \frac{3 a-8 b}{3 a+5 b}=\frac{3 ec-8 d}{2 c+8 d} \end{array} (applying invertendo) In this article, we have seen Componendo and Dividendo proof and example. Hope you like it, and please share with your maths nerds and helped them out to clear their doubts.
# What is a function Much of this course is devoted to the study of properties of real-valued functions of a real variable. Such a function ${\displaystyle f}$ assigns to each element ${\displaystyle x}$ of a given set of real numbers exactly one real number y, called the value of the function ${\displaystyle f}$ at ${\displaystyle x}$. Students can think about function as a process in which when one input a number(e.g. ${\displaystyle x}$), a number(e.g. ${\displaystyle y}$) can be generated. In other word, a function is simply a relation where for every independent variable(e.g. ${\displaystyle x}$), there is only one dependent variable(e.g. ${\displaystyle y}$). The dependence of ${\displaystyle y}$ on ${\displaystyle f}$ and on ${\displaystyle x}$ is made explicit by using the notation ${\displaystyle f(x)}$ to mean the value of ${\displaystyle f}$ at ${\displaystyle x}$. Example: ${\displaystyle f(x)=x^{2}}$ For Every value of ${\displaystyle x}$, ${\displaystyle f(x)}$ only generate a single value. This can be shown through the table of values. A graph of ${\displaystyle y=x^{2}}$. ${\displaystyle x}$ 0 1 2 3 4 ${\displaystyle f(x)}$ 0 1 4 9 16 The function y = ${\displaystyle f(x)}$may be represented pictorially by its graph, which is the set of points (${\displaystyle x}$,${\displaystyle f(x)}$) for each ${\displaystyle x}$ in the domain of ${\displaystyle f}$, indicated with respect to cartesian coordinate axes x,y. Notice that a vertical line only cuts the graph once, thus making the graph a function. A relation, however, is different to a function. A relation is a set of operations that links the independent variables to a set of dependent variables. In order words, for a single ${\displaystyle x}$ value. there can be multiple y values. Example: ${\displaystyle x^{2}+y^{2}=1}$ This is the formula for a circle. Notice that a vertical line may cross the circle more than once, thus for a given x value, there may be more than 1 y value exist. This can be illustrated pictorially or algebraically. ${\displaystyle x^{2}+y^{2}=1}$ ${\displaystyle y^{2}=1-x^{2}}$ ${\displaystyle y=\pm {\sqrt[{}]{1-x^{2}}}}$ # Domain The domain of a function is simply a set of possible ${\displaystyle x}$ values for which it's corresponding ${\displaystyle f(x)}$ exists. Example: ${\displaystyle f(x)={\sqrt[{}]{1-x^{2}}}}$ As we know, in real numbers, we can not square root a negative number. Thus, by solving the following inequality, we can find out the this function's domain. ${\displaystyle 1-x^{2}\geq 0}$ ${\displaystyle 1\geq x^{2}}$ ${\displaystyle x^{2}\leq 1}$ ∴ the domain of this function is all real x where ${\displaystyle 1\leq x\leq 1}$. Note that a lot of questions test this topic around rational functions. It is important to remember that when the denominator is zero, the number is undefined. Example: ${\displaystyle f(x)={\frac {1}{x^{2}+5x+6}}}$ Since the number is undefined when the denominator is zero, we can equate the denominator to zero to find out exactly when this function is undefined. ${\displaystyle x^{2}+5x+6=0}$ ${\displaystyle (x+2)(x+3)=0}$ ${\displaystyle (x+2)=0}$ OR ${\displaystyle (x+3)=0}$ ${\displaystyle x=2,3}$ ∴ the domain of this function is all real x where ${\displaystyle x\neq 2,3}$. # Range A graph of ${\displaystyle y=x^{2}}$. The domain of a function is simply a set of possible ${\displaystyle f(x)}$ values for which it's corresponding ${\displaystyle x}$ exists. This is very similar to Domain, but concerning a different axis. Example: ${\displaystyle f(x)=x^{2}}$ The best way to approach this problem is by drawing a quick sketch of the function in question. As you can see, this function is simply a parabola. We know that all parabolas have their two arms extending to either positive or negative infinity depending on the signs of the parabola. In this parabola, we can see that its minimum point(minima) is at 0 with no ${\displaystyle f(x)}$ value exisiting below it. Thus, we know that all the possible ${\displaystyle f(x)}$ value of this function is above 0 inclusive. ∴the range of this function is all real y where ${\displaystyle y\geq 0}$. # Even/Odd Functions First that's have a brief look of some of the typical even functions. A graph of ${\displaystyle y=x^{2}}$. y= cos(x) As it is clearly shown, an even function is one that is symmetrical with respect to the Y axis. Now, in order to solve problems around even functions, we need to make this geographical description more mathematical. As you can see if you input 1 or -1, the output of this function remains constant. Thus we can deduce that if a function is even, then ${\displaystyle f(x)=f(-x)}$. Example: Show that the function ${\displaystyle f(x)={\frac {x^{4}+1}{x^{2}+x^{4}}}}$ is even. First don't be intimidated by this seemingly complex function, you don't have to graph it. Calculate what ${\displaystyle f(-x)}$; to do this replace all x with (-x). ${\displaystyle f(-x)={\frac {(-x)^{4}+1}{(-x)^{2}+(-x)^{4}}}}$ As we all know if ${\displaystyle (-x)^{2}=x^{2}}$ so that ${\displaystyle ((-x)^{2})^{2}=(x^{2})^{2}}$ hence ${\displaystyle (-x)^{4}=x^{4}}$ In fact any even number power will make what's inside the bracket positive. Thus ${\displaystyle f(-x)={\frac {x^{4}+1}{x^{2}+x^{4}}}}$ ${\displaystyle f(x)=f(-x)}$, f(x) is even. # Regions/Inequality ## Dependent and independent variables. Functional notation. Range and domain. Much of this course is devoted to the study of properties of real-valued functions of a real variable. Such a function f assigns to each element x of a given set of real numbers exactly one real number y, called the value of the function f at x. The dependence of y on f and on x is made explicit by using the notation f(x) to mean the value of f at x. The set of real numbers x on which f is defined is called the domain of f, while the set of values f(x) obtained as x varies over the domain of f is called the range or image of f. x is called the independent variable since it may be chosen freely within the domain of f, while y = f(x) is called the dependent variable since its value depends on the value chosen for x. The functions f studied in this course are usually given by an explicit rule involving calculations to be made on the variable x in order to obtain f(x). For this reason, a function f is often described in a form such as ‘y = f(x)’ with the domain of x specified. It is also common usage to refer to ‘the function f(x)’ where f(x) is prescribed but no domain is given. In such cases, the understanding required to be developed is that the domain of f is the set of real numbers for which the expression f(x) defines a real number. It is important to realise that use of the notation y = f(x) does not imply that the expression corresponding to f(x) is the same for all x. For example, the rule ${\displaystyle f(x)=x,x\geq 0}$ ${\displaystyle f(x)=-x,x<0\;}$ defines a function with domain all real x. The use of x and y is customary and is related to the geometrical representation of a function f by graphing the set of points (x, f(x)) for x in the domain of f, using cartesian (x, y) coordinates. Other symbols for independent and dependent variables occur frequently in practice and students should become familiar with functions defined in terms of other symbols. ## The graph of a function. Simple examples. The pictorial representation of a function is extremely useful and important, as is the idea that algebraic and geometrical descriptions of functions are both helpful in understanding and learning about their properties. The function y = f(x) may be represented pictorially by its graph, which is the set of points (x, f(x)) for each x in the domain of f, indicated with respect to cartesian coordinate axes 0x0y. Denoting the point with coordinates (x, f(x)) by P, the graph of the function (and sometimes the function itself) is often referred to as the ‘set of points P(x, f(x))’. Since f is a function, there is at most one point P of its graph on any ordinate. The graph of y = f(x) is also called the curve y = f(x) and the part of the curve lying between two ordinates is called an arc. Examples of functions y = f(x) should be given which illustrate different types of domain, bounded and unbounded ranges, continuous and discontinuous curves, curves which display simple symmetries, curves with sharp corners and curves with asymptotes. Students are to be encouraged to develop the habit of drawing sketches which indicate the main features of the graphs of any functions presented to them. They should also develop at this stage the habit of checking simple properties of functions and identifying simple features such as: where is the function positive? negative? zero?; where is it increasing? decreasing?; does it have any symmetry properties?; is it bounded?; does it have gaps (jumps) or sharp corners?; is there an asymptote? Knowledge of the symmetries of the graphs of odd and even functions is useful in curve sketching. A function f(x) is even if f(–x) = f(x) for all values of x in the domain. Its graph is symmetric with respect to reflection in the y-axis, i.e. it has line symmetry about the y-axis. A function f(x) is odd if f(–x) = –f(x) for all values of x in the domain. Its graph is symmetric with respect to reflection in the point 0 (the origin or axes), i.e. it has point symmetry about the origin. ## Algebraic representation of geometrical relationships. Locus problems. Some of the work of this section might profitably be discussed in conjunction with Topics 6 and 9. A circle with a given centre C and a given radius r is defined as the set of points in the plane whose distance from C is r. If cartesian coordinate axes 0x0y are set up in the plane so that C is the point with coordinates (a, b), then the distance formula shows that P(x, y) lies on the given circle if and only if x and y satisfy the equation (x – a)2 + (y – b)2 = r2, hence this equation is an algebraic representation corresponding to the geometrical description given above. It should be noted that if this equation is used to express y as a function of x, then two functions are obtained: y = b + √(r2 – (x – a)2) and y = b – √(r2 – (x – a)2), each with domain a – r less than or equal to x less than or equal to a + r. Generally, sets of points satisfying simple conditions stated in geometrical terms can be described in algebraic terms by introducing cartesian coordinates and interpreting the original conditions as conditions relating x and y. The conditions then usually reduce to one or more equations or inequalities. problems involving the determination of the set of points which satisfy a given number of conditions (which may be expressed geometrically or algebraically) are called locus problems and often stated in the form ‘Find the locus of a point P which satisfies …’. The means in practice ‘Find a simple algebraic or geometric description of the set of all points P which satisfy …’. ## Region and inequality. Simple examples. Treatment is to be restricted to regions of the (cartesian x, y–) plane which admit a simple geometrical description — for example, by use of words such as interior, exterior, bounded by, boundary, sector, common to, etc., — and which admit a simple algebraic description using one or more inequalities in x and y. Examples should be simple and involve at most one non-linear inequality, but should include both bounded and unbounded regions. Note that the case of one or more linear inequalities is specifically listed in Topic 6.4. A clear sketch diagram, illustrating the relevant regions, should be drawn for each example. Regions whose algebraic description involves two or more inequalities should be understood to correspond to the common part (intersection) of the regions determined by each separate inequality.
## Permutation and Combination Quiz-4 As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background. Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. Q1. In an examination of 9 papers a candidate has to pass in more papers, then the number of papers in which he fails in order to be successful. The number of ways in which he can be unsuccessful, is • 255 • 193 • 319 • 256 Solution (d) ∵ The candidate is unsuccessful, if he fails in 9 or 8 or 7 or 6 or 5 papers. ∴ Numbers of ways to be unsuccessful =9C9 +9C8 +9C7 +9C6 +9C5 = 9C0+ 9C1 +9C2 +9C3 +9C4 =1/2( 9C0+9C1+.....+9C9 =1/2 (29 )=28=256 Q2.If the letters of the word KRISNA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word KRISNA is • 359 • 324 • 341 • None of these Solution (b) The number of words starting from A are =5!=120 The number of words starting from I are =5!=120 The number of words starting from KA are =4!=24 The number of words starting from KI are =4!=24 The number of words starting from KN are =4!=24 The number of words starting from KRA are =3!=6 The number of words starting from KRIA are =2!=2 The number of words starting from KRIN are =2!=2 The number of words starting from KRISA are=1!=1 The number of words starting from KRISNA are=1!=1 Hence, rank of the word KRISNA =2(120)+3(24)+6+2(2)+2(1)=324 Q3. There is a set of m parallel lines intersecting a set of another n parallel lines in a plane. The number of parallelograms formed, is • m-1C2.n-1C2 • m-1C2.nC2 • mC2.n-1C2 • mC2.nC2 Solution (d)Since, a set of m parallel lines intersecting a set of another n parallel lines in a plane, then the number of parallelograms formed is mC2.nC2. Q4. From 12 books, the difference between number of ways a selection of 5 books when one specified book is always excluded and one specified book is always included, is • 64 • 118 • 330 • 132 Solution (d) Required number of ways = 11C5- 11C4 =11!/5!6!=11!/4!7!=132 Q5. The exponent of 3 in 100 !, is • 48 • 44 • 52 • 33 Solution (a) We have, E3 (100 !)=[100/3]+[100/32 ]+[100/33 ][100/34 ]=33+11+3+1=48 Q6. The number of all the possible selections which a student can make for answering one or more questions out of eight given questions in a paper, when each question has an alternative is • 256 • 6560 • 6561 • None of these Solution (b) Each question can be omitted or one of the two parts can be attempted i.e. it can be taken in 3 ways. So, 8 questions can be attempted in 38-1=6560 ways Q7. The number of times the digit 5 will be written when listing the integers from 1 to 1000, is • 271 • 272 • 300 • None of these Solution (c) Since, 5 does not occur in 1000, we have to count the number of times 5 occurs when we list the integers from 1 to 999. Any number between 1 and 999 is of the form xyz,0≤x,y,z≤9 The number in which 5 occurs exactly once =(3C1 )9×9=243 The number in which 5 occurs exactly twice=(3C2.9)=27 The number in which 5 occurs in all three digits =1 Hence, the number of times 5 occurs =1×243+2×27+3×1=300 Q8. If the letters of the word LATE be permuted and the words so formed be arranged as in a dictionary. Then, the rank of LATE is • 12 • 14 • 13 • 15 Solution (b) 14 Q9. A student is to answer 10out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is • 140 • 280 • 346 • 196 Solution (d) Total number of ways=5C4×8C6×5C5×8C4 =5!/(4!×1!)×8!/(2!×6!)+8!/(5!×3!) =(5×8×7)/2+(8×7×6)/6 =140+56=196 Q10. A lady gives a dinner party to 5 guests to be selected from nine friends. The number of ways of forming the party of 5, given that two of the friends will not attend the party together is • 56 • 91 • 126 • None of these Solution (b) Required number = 9 C5- 7 C3=91 #### Written by: AUTHORNAME AUTHORDESCRIPTION ## Want to know more Please fill in the details below: ## Latest NEET Articles\$type=three\$c=3\$author=hide\$comment=hide\$rm=hide\$date=hide\$snippet=hide Name ltr item BEST NEET COACHING CENTER \| BEST IIT JEE COACHING INSTITUTE \| BEST NEET, IIT JEE COACHING INSTITUTE: Permutation-and-combination-quiz-4 Permutation-and-combination-quiz-4 https://1.bp.blogspot.com/-wsuLBGivsiY/X3iv4Kp0R6I/AAAAAAAAAY8/A6ZvHiP4OwAHEfjN7tVTHUWXu2rW-4HkQCLcBGAsYHQ/w640-h336/Quiz%2BImage%2BTemplate%2B%25283%2529.jpg https://1.bp.blogspot.com/-wsuLBGivsiY/X3iv4Kp0R6I/AAAAAAAAAY8/A6ZvHiP4OwAHEfjN7tVTHUWXu2rW-4HkQCLcBGAsYHQ/s72-w640-c-h336/Quiz%2BImage%2BTemplate%2B%25283%2529.jpg BEST NEET COACHING CENTER \| BEST IIT JEE COACHING INSTITUTE \| BEST NEET, IIT JEE COACHING INSTITUTE https://www.cleariitmedical.com/2020/10/permutation-and-combination-quiz-4.html https://www.cleariitmedical.com/ https://www.cleariitmedical.com/ https://www.cleariitmedical.com/2020/10/permutation-and-combination-quiz-4.html true 7783647550433378923 UTF-8
# 445 in words 445 in words is written as Four Hundred and Forty Five. 445 represents the count or value. The article on Counting Numbers can give you an idea about count or counting. The number 445 is a 3 digit number that is used in expressions related to money, days, distance, length, weight and so on. Let us consider an example for 445. “The living room area is Four Hundred and Forty Five square meters”. Another example is “I have a collection of Four Hundred and Forty Five mickey mouse items.” 445 in words Four Hundred and Forty Five Four Hundred and Forty Five in Numbers 445 ## How to Write 445 in Words? We can convert 445 to words using a place value chart. The number 445 has 3 digits, so let’s make a chart that shows the place value up to 3 digits. Hundreds Tens Ones 4 4 5 Thus, we can write the expanded form as: 4 × Hundred + 4 × Ten + 5 × One = 4 × 100 + 4 × 10 + 5 × 1 = 445 = Four Hundred and Forty Five. 445 is the natural number that is succeeded by 444 and preceded by 446. 445 in words – Four Hundred and Forty Five. Is 445 an odd number? – Yes. Is 445 an even number? – No. Is 445 a perfect square number? – No. Is 445 a perfect cube number? – No. Is 445 a prime number? – No. Is 445 a composite number? – Yes. ## Solved Example 1. Write the number 445 in expanded form Solution: 4 × 100 + 4 × 10 + 5 × 1 We can write 445 = 400 + 40 + 5 = 4 × 100 + 4 × 10 + 5 × 1. ## Frequently Asked Questions on 445 in words Q1 ### How to write the number 445 in words? 445 in words is written as Four Hundred and Forty Five. Q2 ### State True or False. 445 is divisible by 3? False. 445 is not divisible by 3. Q3 ### Is 445 a perfect square number? No. 445 is not a perfect square number.
Suggested languages for you: Americas Europe Problem 427 # Find the general solution to each of the following differential equations: $$[1] \mathrm{y}^{\prime}-\mathrm{y}=0 .$$ [2] $$\mathrm{y}^{\prime \prime}+4 \mathrm{y}^{\prime}-\mathrm{y}=0$$. [3] $$\mathrm{y}^{\prime \prime}+8 \mathrm{y}^{\prime}+16 \mathrm{y}=0$$ Expert verified The general solutions for the given differential equations are: 1. $$y = Ce^x$$ 2. $$y(x) = A e^{(-2 + \sqrt{5})x} + B e^{(-2 - \sqrt{5})x}$$ 3. $$y(x) = (A + Bx)e^{-4x}$$ See the step by step solution ## Step 1: Solution for equation [1] To solve the first-order linear ODE, we can use the method of separation of variables. We have: $y' - y = 0$ Rearranging the equation, we get: $y' = y$ Now, we can separate the variables: $\frac{dy}{y} = dx$ Next, we will integrate both sides with respect to their respective variables: $\int \frac{dy}{y} = \int dx$ $\ln \|y\| = x + C_1$ Now, we will use exponentiation to solve for y: $y = e^{x + C_1}$ $y = e^{C_1}e^x$ Since $$e^{C_1}$$ is an arbitrary constant, we can write the general solution for equation [1] as: $y = Ce^x$ ## Step 2: Solution for equation [2] To solve the second-order linear homogeneous ODE, we will use the method of solving the characteristic equation. We start with the given equation: $y'' + 4y' - y = 0$ The characteristic equation of the given ODE is: $r^2 + 4r - 1 = 0$ Solving the quadratic equation, we get: $r = \frac{-4\pm \sqrt{4^2 - 4(-1)}}{2}$ $r = -2 \pm \sqrt{5}$ As the roots are real and distinct, the general solution for equation [2] is: $y(x) = A e^{(-2 + \sqrt{5})x} + B e^{(-2 - \sqrt{5})x}$ ## Step 3: Solution for equation [3] To solve the second-order linear homogeneous ODE, we will also use the method of solving the characteristic equation. We start with the given equation: $y'' + 8y' + 16y = 0$ The characteristic equation for this ODE is: $r^2 + 8r + 16 = 0$ Solving the quadratic equation, we get: $r = \frac{-8 \pm \sqrt{8^2 - 4(16)}}{2}$ $r = -4$ As the roots are real and repeated, the general solution for equation [3] is: $y(x) = (A + Bx)e^{-4x}$ In conclusion, the general solutions for the given differential equations are: 1. $$y = Ce^x$$ 2. $$y(x) = A e^{(-2 + \sqrt{5})x} + B e^{(-2 - \sqrt{5})x}$$ 3. $$y(x) = (A + Bx)e^{-4x}$$ We value your feedback to improve our textbook solutions. ## Access millions of textbook solutions in one place • Access over 3 million high quality textbook solutions • Access our popular flashcard, quiz, mock-exam and notes features ## Join over 22 million students in learning with our Vaia App The first learning app that truly has everything you need to ace your exams in one place. • Flashcards & Quizzes • AI Study Assistant • Smart Note-Taking • Mock-Exams • Study Planner
# Tessellations in geometry A couple of examples of tessellations in geometry are shown below. Basically, whenever you place a polygon together repeatedly without any gaps or overlaps, the resulting figure is a tessellation. A tessellation is also called tiling. Of course, tiles in your house is a real-life example of tessellation. Tessellation with rectangles Tessellation with equilateral triangles ## Tessellations in geometry: How to quickly check if a geometric figure will tessellate. What makes the above figures tessellations? 1) The rectangles or triangles are repeated to cover a flat surface. 2) No gaps, or overlaps between the rectangles or the triangles. Not all figures will form tessellations in geometry. When a figure can form a tessellation, the figure is said to tessellate. Every triangle tessellates Why can we say with confidence that the above 2 statements are true? Well, since there are no gaps and no overlaps, the sum of the measures of the angles around any vertex must be equal to 360 degrees as seen below with red circles. Therefore, if the measure of an angle of a figure is not a factor of 360, it will not tessellate. This is a very important concept since it will help determine when a figure can tessellate. We can use then the formula to find the interior angle of a regular polygon to check if a figure will tessellate. Interior angle of a regular polygon = [180 × (n-2)] / n ## A couple of examples showing how to use the formula above to determine which figures will tessellate. Example #1 Determine whether a regular pentagon will tessellate A pentagon has 5 sides, so n = 5. Interior angle of the pentagon = [180 × (5-2)] / 5 Interior angle of the pentagon = [180 × 3] / 5 Interior angle of the pentagon = 540 / 5 Interior angle of the pentagon = 108 degrees There is no way to make 360 with 108 since 108 + 108 + 108 = 324 and 108 + 108 + 108 + 108 = 432 Therefore the pentagon will not tessellate as you can see below: The gap is shown with a red arrow! Example #2 Determine whether a regular hexagon will tessellate A hexagon has 6 sides, so n = 6. Interior angle of the hexagon = [180 × (6-2)] / 6 Interior angle of the hexagon = [180 × 4] / 6 Interior angle of the hexagon = 720 / 6 Interior angle of the hexagon = 120 degrees 120 is a factor of 360, so the hexagon will tessellate. Example #3 Determine whether a regular 16-gon will tessellate A 16-gon has 16 sides, so n = 16. Interior angle of the 16-gon = [180 × (16-2)] / 16 Interior angle of the 16-gon = [180 × 14] / 16 Interior angle of the 16-gon = 2520 / 16 Interior angle of the 16-gon = 157.5 degrees 157.5 is not a factor of 360, so the 16-gon will not tessellate. Tessellations can happen with translations, rotations, and reflections and it can also happen with irregular figures. ## Types of tessellations The most common types of tessellations are regular tessellations and semi-regular tessellations. ### Regular tessellations Regular tessellations are made using only regular polygons. For example, you can only use an equilateral triangle, a square, octagon, or any other regular polygon. However, a regular tessellation can only be formed with an equilateral triangle, a square, and a hexagon. If you use any other regular polygon, the interior angle of the regular polygon will never be a factor of 360. The following figure shows a regular tessellation made with hexagons. ### Semi-regular tessellations A semi-regular tessellation is put together with two or more regular polygons. The following semi-regular tessellation is made up of triangles and octagons. Eight types of semi-regular tessellations can be formed • Using squares and octagons like the one shown above • Using triangles and squares (two different patterns) • Using hexagons and triangles (two different patterns) • Using dodecagons and triangles • Using dodecagons, squares, and hexagons • Using hexagons, triangles, and squares 100 Tough Algebra Word Problems. If you can solve these problems with no help, you must be a genius! Recommended
Architecture # How Do You Write 29/9 As A Mixed Number As A Mixed Number? How Do You Write 29/9 As A Mixed Number Trying to find out how to convert 29/9 into a mixed number or fraction? Have I got the answer for you! In this guide, we”ll walk you through the step-by-step process of converting an improper fraction, in this case 29/9, to a mixed number. Read on! Want to quickly learn or show students how to convert 29/9 to a mixed number? Play this very quick and fun video now! Before we begin, let”s revisit some basic fraction terms so you understand exactly what we”re dealing with here: Numerator. This is the number above the fraction line. For 29/9, the numerator is 29.Denominator. This is the number below the fraction line. For 29/9, the denominator is 9.Improper fraction. This is a fraction where the numerator is greater than the denominator.Mixed number. This is a way of expressing an improper fraction by simplifying it to whole units and a smaller overall fraction. It”s an integer (whole number) and a proper fraction. You are watching: 29/9 as a mixed number Now let”s go through the steps needed to convert 29/9 to a mixed number. ## Step 1: Find the whole number We first want to find the whole number, and to do this we divide the numerator by the denominator. Since we are only interested in whole numbers, we ignore any numbers to the right of the decimal point. 29/9= 3.2222222222222 = 3 Now that we have our whole number for the mixed fraction, we need to find our new numerator for the fraction part of the mixed number. ## Step 2: Get the new numerator To work this out we”ll use the whole number we calculated in step one (3) and multiply it by the original denominator (9). The result of that multiplication is then subtracted from the original numerator: 29 – (9 x 3) = 2 ## Step 3: Our mixed fraction We”ve now simplified 29/9 to a mixed number. To see it, we just need to put the whole number together with our new numerator and original denominator: 3 2/9 ## Step 4: Simplifying our fraction In this case, our fraction (2/9) can be simplified down further. In order to do that, we need to calculate the GCF (greatest common factor) of those two numbers. You can use our handy GCF calculator to work this out yourself if you want to. We already did that, and the GCF of 2 and 9 is 1. We can now divide both the new numerator and the denominator by 1 to simplify this fraction down to its lowest terms. 2/1 = 2 9/1 = 9 When we put that together, we can see that our complete answer is: 3 2/9 Hopefully this tutorial has helped you to understand how to convert any improper fraction you have into a mixed fraction, complete with a whole number and a proper fraction. You”re free to use our calculator below to work out more, but do try and learn how to do it yourself. It”s more fun than it seems, I promise! ## Improper Fraction to Mixed Number Enter an improper fraction numerator and denominator
## 21 To 25 Table #### In which table 25 comes in? Table of 25 – Learn 25 times table \| Table of Twenty-Five PDF The table of 25 is the multiplication table that includes the multiples of the number 25 when multiplied by a set of whole numbers. Table of 25 represents the repeated addition of 25 with itself. 1. For example, 3 rows of 25 soldiers each gives the sum of 25 + 25 + 25 = 75. 2. Therefore, the total number of soldiers is 75. 3. Or we can also write it as, 3 x 25 = 75, where 3 is the number of rows and 25 is the number of soldiers standing in each row. 4. Table of 25 is an easy table to learn and memorise. 5. In schools, from the very primary classes, we have learned about Maths tables. These maths tables result from repeated addition of any number, repeated for a certain number of times, In this article, you will learn what is 25 times table, how to write the table of 25 using multiplication along with chart and downloadable PDF. ### Is 21 in the 7 times table? 7 × 1 = 7.7 × 2 = 14 (7 + 7 = 14) 7 × 3 = 21 (7 + 7 + 7 = 21) 7 × 4 = 28 (7 + 7 + 7 + 7 = 28) ## What numbers get to 25? Examples – Example 1: Find the factors of 25 and 24. Solution: The factors of 25 are 1, 5 and 25. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. Thus, the common factor of 25 and 24 is 1. Example 2: Find the factors of 25 and 26. Solution: Factors of 25 = 1, 5 and 25. Factors of 26 = 1, 2, 13 and 26. Hence, the common factor of 25 and 26 is 1. Example 3: Find the factors of 25 and 13. Solution: The factors of 25 are 1, 5 and 25. The factors of 13 are 1 and 13. As 13 is a prime number, the common factor of 25 and 13 is 1. Learn Maths topics, factors and prime factors here with us and also download BYJU’S – The Learning App for interactive videos. You might be interested: 11Th Public Time Table 2023 The factors of 25 are 1, 5 and 25 and its negative factors are -1, -5 and -25. The prime factorization of 25 is 5 × 5 or 5 2, The positive factors of 25 are (1, 25) and (5, 5) The negative factors of 25 are (-1, -25) and (-5, -5) The sum of factors of 25 is 31. ## How do you write 25 in words? How to Write 25 in Words? – The number 25 can be expressed in words with the help of a place value chart. Therefore, for any number, first, we have to find the place value of each digit of the given number. For the number 25, the digit in one’s place is 5 and the digit in ten’s place is 2. Since 25 is a two-digit number, there are no hundreds and thousands. Hence, the place value chart for the number 25 is:Therefore, we can write the expanded form as:2 × Ten + 5 × One= 2 × 10 + 5 × 1= 20 + 5= 25= Twenty-FiveThus, 25 in words is written as Twenty-Five.Learn more about here Interesting way of writing 25 in words 2 = Two5 = Five20 = Twenty20 + 5 = 25Twenty + five = Twenty-FiveTherefore, the word form of the number 25 is Twenty-Five.
# Star Problem: Ratio of girls and boys in a class ## Problem: There are 40 pupils in a class. Each girl is given 5 books and each boy is given 3 books. The girls have 128 more books than the boys. How many girls are there in the class? ### Given: 40 pupils 5 books each girl = 5 books / girl 3 books each boy = 3 books / boy Girls have 128 more books than the boys How many girls are there in the class? ### Solution: LET: Gnumber of girls in the classBnumber of boys in the class Total books of the girls = number of books each girl x number of girls in the class NS. 1 Total books of the girls = ( 5 books / girl ) x G Total books of the girls = 5G Eq. 1 Total books of the boys = number of books each boy x number of boys in the class NS. 2 Total books of the boys = ( 3 books / boy ) x B Total books of the boys = 3B Eq. 2 It was given that, the Girls have 128 more books than the boys; using comparing entities, we have the following Number Sentence (NS) : Total books of the girls = Total books of the boys + 128 NS. 3 Substituting, Equations (Eq.) 1 & 2 on NS. 3, we have: 5G = 3B + 128 Eq. 3 The problem hinted us that there were 40 pupils; the pupils consists of girls and boys, hence, we have the following Number Sentence (NS): Number of girls in the class + Number of boys in the class = 40 G + B = 40 Eq. 4 Manipulating Eq. 4, we get: B = 40 – G Eq. 4a Substituting Eq. 4a on Eq. 3, we get: 5G = 3(40 – G) + 128 Simplifying: 5G = 120 – 3G + 128 5G + 3G = 120 + 128 8G = 248 8G / 8 = 248 / 8 G = 31 Ans. There are 31 girls in the class Using Eq. 4a, we get the number of boys in the class B = 40 – G B = 40 – 31 B = 9 Ans. There are 9 boys in the class ### Checking: Using Eq. 1, Total books of the girls = 5G, we get : 5G = (5 books/girl) x 31 girls = 155 books Using Eq. 2, Total books of the boys = 3G, we get : 3G = (3 books/boy) x 9 boys = 27 books Using NS. 3, Total books of the girls = Total books of the boys + 128 Total books of the girls = 27 books + 128 Total books of the girls = 155 books ## 2 Replies to “Star Problem: Ratio of girls and boys in a class” 1. Missy Torralba says: My heartfelt thanks for the quick reply! My daughter is happy she has completed her homework before going to school. Homeworkz now will be our bestfriend in the coming days! 2. estong says: hi missy, you’re welcome 🙂 sorry for the late reply, i’ve been busy lately. but anyhow, please feel free to post your questions & i’ll try my best to reply as soon as i can. thanks
# How do you write an equation of a line with slope of -3 and passing through (-2,4)? Mar 9, 2018 $y = - 3 x - 2$ #### Explanation: Given - Slope of the line $- 3$ Point $\left(- 2 , 4\right)$ Use the formula - $m x + c = y$ Where - $m$ slope of the line $x , y$ x and y coordinates, through which the line passes in our case - $m = - 3$ $x = - 2$ $y = 4$ $\left(- 3\right) \left(- 2\right) + C = 4$ $6 + c = 4$ $c = 4 - 6 = - 2$ $- 3 x - 2 = y$ The equation of the required line is $y = - 3 x - 2$ Mar 9, 2018 $y = - 3 x - 2$ #### Explanation: There are three ways to write the equation of a line: slope intercept form, point slope form, and standard (general) form. Slope Intercept Form: $y = m x + b$ where m is the slope of the line $\left(\frac{\Delta y}{\Delta x}\right)$ and b is the y-intercept. For a line with slope -3 and point (-2,4), plug -3 in for m, -2 for x, 4 for y, and solve for b. $4 = \left(- 3\right) \cdot \left(- 2\right) + b$ $4 = 6 + b$ $- 2 = b$ The equation in slope intercept form is $y = - 3 x - 2$ Point Slope Form: $\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$ Where m is the slope of the line $\left(\frac{\Delta y}{\Delta x}\right)$, ${y}_{1}$ is the y coordinate of a point, and ${x}_{1}$ is the x coordinate of a point. For a line with slope -3 and point (-2,4), plug -3 in for m, -2 for ${x}_{1}$, 4 for ${y}_{1}$. $\left(y - 4\right) = - 3 \left(x + 2\right)$ Standard Form: Ax + By = C Where A, B, and C are integers. To write an equation in standard form, rewrite the equation in point slope form so that it fits the formula for standard form. $\left(y - 4\right) = - 3 \left(x + 2\right)$ $y - 4 = - 3 x - 6$ $y + 3 x - 4 = - 6$ $y + 3 x = - 2$
# Difference between revisions of "Tessellations by Polygons" Oberkapfenberg Castle, Syria. Relevant examples from Escher's work: Recall that a polygon is a closed plane figure made by joining line segments. You might want to review the relevant material in Fundamental Concepts concerning polygons before reading this section. ## Some Basic Tessellations The fundamental question we will discuss in this section is: Which polygons tessellate? More precisely, which polygons can be used as the only tile in a monohedral tessellation of the plane? Before moving on, you may want to do the Tessellation Exploration: The Basics The most common and simplest tessellation uses a square. Squares easily form horizontal strips: Stacks of these strips cover a rectangular region and the pattern can clearly be extended to cover the entire plane. The same technique works with parallelograms, and so: All parallelograms tessellate. Special parallelograms such as rectangles, and rhombuses also tessellate. Looking for other tessellating polygons is a complex problem, so we will organize the question by the number of sides in the polygon. The simplest polygons have three sides, so we begin with triangles: All triangles tessellate. To see this, take an arbitrary triangle and rotate it about the midpoint of one of its sides. The resulting parallelogram tessellates: The picture works because all three corners (A, B, and C) of the triangle come together to make a 180° angle - a straight line. This property of triangles will be the foundation of our study of polygon tessellations, so we state it here: The sum of angles of any triangle is 180°. Moving up from triangles, we turn to four sided polygons, the quadrilaterals. Before continuing, try the Quadrilateral Tessellation Exploration. Recall that a quadrilateral is a polygon with four sides. The sum of angles in any quadrilateral is 360° To prove, divide a quadrilateral into two triangles as shown: Since the angle sum of any triangle is 180°, and there are two triangles, the angle sum of the quadrilateral is 180° + 180° = 360°. Taking a little more care with the argument, we have: $\alpha_1 + \delta_1 + \gamma = 180^\circ$ and $\alpha_2 + \delta_2 + \beta = 180^\circ$. Then $360^\circ = \alpha_1 + \delta_1 + \gamma + \alpha_2 + \delta_2 + \beta = \alpha_1 + \alpha_2 + \beta + \gamma + \delta_1 + \delta_2 = \angle A + \angle B + \angle C + \angle D$. This division into triangles does not calculate the angle sum of the quadrilateral. The point of all the letters is that the angles of the triangles make the angles of the quadrilateral, which would not work if the quadrilateral was divided as shown on the right. We now turn to the main result of this section: Begin with an arbitrary quadrilateral ABCD. Rotate by 180° about the midpoint of one of its sides, and then repeat using the midpoints of other sides to build up a tessellation. The angles around each vertex are exactly the four angles of the original quadrilateral. Since the angle sum of the quadrilateral is 360°, the angles close up, the pattern has no gaps or overlaps, and the quadrilateral tessellates. Recall from Fundamental Concepts that a convex shape has no dents. All triangles are convex, but there are non-convex quadrilaterals. The technique for tessellating with quadrilaterals works just as well for non-convex quadrilaterals: It is worth noting that the general quadrilateral tessellation results in a wallpaper pattern with p2 symmetry group. ## Tessellations by Convex Polygons Every shape of triangle can be used to tessellate the plane. Every shape of quadrilateral can be used to tessellate the plane. In both cases, the angle sum of the shape plays a key role. Since triangles have angle sum 180° and quadrilaterals have angle sum 360°, copies of one tile can fill out the 360° surrounding a vertex of the tessellation. The next simplest shape after the three and four sided polygon is the five sided polygon: the pentagon. The angle sum of any pentagon is 540°, because we can divide the pentagon into three triangles: Since each triangle has angle sum 180° the angle sum of the pentagon is 180° + 180° + 180° = 540°. Rather than repeat the angle sum calculation for every possible number of sides, we look for a pattern. The angle sum of a triangle (3-gon) is 180°, the angle sum of a quadrilateral (4-gon) is 2x180°, and the angle sum of a pentagon is 3x180°. A general polygon with $n$ sides can be cut into $n-2$ triangles and so we have: The sum of the angles of an $n$-gon is $(n-2)\times 180^\circ$. Unlike the triangle and quadrilateral case, the pentagon's angle sum of 540° is not helpful when trying to fit a bunch of pentagons around a vertex. In fact, there are pentagons which do not tessellate the plane. For example, the regular pentagon has five equal angles summing to 540°, so each angle of the regular pentagon is $\frac{540^\circ}{5} = 108^\circ$. Attempting to fit regular polygons together leads to one of the two pictures below: Both situations have wedge shaped gaps that are too narrow to fit another regular pentagon. Thus, not every pentagon tessellates. On the other hand, some pentagons do tessellate, for example this house shaped pentagon: The house pentagon has two right angles. Because those two angles sum to 180° they can fit along a line, and the other three angles sum to 360° (= 540° - 180°) and fit around a vertex. Thus, some pentagons tessellate and some do not. The situation is the same for hexagons, but for polygons with more than six sides there is the following: No convex polygon with seven or more sides can tessellate. This remarkable fact is difficult to prove, but just within the scope of this book. However, the proof must wait until we develop a counting formula called the Euler characteristic, which will arise in our chapter on Non-Euclidean Geometry. Nobody has seriously attempted to classify non-convex polygons which tessellate, because the list is quite likely to be too long and messy to describe by hand. However, there has been quite a lot of work towards classifying convex polygons which tessellate. Because we understand triangles and quadrilaterals, and know that above six sides there is no hope, the classification of convex polygons which tessellate comes down to two questions: • Which convex pentagons tessellate? • Which convex hexagons tessellate? Question 2 was completely answered in 1918 by K. Reinhardt.[1] Reinhardt showed that there are only three types of convex hexagons which tessellate: Reinhardt also addressed Question 1 and gave five types of pentagon which tessellate. In 1968, R. Kershner[2] found three new types, and claimed a proof that the eight known types were the complete list. A 1975 article by Martin Gardner[3] in Scientific American popularized the topic, and led to a surprising turn of events. In fact Kershner's "proof" was incorrect. After reading the Scientific American article, a computer scientist, Richard James III, found a ninth type of convex pentagon that tessellates. Not long after that, Marjorie Rice, a San Diego homemaker with only a high school mathematics background, discovered four more types, and then a German mathematics student, Rolf Stein, discovered a fourteenth type in 1985. Since 1985, no new types have been discovered, and many mathematicians believe that the list is finally complete. However, there is no well accepted proof of the classification, so it remains possible that there is a fifteenth or even many more types of convex pentagons that tessellate. Today, question 1 is an open problem, a problem whose solution is unknown. Summary of Polygon Tessellations Sides Angle Sum Tessellates? 3 180° Yes. All triangles tessellate. 4 360° Yes. All quadrilaterals tessellate. 5 540° Sometimes. There is a list of 14 types of convex pentagons which tessellate, but nobody knows if the list is complete. 6 720° Sometimes. There is a list of 3 types of convex hexagons which tessellate, and these are the only three. 7 & up 900°, 1080°, ... No convex $n$-gon tessellates for $n \geq 7$ ## Tessellations by Regular Polygons Recall that a regular polygon is a polygon whose sides are all the same length and whose angles all have the same measure. A regular $n$-gon has $n$ equal angles that sum to $(n-2)180^\circ$, so: The corner angle of a regular $n$-gon is $\frac{(n-2)180^\circ}{n}$. The table shows the corner angles for the first few regular polygons: Number of Sides Corner angle 3 4 5 6 7 8 9 10 11 12 60° 90° 108° 120° ~128.6° 135° 140° 144° ~147.3° 150° ### Regular Tessellations Regular tessellation A tessellation using one regular polygon tile, arranged so that edges match up. Corners of the tiles need to fit together around a point, which means the corner angle of the regular polygon must evenly divide 360°. Since $6 \times 60^\circ = 360^\circ$, there is a regular tessellation using six triangles around each vertex. Since $4 \times 90^\circ = 360^\circ$, there is a regular tessellation using four squares around each vertex. And since $3 \times 120^\circ = 360^\circ$ there is a regular tessellation using three hexagons around each vertex. We have already seen that the regular pentagon does not tessellate. A regular polygon with more than six sides has a corner angle larger than 120° (which is 360°/3) and smaller than 180° (which is 360°/2) so it cannot evenly divide 360°. We conclude: There are three regular tessellations of the plane: by triangles, by squares, by hexagons. A major goal of this book is to classify all possible regular tessellations. Apparently, the list of three regular tessellations of the plane is the complete answer. However, these three regular tessellations fit nicely into a much richer picture that only appears later when we study Non-Euclidean Geometry. Tessellations using different kinds of regular polygon tiles are fascinating, and lend themselves to puzzles, games, and certainly tile flooring. Try the Pattern Block Exploration. ### Archimedean tessellations This section is unfinished. An archimedean tessellation using regular triangles and squares ## Notes 1. K. Reinhardt, Über die Zerlegung der Ebene in Polygone. (Inaugural-Disstertation, Univ. Frankfurt a.M.) R. Noske, Borna and Leipzig, 1918. 2. R. Kershner, On Paving the Plane, The American Mathematical Monthly 75, October 1968, pg. 839-844 3. Martin Gardner, Time Travel and Other Mathematical Bewilderments Ch. 13. W.H. Freeman, 1988
• Share Send to a Friend via Email Your suggestion is on its way! An email with a link to: was emailed to: Thanks for sharing About.com with others! Hypothesis Testing Using One-Sample t-Tests Hypothesis Testing Using One-Sample t-Tests First we’ll consider our hypothesis that the intercept variable equals one. The idea behind this is explained quite well in Gujarati’s Essentials of Econometrics. On page 105 Gujarati describes hypothesis testing: “[S]uppose we hypothesize that the true B1 takes a particular numerical value, e.g., B1 = 1. Our task now is to “test” this hypothesis.” “In the language of hypothesis testing a hypothesis such as B1 = 1 is called the null hypothesis and is generally denoted by the symbol H0. Thus H0: B1 = 1. The null hypothesis is usually tested against an alternative hypothesis, denoted by the symbol H1. The alternative hypothesis can take one of three forms: H1: B1 > 1, which is called a one-sided alternative hypothesis, or H1: B1 < 1, also a one-sided alternative hypothesis, or H1: B1 not equal 1, which is called a two-sided alternative hypothesis. That is the true value is either greater or less than 1.” In the above I’ve substituted in our hypothesis for Gujarati’s to make it easier to follow. In our case we want a two-sided alternative hypothesis, as we’re interested in knowing if B1 is equal to 1 or not equal to 1. The first thing we need to do to test our hypothesis is to calculate at t-Test statistic. The theory behind the statistic is beyond the scope of this article. Essentially what we are doing is calculating a statistic which can be tested against a t distribution to determine how probable it is that the true value of the coefficient is equal to some hypothesized value. When our hypothesis is B1 = 1 we denote our t-Statistic as t1(B1=1) and it can be calculated by the formula: t1(B1=1) = (b1 - B1 / se1) Let’s try this for our intercept data. Recall we had the following data: Intercept b1 = 0.47 se1 = 0.23 Our t-Statistic for the hypothesis that B1 = 1 is simply: t1(B1=1) = (0.47 – 1) / 0.23 = 2.0435 So t1(B1=1) is 2.0435. We can also calculate our t-test for the hypothesis that the slope variable is equal to -0.4: X Variable b2 = -0.31 se2 = 0.03 Our t-Statistic for the hypothesis that B2 = -0.4 is simply: t2(B2= -0.4) = ((-0.31) – (-0.4)) / 0.23 = 3.0000 So t2(B2= -0.4) is 3.0000. Next we have to convert these into p-values. The p-value "may be defined as the lowest significance level at which a null hypothesis can be rejected...As a rule, the smaller the p value, the stronger is the evidence against the null hypothesis." (Gujarati, 113) As a standard rule of thumb, if the p-value is lower than 0.05, we reject the null hypothesis and accept the alternative hypothesis. This means that if the p-value associated with the test t1(B1=1) is less than 0.05 we reject the hypothesis that B1=1 and accept the hypothesis that B1 not equal to 1. If the associated p-value is equal to or greater than 0.05, we do just the opposite, that is we accept the null hypothesis that B1=1. Calculating the p-value Unfortunately, you cannot calculate the p-value. To obtain a p-value, you generally have to look it up in a chart. Most standard statistics and econometrics books contain a p-value chart in the back of the book. Fortunately with the advent of the internet, there’s a much simpler way of obtaining p-values. The site Graphpad Quickcalcs: One sample t test allows you to quickly and easily obtain p-values. Using this site, here’s how you obtain a p-value for each test. Steps Needed to Estimate a p-value for B1=1 • Click on the radio box containing “Enter mean, SEM and N.” Mean is the parameter value we estimated, SEM is the standard error, and N is the number of observations. • Enter 0.47 in the box labelled “Mean:”. • Enter 0.23 in the box labelled “SEM:” • Enter 219 in the box labelled “N:”, as this is the number of observations we had. • Under " 3. Specify the hypothetical mean value" click on the radio button beside the blank box. In that box enter 1, as that is our hypothesis. • Click “Calculate Now” You should get an output page. On the top of the output page you should see the following information: P value and statistical significance: The two-tailed P value equals 0.0221 By conventional criteria, this difference is considered to be statistically significant. So our p-value is 0.0221 which is less than 0.05. In this case we reject our null hypothesis and accept our alternative hypothesis. In our words, for this parameter, our theory did not match the data. Be Sure to Continue to Page 3 of "Hypothesis Testing Using One-Sample t-Tests".
Oberlaudacu 2022-01-11 If a frictional force of 50 N is applied to each side of the tires, determine the average shear strain in the rubber. Each pad has cross-sectional dimensions of 20 mm and 50 mm. ${G}_{r}=0.20MPa$ Jenny Sheppard The base is 20 mm and 50 mm in size. The strain in the tire must be ascertained.Friction force ${F}_{t}=50N$ Using the expressed shear stress, we will determine T. Surface cross-sectional area: $A=50\cdot 20$ $=1000m{m}^{2}$ $\tau =\frac{{F}_{t}}{A}$ $=\frac{50}{1000}$ $\tau =0.05MPa$ We can also calculate the shear stress in the following way. $\tau =\gamma \cdot G$ Where: $\gamma$- shear strain $\tau$- shear strain $⇒\gamma =\frac{\tau }{G}$ $=\frac{0.05}{0.20}$ $\gamma =0.25rad$ The solution is: $\gamma =0.25rad$ Shannon Hodgkinson Given that $⇒$ Area of cross section $\left(Acs\right)=50×20$ $⇒Acs=1000m{m}^{2}$ Forse $\left(F\right)=50N$, $G=0\cdot 2MPa$ $\because ShearstressPSK\left(\tau \right)=\frac{F}{Acs}=\frac{50}{1000}=0.05MPa$ As we know $⇒Shearstra\in =\frac{Shearstress}{Shear\text{mod}\underset{―}{e}s}$ $⇒\gamma =\frac{\tau }{4}$ $⇒\gamma =\frac{0.05}{0.2}=0.25rad$ nick1337 $\gamma =0.00025$ Explanation: $\gamma =\frac{\Delta x}{L}$ where: $\gamma$ is the shear strain, $\Delta x$ is the change in length, and $L$ is the original length. Given that the pads have cross-sectional dimensions of 20 mm and 50 mm, we can assume that the original length $L$ is 20 mm. Since the pads are made of rubber, we can assume that the shear modulus of rubber (${G}_{r}$) is 0.20 MPa. The shear stress ($\tau$) is given by: $\tau =\frac{F}{A}$ where: $\tau$ is the shear stress, $F$ is the applied force, and $A$ is the cross-sectional area. In this case, the applied force ($F$) is 50 N, and the cross-sectional area ($A$) can be calculated as follows: $A=\text{{width}\right\}×\text{{height}\right\}$ Plugging in the values, we have: $A=20\phantom{\rule{0.167em}{0ex}}\text{mm}×50\phantom{\rule{0.167em}{0ex}}\text{mm}$ Now, we can substitute the values into the formula for shear stress to find the shear strain: $\tau ={G}_{r}·\gamma$ Solving for $\gamma$, we get: $\gamma =\frac{\tau }{{G}_{r}}$ Plugging in the values for $\tau$ and ${G}_{r}$, we have: $\gamma =\frac{50\phantom{\rule{0.167em}{0ex}}\text{N}}{20\phantom{\rule{0.167em}{0ex}}\text{mm}×50\phantom{\rule{0.167em}{0ex}}\text{mm}}·\frac{1\phantom{\rule{0.167em}{0ex}}\text{MPa}}{0.20\phantom{\rule{0.167em}{0ex}}\text{MPa}}$ Now, we can simplify the equation: $\gamma =\frac{50\phantom{\rule{0.167em}{0ex}}\text{N}}{1000\phantom{\rule{0.167em}{0ex}}{\text{mm}}^{2}}·5$ Converting the units to meters, we have: $\gamma =\frac{50\phantom{\rule{0.167em}{0ex}}\text{N}}{1000000\phantom{\rule{0.167em}{0ex}}{\text{mm}}^{2}}·5$ Simplifying further: $\gamma =\frac{50×5}{1000000}$ Thus, the average shear strain in the rubber is $0.00025$. Therefore, the average shear strain in the rubber is $\gamma =0.00025$. Vasquez To solve the problem, we can use the equation for shear strain, which is given by: Given that the shear stress is equal to the frictional force applied on each side of the tire (50 N), and the shear modulus is given as G_r = 0.20 MPa, we can substitute these values into the equation to find the shear strain. First, let's convert the shear modulus from MPa to N/m²: ${G}_{r}=0.20×{10}^{6}\phantom{\rule{0.167em}{0ex}}\text{Pa}$ The cross-sectional area of each pad can be calculated as follows: $A=\text{{width}\right\}×\text{{height}\right\}=\left(20\phantom{\rule{0.167em}{0ex}}\text{mm}\right)×\left(50\phantom{\rule{0.167em}{0ex}}\text{mm}\right)$ Next, we can calculate the shear strain using the formula mentioned earlier: Finally, we can simplify the units by noting that 1 mm = 0.001 m: $\gamma =\frac{50}{0.20×{10}^{6}}×\frac{1}{0.001}=\frac{50}{0.20×{10}^{6}×0.001}\phantom{\rule{0.167em}{0ex}}\text{m/m}$ Therefore, the average shear strain in the rubber is $\frac{50}{0.20×{10}^{6}×0.001}\phantom{\rule{0.167em}{0ex}}\text{m/m}$. RizerMix To determine the average shear strain in the rubber brake pads of a bicycle tire, we can use the formula: Given: - Shear stress, $\tau =50\phantom{\rule{0.167em}{0ex}}\text{N}$ - Cross-sectional dimensions: ${d}_{1}=20\phantom{\rule{0.167em}{0ex}}\text{mm}$ (width) and ${d}_{2}=50\phantom{\rule{0.167em}{0ex}}\text{mm}$ (height) - Shear modulus, ${G}_{r}=0.20\phantom{\rule{0.167em}{0ex}}\text{MPa}=0.20×{10}^{6}\phantom{\rule{0.167em}{0ex}}{\text{N/m}}^{2}$ First, let's convert the dimensions to meters: ${d}_{1}=20\phantom{\rule{0.167em}{0ex}}\text{mm}=20×{10}^{-3}\phantom{\rule{0.167em}{0ex}}\text{m}$ ${d}_{2}=50\phantom{\rule{0.167em}{0ex}}\text{mm}=50×{10}^{-3}\phantom{\rule{0.167em}{0ex}}\text{m}$ Now, we can calculate the average shear strain: Substituting the given values: $\text{{Shearstrain}\right\}=\frac{50\phantom{\rule{0.167em}{0ex}}\text{N}}{0.20×{10}^{6}\phantom{\rule{0.167em}{0ex}}{\text{N/m}}^{2}}$ Simplifying the expression: $\text{{Shearstrain}\right\}=\frac{50}{0.20×{10}^{6}}=0.25×{10}^{-3}$ Therefore, the average shear strain in the rubber brake pads is $0.25×{10}^{-3}$ rad. 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Presentation is loading. Please wait. # Learn to work with rates and ratios. ## Presentation on theme: "Learn to work with rates and ratios."— Presentation transcript: Learn to work with rates and ratios. Vocabulary rate unit rate unit price Movie and television screens range in shape from almost perfect squares to wide rectangles. An aspect ratio describes a screen by comparing its width to its height. Common aspect ratios are 4:3, 37:20, 16:9, and 47:20. Additional Example 1A: Ordering Ratios A. Order the ratios 4:3, 23:10, 13:9, and 47:20 from the least to greatest. 4 3 4 3 Divide. = 1.3 1 4:3 = = 1.3 23 10 23:10 = = 2.3 13 9 13:9 = = 1.4 47 20 47:20 = = 2.35 The decimals in order are 1.3, 1.4, 2.3, and 2.35. The ratios in order from least to greatest are 4:3, 13:9, 23:10, and 47:20. Additional Example 1B: Ordering Ratios B. A television has screen width 20 in. and height 15 in. What is the aspect ratio of this screen? The ratio of the width to the height is 20:15. The ratio can be simplified: 20 15 5(4) 5(3) 4 3 = = . The screen has the aspect ratio 4:3. Try This: Example 1A A. Order the ratios 2:3, 35:14, 5:3, and 49:20 from the least to greatest. 2 3 2 3 Divide. = 0.6 1 2:3 = = 0.6 35 14 35:14 = = 2.5 5 3 5:3 = = 1.6 49 20 49:20 = = 2.45 The decimals in order are 0.6, 1.6, 2.45, and 2.5. The ratios in order from least to greatest are 2:3, 5:3, 49:20, 35:14. Try This: Example 1B B. A movie theater has a screen width 36 ft. and height 20 ft. What is the aspect ratio of this screen? The ratio of the width to the height is 36:20. The ratio can be simplified: 36 20 4(9) 4(5) 9 5 = = . The screen has the aspect ratio 9:5. A ratio is a comparison of two quantities. A rate is a comparison of two quantities that have different units. 90 3 Ratio: Read as “90 miles per 3 hours.” 90 miles 3 hours Rate: Unit rates are rates in which the second quantity is 1. The ratio 90 3 can be simplified by dividing: 90 3 30 1 = 30 miles, 1 hour unit rate: or 30 mi/h Additional Example 2: Using a Bar Graph to Determine Rates Use the bar graph to find the number of acres, to the nearest acre, destroyed in Nevada and Alaska per week. 640,000 acres 52 weeks Nevada = 12,308 acres 1 week Additional Example 2 Continued Use the bar graph to find the number of acres, to the nearest acre, destroyed in Nevada and Alaska per week. 750,000 acres 52 weeks Alaska = 14,423 acres 1 week 950,000 acres Montana = 52 weeks 18,269 acres 1 week Try This: Example 2 Use the bar graph to find the number of acres, to the nearest acre, destroyed in Montana and Idaho per week. 950,000 acres 52 weeks Montana = 18,269 acres 1 week Try This: Example 2 Continued Use the bar graph to find the number of acres, to the nearest acre, destroyed in Montana and Idaho per week. 1,400,000 acres 52 weeks Idaho = 26,923 acres 1 week Unit price is a unit rate used to compare costs per item. Additional Example 3A: Finding Unit Prices to Compare Costs Pens can be purchased in a 5-pack for \$1.95 or a 15-pack for \$ Which is the better buy? Divide the price by the number of pens. price for package number of pens \$1.95 5 = \$0.39 price for package number of pens \$6.20 15 = \$0.41 The better buy is the 5-pack for \$1.95. Additional Example 3B: Finding Unit Prices to Compare Costs Jamie can buy a 15-oz jar peanut butter for \$2.19 or a 20-oz jar for \$ Which is the better buy? Divide the price by the number of ounces. price for jar number of ounces \$2.19 15 = \$0.15 price for jar number of ounces \$2.78 20 = \$0.14 The better buy is the 20-oz jar for \$2.78. Try This: Example 3A Golf balls can be purchased in a 3-pack for \$4.95 or a 12-pack for \$ Which is the better buy? Divide the price by the number of balls. price for package number of balls \$4.95 3 = \$1.65 price for package number of balls \$18.95 12 = \$1.58 The better buy is the 12-pack for \$18.95. Try This: Example 3B John can buy a 24 oz bottle of ketchup for \$2.19 or a 36 oz bottle for \$ Which is the better buy? Divide the price by the number of ounces. price for bottle number of ounces \$2.19 24 = \$0.09 price for bottles number of ounces \$3.79 36 = \$0.11 The better buy is the 24-oz jar for \$2.19. Lesson Quiz 1. At a family golf outing, a father drove the ball 285 ft. His daughter drove the ball 95 ft. Express the ratio of the father’s distance to his daughter’s in simplest terms. 2. Find the unit price of 6 stamps for \$2.22. 3. Find the unit rate of 8 heartbeats in 6 seconds. 4. What is the better buy, a half dozen carnations for \$4.75 or a dozen for \$9.24? 5. Which is the better buy, four pens for \$5.16 or a ten-pack for \$12.90? 3:1 \$0.37 per stamp  1.3 beats/s a dozen They cost the same. Download ppt "Learn to work with rates and ratios." Similar presentations Ads by Google
# Inter 1st Year Maths 1B The Plane Solutions Ex 7(a) Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B The Plane Solutions Exercise 7(a) will help students to clear their doubts quickly. ## Intermediate 1st Year Maths 1B The Plane Solutions Exercise 7(a) I. Question 1. Find the equation of the plane if the foot of the perpendicular from origin to the plane is (1, 3, -5). Solution: OP is the normal to the plane D. Rs of op are 1, 3, -5 The plane passes through P( 1, 3, -5) equation of the plane is 1(x – 1) + 3(y – 3) – 5(z + 5) = 0 x – 1 + 3y – 9 – 5z – 25 = 0 x + 3y – 5z – 35 = 0 Question 2. Reduce the equation x + 2y – 3z – 6 = 0 of the plane to the normal form. Solution: Equation of the plane is x + 2y – 3z – 6 = 0 i.e., x + 2y – 3z = 6 Dividing in the $$\sqrt{1^{2}+2^{2}+(-3)^{2}}=\sqrt{1+4+9}=\sqrt{14}$$ The equation of the plane in the normal form is Question 3. Find the equation of the plane. Whose intercepts on X, Y, Z – axis are 1, 2, 4 respectively. Solution: Equation of the plane in the intercept form is $$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}$$ = 1 Given a = 1, b = 2, c = 4. Equation of the required plane in the intercept form is $$\frac{x}{1}+\frac{y}{2}+\frac{z}{4}$$ = 1 Multiplying with 4, we get 4x + 2y + z = 4 Question 4. Find the intercepts of the plane 4x + 3y – 2z + 2 = 0 on the co-ordinate axes. Solution: Equation of the plane is 4x + 3y – 2z + 2 = 0 – 4x – 3y + 2z = 2 x – intercept = -1/2, y – intercept = -2/3, z intercept =1. Question 5. Find the d.c.’s of the normal to the plane x + 2y + 2z – 4 = 0. Solution: Equation of the plane is x + 2y + z- 4 = 0 d.r.’s of the normal are (1, 2, 2) Dividing with $$\sqrt{1+4+4}$$ = 3, d.c.’s of the normal to the plane are ($$\frac{1}{3}, \frac{2}{3}, \frac{2}{3}$$) Question 6. Find the equation of the plane passing through the point (-2, 1, 3) and having (3, -5, 4) as d.r.’s of its normal. Solution: d.r.’s of the normal are (3, -5, 4) and the plane passes through (-2, 1, 3). Equation of the plane is 3(x + 2) – 5(y – 1) + 4(z – 3) = 0 3x + 6 – 5y + 5 + 4z – 12 = 0 3x -5y+ 4z – 1 = 0. Question 7. Write the equation of the plane 4x – 4y + 2z + 5 = 0 in the intercept form. Solution: Equation of the plane is 4x – 4y + 2z + 5 = 0 – 4x + 4y – 2z = 5 $$-\frac{4x}{5}, \frac{4y}{5}, \frac{2z}{5}$$ Intercept form is x – intercept = $$\frac{5}{4}$$, y – intercept = $$\frac{5}{4}$$, z – intercept = –$$\frac{5}{2}$$ Question 8. Find the angle between the planes x + 2y + 2z-5 = 0 and 3x + 3y + 2z – 8 = 0. Solution: Equation of the planes are x + 2y + 2z – 5 = 0 3x + 3y + 2z – 8 = 0 II. Question 1. Find the equation of the plane passing through the point (1,1,1) and parallel to the plane x +.2y + 3z – 7 = 0. Solution: Equation of the given plane is x + 2y + 3z – 7 = 0. Equation of the parallel plane is x + 2y + 3z = k. This plane passes through the point P (1, 1, 1) 1 + 2 + 3 = k ⇒ k = -6 Equation of the required plane is x + 2y + 3z = 6 Question 2. Find the equation of the plane passing through (2, 3, 4) and perpendicular to x – axis. Solution: The plane is perpendicular to x – axis ∴ x – axis is the normal to the plane d.c.’s of x -axis are 1, 0, 0 Equation of the required plane is x = k This plane passes through the point P(2, 3, 4) ∴ 2 = k Equation of the required plane is x = 2. Question 3. Show that 2x + 3y + 7 = 0 represents a plane perpendicular to XY – plane. Solution: Equation of the given plane is 2x + 3y + 7 = 0 Equation of xy – plane is z = 0 a1a2 + b1b2 + c1c2 = 2.0 + 3.0 + 0.1 = 0 +0+0=0 The plane 2x + 3y + 7 = 0 is perpendicular to XY – plane. Question 4. Find the constant k so that the planes x – 2y + kz = 0 and 2x + 5y – z = 0 are at right angles. Find the equation of the plane through (1, -1, -1) and perpen-dicular to these planes. Solution: Equations of the given planes are x – 2y + kz = 0 and 2x + 5y – z = 0 These the planes are perpendicular 1.2 – 2.5 + k (-1) = 0 2 – 10 = k ⇒ k = -8 Equation of the planes x – 2y – 8z = 0 ………. (1) 2x + 5y – z = 0 ……….. (2) The required plane passes through (1, -1, -1) ∴ Equation of the plane can be taken as a(x – 1) + b(y + 1) + c(z + 1) = 0 ………… (3) This plane is perpendicular to the planes (1) and (2) a – 2b – 8c = 0 2a + 5b – c = 0 Substituting in (3), equation of the required planes 42 (x – 1) – 15(y + 1) + 9(z + 1) = 0 42x – 42 – 15y – 15 + 9z + 9 = 0 42x – 15y + 9z – 48 = 0. Question 5. Find the equation of the plane through (-1, 6, 2) and perpendicular to the join of (1, 2, 3) and (-2, 3, 4). Solution: The plane is perpendicular to the line joining A(1, 2, 3) and B(-2, 3, 4). d.r.’s of AB are 1 + 2, 2 – 3, 3 – 4 i.e., 3, -1, -1 AB is normal to the plane and the plane passes through the point P(-1, 6, 2) Equation of the required plane is 3(x + 1) – 1(y – 6) -1(z – 2) = 0 3x + 3 – y + 6- z + 2 = 0 3x – y – z + 11 = 0 Question 6. Find the equation of the plane bisecting the line segment joining (2, 0, 6) and (-6, 2, 4) and perpendicular to it. Solution: A (2, 0, 1), B(-6, 2, 4) are the given points ‘o’ is the mid point of AB Co-ordinates of O are $$\left(\frac{2-6}{2}, \frac{0+2}{2}, \frac{6+4}{2}\right)$$ = (-2, 1, 5) The plane is perpendicular to AB d.r.’s of the normal to the plane are 2 + 6, 0 – 2, 6 – 4 8, -2, 2 Equation of the required plane is +8 (x + 2) – 2(y – 1) + 2 (z – 5) = 0 8x + 16 – 2y + 2 + 2z – 10 = 0 8x – 2y + 2z + 8 = 0 Question 7. Find the equation of the plane passing through (0, 0, -4) and perpendicular to the line joining the points (1, -2, 2) and (-3, 1, -2). Solution: A(1, -2, 2), B(-3, 1, -2) are the given points d.r.’s of AB are 1 + 3, -2 -1, 2 + 2 i.e., 4, -3, 4 AB is normal to the plane and the plane passes through the point P(0, 0, -4). Equation of the required plane is 4(x – 0) – 3(y – 0) + 4(z + 4) = 0 4x – 3y + 4z + 16 = 0 Question 8. Find the equation of the plane through (4, 4, 0) and perpendicular to the planes 2x + y + 2 z + 3 = 0 and 3x + 3y + 2z – 8 = 0. Solution: Equation of the plane passing through P(4, 4, 0) is a(x – 4) + b(y – 4) + c(z – 0) = 0 ………. (1) This plane is perpendicular to 2x + y + 2z – 3 = 0 3x + 3y + 2z – 8 = 0 ∴ 2a + b + 2c = 0 ………… (2) 3a + 3b + 2c = 0 ………….(3) Substituting in (1), equation of the required plane is -4 (x – 4) + 2(y – 4) + 3(z – 0) = 0 -4x + 16 + 2y – 8 + 3z = 0 -4x + 2y + 3z + 8 = 0 III. Question 1. Find the equation of the plane through the points (2, 2, -1), (3, 4, 2), (7, 0, 6). Solution: A(2, 2, -1), B(3, 4, 2), C(7, 0, 6) are the given points. Equation of the plane passing through A(2, 2, -1) is a(x – 2) + b(y – 2) + c(z + 1) = 0 This plane passes through B(3, 4, 2) and C(7, 0, 6) a(3 – 2) + b(4 – 2) +c(2 + 1) = 0 a + 2b + 3c = 0 ……….. (2) a(7 – 2) + b(0 – 2) + c(6 + 1) = 0 5a – 2b + 7c = 0 ……….. (3) From (2) and (3) Substituting in (1) equation of the required plane is 5(x- 2) + 2(y- 2) – 3(z +1) = 0 5x – 10 + 2y – 4 – 3z – 3 = 0 5x + 2y – 3z -17 = 0 or 5x + 2y- 3z = 17 Question 2. Show that the points (0, -1, 0), (2, 1, -1), (1, 1, 1), (3, 3, 0) are coplanar. Solution: Equation of the plane through A(0, -1, 0) is ax + b(y + 1) + cz = 0 …….. (1) This plane passes through B(2, 1, -1) and C(1, 1, 1) 2a + 2b – c = 0 …………. (2) a + 2b + c = 0 ………….. (3) (2) – (3) gives a – 2c = 0 ⇒ a = 2c ⇒ $$\frac{a}{2}=\frac{c}{1}$$ (2) + (3) gives 3a + 4b = 0 ⇒ 3a = – 4b ⇒ $$\frac{a}{4}=\frac{b}{-3}$$ ∴ $$\frac{a}{4}=\frac{b}{-3}=\frac{c}{2}$$ Substitutes in (1) equation of the plane ABC is 4x – 3(y + 1) + 2(z – 0) = 0 4x – 3y + 2z – 3 = 0 4x – 3y + 2z – 3 = 4.3 – 3.3. + 0.3 = 12 – 9 – 3 = 0 The given points A, B, C, D are coplanar. Question 3. Find the equation of the plane through (6, – 4, 3), (0, 4, -3) and cutting of inter-cepts whose sum is zero. Solution: Suppose a, b, c are the intercepts of the plane. Equation of the plane is $$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}$$ = 1 Given a + b + c = 0 c = – (a + b) The plane passes through P(6, -4, 3), Q(0, 4, -3) c = -a – b = -3 – b 4(-3 – b) – 3b = b(-3 – b) -12 – 4b – 3b = – 3b – b² b² – 4b – 12 = 0 (b – 6) (b + 2) = 0 ⇒ b = 6 – 2 Case i) b = 6 c = -3 – b = -3 – 6 = -9 Equation of the plane is $$\frac{x}{3}+\frac{y}{6}+\frac{z}{9}$$ = 1 6x + 3y – 2z = 18 Case ii) b = -2 c = -3 – b = -3 + 2 = -1 Equation of the plane is $$\frac{x}{3}+\frac{y}{-2}+\frac{z}{-1}$$ = 1 Question 4. A plane meets the co-ordinate axes in A, B, C. If the centroid of ∆ABC is (a, b, c). Show that the equation of the plane is $$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}$$ = 1 Solution: Suppose α, β, γ are the intercepts of the plane ABC. Equation of the plane is the intercept form is $$\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=1$$ ………… (1) Co-ordinates of A are (α, 0, 0), B are (0, β, 0) and C are (0, 0, γ) G is the centroid of ∆ABC Co-ordinates of Gare = α = 3a, β = 3b, γ = 3c Substituting in (1), equation of the plane ABC is Question 5. Show that the plane through (1, 1, 1) , (1, -1, 1) and (-7, -3, -5) is parallel . to Y-axis. Solution: Equation of the plane through A (1, 1, 1) can be taken as a(x – 1) +b(y – 1) + c(z – 1) = 0 ……….. (1) This plane passes through B(1, -1, 1) and C(-7, -3, -5) 0 – 2b + 0 = 0 ⇒ b – 0 Equation of zy – plane is y = 0 0.x + 1.y + 0.z = 0 a.0 + 0.1 + c.0 = 0 The required plane is perpendicular to zx – plane hence it is parallel to Y – axis. Question 6. Show that the equations ax + by + r =0, by + cz + p = 0, cz + ax + q = 0 represent planes perpendicular to xy,yz, zx planes respectively. Solution: Equation of the given plane is ax + by + c = Q d.rs of the normal are (a, b, c) Equation of XY – plane is z = 0 d.rs of the normal are (0, 0, 1) a.0 + b.0 + 0.1 = 0 ∴ ax + by + r = 0 is perpendicular to xy – plane. Similarly we can show that by + cz + p = 0 is perpendicular to yz – plane and cz + ax + q – 0 is perpendicular to zx – plane.
# How is the modulus calculated? ### How is the modulus calculated? Method 2: Perform integer division and calculate the difference value. Example : Calculation of A=123 module N=4, divide: 123/4=30.75 123/4 = 30.75. Retrieve the integer part: 30 , the multiple by N=4: 30×4=120 . ### How to calculate the IBAN control key? The IBAN key and his method of calculation Move the first 4 characters of theIBAN to the right of the number. Convert letters to numbers according to the conversion table below. Apply MOD 97-10 (see ISO 7604). ### How to change mod? In ADDITION or MULTIPLICATION (mod m), we can replace a number by another equal module Mr. You can add or subtract or multiply by the SAME NUMBER on each side of an equality mod Mr. We can raise to a POWER on each side of an equality mod m (but not vice versa: false reciprocal). ### What is Modulo in Mathematics? In some computer languages, the modulo is represented by a percent sign “%”. We will write a mod n to represent the remainder of the division of a by n. A modulo is therefore equivalent to the difference between a and the multiplication of the truncated value of the quotient of a by n. In mathematics, we sometimes note this: ### How to calculate the modular power? Modular power calculation a^b mod n. Calculation tool for the modulo operation. The modulo is the operation of calculating the remainder of the Euclidean division. The modulo % calculator returns the remainder of integer division. ### What is modulo n? Given two numbers, a (the dividend) and n (the divisor), a modulo n (abbreviated a mod i>n) is the remainder of dividing a by n. For example, the expression “7 mod 5” would evaluate to 2 because 7 divided by 5 leaves a remainder of 2, while “10 mod 5” would evaluate to 0 because dividing 10 by 5 leaves a remainder of 0. ### What is the difference between modulo and integer division? As a first approach, we can say that the modulo is the remainder of the integer division. Thus 9 modulo 4 gives 1 (9 divided by 4 gives a quotient of 2 and a remainder of 1), while 9 modulo 3 gives 0 (9 divided by 3 gives a quotient of 3 and a remainder of 0). In mathematics and computer programming,…
If you're reading this guide now, you've probably dealt with functions in great detail already, so I'll just include some brief highlights you'll need to get started with calculus. Much of this should be review, so feel free to skip sections you feel comfortable with. Definition of a Function A function is a rule that assigns to each element x from a set known as the "domain" a single element y from a set known as the "range". For example, the function y = x2 + 2 assigns the value y = 3 to x = 1, y = 6 to x = 2, and y = 11 to x = 3. Using this function, we can generate a set of ordered pairs of (x, y) including (1, 3),(2, 6), and (3, 11). We can also represent this function graphically, as shown below. The Vertical Line Test Note that in the graph above, each element x is assigned a single value y. If a rule assigned more than one value y to a single element x, that rule could not be considered a function. As you may recall from precalc, we can test for this property using the vertical line test, where we see whether we can draw a vertical line that passes through more than one point on the graph: Because any vertical line would pass through only one point, y = x2 + 2 must be assigning only one y value to each x value, and it therefore passes the vertical line test. Thus, y = x2 + 2 can rightfully be considered a function. The Horizontal Line Test Although a function can only assign one y value to each element x, it is allowed to assign more than one x value to each y. This is the case with our function y = x2 + 2. The value x = 4 is mapped to the single value y = 18, but the value y = 18 is mapped to both x = 4 and x = - 4. A one-to-one function is a special type of function that maps a unique x value to each element y. So, each element x maps to one and only one element y, and each element y maps to one and only one element x. An example of this is the function x3:
# Volume of a Triangular Pyramid – Formulas and Examples The volume of a hexagonal prism is the total space occupied by the prism in three-dimensional space. We can calculate the volume of these prisms by multiplying the area of the base by the height of the prism. The area of the base is equal to the area of a hexagon. The difference between a hexagonal prism and other prisms is the shape of its cross-section. Here, we will learn about the formula that we can use to calculate the volume of hexagonal prisms. Also, we will use this formula to solve some practice problems. ##### GEOMETRY Relevant for Learning about the volume of a triangular pyramid. See examples ##### GEOMETRY Relevant for Learning about the volume of a triangular pyramid. See examples ## Formula to find the volume of a triangular prism The volume of any pyramid is equal to the area of the base times the height of the pyramid divided by three. Therefore, the following is the formula for the volume of a pyramid: In turn, we know that the base of a triangular pyramid is a triangle and the area of any triangle is found by multiplying the length of its base by its height and dividing by two. Therefore, we have: where b is the length of the base of the triangle, a is the length of the height of the triangle, and h is the height of the pyramid. ## Volume of a triangular pyramid – Examples with answers The formula for the volume of triangular pyramids is used to solve the following examples. Each example has its respective solution, but it is recommended that you try to solve the problems yourself before looking at the answer. ### EXAMPLE 1 A pyramid has a height of 5 m and a triangular base with a base of length 4 m and a height of 3 m. What is its volume? We have the following information: • Pyramid height, • Base of the triangle, • Height of the triangle, We use these values in the volume formula: The volume is equal to 10 m³. ### EXAMPLE 2 What is the volume of a pyramid that has a height of 8 m and a triangular base with a base of 5 m and a height of 6 m? We have the following values: • Pyramid height, • Base of the triangle, • Triangle height, Using this information in the volume formula, we have: The volume is equal to 40 m³. ### EXAMPLE 3 A pyramid has a height of 11 m and a triangular base with a base of length 7 m and a height of 8 m. What is its volume? From the question, we get the following values: • Pyramid height, • Base of the triangle, • Triangle height, We use the volume formula with these values: The volume is equal to 102.7 m³. ### EXAMPLE 4 What is the volume of a pyramid with a height of 9 m and a triangular base of a base of 7 m and a height of 9 m? We have the following values: • Pyramid height, • Base of the triangle, • Height of the triangle, We substitute these values in the volume formula: The volume is equal to 94.5 m³. ## Volume of a triangular pyramid – Practice problems Practice using the formula for the volume of triangular pyramids and solve the following problems. If you need help with this, you can look at the solved examples above.
# Angle bisector theorem Angle bisector theorem In this diagram, BD:DC = AB:AC. In geometry, the angle bisector theorem is concerned with the relative lengths of the two segments that a triangle's side is divided into by a line that bisects the opposite angle. It equates their relative lengths to the relative lengths of the other two sides of the triangle. Contents 1 Theorem 2 Proofs 2.1 Proof 1 2.2 Proof 2 2.3 Proof 3 3 Exterior angle bisectors 4 History 5 Applications 6 References 7 Further reading 8 External links Theorem Consider a triangle ABC. Let the angle bisector of angle A intersect side BC at a point D between B and C. The angle bisector theorem states that the ratio of the length of the line segment BD to the length of segment CD is equal to the ratio of the length of side AB to the length of side AC: {displaystyle {frac {\|BD\|}{\|CD\|}}={frac {\|AB\|}{\|AC\|}},} and conversely, if a point D on the side BC of triangle ABC divides BC in the same ratio as the sides AB and AC, then AD is the angle bisector of angle ∠ A. The generalized angle bisector theorem states that if D lies on the line BC, then {displaystyle {frac {\|BD\|}{\|CD\|}}={frac {\|AB\|sin angle DAB}{\|AC\|sin angle DAC}}.} This reduces to the previous version if AD is the bisector of ∠ BAC. When D is external to the segment BC, directed line segments and directed angles must be used in the calculation. The angle bisector theorem is commonly used when the angle bisectors and side lengths are known. It can be used in a calculation or in a proof. An immediate consequence of the theorem is that the angle bisector of the vertex angle of an isosceles triangle will also bisect the opposite side. Proofs Proof 1 In the above diagram, use the law of sines on triangles ABD and ACD: {displaystyle {frac {\|AB\|}{\|BD\|}}={frac {sin angle ADB}{sin angle DAB}}} (1) {displaystyle {frac {\|AC\|}{\|CD\|}}={frac {sin angle ADC}{sin angle DAC}}} (2) Angles ∠ ADB and ∠ ADC form a linear pair, that is, they are adjacent supplementary angles. Since supplementary angles have equal sines, {displaystyle {sin angle ADB}={sin angle ADC}.} Angles ∠ DAB and ∠ DAC are equal. Therefore, the right hand sides of equations (1) and (2) are equal, so their left hand sides must also be equal. {displaystyle {frac {\|BD\|}{\|CD\|}}={frac {\|AB\|}{\|AC\|}},} which is the angle bisector theorem. If angles ∠ DAB and ∠ DAC are unequal, equations (1) and (2) can be re-written as: {displaystyle {{frac {\|AB\|}{\|BD\|}}sin angle DAB=sin angle ADB},} {displaystyle {{frac {\|AC\|}{\|CD\|}}sin angle DAC=sin angle ADC}.} Angles ∠ ADB and ∠ ADC are still supplementary, so the right hand sides of these equations are still equal, so we obtain: {displaystyle {{frac {\|AB\|}{\|BD\|}}sin angle DAB={frac {\|AC\|}{\|CD\|}}sin angle DAC},} which rearranges to the "generalized" version of the theorem. Proof 2 Let D be a point on the line BC, not equal to B or C and such that AD is not an altitude of triangle ABC. Let B1 be the base (foot) of the altitude in the triangle ABD through B and let C1 be the base of the altitude in the triangle ACD through C. Then, if D is strictly between B and C, one and only one of B1 or C1 lies inside triangle ABC and it can be assumed without loss of generality that B1 does. This case is depicted in the adjacent diagram. If D lies outside of segment BC, then neither B1 nor C1 lies inside the triangle. ∠ DB1B and ∠ DC1C are right angles, while the angles ∠ B1DB and ∠ C1DC are congruent if D lies on the segment BC (that is, between B and C) and they are identical in the other cases being considered, so the triangles DB1B and DC1C are similar (AAA), which implies that: {displaystyle {frac {\|BD\|}{\|CD\|}}={frac {\|BB_{1}\|}{\|CC_{1}\|}}={frac {\|AB\|sin angle BAD}{\|AC\|sin angle CAD}}.} If D is the foot of an altitude, then, {displaystyle {frac {\|BD\|}{\|AB\|}}=sin angle BAD{text{ and }}{frac {\|CD\|}{\|AC\|}}=sin angle DAC,} and the generalized form follows. Proof 3 {displaystyle alpha ={tfrac {angle BAC}{2}}=angle BAD=angle CAD} A quick proof can be obtained by looking at the ratio of the areas of the two triangles {displaystyle triangle BAD} and {displaystyle triangle CAD} , which are created by the angle bisector in {displaystyle A} . Computing those areas twice using different formulas, that is {displaystyle {tfrac {1}{2}}gh} with base {displaystyle g} and altitude {displaystyle h} and {displaystyle {tfrac {1}{2}}absin(gamma )} with sides {displaystyle a} , {displaystyle b} and their enclosed angle {displaystyle gamma } , will yield the desired result.
# A Math Problem Where Two or More Factors Must Be Taken Into Account A Math Problem Where Two or More Factors Must Be Taken Into Account Mathematics is a subject that often requires careful consideration of multiple factors to arrive at the correct solution. One such problem that exemplifies this concept is the calculation of compound interest. Compound interest is the interest earned on both the initial principal amount and the accumulated interest from previous periods. It is a crucial concept in finance and investment planning, as it determines how much money will be earned or owed over a given period. To understand compound interest, let’s consider a hypothetical scenario. Suppose you deposit \$1,000 into a savings account that offers an annual interest rate of 5%. This interest rate is compounded annually, meaning that at the end of each year, the interest earned is added to the initial principal amount, and the next year’s interest is calculated based on the new total. To calculate the compound interest, we use the formula: A = P(1 + r/n)^(nt) Where: A = the final amount including interest P = the principal amount (initial deposit) r = the annual interest rate (expressed as a decimal) n = the number of times that interest is compounded per year t = the number of years See also How Do I Get a Medical Marijuana Card in Kentucky Let’s plug in the values from the scenario above: A = 1000(1 + 0.05/1)^(1*1) = 1000(1.05)^1 = 1000(1.05) = \$1,050 After one year, your initial deposit of \$1,000 would grow to \$1,050. To calculate the compound interest earned, subtract the principal amount from the final amount: Compound Interest = A – P = 1050 – 1000 = \$50 Now, let’s answer some common questions related to compound interest: 1. What is the formula for calculating compound interest? – The formula is A = P(1 + r/n)^(nt), where A is the final amount, P is the principal amount, r is the annual interest rate, n is the number of times that interest is compounded per year, and t is the number of years. 2. How is compound interest different from simple interest? – Compound interest takes into account the accumulated interest from previous periods, while simple interest only considers the initial principal amount. 3. Is compound interest always calculated annually? – No, compound interest can be calculated annually, semi-annually, quarterly, monthly, or even daily, depending on the terms of the investment. 4. What happens to compound interest if the interest rate increases? – As the interest rate increases, the compound interest earned on the principal amount also increases. 5. Does the frequency of compounding affect the final amount? – Yes, a higher compounding frequency leads to a higher final amount, as the interest is added more frequently. 6. Can compound interest work against you? – Yes, compound interest can work against you if you have a loan or debt, as it will accumulate over time and increase the amount owed. – Compound interest can be advantageous when investing, as it allows your money to grow exponentially over time. 8. Can compound interest be negative? – No, compound interest cannot be negative. It only represents the growth in value or debt owed. 9. Is compound interest used in any other fields besides finance? – Yes, compound interest is also used in fields such as population growth, scientific research, and exponential decay.
# 11 Trigonometric Functions of Acute Angles Save this PDF as: Size: px Start display at page: ## Transcription 1 Arkansas Tech University MATH 10: Trigonometry Dr. Marcel B. Finan 11 Trigonometric Functions of Acute Angles In this section you will learn (1) how to find the trigonometric functions using right triangles, () compute the values of these functions for some special angles, and () solve model problems involving the trigonometric functions. First, let s review some of the features of right triangles. A triangle in which one angle is 90 is called a right triangle. The side opposite to the right angle is called the hypotenuse and the remaining sides are called the legs of the triangle. Suppose that we are given an acute angle θ as shown in Figure Note that a 0 and b 0. Figure 11.1 Associated with θ are three lengths, the hypotenuse, the opposite side, and the adjacent side. We define the values of the trigonometric functions of θ as ratios of the sides of a right triangle: sin θ = opposite hypotenuse = b r cos θ = adjacent hypotenuse = a r tan θ = opposite adjacent = b a csc θ = hypotenuse opposite = r b sec θ = hypotenuse adjacent = r a cot θ = adjacent opposite = a b where r = a + b (Pythagorean formula). The symbols sin, cos, sec, csc, tan, and cot are abbreviations of sine, cosine, secant, cosecant, tangent, and cotangent. The above ratios are the same regardless of the size of the triangle. That is, the trigonometric functions defined above depend only on the angle θ. To see this, consider 1 2 Figure 11.. Figure 11. The triangles ABC and AB C are similar. Thus, AB For example, using the cosine function we have cos θ = AB AC = AB AC. AB = AC AC = BC B C. The following identities, known as reciprocal identities, follow from the definition given above. sin θ = 1 csc θ, cos θ = 1 sec θ, tan θ = 1 csc θ = 1 sin θ, sec θ = 1 cos θ, cot θ = 1 cot θ, tan θ. Example 11.1 Find the exact value of the six trigonometric functions of the angle θ shown in Figure 11.. Figure 11. By the Pythagorean Theorem, the length of the hypotenuse is = 169 = 1. Thus, sin θ = 1 1 cos θ = 5 1 tan θ = 1 5 csc θ = 1 1 sec θ = 1 5 cot θ = 5 1 Given the value of one trigonometric function, it is possible to find the values of the remaining trigonometric functions of that angle. 3 Example 11. Suppose that θ is an acute angle for which cos θ = 5 7. Determine the values of the other five trigonometric functions. Since cos θ = 5 7, the adjacent side has length 5 and the hypotenuse has length 7. See Figure Using the Pythagorean Theorem, the opposite side has length 49 5 = 6. Thus, sin θ = 6 7 ; cos θ = 5 7 sec θ = 7 5 ; tan θ = 6 5 csc θ = ; cot θ = 5 6 = 7 = Figure 11.4 Example 11. Solve for y given that tan 0 =.(See Figure 8.5). Figure 11.5 According to Figure 11.5, y = 75 tan 0 = 75( ) = 5 Trigonometric Functions of Special Angles Next, we compute the trigonometric functions of some special angles. It s useful to remember these special trigonometric ratios because they occur often. 4 Example 11.4 Determine the values of the six trigonometric functions of the angle 45. See Figure Figure 11.6 Using Figure 11.6, the triangle OAP is a right isosceless triangle. By the Pythagorean Theorem we find that r = a or r = a. Thus, sin 45 = ; csc 45 = cos 45 = ; sec 45 = tan 45 = 1 ; cot 45 = 1 Example 11.5 Determine the trigonometric functions of the angles (a) θ = 0 (b) θ = 60. (a) Let ABC be an equilateral triangle with side of length a. Let P be the midpoint of the side AC and h the height of the triangle. See Figure Using the Pythagorean Theorem we find h = a. Figure 5 Thus, (b) Similarly, sin 0 = 1 ; cos 0 = ; tan 0 = csc 0 = ; sec 0 = ; cot 0 =. sin 60 = ; cos 60 = 1 ; tan 60 = csc 60 = ; sec 60 = ; cot 60 = Example 11.6 Find the exact value of sin 60 sec 45 tan 60. Using the results of the previous two problems we find that sin 60 sec 45 tan 60 = ( ) = 6. We follow the convention that when we write a trigonometric function, such as sin t, then it is assumed that t is in radians. If we want to evaluate the trigonometric function of an angle measured in degrees we will use the degree notation such as cos 0. Angles of Elevation and Depression If an observer is looking at an object, then the line from the observer s eye to the object is known as the line of sight. If the object is above the horizontal then the angle between the line of sight and the horizontal is called the angle of elevation. If the object is below the horizontal then the angle between the line of sight and the horizontal is called the angle of depression. See Figure Figure 6 Example 11.7 From a point 115 feet from the base of a redwood tree, the angle of elevation to the top of the tree is 64.. Find the height of the tree to the nearest foot. See Figure 8.9. Figure 11.9 According to Figure 11.9, we use the tangent function to find the height h of the tree: tan 64. = h 115 so that h = 115 tan ft. Evaluating trigonometric functions with a calculator When evaluating trigonometric functions using a calculator, you need to set the calculator to the desired mode of measurement (degrees or radians). The functions sine, cosine, and tangent have a key in a standard scientific calculator. For the remaining three trigonometric functions the key x 1 is used in the process. For example, to evaluate sec π 8, set the calculator to radian mode, then apply the following sequence of keystrokes: π,, 8, cos, x 1 and enter to obtain sec π ### Right Triangle Trigonometry Right Triangle Trigonometry MATH 160, Precalculus J. Robert Buchanan Department of Mathematics Fall 2011 Objectives In this lesson we will learn to: evaluate trigonometric functions of acute angles, use ### RIGHT TRIANGLE TRIGONOMETRY RIGHT TRIANGLE TRIGONOMETRY The word Trigonometry can be broken into the parts Tri, gon, and metry, which means Three angle measurement, or equivalently Triangle measurement. Throughout this unit, we will ### Section 9.4 Trigonometric Functions of any Angle Section 9. Trigonometric Functions of any Angle So far we have only really looked at trigonometric functions of acute (less than 90º) angles. We would like to be able to find the trigonometric functions ### y = rsin! (opp) x = z cos! (adj) sin! = y z = The Other Trig Functions MATH 7 Right Triangle Trig Dr. Neal, WKU Previously, we have seen the right triangle formulas x = r cos and y = rsin where the hypotenuse r comes from the radius of a circle, and x is adjacent to and y ### Unit 2: Right Triangle Trigonometry RIGHT TRIANGLE RELATIONSHIPS Unit 2: Right Triangle Trigonometry This unit investigates the properties of right triangles. The trigonometric ratios sine, cosine, and tangent along with the Pythagorean Theorem are used to solve right ### Right Triangle Trigonometry Section 6.4 OBJECTIVE : Right Triangle Trigonometry Understanding the Right Triangle Definitions of the Trigonometric Functions otenuse osite side otenuse acent side acent side osite side We will be concerned ### Trigonometry (Chapters 4 5) Sample Test #1 First, a couple of things to help out: First, a couple of things to help out: Page 1 of 24 Use periodic properties of the trigonometric functions to find the exact value of the expression. 1. cos 2. sin cos sin 2cos 4sin 3. cot cot 2 cot Sin ### 4-1 Right Triangle Trigonometry Find the exact values of the six trigonometric functions of θ. 1. The length of the side opposite θ is 8 is 18., the length of the side adjacent to θ is 14, and the length of the hypotenuse 3. The length ### Pre-Calculus II. where 1 is the radius of the circle and t is the radian measure of the central angle. Pre-Calculus II 4.2 Trigonometric Functions: The Unit Circle The unit circle is a circle of radius 1, with its center at the origin of a rectangular coordinate system. The equation of this unit circle ### Types of Angles acute right obtuse straight Types of Triangles acute right obtuse hypotenuse legs MTH 065 Class Notes Lecture 18 (4.5 and 4.6) Lesson 4.5: Triangles and the Pythagorean Theorem Types of Triangles Triangles can be classified either by their sides or by their angles. Types of Angles An ### 4.3 & 4.8 Right Triangle Trigonometry. Anatomy of Right Triangles 4.3 & 4.8 Right Triangle Trigonometry Anatomy of Right Triangles The right triangle shown at the right uses lower case a, b and c for its sides with c being the hypotenuse. The sides a and b are referred ### 4-1 Right Triangle Trigonometry Find the exact values of the six trigonometric functions of θ. 3. The length of the side opposite θ is 9, the length of the side adjacent to θ is 4, and the length of the hypotenuse is. 7. The length of ### 4-1 Right Triangle Trigonometry Find the measure of angle θ. Round to the nearest degree, if necessary. 31. Because the lengths of the sides opposite θ and the hypotenuse are given, the sine function can be used to find θ. 35. Because ### MA Lesson 19 Summer 2016 Angles and Trigonometric Functions DEFINITIONS: An angle is defined as the set of points determined by two rays, or half-lines, l 1 and l having the same end point O. An angle can also be considered as two finite line segments with a common ### Trigonometry. Week 1 Right Triangle Trigonometry Trigonometry Introduction Trigonometry is the study of triangle measurement, but it has expanded far beyond that. It is not an independent subject of mathematics. In fact, it depends on your knowledge ### You can solve a right triangle if you know either of the following: Two side lengths One side length and one acute angle measure Solving a Right Triangle A trigonometric ratio is a ratio of the lengths of two sides of a right triangle. Every right triangle has one right angle, two acute angles, one hypotenuse, and two legs. To solve ### Right Triangles A right triangle, as the one shown in Figure 5, is a triangle that has one angle measuring Page 1 9 Trigonometry of Right Triangles Right Triangles A right triangle, as the one shown in Figure 5, is a triangle that has one angle measuring 90. The side opposite to the right angle is the longest ### 2. Right Triangle Trigonometry 2. Right Triangle Trigonometry 2.1 Definition II: Right Triangle Trigonometry 2.2 Calculators and Trigonometric Functions of an Acute Angle 2.3 Solving Right Triangles 2.4 Applications 2.5 Vectors: A Geometric ### MATH 150 TOPIC 9 RIGHT TRIANGLE TRIGONOMETRY. 9a. Right Triangle Definitions of the Trigonometric Functions Math 50 T9a-Right Triangle Trigonometry Review Page MTH 50 TOPIC 9 RIGHT TRINGLE TRIGONOMETRY 9a. Right Triangle Definitions of the Trigonometric Functions 9a. Practice Problems 9b. 5 5 90 and 0 60 90 ### Intermediate Algebra with Trigonometry. J. Avery 4/99 (last revised 11/03) Intermediate lgebra with Trigonometry J. very 4/99 (last revised 11/0) TOPIC PGE TRIGONOMETRIC FUNCTIONS OF CUTE NGLES.................. SPECIL TRINGLES............................................ 6 FINDING ### 6.1 Basic Right Triangle Trigonometry 6.1 Basic Right Triangle Trigonometry MEASURING ANGLES IN RADIANS First, let s introduce the units you will be using to measure angles, radians. A radian is a unit of measurement defined as the angle at ### 18 Verifying Trigonometric Identities Arkansas Tech University MATH 1203: Trigonometry Dr. Marcel B. Finan 18 Verifying Trigonometric Identities In this section, you will learn how to use trigonometric identities to simplify trigonometric ### Right Triangle Trigonometry Test Review Class: Date: Right Triangle Trigonometry Test Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Find the length of the missing side. Leave your answer ### CRASH COURSE IN PRECALCULUS CRASH COURSE IN PRECALCULUS Shiah-Sen Wang The graphs are prepared by Chien-Lun Lai Based on : Precalculus: Mathematics for Calculus by J. Stuwart, L. Redin & S. Watson, 6th edition, 2012, Brooks/Cole ### opposite side adjacent side sec A = hypotenuse opposite side adjacent side = a b Trigonometry Angles & Circular Functions: Solving Right Triangles Trigonometric Functions in a Right Triangle We have already looked at the trigonometric functions from the perspective of the unit circle. ### Geometry Mathematics Curriculum Guide Unit 6 Trig & Spec. Right Triangles 2016 2017 Unit 6: Trigonometry and Special Right Time Frame: 14 Days Primary Focus This topic extends the idea of triangle similarity to indirect measurements. Students develop properties of special right triangles, ### 6.3 Inverse Trigonometric Functions Chapter 6 Periodic Functions 863 6.3 Inverse Trigonometric Functions In this section, you will: Learning Objectives 6.3.1 Understand and use the inverse sine, cosine, and tangent functions. 6.3. Find the ### CHAPTER 38 INTRODUCTION TO TRIGONOMETRY CHAPTER 38 INTRODUCTION TO TRIGONOMETRY EXERCISE 58 Page 47. Find the length of side x. By Pythagoras s theorem, 4 = x + 40 from which, x = 4 40 and x = 4 40 = 9 cm. Find the length of side x. By Pythagoras ### Any two right triangles, with one other angle congruent, are similar by AA Similarity. This means that their side lengths are. Lesson 1 Trigonometric Functions 1. I CAN state the trig ratios of a right triangle 2. 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Laval Kennesaw STate University August 27, 2010 Abstract This handout defines the trigonometric function of angles and discusses the relationship between ### Mid-Chapter Quiz: Lessons 4-1 through 4-4 Find the exact values of the six trigonometric functions of θ. Find the value of x. Round to the nearest tenth if necessary. 1. The length of the side opposite is 24, the length of the side adjacent to ### Definition III: Circular Functions SECTION 3.3 Definition III: Circular Functions Copyright Cengage Learning. All rights reserved. Learning Objectives 1 2 3 4 Evaluate a trigonometric function using the unit circle. 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MATH19730 Part 1 Trigonometry Section2 Trigonometry provides methods to relate angles and lengths but the functions we define have many other applications in mathematics. An angle can be measured in degrees ### Inverse Trigonometric Functions SECTION 4.7 Inverse Trigonometric Functions Copyright Cengage Learning. All rights reserved. Learning Objectives 1 2 3 4 Find the exact value of an inverse trigonometric function. Use a calculator to approximate ### 4.1 Radian and Degree Measure Date: 4.1 Radian and Degree Measure Syllabus Objective: 3.1 The student will solve problems using the unit circle. Trigonometry means the measure of triangles. Terminal side Initial side Standard Position ### Right Triangles and SOHCAHTOA: Finding the Measure of an Angle Given any Two Sides (ONLY for ACUTE TRIANGLES Why?) Name Period Date Right Triangles and SOHCAHTOA: Finding the Measure of an Angle Given any Two Sides (ONLY for ACUTE TRIANGLES Why?) 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Cliffhanger Tools Cliff Howard REPEATABLE CONDUIT BENDING Cliffhanger Tools Cliff Howard http://www.cliffhangertools.com Purpose for this Training Repeatable Bending Review theory Learn how to gather the facts Understanding the Cliffhanger ### 4.4 Right Triangle Trigonometry 4.4 Right Triangle Trigonometry Trigonometry is introduced to students in two different forms, as functions on the unit circle and as functions on a right triangle. The unit circle approach is the most ### Section 0.4 Inverse Trigonometric Functions Section 0.4 Inverse Trigonometric Functions Short Recall from Trigonometry Definition: A function f is periodic of period T if f(x + T ) = f(x) for all x such that x and x+t are in the domain of f. The ### Campus Academic Resource Program Double Angle and Half Angle Formulae This handout is an overview of double angle and half angle formulae found in pre-calculus. In particular, this handout will: Discuss a strategy to find an expression for the double-angle formula for cosine. ### G E O M E T R Y CHAPTER 9 RIGHT TRIANGLES AND TRIGONOMETRY. Notes & Study Guide G E O M E T R Y CHAPTER 9 RIGHT TRIANGLES AND TRIGONOMETRY Notes & Study Guide 2 TABLE OF CONTENTS SIMILAR RIGHT TRIANGLES... 3 THE PYTHAGOREAN THEOREM... 4 SPECIAL RIGHT TRIANGLES... 5 TRIGONOMETRIC RATIOS... ### 1. The Six Trigonometric Functions 1.1 Angles, Degrees, and Special Triangles 1.2 The Rectangular Coordinate System 1.3 Definition I: Trigonometric 1. The Si Trigonometric Functions 1.1 Angles, Degrees, and Special Triangles 1. The Rectangular Coordinate Sstem 1.3 Definition I: Trigonometric Functions 1.4 Introduction to Identities 1.5 More on Identities ### Parallel and Perpendicular. We show a small box in one of the angles to show that the lines are perpendicular. CONDENSED L E S S O N. Parallel and Perpendicular In this lesson you will learn the meaning of parallel and perpendicular discover how the slopes of parallel and perpendicular lines are related use slopes ### as a fraction and as a decimal to the nearest hundredth. Express each ratio as a fraction and as a decimal to the nearest hundredth. 1. sin A The sine of an angle is defined as the ratio of the opposite side to the hypotenuse. So, 2. tan C The tangent of an ### Algebra 2/Trigonometry Practice Test Algebra 2/Trigonometry Practice Test Part I Answer all 27 questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. For each question, write on the separate ### Math Placement Test Practice Problems Math Placement Test Practice Problems The following problems cover material that is used on the math placement test to place students into Math 1111 College Algebra, Math 1113 Precalculus, and Math 2211 ### BASICS ANGLES. θ x. Figure 1. Trigonometric measurement of angle θ. trigonometry, page 1 W. F. Long, 1989 TRIGONOMETRY BASICS Trigonometry deals with the relation between the magnitudes of the sides and angles of a triangle. In a typical trigonometry problem, two angles and a side of a triangle would be given ### NCERT. There is perhaps nothing which so occupies the middle position of mathematics as trigonometry. J.F. Herbart (1890) INTRODUCTION TO TRIGONOMETRY 7 8. Introduction There is perhaps nothing which so occupies the middle position of mathematics as trigonometry. J.F. Herbart (890) 8 You have already studied about triangles, ### Verifying Trigonometric Identities. Introduction. is true for all real numbers x. So, it is an identity. Verifying Trigonometric Identities 333202_0502.qxd 382 2/5/05 Chapter 5 5.2 9:0 AM Page 382 Analytic Trigonometry Verifying Trigonometric Identities What you should learn Verify trigonometric identities. Why you should learn it You can ### Trigonometric Functions Trigonometric Functions MATH 10, Precalculus J. Robert Buchanan Department of Mathematics Fall 011 Objectives In this lesson we will learn to: identify a unit circle and describe its relationship to real ### Section 7.5 Unit Circle Approach; Properties of the Trigonometric Functions. cosθ a. = cosθ a. Section 7.5 Unit Circle Approach; Properties of the Trigonometric Functions A unit circle is a circle with radius = 1 whose center is at the origin. Since we know that the formula for the circumference ### 1. Introduction circular deﬁnition Remark 1 inverse trigonometric functions 1. Introduction In Lesson 2 the six trigonometric functions were defined using angles determined by points on the unit circle. This is frequently referred to as the circular definition of the trigonometric ### Law of Cosines. If the included angle is a right angle then the Law of Cosines is the same as the Pythagorean Theorem. 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Right Triangle Trig Solving right triangles is possible using trigonometry so long as two sides or two angles and a side are known. The right angle is going to be 90 degrees, so there’s always at least one angle that is already known. When a person needs to determine all the sides and angles of the triangle, there are a few formulas they can use depending on the information in the problem. Knowing Two Angles and One Side, Solve the Triangle The right angle is always 90 degrees. If another angle is known, it’s easy to find the measurement of the third angle. Simply subtract the second angle from 90 degrees to get the third angle. For instance, if the second angle is 32, the three angles for the triangle would be 90, 32 and 58. When one side is known, it’s possible to determine the remaining sides using a specific formula. The unknown side divided by the known sign is going to be sin of the angle across from the unknown side. For example, angle A is 38 degrees and side b is 10. To find side a, the equation would look like: a/10=sin 38. At this point, the student can use the three-place trigonometric table to determine sin 38 or a calculator. The equation is then a/10=0.616. Solve for a, which is 6.16. Now that two sides are known, it’s possible to determine the third side of the triangle using the Pythagorean theorem, a^2+b^2=c^2. Solving this equation, the three sides are 10, 6.16, and 11.75 Knowing Two Sides When two sides are known, the third one can be found using the Pythagorean theorem as shown above. After this, it’s important to determine the angles. One of the angles is going to be 90 degrees, so only two more are needed. Assume for the following example that angle C is 90. Sides are a=10, b=6, and c=8. To find the second angle, use cos A=6/10 (the 8 and 10 are because these are the sides connected to the angle). This means cos A=0.6, but an angle is needed. Use the three-place trigonometric table to determine the angle for A, which is 37 degrees. Once the second angle is known, the third one can be determined by subtracting the second one from 90. In this example, the second angle would be 53. Solving for a right triangle is made easier because one of the angles is always known. So long as a second angle and a side or two sides are known, it’s possible to solve all the angles and sides. In more advanced problems, knowing an angle outside of the triangle can help a person solve all of the angles inside it as well. Practice makes it easier to remember what formulas are needed to solve for what the person needs and for the student to learn how to easily use the three-place trigonometric table. Try a few problems today to get a little more practice solving all parts of these triangles.
# Standard deviation of a discrete random variable The standard deviation of a discrete random variable is denoted by σ and the formula to use to compute the standard deviation is the one you see below. We can use the example in the previous lesson about the number of people going to the movie theater each week to look for the standard deviation. We will show you how to use both formulas above. ## Calculating the standard deviation of a discrete random variable In the lesson about mean of a discrete random variable we have the probability distribution table shown below. x P(x) 0 0.5 1 0.25 2 0.15 3 0.09 4 0.01 ΣP(x) = 1 We have already looked for the mean in the lesson about mean of a discrete random variable and we found that E(x) = 0.86. Here is a table showing how to compute the standard deviation using this formula. $$σ = \sqrt{Σ[(x-µ)^2 × P(x)]}$$ x x - μ (x - μ)2 P(x) (x - μ)2× P(x) 0 -0.86 0.7396 0.5 0.3698 1 0.14 0.0196 0.25 0.0049 2 1.14 1.2996 0.15 0.19494 3 2.14 4.5796 0.09 0.412164 4 3.14 9.8596 0.01 0.098596 ∑[(x - μ)2× P(x)] = 1.0804 1.0804 was of course found by adding all the numbers in the last column. This number is called variance, more specifically variance of a discrete random variable. Variance of a discrete random variable = ∑[(x - μ)2× P(x)] = 1.0804 The standard deviation is the square root of the variance or σ = √(1.0804) = 1.039422 Here is a table showing how to compute the standard deviation using the other formula. $$σ = \sqrt{Σ[x^2 × P(x)] - µ^2}$$ x x2 P(x) x2 × P(x) 0 0 0.5 0 1 1 0.25 0.25 2 4 0.15 0.6 3 9 0.09 0.81 4 16 0.01 0.16 ∑[(x2 × P(x)] = 1.82 1.82 is found by adding all the numbers in the fourth column. μ = 0.86, so μ2 = 0.7396 ∑[(x2 × P(x)] - μ2 = 1.82 - 0.7396 = 1.0804 Again, this number is the variance of the discrete random variable. Just get the square root of this number to get the standard deviation. Since the number is the same as the one before, the answer is the same! 100 Tough Algebra Word Problems. If you can solve these problems with no help, you must be a genius! Recommended
Tuesday, February 12, 2013 The power set of X If you are in Hong Kong and you need help for university mathematics courses, please visit www.all-r-math.com. Suppose X and Y are sets. What is YX? YX is the set of all functions from X to Y. From the last post, we see that if X has n elements and Y has m elements, then YX has mn elements. What is 2X? If we write like that, we identify 2 with a set with two elements, in particular we can consider 2={0,1}. Therefore 2X is simply a set of functions that for each x∈X, f(x) takes up values as 0 or 1. How do we describe a function in 2X? Since f(x) can only have two values, we can describe the function f using the preimage of 0, i.e. f-1(0)={x∈X : f(x)=0}. In other words, each function in 2X corresponds to exactly one subset of X. 2X is sometimes used to denote the power set of X, right? A power set of X, usually denoted by P(X), is the set of all subsets of X. Because there exists a correspondence between the functions in 2X and the subsets in P(X), so we sometimes really consider them identical. For example, if X={a,b,c}, then P(X)={∅, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}} and the eight functions in 2X are f(a)=f(b)=f(c)=1; f(a)=0, f(b)=f(c)=1; f(b)=0, f(a)=f(c)=1; f(c)=0, f(a)=f(b)=1; f(a)=f(b)=0, f(c)=1; f(a)=f(c)=0, f(b)=1; f(b)=f(c)=0, f(a)=1; f(a)=f(b)=f(c)=0.
# Trigonometry/The most Difficult Triangles to Solve ## Given SSA (Two Sides and an Angle Not Included by the Two Sides - the Ambiguous Case)Edit In the diagram below, the red arcs indicate some points at the same distance from the bottom left corner of the triangle. These are points that make one of the side lengths right. The missing corner of the triangle must be on one of the points where the red arcs intersect the sloping dotted line. Look at the diagram above. We run into trouble. There are two possible solutions. The given information may not uniquely determine a triangle. It's worse than that. There may not be a solution at all, if angle $\displaystyle \theta$ is too large. Here's how we proceed. One of the two given sides, side $\displaystyle a$ is opposite the given angle, so we can apply the Law of Sines to try to find the angle opposite the other given side. When we do this (given sides $\displaystyle a$ and $\displaystyle b$ and angle $\displaystyle A$), we'll have $\displaystyle \sin B = \frac{b\sin \theta}{a}$ and there are three possibilities: $\displaystyle \text{Either } \sin B >1 \text{,} \sin B =1 \text{, or} \sin B <1 \text{.}$ • $\displaystyle \text{If } \sin B >1$, then there is no angle $\displaystyle B$ that meets the given information, so no triangle can be formed with the given sides and angle. This is because the side is too short to reach the dotted line, whatever position it is in. • $\displaystyle \text{If } \sin B =1$, then we have a right triangle with right angle at $\displaystyle B$ and we can proceed as above for right triangles. • $\displaystyle \text{If } \sin B <1$ and the side opposite the given angle is shorter than the other given side, then there are two possible measures for angle $\displaystyle B$, one acute (the inverse sine of the value of $\displaystyle \sin B$) and one obtuse (the supplement of the acute one, i.e. 180º less that angle). Whichever of the two angles we choose, we can find the missing information as above now that we have two angles. Thus there are two possible solutions. • $\displaystyle \text{If } \sin B <1$ and the side opposite the given angle is longer than the other given side, then again there may appear to be two solutions, but one is invalid since we will find that two angles add to more than 180º so the third angle would be negastive. Thus there is only one possible solution. (If the two given sides are of equal length, the second solution would be a triangle of zero area where these two sides coincide and the third side is of zero length.) ## The case Sin B = 1Edit Thanks to the Pythagorean Theorem, it is possible to prove that two right triangles are congruent if they have equal hypotenuses and also one other pair of sides is equal. (However, Euclid proves this theorem without needing Pythagoras' Theorem.) ### ProofEdit Okay, we already know the theorem: $\displaystyle a^2+b^2=c^2$. Right? As a result, you see that it can be turned around to yield the formula $\displaystyle c^2-b^2=a^2$. This means that, if you know the hypotenuse and one leg of a triangle, you can calculate the length of the second leg. This is not only important in and of itself, but also because it means that you know the lengths of all three sides of the triangle, and there is only one unique triangle with three particular side lengths, which means that if you know that the hypotenuse and one leg of two triangles are congruent, you also know that the two triangles are congruent!
You are on page 1of 20 CONTINUTITY AND DIFFERENTIABILITY 1 CONTINUITY AND DIFFERENTIABILITY 3.1 BASIC CONCEPTS AND IMPORTANT RESULTS (a) Continuity of a real function at a point A function f is said to be left continuous or continuous from the left at x = c iff (i) f(c) exists (ii) Lt f(x) exists and (iii) x ®c - Lt f(x) = f(c). x ®c - A function f is said to be right continuous or continuous from the right at x = c iff (i) f(c) exists (ii) Lt x ®c + f(x) exists and (iii) Lt x ®c + f(x) = f(c). A function f is said to be continuous at x = c iff (i) f(c) exists (ii) Lt f(x) exists and (iii) x ®c Lt f(x) = f(c). x ®c Hence, a function is continuous at x = c iff it is both left as well as right continuous at x = c. When xLt f(x) exists but either f(c) does not exist or ®c Lt f(x) ¹ f(c), we say that f x ®c has a removable discontinuity; otherwise, we say that f has non-removable discontinuity. (b) Continuity of a function in an interval A function f is said to be continuous in an open interval (a, b) iff f is continuous at every point of the interval (a, b) ; and f is said to be continuous in the closed interval [a, b] iff f is continuous in the open interval (a, b) and it is continuous at a from the right and at b from the left. Continuous function. A function is said to be a continuous function iff it is continuous at every point of its domain. In particular, if the domain is a closed interval, say [a, b], then f must be continuous in (a, b) and right continuous at a and left continuous at b. The set of all point where the function is continuous is called its domain of continuity. The domain of continuity of a function may be a proper subset of the domain of the function. 3.2 PROPERTIES OF CONTINUOUS FUNCTIONS Property 1. Let f, g be two functions continuous at x = c, then (i) af is continuous at x = c, " a Î R (ii) f + g is continuous at x = c (iii) f – g is continuous at x = c (iv) fg is continuous at x = c (v) f is continuous at x = c, provided g(c) ¹ 0. g Property 2. Let D1 and D2 be the domains of continuity of the functions f and g respectively then (i) af is continuous on D1 for all a Î R (ii) f + g is continuous on D1 Ç D2 (iii) f – g is continuous on D1 Ç D2 (iv) fg is continuous on D1 Ç D2 (v) f is continuous on D1 Ç D2 except those points where g(x) = 0. g Property 3. A polynomial function is continuous everywhere. In particular, every constant function and every identity function is continuous. Property 4. A rational function is continuous at every point of its domain. Property 5. If f is continuous at c, then \| f \| is also continuous at x = c. www.thinkiit.in CONTINUTITY AND DIFFERENTIABILITY 2 In particular, the function \| x \| is continuous for every x Î R. Property 6. Let f be a continuous one-one function defined on [a, b] with range [c, d], then the inverse function f–1 : [c, d] ® [a, b] is continuous on [c, d] Property 7. If f is continuous at c and g is continuous at f(c), then gof is continuous at c. Property 8. All the basic trigonometric functions i.e. sin x, cos x, tan x, cot x, sec x and cosec x are continuous. Property 9. All basic inverse trigonometric functions i.e. sin–1 x, cos–1 x, tan–1 x, cot–1 x, sec– 1 x, cosec–1 x are continuous (in their respective domains). Property 10. Theorem. If a function is differentiable at any point, it is necessarily continuous at that point. The converse of the above theorem may not be true i.e. a function may be continuous at a point but may not be derivable at that point. 3.3 DERIVATIVE OF VARIOUS FUNCTIONS (a) Derivative of composite functions Theorem. If u = g(x) is di fferentiabl e at x and y = f(u) is differentiable at u, then y is differentiable at x and dy dy du = . . dx du dx If g is differentiable at x and f is differentiable at g(x), then the composite function h(x) = f(g(x)) is differentiable at x and h’(x) = f’(g(x)). g’(x). Chain Rule. The above rule is called the chain rule of differentiation, since determining the derivative of y = f(g(x)) at x involves the following chain of steps : (b) (i) First, find the derivative of the outer function f at g(x). (ii) Second, find the derivative of the inner function g at x. (iii) The product of these two derivatives gives the required derivative of the composite function fog at x . (i) dy dy dx dt = dx , provided ¹ 0. dx dt dt (ii) (iii) dy dx . =1 dx dy (iv) dx 1 dy = ¹ 0. , provided dx dy dx dy x d (\| x \|) = , x ¹ 0. \|x\| dx Derivatives of inverse trigonometric functions 1 (i) d (sin–1 x) = dx (ii) 1 d (cos–1 x) = – , x Î (–1, 1) i.e. \| x \| < 1 dx 1 - x2 (iii) 1 d (tan–1 x) = , for all x Î R dx 1+ x2 (iv) 1 d (cot–1 x) = – , for all x Î R dx 1+ x2 1 - x2 , x Î (–1, 1) i.e. \| x \| < 1 www.thinkiit.in CONTINUTITY AND DIFFERENTIABILITY 3 (c) (v) 1 d ,x>1 (sec–1 x) = dx x x2 - 1 (vi) 1 d (cosec–1 x) = – ,x>1 dx x x2 - 1 Derivatives of algebraic and trigonometric functions (i) d (xn) = nxn – 1 dx (ii) d (xx) = xx log ex dx (iii) d (sin x) = cos x dx (iv) d (cos x) = – sin x dx (v) d (tan x) = sec2 x dx (vi) d (cot x) = – cosec2 x dx d (cosec x) = – cosec x cot x. dx Derivatives of exponential and logarithmic functions (vii) (d) (e) (i) d (ex) = ex , for all x Î R dx (vi) 1 d (loga \| x \|) = , x ¹ 0, a > 0, a ¹ 1. x log a dx (ii) d (ax) = ax log a, a > 0, a ¹ 1, x Î R dx (v) d 1 (log \| x \|) = , x ¹ 0 dx x (iii) d 1 (log x) = , x > 0 dx x (iv) 1 d (loga x) = , x > 0, a > 0, a ¹ 1 x log a dx Logarithmic differentiation If u, n are differentiable functions of x, then (f) d d (un) = un (n log u). dx dx Derivatives of functions in parametric form If x and y are two variables such that both are explicitly expressed in terms of a third variable, say t, i.e. if x = f(t) and y = g(t), then such functions are called parametric functions and dy dy dx dt = dx , provided ¹ 0. dx dt dt www.thinkiit.in CONTINUTITY AND DIFFERENTIABILITY 4 (g) Derivative of second order If a function f is differentiable at a point x, then its derivative f’ is called the first derivative or derivative of first order of the function f. If f’ is further differentiable at the same point x, then its derivative is called the second derivative or derivative of the second order of f at that point and is denoted by f’’. If the function f is defined by y = f(x), then its first and second derivatives are denoted by f ’(x) and f’’(x) or by d2 y dy and or by y1 and y2 or by y’ and y’’ respectively.. dx dx2 3.4 ROLLE’S THEOREM AND LAGRANGE’S MEAN VALUE THEOREM (i) Rolle’s theorem If a function f is (i) continuous in the closed interval [a, b] (ii) derivable in the open interval (a, b) and (iii) f(a) = f(b), then there exists atleast one real number c in (a, b) such that f’(c) = 0. Thus converse of Rolle’s theorem may not be true. (ii) Lagrange’s mean value theorem If a function f is (i) continuous in the closed interval [a, b] and (ii) derivable in the open interval (a, b), then there exists atleast one real number c in (a, b) such that f ’(c) = The converse of Lagrange’s mean value theorem may not be true. www.thinkiit.in f (b) - f (a) b-a CONTINUTITY AND DIFFERENTIABILITY 5 SOLVED PROBLESM Ex.1 Is the function defined by f(x) = x2 – sin x + 5 continuous at x = p ? Sol. Here, f(p) = (p)2 – sin p – 5 = p2 – 5 lim f(x)= lim f(p+h)= lim [(p+h)2 –sin(p+h)–5] h®0 h®0 x ®p + 2 2 = hlim ® 0 [(p + h) + sin h – 5] = p – 5 and lim x ® p- lim 2 f(x)= hlim ® 0 f(p–h) = h ® 0 [(p–h) –sin (p–h)–5] 2 2 = hlim ® 0 [(p – h) – sin h – 5] = p – 5 lim f(x) = x ® p - f(x) = f(p), Since, xlim ®p + the function f is continuous at x = p. Ex.2 Discuss the continuity of the cosine, cosecant, secant and cotangent functions. Sol. Continuity of f(x) = cos x Let a be an arbitrary point of the domain of the function f(x) = cos x. Then, f (a) = cos a lim x ®a + lim f(x) = hlim ® 0 f(a + h) = h ® 0 cos (a + h) = hlim ® 0 [cos a cos h – sin a sin h] = cos a × 1 – sin a × 0 = cos a and lim x ®a - lim f(x) = hlim ® 0 f(a – h) = h ® 0 cos (a – h) = hlim ® 0 [cos a cos h + sin a sin h] = cos a × 1 + sin a × 0 = cos a Since, lim x ®a + lim f(x) = x ®a - f(x)=f(a), the function is continuous at x = a. As a is an arbitrary point of the domain, the function is continuous on the domain of the functions, Proceed as above and prove yourself the continuity of other trigonometric Ex.3 Find all points of discontinuity of f, where ì sinx ï , if x < 0 f(x) = í x ïî x + 1 , if x ³ 0 Sol. The point of discontinuity of f can at most be x = 0. Let us examine the continuity of f at x = 0. lim lim Here, x ®0+ f(x)= hlim ® 0 f(0+h)= h ® 0 [(0+h)+1]=1 www.thinkiit.in CONTINUTITY AND DIFFERENTIABILITY 6 lim lim sin( 0 - h) = lim - sinh = 1 f(x) = hlim ® 0 f(0 – h) = h ® 0 h ®0 0-h -h and x ®0 - Also, f(0) = 0 + 1 = 1 Since, x ®0 + lim f(x) = lim x ®0 - f(x) = f(0), f is continuous at x = 0. Hence, there is no point of discontinuity of f. Ex.4 Determine if f defined by 1 ì 2 ïx sin , if x ¹ 0 f(x) = í x ïî 0, if x = 0 is a continuous function. Sol. It is sufficient to examine the continuity of the function f at x = 0. Here f (0) = 0 Also, lim x ®0 + f(x) = hlim ® 0 f(0 + h) 1 ù é é 2 (0 + h)2 sin h sin ú = 0 = hlim = hlim ® 0 êë ® 0 êë 0 + h úû and lim x ®0 - f(x) = hlim ® 0 f(0 – h) é 2 1 ù æ 1 öù é é ù 1 (0 - h)2 sin = hlim = lim êh sin ç - h ÷ ú = 0 êQ sin £ 1ú ® 0 êë 0 - h úû h ® 0 ë øû è h ë û Hence, lim x ®0 + f(x) = lim x ®0 - f(x) = f(0) So, f is continuous at x = 0 This implies that f is a continuous function at all x Î R. Ex.5 Examine the continuity of f, where f is defined by Sol. ìsin x - cos x , if x ¹ 0 f(x) = í - 1, if x = 0 î Here, f(0) = –1 lim lim Also, x ®0+ f(x)= hlim ® 0 f(0+h)= h ® 0 [sin(0+h)–cos(0+h)] = hlim ® 0 [sin h – cos h] = –1 lim lim and x ®0 - f(x)= hlim ® 0 f(0–h)= h ® 0 [sin(0–h)–cos(0–h)] = hlim ® 0 [–sin h – cos h] [Q sin (–h) = –sin h] = –0 – 1 = –1 and cos (–h) = cos h] lim lim Hence, x ®0+ f(x) = x ®0- f(x) = f(0) So, f is continuous at x = 0; and hence continuous at all x Î R. www.thinkiit.in CONTINUTITY AND DIFFERENTIABILITY 7 Ex.6 Sol. Find the value of k so that the following function f is continuous at the indicated point: (i) ì kx + 1 , if x £ 5 f(x) = í3 x - 5 , if x > 5 at x = 5 î (ii) ì kx2 , if x £ 2 f(x) = í at x = 2 if x > 2 î3 , (i) Since f is given to be continuous at x = 5, we have lim x ®5 + lim + h) = hlim ® 0 f(5 – h) = f(5) Þ h ® 0 f(5 Þ h ® 0 [3(5+h)–5]= h ® 0 Þ 10 = 5k + 1 (ii) Since f is given to be continuous at x = 2, we have lim lim x ® 2+ Ex.7 lim f(x) = x ®5- f(x) = f(5) lim [k(5–h)+1] = 5k + 1 f(x) = 9 5 Þ k= lim x ® 2- f(x) = f(2) lim + h) = hlim ® 0 f(2 – h) = f(2) lim 2 = hlim ® 0 [k(2 – h) ] = 4k Þ h ® 0 f(2 Þ h ® 0 (3) Þ 3 = 4k Þ k= 3 4 Find the values of a and b such that the function defined by if x £ 2 ì5, ï f(x) = íax + b, if 2 < x £ 10 ïî 21, if x ³ 10 is a continuous function. Sol. Since the function f is continuous, it is continuous at x = 2 as well as at x = 10. lim lim f(x) = x ®2- f(x) = f(2) So, x ®2+ i.e., h ® 0 f(2 i.e., 2a + b = 5 and x ®10 + i.e., h ® 0 f(10 i.e., 21 = 10a + b lim lim lim + h) = hlim ® 0 f(2 – h) = f(2) f(x) = (......1) lim x ®10 - f(x) = f(10) + h) = hlim ® 0 f(10 – h) = f(10) (......2) From (1) and (2), we find that a = 2 and b = 1 www.thinkiit.in CONTINUTITY AND DIFFERENTIABILITY 8 Ex.8 Sol. Show that the function defined by g(x) = x – [x] is discontinuous at all integral points. Here, [x] denotes the greatest integer less than or equal to x. The function f(x) = x – [x] can be written as ì x - (k - 1), if k - 1 < x < k f(x) = í x - k, if k < x < k + 1, where k is an arbitrary integer.. î lim lim f(x)= hlim ® 0 f(k + h)= h ® 0 [(k+h) –k]=0 lim lim f(x)= hlim ® 0 f(k–h)= h ® 0 [(k–h)– (k–1)]=1 lim f(x) ¹ Now, x ®k + and x ®k - Since, x ®k + lim x ®k - f(x), the function f is not continuous at x = k. Since k is an arbitrary integer, we can easily conclude that the function is discontinuous at all integral points. Ex.9 Verify LMV Theorem for the function ìx 3 + 2, when x £ 1 on [ -1, 2 ]. f(x) = í î 3 x , when x > 1 Sol. Both x3 + 2 and 3x are polynomial functions. So, f (x) is continuous and differentiable everywhere except at x = 1. Here, lim f ( x ) = 3.1 = 3 x ®1+ lim f ( x ) = 13 + 2 = 3 x ® 1- As lim f(x) = lim f(x) =f(1),f(x) iscontinuous at x =1. x®1+ x®1- Obviously, then f(x) is continuous on [–1, 2]. Again to test the differentiability of f(x) at x = 1, we have L f ‘(1) = 3 3 lim f ( x ) - f (1) lim ( x + 2) - (1 + 2) = x ®1 x ®1 x -1 x -1 lim x3 - 1 lim = x ®1= x ®1- (x2 + x + 1) = 3 x -1 lim f ( x ) - f (1) R f ‘(1) = x ®1+ x -1 = lim 3 x - 3.1 x -1 x ®1+ = lim x ®1+ (3) = 3 As L f ‘(1) = R f ‘(1), the function f (x) is differentiable at x = 1. Hence, f is differentiable in (– 1, 2). Thus, both the conditions required for the applicability of the LMV Theorem are satisfied and hence, there exists at least one c Î (–1, 2) such that f ‘(c) = f (2) - f (-1) 6 -1 5 = Þ f ‘(c) = 2 - ( -1) 3 3 www.thinkiit.in CONTINUTITY AND DIFFERENTIABILITY 9 Now, in x > 1, f ‘(x) = 3. So, f ‘(c) cannot be In x £ 1. f ‘(x) = 3x2 Þ f ‘(c) = 3c2 Obviously, 3c2 = 5 5 5 gives c2 = or c = ± 3 9 3 Both 5 in this interval. 3 5 5 and – lie in (–1, 2). Thus, LMV is verified for f(x) and [–1, 2]. 3 3 Ex.10 Verify Rolle’s theorem for the function f (x) = x (x – 3)2 in the closed interval 0 £ x £ 3. Sol. (i) Here, f(x) = x (x – 3)2 = x (x2 – 6x + 9) = x3 – 6x2 + 9x Since f(x) is a polynomial function of x, it is continuous in [0, 3] (ii) f ‘(x) = 3x2 – 12x + 9 exists uniquely in the open interval (0, 3) (iii) f(0) = (0)3 – 6(0)2 + 9(0) =0–0+0=0 f(3) = (3)2 – 6(3)2 + 9(3) = 27 – 54 + 27 = 0 f(0) = f(3) Thus, all the three conditions are satisfied, Hence, Rolle’s Theorem is applicable. Let us now solve i.e. f ‘(c) = 0 3c – 12c + 9 = 0 2 3(c2 – 4c + 3) = 0 (c – 3) (c – 1) = 0 c = 3, 1 SInce, c = 1 Î (0, 3), the Rolle’s Theorem is verified for the function. f(x) = x(x – 3)2 in the closed interval [0, 3]. Ex.11 Verify Rolle’s Theorem for the function f(x) = (x – a)m (x – b)n in [a, b] ; m, n being positive integers. Sol. Here, f(x) is a polynomial function of degree (m + n). So, it is a continuous function in [a, b]. f ‘(x) = (x – a)m – 1 (x – b)n – 1 [m (x – b) + n (x – a)] exists uniquely in (a, b). So, it is derivable m (a, b). Further, f(a) = 0 and f(b) = 0. So, f(a) = f(b) Thus, all the three conditions of Rolle’s Theorem are satisfied. Hence, Rolle’s Theorem is applicable. Let us now solve f ‘(c) = 0 Þ (c–a)m – 1 (c – b)n – 1 [m (c – b) + n (c–a)] = 0 Þ c = a or c = b or c = Since c = mb + na m+n mb + na Î (a, b), the Rolle’s Theorem is verified. m+n www.thinkiit.in CONTINUTITY AND DIFFERENTIABILITY 10 ì1 + x, if x £ 2 Ex.12 Show that the function f(x) = í5 - x , if x > 2 is continuous at x = 2, but not differentiable î at x = 2. Sol. At x = 2, lim lim f(x) = hlim ® 0 f(2 + h) = h ® 0 [5–(2+ h)] = 3 lim lim f(x) = hlim ® 0 f(2 – h) = h ® 0 [1+(2–h)] = 3 x ®2+ x ®2- Also, f(2) = 1 + 2 = 3 Since, x ® 2+ lim f(x) = lim x ® 2- f(x) = f(2), f(x) is continuous at x = 2. f (2 - h) - f (2) Next, Lf ‘(2) = hlim ®0 -h (1 + 2 - h) - (1 + 2) = hlim =1 ®0 = -h f (2 + h ) - f ( 2 ) 5 - (2 + h) - (1 + 2) = hlim = –1 Rf ‘(2)= hlim ®0 = ®0 h h Since, Lf ‘(2) ¹ Rf ‘(2), the function f is not differentiable at x = 2. ì1 - x, if x < 1 Ex.13 Show that the function f(x) = í x2 - 1, if x ³ 1 is continuous at x = 1, but not differentiable î thereat. Sol. The function is continuous at x = 1, because lim x ®1+ f(x) = lim x ®1- f(x) = f(1) as shown below : lim lim lim 2 2 f(x) = hlim ® 0 f(1 + h) = h ® 0 [(1 + h) – 1] = h ® 0 (h + 2h) = 0 ; lim lim lim f(x) = hlim ® 0 f(1 – h) = h ® 0 [1 – (1 – h)] = h ® 0 (h) = 0 x ®1+ x ®1- and f(1) = (1)2 – 1 = 1 – 1 = 0 Further, f (1 + h) - f (1) Rf ‘(1) = hlim ®0 h [(1 + h ) 2 - 1] - [0 ] = hlim =2 ®0 h f(1 - h) - f(1) [(1 - h ) 2 - 1] - [0 ] Lf ‘(1) = hlim = hlim ® 0 ®0 h h æ h ö lim ç ÷ = hlim ® 0 è - h ø = h ® 0 (–1) = –1 Since, Rf ‘(1) ¹ Lf ‘(1), the function is not differentiable at x = 1. www.thinkiit.in CONTINUTITY AND DIFFERENTIABILITY 11 Ex.14 Show that the function f defined as ì 3x - 2, if 0 < x £ 1 ï 2 f(x) = í2 x - x , if 1 < x £ 1 ï 5 x - 4, if x > 2 î is continuous at x = 2, but not differentiable thereat. Sol. lim lim At x=2, x ®2+ f(x)= hlim ® 0 f(2+h)= h ® 0 [5(2+h)–4]=6 lim x ®2- lim 2 f(x) = hlim ® 0 f(2–h)= h ® 0 [2(2–h) –(2–h)] 2 = hlim ® 0 [2(4 – 4h + h ) – (2 – h)] 2 = hlim ® 0 [6 – 7h + 2h ] = 6 and f(2) = 2 (2)2 – 2 = 8 – 2 = 6 Since x ®2+ lim lim f(x) = x ®2- f(x)=f(2), the function f is continuous at x = 2. f (2 - h) - f (2) Next, Lf ‘(2) = hlim ®0 -h 2(2 - h)2 - ( 2 - h) - [5(2) - 4] = hlim ®0 -h = 6 - 7h + 2h2 - 6 =7 -h f (2 + h) - f (2) f ‘(2) = hlim ®0 h [5(2 + h) - 4] - [5(2) - 4] = hlim ®0 h [5(2 + h) - 4] - [5(2) - 4] = hlim ®0 h é 6 + 5h - 6 ù = hlim ú =5 ® 0 êë h û Since, Lf ’(2) = Rf ‘(2), the function f is not differentiable at x = 2. www.thinkiit.in CONTINUTITY AND DIFFERENTIABILITY 12 UNSOLVED PROBLEMS EXERCISE – I Q.1 ì 1 ï e x -1 ï , when x ¹ 0 Show that the function f(x) = í 1 ï ex +1 ï 0 , when x = 0 î is discontinuous at x = 0. Q.2 1 ì ï x sin , when x ¹ 0 Show that the function f(x) = í x ïî 0 , when x = 0 is continuous at x = 0 Q.3 Is the following function continuous at the origin ? f(x) = f(0) = Q.4 cos ax - cos bx x2 , when x ¹ 0 b2 - a2 , when x = 0 2 ì ex - 1 ï , when x ¹ 0 If the function defined by f(x) = í log(1 + 2x) ï k, when x = 0 î is continuous at x = 0, find the value of k. Q.5 Q.6 ì cos2 x - sin 2 x - 1 , ï if f(x) = í x2 + 1 - 1 ï k , î when x ¹ 0 is continuous at x = 0, find k. when x = 0 ì ï 1 - cos 4x , when x < 0 ï x2 ï k , when x = 0 is continuous Determine the value of k so that the function f(x) = í ï x , when x > 0 ï ïî 16 + x - 4 at x = 0. Q.7 ì x- \| x \| ï , when x ¹ 0 Discuss the continuity of the function f(x) at x = 0 if f(x) = í 2 ïî 2 , when x = 0 Q.8 ì 2, if x £ 3 ï Let f(x) = íax + b, if 3 < x < 5 ïî a , if x ³ 5 find the value of a and b, so that f(x) is continuous. www.thinkiit.in CONTINUTITY AND DIFFERENTIABILITY 13 Q.9 Q.10 Q.11 Q.12 Q.13 x -1 ì , x ¹1 ïï 2 + 2 x 7 x 5 Find the derivative of f(x) = í at x = 1 -1 ï , x =1 ïî 3 ì x 2 + 3x + a , if x £ 1 Find the value of ‘a’ and ‘b’, so that the function f(x) = í is differentiable at , if x > 1 îbx + 2 each x Î R. Differentiate the following w.r.t x : 1 - cos x 1 + cos x æ ö loga ç x + x 2 + a2 ÷ è ø (iii) log (sec x + tan x) (vi) æ 2 ö 2 ç x +a +x÷ log ç 2 ç x + a2 - x ÷÷ è ø (i) log (iv) x æ log ç sin + cos ÷ 2 2 è ø (i) If y = (ii) æ x +1+ x -1ö x + x2 - 1 ÷ , prove that dy = If y = çç ÷ dx x2 - 1 è x +1 - x -1 ø (iii) If y = (iv) If y = sec x + tan x dy = sec x (sec x + tan x) , show that sec x - tan x dx (i) If y = x + (ii) (v) æ 1 + x sin x ö ÷ log ç è 1 - x sin x ø 1- x dy +y=0 , prove that (1 – x2) 1+ x dx cos x + sin x dy æ , show that = sec2 ç x + ÷ cos x - sin x dx è 1 x , show that 2x dy + y = 2 x (ii) dx If y = x sin y, prove that x dy = dx y (1 - x cos y ) (iii) -1 dy ,x¹y If x 1+ y + y 1+ x = 0, prove that = ( 1 + x )2 dx x + x + x + ........ , prove that Q.14 If y = Q.15 Given that cos cosec2x – 1 dy = ( 2 y - 1) dx 1 1 x x x sin x x x . cos . cos ......= , prove that 2 sec2 + 4 sec2 + ......= 2 4 8 x 2 2 4 2 1 x2 www.thinkiit.in CONTINUTITY AND DIFFERENTIABILITY 14 2t Q.16 If x = tan–1 Q.17 Differentiate : (i) (iii) 1- t 2 and y = sin–1 2t 1+ t 2 , show that dy = 1. dx ì ü sin–1 í x 1 - x + x 1 - x 2 ý î þ æ 1 ö ÷÷ + tan–1 tan–1 çç 2 è x + x + 1ø æ 1 ö çç 2 ÷÷ + tan–1 è x + 3x + 3 ø æ x ç tan–1 ç 2 è 1- x ö ÷ –1 ÷ + tan ø (ii) æ 2x ö ÷ tan–1 çç è 1 + 15 x ø æ 1 ö çç 2 ÷÷ + ............ to n terms. è x + 5x + 7 ø æ x ç ç 2 è 1+ 1- x ö ÷ ÷ (ii) ø æ 3 cos x - 4 sin x ö ÷ cos–1 ç 5 è ø Q.18 Differentiate : (i) Q.19 æ 1 Differentiate : sin–1 çç 2 è 1+ x Q.20 Differentiate : (i) Q.21 ì2x - 1, if x < 0 Discuss the continuity of the function f(x) = í2x + 1, if x ³ 0 î Q.22 ì\| x -4\| ï , x¹4 If a function f(x) is defined as f(x) = í x - 4 show that f is everywhere continuous except ïî 0 , x =4 ö ÷ + tan–1 ÷ ø æ ö 2 ç 1 + x - 1÷ çç ÷÷ . x è ø æ 1 + x + 1- x sin–1 çç 2 è ö ÷ ÷ ø (ii) æ x - x -1 ö ÷ cos–1 çç -1 ÷ èx+x ø at x = 4. Q.23 Discuss the continuity of the function f(x) = \| x \| + \|x – 1\| in the interval [–1, 2] Q.24 Show that f(x) = \| x \| is not differential at x = 1. Q.25 ì2 + x , if x ³ 0 Let f(x) = í2 - x , if x < 0 , show that f(x) is not derivable at x = 0. î Q.26 ì 2 æ 1ö ï x sinç ÷ , if x ¹ 0 Show that the function f(x) = í is differential at x = 0 and f ‘(0) = 0. èxø ï 0 , if x = 0 î www.thinkiit.in CONTINUTITY AND DIFFERENTIABILITY 15 BOARD PROBLES EXERCISE – II log x dy . = dx (1 + log x )2 Q.1 If xy = ex – y, prove that Q.2 If xp yq = (x + y)p + q, prove that Q.3 æ x2 ö ç ÷ Find 2 when y = log ç e x ÷ . dx è ø Q.4 If y = ae2x + be–x, prove that Q.5 If y = A cos nx + B sin nx show that Q.6 ì2 x - 1 , x < 0 Discuss the continuity of the function f(x) at x = 0 if f(x) = í2x + 1 , x ³ 0 î [C.B.S.E. 2002] Q.7 Show that the function f(x) = 2x – \| x \| is continuous at x = 0. [C.B.S.E. 2002] Q.8 ì 3ax + b , x > 1 ï , x = 1 is continuous. at x = 1, find the values of a and b. If the function f(x) = í 11 ïî 5ax - 2b , x < 1 [C.B.S.E. 2000] dy y = . dx x [C.B.S.E. 2000] d2 y [C.B.S.E. 2000] d2 y dx2 dy – 2y = 0. dx d2 y dx2 [C.B.S.E. 2000] + n2y = 0. [C.B.S.E. 2001] [C.B.S.E. 2002] æp ö 1 - sin 2x dy , prove that + sec2 ç - x ÷ = 0. 1 + sin 2x dx è4 ø [C.B.S.E. 2002] Q.9 If y = Q.10 dy æ æ p x öö If y = log çç tanç + ÷ ÷÷ , show that – sec x = 0. dx è 4 2 øø è Q.11 Verify Lagrange’s mean value theorem for the following functions in the given intervals. [C.B.S.E. 2002] [C.B.S.E. 2002] Also find ‘c’ of this theorem : (i) f(x)=x2 +x–1 in [0, 4] Q.12 If y = ex (sin x + cos x), prove that Q.13 Differentiate the following w.r.t. x æ 1 - cos x (i) log çç è 1 + cos x ö ÷ (ii) log (x + ÷ ø d2 y dx2 –2 dy + 2y = 0. dx 1+ x 2 ) www.thinkiit.in (ii) f(x)= x 2 - 4 on [2, 4] [C.B.S.E. 2002] [C.B.S.E. 2003] CONTINUTITY AND DIFFERENTIABILITY 16 dy p at t = . dx 2 Q.14 If x = a(t + sin t), y = a(1 – cos t), find Q.15 Differentiate the following functions w.r.t x : æ ö (i) 1 + sin x tan–1 çç 1 - sin x ÷÷ . è ø (iv) æ 2 ç 5x + 12 1 - x sin ç 13 ç è (vi) æ 1+ x - 1 - x ö ç ÷ tan–1 ç ÷ è 1 + x + 1- x ø –1 d dx (ii) ö ÷ ÷÷ (v) ø æ 1 - sin x ç cot–1 ç 1 + sin x è [C.B.S.E. 2004] ö ÷ ÷ ø æ 2 2 ç 1+ x - 1 - x tan çç 2 2 è 1 + x + 1- x –1 æx a2 ç sin -1 ÷ = a2 - x 2 + ç2 ÷ 2 2 è ø [C.B.S.E. 2003] (iii) ö æ 1 + x 2 - 1÷ tan–1 çç ÷÷ x ç ø è ö ÷ ÷÷ ø a2 - x 2 . Q.16 Prove that Q.17 If y = (sin x)x + (cos x)tan x, find Q.18 Find Q.19 æ 2x ö ÷÷ w.r.t. sin–1 Differentiate tan–1 çç è 1 - x2 ø Q.20 æ3+ xö ÷ If f(x) = ç è 1+ x ø Q.21 Find Q.22 2 æ x ö d2 y 1 æ a ö ç ÷ ç ÷ If y = x log , prove that . = è a + bx ø x è a + bx ø dx2 Q.23 ì ï 1 - cos 4x , x<0 ï x2 ï a , x = 0 is continuous at x = 0, find the value If the function f defined by f(x) = í ï x ï , x>0 ï 16 + x - 4 î dy dx [C.B.S.E. 2004] [C.B.S.E. 2004] 2bt 1- t2 dy , when x = a ,y= 2 dx 1+ t2 1+ t [C.B.S.E. 2004] æ 2x ö çç ÷÷ . è 1 + x2 ø [C.B.S.E. 2004] 2 + 3x , find f ’(0). æ 1 + t2 ö 2t dy ÷ if : x = a çç 2 ÷, y = dx 1- t2 è 1- t ø of a. [C.B.S.E. 2005] [C.B.S.E. 2005] [C.B.S.E. 2005] [C.B.S.E. 2006] www.thinkiit.in CONTINUTITY AND DIFFERENTIABILITY 17 1 dy +y=2 x. dx Q.24 If y = Q.25 æ 1 + sin x + 1 - sin x Differentiate w.r.t. x : tan–1 çç è 1 + sin x - 1 - sin x Q.26 If y x 2 + 1 = log ( x 2 + 1 – x), prove that (x2 + 1) Q.27 b æ dy ö = . If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t), show that ç ÷ p a è dx ø t = x + x , then show that 2x ö ÷ ÷ ø [C.B.S.E. 2006] [C.B.S.E. 2006] dy + xy + 1 = 0. dx [C.B.S.E. 2006] 4 [C.B.S.E. 2006] d2 y Q.28 If y = cosec x + cot x, show that sin x . Q.29 Verify LMV ; find ‘c’ f(x) = x 2 + 2x + 3 in [4, 6] [C.B.S.E. 2006] Q.30 ì x 2 - 25 ï If f(x) = í x - 5 , x ¹ 5 is continuous at x = 5, find the value of k. ïî k , x =5 [C.B.S.E. 2007] Q.31 If y = 3e2x + 2e3x, prove that Q.32 If y = A emx + B enx, prove that Q.33 If y = sin (log x), prove that x 2 Q.34 For what value of k is the following function continuous at x = 2 ? d2 y dx 2 –5 d2 y dx2 d2 y dx 2 dx2 = y2. dy + 6y = 0. dx – (m + n) +x dy + mny = 0. dx dy + y = 0. dx ì 2x + 1 , x < 2 ï f(x) = í k , x = 2 ïî 3 x - 1 , x > 2 Q.35 [C.B.S.E. 2006] [C.B.S.E. 2007] [C.B.S.E. 2007] [C.B.S.E. 2007] [C.B.S.E. 2008] Discuss the continuity of the following function at x = 0 : ì x 4 + 2x3 + x 2 ï , x¹0 f(x) = í . tan-1 x ï 0 , x =0 î www.thinkiit.in [C.B.S.E. 2008] CONTINUTITY AND DIFFERENTIABILITY 18 Q.36 ì 1 - sin 3 x ï ï 3 cos2 x ï a Let f(x) = í ï ï b(1 - sin x ) ï ( p - 2x )2 î Q.37 If f(x), defined by the following, is continuous at x = 0, find the values of a, b and c. p 2 p p , if x = . If f(x) be a continuous function at x = , find a and b. 2 2 p , if x > 2 , if x < [C.B.S.E. 2008] ì ï sin( a + 1)x + sin x , if x < 0 ï x ï c , if x = 0 f(x) = í ï 2 ï x + bx - x , if x > 0 ï bx 3 / 2 î [C.B.S.E. 2008] dy n x 2 + a 2 ) , prove that dx = ny . [C.B.S.E. 2008] Q.38 If y = (x + Q.39 If x 1 + y + y 1 + x = 0 , find Q.40 If y = Q.41 dy p æ at q = . If x = a ç cos q + log tan ÷ and y = a sin q, find the value of dx 4 è Q.42 If y = (log (x + Q.43 If sin y = x sin (a + y), prove that Q.44 If (cos x)y = (sin y)x, find Q.45 If y = Q.46 Differentiate the following function w.r.t. x : xsin x + (sin x)cos x Q.47 x sin x 2 x + a2 dy . dx [C.B.S.E. 2008] æ1 1 ö÷ dy ç . x 2 + 1 – log ç x + 1 + 2 ÷ , find dx x è ø sin -1 x 1- x 2 [C.B.S.E. 2008] x 2 + 1 ))2, show that (1 + x2) d2 y dx2 +x dy – 2 = 0. dx dy sin 2 (a + y ) = . dx sin a dy . dx , show that (1 – x2) [C.B.S.E. 2008] [C.B.S.E. 2008] [C.B.S.E. 2009] [C.B.S.E. 2009] d2 y dx 2 – 3x dy –y=0 dx æ sin x ö + cos x log x ÷ + (sin x)cos x (cos x cot x – sin x logsin x) ç è x ø www.thinkiit.in [C.B.S.E. 2009] [C.B.S.E. 2009] CONTINUTITY AND DIFFERENTIABILITY 19 dy if (x2 + y2)2 = xy.. dx Q.48 Find Q.49 If y = 3 cos (log x) + 4 sin (log x), then show that x 2 . æ 3x + 4 1 - x 2 ç çç 5 è [C.B.S.E. 2009] d2 y dx 2 +x dy + y = 0 [C.B.S.E. 2009] dx ö dy ÷ ÷÷ , find dx . ø Q.50 If y = cos–1 Q.51 If y = cosec–1 x, x > 1, then show that x (x 2 – 1) Q.52 If xy = ex – y, show that Q.53 d2y dy æ1 ö =0 If x = tan ç log y ÷ , show that (1 + x2) + (2x – a) 2 dx dx èa ø Q.54 If x = dy y =- . dx x [C.B.S.E. 2012] Q.55 é 1 + x2 - 1 ù ú with respect to x. Differentiate tan–1 ê x úû ëê [C.B.S.E. 2012] Q.56 If x = a (cos t + t sin t) and y = a (sin t – t cos t), 0 < t < -1 asin t ,y= [C.B.S.E. 2010] d2 y dx2 + (2x2 – 1) dy = 0 [C.B.S.E. 2010] dx log x dy = . {log (xe)}2 dx -1 acos t , show that [C.B.S.E. 2011] [C.B.S.E. 2011] 2 2 2 dy dx dy p , 2 and , find 2 2 . 2 dt dt dx [C.B.S.E. 2012] dy (1 + log)2 = . dx log y Q.57 If yx = ey – x, prove that Q.58 Differentiate the following with respect to x : æ 2 x +1.3x sin–1 çç x è 1 + (36) Q.59 [C.B.S.E. 2013] [C.B.S.E. 2013] ö ÷ ÷ ø Find the value of k, for which [C.B.S.E. 2013] ì 1 + kx - 1 - kx , if - 1 £ x < 0 ïï x í f(x) = ï 2x + 1 , if 0 £ x < 1 x -1 îï is continuous at x = 0. OR If x = a cos3 q and y = a sin3 q, then find the value of d2 y dx 2 at q = www.thinkiit.in p . 6 CONTINUTITY AND DIFFERENTIABILITY 20 EXERCISE – 1 (UNSOLVED PROBLEMS) 3. yes 4. 1/2 9. –2/9 10. 3 and 5 11. (i) cosec x (ii) (v) (iii) 2( x cos x + sin x) 2 (vi) 2 1 - x sin x 1 2 1+ (x + x) 18. (i) 3 2 1- x 2 5. –4 2 2 x +a 6. 8 1 17. (i) 2 7. discontinuous 1- x 1 2 log a x + a + 2 1 2 x-x 2 (ii) 2 8. 7/5 and –17/2 (iii) sec x (iv) 5 1 + 25 x 2 æp xö 1 tan ç 4 - 2 ÷ è ø 2 3 1 + 9x2 1 1+ x2 (ii) 1 19. -1 2 2(1 + x ) 20. (i) -1 2 1- x (ii) 2 -2 1+ x2 EXERCISE – 2 (BOARD PROBLEMS) 3. -2 6. Discontinuous x2 13. (i) cosec x (ii) 15. (i) 18. 1 2 (ii) 1 2 8. 3,2 1 6 14. 1 1+ x2 (iii) 11. (i) 2 (ii) 1 1 (iv) 2 2(1 + x ) 1- x (v) 2 x 1- x 4 (vi) -1 2 1 - x2 - b(1 - t 2 ) 2at 19. 1 21. 1+ t2 2at 23. 8 25. –1/2 29. 5 30. 10 32. 5/2 34. 5 35. continuous 36. 1 ,4 2 37. – -1 3 1 , any real number, 39. (1 + x )2 2 2 41. 1 44. log sin y + y tan x log cos x - cot y 56. at cos t, at sin t and 48. y - 4x 3 - 4xy 2 2 3 4x y + 4y - x 50. -1 1- x 2 55. 1 2(1 + x2 ) sec3 t at www.thinkiit.in 40. x2 + 1 x
# Finding a percentage of an amount using a multiplier ## Home learning focus In this lesson, you will learn how to find percentages of amounts using decimal multipliers. You will need a calculator. This lesson includes: • a learning summary • two activities Created in partnership with Numerise. # Learn ### Recap: Finding percentages of amounts by hand You should have already learnt how to find percentages of amounts by hand (without a calculator), but here is a quick recap. If you needed to find 15% of 400 by hand, then there are a couple of different methods you could use: Method 1: Find 1% of 400 and then multiply by 15 To find 1% of a number, you divide by 100, so: 1% of 400 400 ÷ 100 = 4 1% of 400 = 4 15% is 15 lots of 1%, so: 15% of 400 15 × 4 = 60 15% of 400 = 60 Method 2: Add together 10% and 5% of 400 You can break 15% into smaller percentages, for example: 15% = 10% + 5% To find 10% of a number, divide by 10, so: 10% of 400 400 ÷ 10 = 40 5% is half of 10%, so 5% of 400 is half of 40, which is 20 15% of 400 = 10% of 400 + 5% of 400 15% of 400 = 40 + 20 15% of 400 = 60 ## Finding percentages of amounts using a decimal multiplier Sometimes Method 1 and Method 2 can be slow, or the percentage may be hard to break down into smaller chunks. In these cases, you can use decimals. Remember, percentages can be written as decimals. Look at the diagram below. For our 15% of 400 example above, 15% written as a decimal is 0.15 You can use the decimal equivalent of a percentage to help us find percentages of amounts - you just need to multiply the amount by the decimal. So to find 15% of 400, you would multiply 0.15 by 400. You can use a calculator. 0.15 × 400 = 60 ## Example: Find 27% of 300 First, rewrite 27% as a decimal. 27 ÷ 100 = 0.27 27% = 0.27 Then, use a calculator to multiply the original amount by the decimal multiplier. 0.27 × 300 = 81 So, 27% of 300 is 81. ## Example: Find 36.2% of 640 First, rewrite 36.2% as a decimal. 36.2 ÷ 100 = 0.362 36.2% = 0.362 Then, use a calculator to multiply the original amount by the decimal multiplier. 0.362 × 640 = 231.68 So, 36.2% of 640 is 231.68. ## Example: Find 112% of 86 Here the percentage is greater than 100%, but you can still use the same method. First, rewrite 112% as a decimal. 112 ÷ 100 = 1.12 112% = 1.12 Then, use a calculator to multiply the original amount by the decimal multiplier. 1.12 × 86 = 96.32 So, 112% of 86 is 96.32. # Practise ## Activity 1 Worksheet: Decimal multipliers Practise finding percentages of amounts using this worksheet from Numerise. You can print it out or write your answers on a piece of paper. You will need to use a calculator, but don’t forget to write down your workings! ## Activity 2 Worksheet: Finding the percentage of an amount using a multiplier Have a go at this worksheet from Beyond where you have to find the percentage of amounts using a multiplier. You can print it out or write your answers on a piece of paper.
16 Pages Perimeter and Area Class 5 Worksheets with Answers (PDF) Looking for helpful worksheets to help your 5th graders understand perimeter and area? We've got you covered! Check out our collection of 16 printable PDF pages of fun and informative worksheets with answers to get them understanding this math concept quickly. Calculate the area and perimeter of different polygons. Students can practice the concept of area and perimeter by calculating the area and perimeter of different polygons. These worksheets include questions that require students to calculate the areas and perimeters of regular or irregular shapes such as rectangles, triangles, quadrilaterals, pentagons, hexagons, octagons, circles and more! They are great resources to help reinforce this important math concept. Determine area and perimeter for triangles and rectangles. While working with triangles and rectangles, students need to calculate the length of each side in order to determine the perimeter. The formula for area for a triangle is A = ½bh, where b is the base and h is the height of the triangle. To find the area of a rectangle or square, multiply the length by width (A=l×w). For perimeter, add together all sides of a shape. As an example, for a rectangle A + B + C + D = perimeter. Learn to calculate circumference and area of circles. When working with circles, use the formula c = 2πr to find circumference of circles. The radius is half of the diameter so that r = d/2. To calculate area of a circle, use the formula A = πr2, where r is the radius of the circle and π (3.14) is a constant available from any scientific calculator. Use these formulas to solve worksheets and prepare for exams! Understand the differences between perimeter and circumference. Knowing the difference between perimeter andcircumference is essential for mastering these topics. Perimeter is the length of a shape and can be calculated by adding up all the sides of a particular shape. In contrast, circumference is the distance around a circle and can be found using the formula c = 2πr, where r is the radius of the circle. Solve real-world problems involving perimeter and area. Solving real-world problems involving perimeter and area will help you apply what you have learnt in a practical way. Start by asking yourself questions like, “How can I use these formulas to determine the boundary of a given shape?” or “What is the formula for finding the area of an ellipse?” This will not only help build your understanding, but also ensure you’re ready for exams. When it comes to teaching perimeter and area class 5, incorporating engaging and comprehensive worksheets is essential for students to develop a solid understanding of these mathematical concepts. Our perimeter and area class 5 worksheets with answers provide a wide variety of questions and exercises, ranging from perimeter area and volume class 5 problems to area and perimeter questions for class 5. These worksheets have been thoughtfully designed to support students in mastering the fundamentals of these topics, ensuring they excel in class 5 maths. One popular resource is the perimeter and area worksheet for class 5, which covers a diverse array of class 5 maths area and perimeter problems. This worksheet is complemented by our perimeter area and volume class 5 worksheets that delve deeper into the concepts of area, perimeter, and volume, offering more challenging exercises for students who are ready to advance their skills. An additional resource, the area perimeter worksheet for class 5, presents a variety of questions on area and perimeter for class 5, ensuring students have ample opportunity to practice and apply their knowledge. Our perimeter and area class 5 worksheet collection is available in PDF format, making it easy for teachers and parents to download and print the materials for offline use. The perimeter and area class 5 worksheets cover a wide range of topics, such as rectangle perimeter, area of a square, and word problems on area and perimeter for class 5. These questions on area and perimeter for class 5 PDF worksheets are designed to engage students while reinforcing their understanding of these crucial mathematical concepts. To accommodate different learning styles and curriculum requirements, our area and perimeter worksheet for class 5 comes in various formats, including the worksheet on perimeter and area, worksheet for perimeter and area, and worksheet on perimeter and area for class 5. These resources are perfect for students who need extra practice with perimeter questions for class 5, area questions for class 5, and perimeter sums for class 5. As an added bonus, we offer a range of complementary materials, such as maths worksheets for class 5th, perimeter and area class 5 worksheets pdf, volume worksheets for class 5, and area and its boundary class 5 extra questions. With these resources, students can explore other related topics, like perimeter and area class 5 PDF materials, area of a square worksheet, and area and perimeter class 5 exercises. Our volume worksheet for class 5, for instance, offers a variety of volume questions for class 5, which can help students gain a deeper understanding of the relationships between perimeter, area, and volume. To further support students in their learning journey, we provide area and its boundary class 5 worksheets, perimeter and area class 6 worksheets pdf with answers, and perimeter worksheet options. These materials are designed to help students transition from class 5 to class 6 seamlessly while strengthening their skills in area, perimeter, and volume. Additionally, we offer area worksheet, perimeter and area mcq class 5, mcq on perimeter and area class 5, and word problems on perimeter and area class 5 resources to help students build confidence in their mathematical abilities. These area and perimeter word problems worksheets pdf documents are especially useful for students who want to practice real-life applications of their newfound knowledge. In conclusion, our comprehensive collection of perimeter and area class 5 worksheets and resources provides the perfect platform for students to develop a strong foundation in these essential mathematical concepts. With a wide variety of exercises and problem-solving opportunities, these worksheets will engage and challenge students, ensuring they excel in their studies and are well-prepared for future mathematical endeavors. Perimeter and area formulas for class 5th Maths 1. Perimeter Formulas: • Perimeter of a Rectangle: P = 2(L + W), where P is the perimeter, L is the length, and W is the width. • Perimeter of a Square: P = 4s, where P is the perimeter and s is the side length of the square. • Perimeter of a Triangle: P = a + b + c, where P is the perimeter, and a, b, and c are the lengths of the three sides of the triangle. 1. Area Formulas: • Area of a Rectangle: A = L × W, where A is the area, L is the length, and W is the width. • Area of a Square: A = s², where A is the area and s is the side length of the square. • Area of a Triangle: A = 0.5 × b × h, where A is the area, b is the base, and h is the height of the triangle. Perimeter and area class 5 word problems 1. John has a rectangular garden with a length of 8 meters and a width of 5 meters. What is the perimeter and area of his garden? 2. Alice has a square-shaped room with a side length of 6 meters. Calculate the perimeter and area of her room. 3. A triangular park has sides measuring 7 meters, 10 meters, and 15 meters. Find the perimeter of the park. If the height corresponding to the base of 10 meters is 6 meters, find the area of the park. 4. The length of a rectangular field is 50 meters, and its width is 30 meters. How much fencing will be required to enclose the entire field? Also, find the area of the field. 5. A square garden has an area of 144 square meters. Calculate the side length and perimeter of the garden. 6. The width of a rectangular swimming pool is 12 meters, and its area is 240 square meters. Determine the length of the swimming pool and its perimeter. 7. A rectangular playground is twice as long as it is wide. If the perimeter of the playground is 60 meters, find the length and width of the playground and its area. FAQs 1. What is the perimeter? The perimeter is the total distance around the edge or boundary of a two-dimensional shape. It is the sum of the lengths of all the sides of the shape. 1. What is the area? The area is the amount of space occupied by a two-dimensional shape. It is measured in square units, such as square meters, square centimeters, or square inches. 1. How do you calculate the perimeter and area of a rectangle? To calculate the perimeter of a rectangle, use the formula P = 2(L + W), where P is the perimeter, L is the length, and W is the width. To calculate the area of a rectangle, use the formula A = L × W, where A is the area. 1. How do you calculate the perimeter and area of a square? To calculate the perimeter of a square, use the formula P = 4s, where P is the perimeter and s is the side length. To calculate the area of a square, use the formula A = s², where A is the area. 1. How do you calculate the perimeter and area of a triangle? To calculate the perimeter of a triangle, add the lengths of its three sides: P = a + b + c, where P is the perimeter, and a, b, and c are the side lengths. To calculate the area of a triangle, use the formula A = 0.5 × b × h, where A is the area, b is the base, and h is the height. 1. Why are perimeter and area important concepts to learn? Understanding perimeter and area is important because these concepts have practical applications in real-life situations, such as measuring the dimensions of rooms, fields, or objects, estimating the amount of materials needed for construction or decoration, and solving problems related to space and size. 1. How can students practice perimeter and area concepts in class 5? Students can practice perimeter and area concepts by working on a variety of worksheets and word problems, participating in class discussions, and applying their knowledge to real-life situations. Teachers can provide engaging and challenging exercises to help students develop a strong foundation in these essential mathematical concepts. • Tags : • Class 5 perimeter and area
# Unit 6: Series AC Circuits In this unit we'll pull together a lot of things from earlier units and use them to analyze series circuits containing AC voltage sources. You'll need to remember what you've learned about AC fundamentals, capacitors, inductors, complex numbers, and phasors. Once you take all of that into account, though, you'll find that to analyze a series AC circuit, you follow the same steps that you follow to analyze a series DC circuit. Also, the same rules that hold for series DC circuits (such as Ohm's law, Kirchhoff's Voltage Law, and the Voltage-Divider Rule) also hold for series AC circuits. But the math is a little more complicated, because each step involves complex numbers instead of real numbers. ##### Units 4 and 5 Review • This unit will build on material that you studied in Unit 4 and Unit 5 . So let's begin by taking these two self-tests to review what you learned in those units. ##### Review of Series DC Circuits • In EET 1150 you learned how to analyze series DC circuits like the one shown below. Let's do a quick review of what you learned there. • You should recall that the basic steps in analyzing a circuit like this one are: 1. Add the resistance values to find the circuit's total resistance, RT. For the circuit shown, this means that RT = R1 + R2 + R3 2. Apply Ohm's law to the entire circuit to find the circuit's total current: IT = VS ÷ RT 3. Recognize that, since we're dealing with a series circuit, each resistor's current is equal to the total current. For the circuit shown: I1 = I2 = I3 = IT 4. Apply Ohm's law to each resistor to find the voltage drops: V1 = I1 × R1 and V2 = I2 × R2 and V3 = I3 × R3 ##### Review: Kirchhoff's Voltage Law in Series DC Circuits • In EET 1150 you also learned that Kirchhoff's Voltage Law (KVL) says that the sum of the voltage drops around any closed loop equals the sum of the voltage rises around that loop. • In terms of a simple series DC circuit like the one you just analyzed, this means that the sum of all the resistor voltage drops must equal the source voltage. ##### Review: Voltage Divider Rule in Series DC Circuits • In EET 1150 you also learned that the voltage-divider rule is a shortcut rule that you can use to find the voltage drop across a resistor in a series circuit. • The rule says that the voltage across any resistance in a series circuit is equal to the ratio of that resistance to the circuit's total resistance, multiplied by the source voltage. • In equation form, this rule is expressed as: Vx = VS×(Rx÷RT) ##### Review: Troubleshooting Series DC Circuits • Troubleshooting a non-working circuit means finding the problem that is preventing the circuit from working correctly. • The two most common types of problems are open circuits and short circuits. • An open circuit, or "open," is a break in a circuit path. • The most important thing to remember about opens is that no current can flow through an open. • Therefore, no current can flow anywhere in a series circuit containing an open. • Since no current flows through an open, you can think of the open as having infinite resistance (R = ∞). • Usually, an open will not have a voltage drop of 0 V. In fact, in a series DC circuit that contains an open, the entire source voltage will appear across the open, and no voltage will appear across any of the other resistors. • So if you measure the voltage between any two points in a series circuit containing an open, you'll measure 0 V if the two points are on the same side of the open, but you'll measure the entire source voltage if the points are on opposite sides of the open. • For example, suppose R3 is open in the circuit shown below. Then there will be 0 V across R1, across R2, and across R4. Also, Vab = 0 V. But there will be 9 V across R3. Also, Vac = 9 V, and Vbc = 9 V. • A short circuit, or "short," is a path of zero resistance connecting two points in a circuit that are not supposed to be connected. • Since a short has zero resistance, the voltage across it must be zero. This follows from Ohm's law, × R. • A component is said to be short-circuited, or "shorted out," when there is a short circuit connected in parallel with it. No current flows through a short-circuited component. Instead, current is diverted through the short itself. • For example, suppose that in the circuit shown below there is a short between points a and b, perhaps caused by a loose wire clipping that connects these two points. Then R2 is short-circuited. No current will flow through R2; instead, current will follow the path of zero resistance through the short itself (the wire clipping). • A short in a series DC circuit reduces the circuit's total resistance, causing more current to flow out of the voltage source. • For example, in the circuit shown above, if R2 is short-circuited by a wire clipping that connects points a and b, then R2's resistance disappears from the circuit, and the circuit's total resistance is equal to R1 + R3 + R4. • That ends our quick review of series DC circuits. If you'd like a more thorough review, go to Unit 7 of EET 1150. Now let's get back to AC circuits. ##### Sinusoidal Response of AC Circuits • Here's an important point that we've mentioned a couple of times before and that is worth repeating: When a sinusoidal voltage is applied to any circuit containing resistors, capacitors, and inductors, all of the circuit's current waveforms and voltage waveforms are sinusoids and have the same frequency as the source voltage. • So, for example, if you're given the circuit shown below, and if you're told that the source voltage is a sinusoid having a frequency of 5 kHz, then you can say immediately that the current through every component is a 5-kHz sinusoidal current, and the voltage drop across every component is a 5-kHz sinusoidal voltage. • Where things get a bit tricky is figuring out the peak values and phase shifts of these current and voltage waveforms. But we'll be able to do it, thanks to complex numbers. ##### Boldface Notation for Phasors • Up to now we have used italicized, non-boldface letters to represent voltage and current. In particular: • V represents voltage. • I represents current. • From this point onward, we will usually treat voltage and current as phasors, which means we'll treat them as complex numbers with both a magnitude and an angle. We'll use the italic letters listed above to denote the magnitude of the phasor, and we'll use boldface letters to represent the total phasor quantity (which includes both the magnitude and the angle). In particular: • V represents a phasor voltage, which has both a magnitude (V) and an angle. • I represents a phasor current, which has both a magnitude (I) and an angle. ##### Impedance • In AC circuits, every component or combination of components has a quantity called an impedance, which we denote with the boldface letter Z. • If we know a component's voltage V and its current I, we can use the following equation to find the component's impedance: Z = ÷ I • According to this equation, impedance Z is equal to the ratio of two complex numbers, V and I. Therefore impedance itself is a complex number. (That's why we denote it with a boldface letter.) • Like any complex quantity, impedance can be expressed in either polar form or rectangular form. When we express it in polar form, we specify its magnitude and its angle. • Because impedance is the ratio of a voltage to a current, impedance is measured in ohms. • For example, suppose that dividing a component's voltage by its current gives the result ÷ I = 736 ∠26.5° Ω Then the component's impedance Z has a magnitude of 736 Ω and an angle of 26.5°. At times we might be interested in talking just about this impedance's magnitude, in which case we would write Z = 736 Ω Note that in this last equation, Z is not written in boldface, because it does not denote a complex number. Rather it just denotes a complex number's magnitude. ##### Ohm's Law for AC Circuits • Look again at the equation Z = ÷ I • This equation, which applies to components in AC circuits, is similar to Ohm's law, which you know from your study of DC circuits: R = V ÷ I • In fact, we'll refer to the equation Z = ÷ I as Ohm's law for AC circuits. In AC circuits, impedance plays a role very similar to the role played by resistance in DC circuits. But you must keep in mind that impedance is a complex quantity, which has both a magnitude and an angle. • Of course, you can also rearrange this equation to solve for voltage if you know current and impedance, or to solve for current if you know voltage and impedance: V = × Z I = ÷ Z • Remember, in each case, all quantities are complex numbers, not real numbers. ##### How is Impedance Related to Resistance and Reactance? • Recall from previous Units that: • Resistance R represents a resistor's opposition to current. • Capacitive reactance XC represents a capacitor's opposition to current. • Inductive reactance XL represents an inductor's opposition to current. • Recall also that each of these quantities is measured in ohms. • Impedance Z, which is also measured in ohms, can be thought of as a generalization of the concepts of resistance and reactance. It represents any component's (or combination of components) opposition to AC current. • Sometimes we attach a subscript to Z to indicate the type of component whose impedance we're talking about: • We might write ZR when discussing a resistor's impedance, which is closely related to its resistance R. • Similarly, we might write ZC when discussing a capacitor's impedance, which is closely related to its reactance XC. • Again, we might write ZL when discussing an inductor's impedance, which is closely related to its reactance XL. • Let's look at each of these cases more closely. ##### Impedance of a Resistor • A resistor's impedance ZR is a complex quantity whose magnitude R is the resistance in ohms and whose angle is 0°. • We use 0° because voltage and current are in phase in resistors. (In other words, there is a 0° phase angle between a resistor's current and its voltage.) • So in polar notation, a resistor's impedance is ZR = R∠0° • We can easily convert this to rectangular notation, to get ZR = R + j0 or simply ZR = R • Thus, ZR for any resistor has a real part but no imaginary part. In the complex plane, a resistor's impedance lies along the positive real axis, as shown in the following diagram representing a resistor whose resistance is 50 Ω: ##### Impedance of a Capacitor • A capacitor's impedance ZC is a complex quantity whose magnitude XC is 1 ÷ (2pfC), and whose angle is −90°. • We use −90° because voltage lags current by 90° in a capacitor. • So in polar notation, a capacitor's impedance is ZC = XC ∠−90° • We can easily convert this to rectangular notation, to get • ZC = 0 − jXC or simply ZC = −jXC • Remember, in each of these equations XC = 1 ÷ (2pfC), which is also equal to 1 ÷ (ωC). • Thus, ZC for any capacitor has a negative imaginary part but no real part. In the complex plane, a capacitor's impedance lies along the negative imaginary axis, as shown in the following diagram representing a capacitor whose reactance is 50 Ω: . ##### Impedance of an Inductor • An inductor's impedance ZL is a complex quantity whose magnitude XL is 2pfL and whose angle is +90°. • We use +90° because voltage leads current by 90° in an inductor. • So in polar notation, an inductor's impedance is ZL = XL ∠90° • We can easily convert this to rectangular notation, to get • ZL = 0 + jXL or simply ZL = jXL • Remember, in each of these equations XL = 2pfL, which is also equal to ωL. • Thus, ZL for any inductor has a positive imaginary part but no real part. In the complex plane, an inductor's impedance lies along the positive imaginary axis, as shown in the following diagram representing an inductor whose reactance is 50 Ω: . ##### Putting It All Together • We've seen that Ohm's law for AC circuits can be written in any of the following forms: I = ÷ Z V = × Z Z = ÷ I • We've also seen how to calculate a resistor's impedance, or a capacitor's impedance, or an inductor's impedance. • Let's look at some problems that require us to combine these pieces. ##### Series Impedance • Suppose we have a resistance in series with a reactance. We'd like to find the total impedance of these two components, but we can't simply add resistances and reactances as real numbers. For example, a 1 kΩ resistance in series with a 2 kΩ capacitive reactance does not add up to 3 kΩ. • Instead, we must add them as complex numbers. So, to find the total impedance of a 1 kΩ resistance in series with a 2 kΩ capacitive reactance, we must add 1∠0° kΩ plus 2∠−90° kΩ, which gives us a total of 2.24∠−63.4° kΩ. • It may seem strange that you can combine a 1 kΩ resistance with a 2 kΩ reactance and come up with a total of only 2.24 kΩ, but that's how it works. ##### Analyzing Series AC Circuits • Now that you know how to treat resistances and reactances as complex quantities, and how to use the phasor form of Ohm's law, and how to use complex numbers to find total impedance, you're ready to analyze any series AC circuit, such as the series RLC circuit shown below. • Here are the steps to follow: 1. Use complex addition to find the circuit's total impedance, ZT. 2. Apply Ohm's law to the entire circuit to find the circuit's total current. 3. Recognize that, since we're dealing with a series circuit, each component's current is equal to the total current. 4. Apply Ohm's law to each component to find the voltage drops. • These are very similar to the steps that you followed in EET 1150 to analyze a simple series DC circuit containing resistors. The big difference is that throughout this procedure, we must now use complex numbers instead of real numbers. • Let's look at each step in more detail. ##### Step 1: Find Total Impedance • For n impedances in series, total impedance is given by ZT = Z1 + Z2 + … + Zn • Each Z on the right-hand side of this equation may be the impedance of a resistor, an inductor, or a capacitor. So this step will require you first to find the reactances of any capacitors or inductors in the circuit. • Remember: we're adding complex numbers here, not real numbers. ##### Step 2: Find Total Current • Knowing the source voltage VS and the total impedance ZT, you can use Ohm's Law to find the current: IT = VS ÷ ZT • Again, remember that we're dividing complex numbers, not real numbers. ##### Step 3: Find Individual Currents • This step is the easiest. It simply requires you to remember that in any series circuit (DC or AC), every component's current is equal to the total current: IT = I1 = I2 = … = In ##### Step 4: Find Voltage Drops • Now that you know the current through each component, use Ohm's Law to find the voltage drop across each component: V1 = I1 × Z1 and V2 = I2 × Z2 and V3 = I3 × Z3 and ... • Remember, Z1 is the first component's impedance, which will have an angle of 0° if the first component is a resistor, or an angle of −90° if the first component is a capacitor, or an angle of 90° if the first component is an inductor. Similarly for Z2, Z3, and so on for however many components the circuit contains. • Again, remember that we're multiplying complex numbers, not real numbers. ##### Practice Problems • Want more practice analyzing series AC circuits? Here are a couple of lessons that will generate as many practice problems as you want, and then let you check your answers against the correct answers. • The first one covers series RC circuits: • And the second one covers series RL circuits: ##### Phasor Diagrams • After you've analyzed a circuit by finding currents and voltage drops, you can draw a phasor diagram that shows in graphical form how these quantities relate to each other. A phasor diagram simply shows each voltage and current as a vector in the complex plane, drawn with the appropriate angle and magnitude. • Here is a simple example showing the current and voltage phasors for a single component: • Here's another example showing the phasors for all voltages and currents in a particular series RC circuit. • From the diagram you can quickly see the relationship between the circuit's current and voltages. • In every phasor diagram for a series circuit, you should find that the current and the resistor's voltage drop have the same angle, since current and voltage in a resistor are always in phase with each other. • Also, you should find that inductor voltage and capacitor voltage are always at a 90° angle to the current, which should make sense. (Remember ELI the ICEman from Unit 4?) ##### Kirchhoff's Voltage Law • As in DC circuits, Kirchhoff's Voltage Law (KVL) says that the sum of the voltage drops around any closed loop equals the sum of the voltage rises around that loop. • As we'll see in later units of this course, KVL applies to all circuits, whether series, parallel, or series-parallel. • In this unit we're restricting our attention to series circuits containing a single voltage source. In these circuits, KVL simplifies to the following form: the source voltage in a series circuit is equal to the sum of the voltage drops across the circuit's resistors, capacitors, and inductors. • Whenever you apply KVL to an AC circuit, you must use complex numbers, not real numbers. If you just add the magnitudes of the voltages, instead of adding the magnitudes along with their angles, you won't get good results. ##### Voltage-Divider Rule • As in DC circuits, this is a shortcut rule for finding voltage drops in a series circuit. • The voltage-divider rule says that the voltage Vx across any impedance Zx in a series circuit with source voltage VS is given by: Vx = (Zx ÷ ZT) × VS • Again, use complex numbers, not real numbers. ##### Troubleshooting Series AC Circuits • Above we reviewed the basics of troubleshooting series DC circuits. Almost all of these same points apply to series AC circuits. (But there's one important difference noted below.) In particular: • No current flows through an open. • Therefore, no current flows anywhere in a series circuit containing an open. • Since no current flows through an open, you can think of the open as having infinite resistance (R = ∞). • Usually, an open will not have a voltage drop of 0 V. In fact, in a series circuit that contains an open, the entire source voltage will appear across the open, and no voltage will appear across any of the other resistors, capacitors, or inductors. • So if you measure the voltage between any two points in a series circuit containing an open, you'll measure 0 V if the two points are on the same side of the open, but you'll measure the entire source voltage if the points are on opposite sides of the open. • A short has zero resistance and zero voltage. • A component is said to be short-circuited, or "shorted out," when there is a short connected in parallel with it. No current flows through a short-circuited component. Instead, current is diverted through the short itself. • So far, everything we've said about opens and shorts in series AC circuits is the same as what we said earlier about opens and shorts in series DC circuits. But here's a difference: • A short in a series DC circuit will always reduce the circuit's total resistance, increasing the circuit's total current. • But in a series AC circuit, a short could either increase or decrease the circuit's total impedance, and therefore could either decrease or increase the total current. • Why this difference between shorts in DC circuits and shorts in AC circuits? It's because of the difference between real numbers and complex numbers. • To find a series DC circuit's total resistance, you add two or more real numbers. If one of the circuit's resistors is shorted out, then you replace one of these numbers with zero, and this will decrease the total. • For example, suppose a series circuit contains a 100 Ω resistor, a 150 Ω resistor, and a 200 Ω resistor. Then the total resistance is 450 Ω. But if the 200 Ω resistor is shorted out, then you replace its resistance with zero, and so the total resistance decreases to 250 Ω. This decrease in total resistance will increase the circuit's total current. • But to find a series AC circuit's total impedance, you add two or more complex numbers. If one of the circuit's components is shorted out, then you replace one of these complex numbers with zero. When you're adding complex numbers, replacing one of them by zero could either decrease or increase the total. • For example, suppose a series AC circuit contains a 100∠0° Ω resistive impedance, a 150∠90° Ω inductive impedance, and a 200∠−90° Ω capacitive impedance. Then the total impedance is 112∠−26.6° Ω. If the capacitor is shorted out, then you replace its impedance with zero, and so the total impedance increases to 180∠56.3° Ω. This increase in total impedance will decrease the circuit's total current. • On the other hand, suppose that in the same circuit the capacitor is okay but the resistor is shorted out. Then the circuit's total impedance is 50∠−90° Ω, which is a decrease from the original total impedance. This decrease in total impedance will increase the circuit's total current. • So a short in a series AC circuit may either increase or decrease the total impedance. ##### "Systems View" of a Circuit • In your electronics courses up to now, you've considered a circuit as a collection of individual components such as resistors, capacitors, and inductors. Most of the circuits you've studied have been very small, with just a few components. • But real-world circuits are usually much more complex, with hundreds or even thousands of components. To study and analyze such circuits, you have to shift your way of looking at circuits. Instead of concentrating on individual components, it's more useful to think of the circuit as a system made up of parts that perform certain functions. The "parts" that I'm referring to here are not individual components. Rather, they are sub-circuits that contain many components connected together to perform some function. • For example, in your later courses you'll study amplifier circuits. These circuits contain transistors as well as resistors, capacitors, and inductors, so we're not ready to understand the details now. But there are a number of standard designs for amplifier circuits, and rather than focusing on the details you might just want to consider the entire amplifier circuit as a "box." This box has two input terminals to which you can connect an input voltage, and two output terminals at which the amplifier's output voltage will appear. Here's a diagram: • The amplifier circuit contains many components (resistors, capacitors, transistors). But in this diagram we're not showing those details. • A more complete circuit might consist of an oscillator connected to an amplifier connected to a power amplifier, as shown here: • In this diagram we have three boxes representing three complicated sub-circuits. We're not showing the details of these sub-circuits, but we are showing how the sub-circuits are connected to each other. (For example, the two lines between the oscillator and the amplifier show that the oscillator's output voltage is also the amplifier's input voltage.) • Next we'll look at a few simple series RC and RL circuits that we can think of as "boxes" that perform a certain function. ##### Lag Circuits and Lead Circuits • In some applications, a designer needs to shift a voltage's phase angle by a certain amount. In such cases the designer uses a circuit that introduces a phase shift between the circuit's output voltage and its input voltage. • There are two basic possibilities here. Either: • The circuit is designed so that its output voltage lags its input voltage, in which case we're dealing with a lag circuit. Or: • The circuit is designed so that its output voltage leads its input voltage, in which case we're dealing with a lead circuit. ##### RC Lag Circuits and Lead Circuits • A simple series RC circuit can serve as either a lag circuit or a lead circuit, depending on whether you take the output voltage across the resistor or across the capacitor. In particular: • In any series RC circuit, the capacitor's voltage lags the source voltage, so you'll have a lag circuit if you take the output voltage across the capacitor, as shown here: • Also, in any series RC circuit, the resistor's voltage leads the source voltage, so you'll have a lead circuit if you take the output voltage across the resistor, as shown here: • If you remember ELI the ICEman, you'll be able to quickly identify circuits like the ones above as either lead circuits or lag circuits. • ICE reminds you that a capacitor's voltage tends to lag everything else in the circuit, so you've got a lag circuit if you're taking the output voltage across the capacitor. • Whenever you encounter circuits like these, you can always use the general techniques you learned above to analyze the circuit and figure out how far the output voltage is shifted from the input voltage. Or you can remember the following formulas. • For an RC lag circuit, the phase angle φ between the input and output is φ = −tan−1(R ÷ XC) • For an RC lead circuit, the phase angle φ between the input and output is φ = tan−1(XC ÷ R) • Using these formulas, and by choosing appropriate values of R and C, you can design a lead circuit or lag circuit to shift the voltage by any desired angle. • Here's a nice memory trick to help you remember those two formulas: Notice that the order of the R and the XC in each formula is the same as the order of the resistor and the capacitor in the corresponding schematic diagram. ##### RL Lag Circuits and Lead Circuits • A simple series RL circuit can also serve as either a lag circuit or a lead circuit, depending on whether you take the output voltage across the resistor or across the inductor. In particular: • In any series RL circuit, the resistor's voltage lags the source voltage, so you'll have a lag circuit if you take the output voltage across the resistor, as shown here: • Also, in any series RL circuit, the inductor's voltage leads the source voltage, so you'll have a lead circuit if you take the output voltage across the inductor, as shown here: • If you remember ELI the ICEman, you'll be able to quickly identify circuits like the ones above as either lead circuits or lag circuits. • ELI reminds you that an inductor's voltage tends to lead everything else in the circuit, so you've got a lead circuit if you're taking the output voltage across the inductor. • Whenever you encounter circuits like these, you can always use the general techniques you learned above to analyze the circuit and figure out how far the output voltage is shifted from the input voltage. Or you can remember the following formulas. • For an RL lag circuit, the phase angle φ between the input and output is φ = −tan−1(XL ÷ R) • For an RL lead circuit, the phase angle φ between the input and output is φ = tan−1(R ÷ XL) • Here's a nice memory trick to help you remember those two formulas: Notice that the order of the R and the XL in each formula is the same as the order of the resistor and the inductor in the corresponding schematic diagram. ##### Unit 6 Review • This e-Lesson has covered several important topics, including: • series AC circuits • phasor diagrams • Kirchhoff's Voltage Law • voltage-divider rule • troubleshooting series AC circuits • lag circuits and lead circuits. • To finish the e-Lesson, take this self-test to check your understanding of these topics. Congratulations! You've completed the e-Lesson for this unit.
# What is the equivalent of 1/2 2/4 ## What is the equivalent of ½? 2/4Answer: The fractions which are equivalent to 1/2 are 2/4, 3/6, 4/8, 6/12 etc. Equivalent fractions have same value in reduced form. Explanation: Equivalent fractions can be written by multiplying or dividing both the numerator and the denominator by the same number. ## What does 1/2 mean as a fraction? The fraction one-half, written in symbols as 1/2, means “one piece, where it takes two pieces to make a whole.” The fraction a half, written in symbols as 1/2, means “one piece, where it takes two pieces to make a whole.” ## What is the equivalent of 1/2 not? To compare 1/2 and 3/7 we would multiply 1/2 by 3/3 to produce 3/6. Since 3/6 is not the same as 3/7, the fractions are not equivalent. Fractions equivalent to 1/2 are 2/4, 3/6, 4/8, 5/10, 6/12 … Fractions equivalent to 1/3 are 2/6, 3/9, 4/12, 5/15, … ## What fraction is equivalent to? Equivalent Fractions ChartUnit FractionEquivalent Fractions1/32/6, 3/9, 4/12..1/42/8, 3/12, 4/16..1/52/10, 3/15, 4/20,..1/62/12, 3/18, 4/24,..4 more rows ## What is 1/2 in a whole number? Also, if rounded up to the nearest whole number, we get 1. Hence, 1/2 as a whole number will be 0 or 1. ## How do you find 1/2 of a number? Halves are calculated by dividing by 2. For example: One half of 10 = ½ of 10 = 10/2 = 5. One half of 34 = ½ of 34 = 34/2 = 17. ## What fraction is more than 1 2? Answer: Yes, 3/4 is bigger than 1/2.3/4=0.751/2=0.5 ## What fraction is smaller than 1 2? The fraction 1/4 is less than 1/2 . ## What is the decimal of 1 2? Answer: 1/2 as a decimal is 0.5. ## What is 1/4 equal to as a fraction? 2/8Answer: The fractions equivalent to 1/4 are 2/8, 3/12, 4/16, etc. Equivalent fractions have the same value in their reduced form. ## What is 1/3 equivalent to as a fraction? 2/6Fractions equivalent to 1/3: 2/6, 3/9, 4/12, 5/15 and so on … ## What is equivalent calculator? Equivalent Expression Calculator is a free online tool that displays the equivalent expressions for the given algebraic expression. BYJU’S online equivalent expression calculator tool makes the calculations and simplification faster and it displays the equivalent expression in a fraction of seconds. ## What is the fraction 3/4 equivalent to? Equivalent fractions of 3/4 : 6/8 , 9/12 , 12/16 , 15/ ## What is 3/5 equal to as a fraction? 6/10So, 3/5 = 6/10 = 9/15 = 12/20. ## What fraction is 4/5 equivalent to? 8/10Decimal and Fraction Conversion ChartFractionEquivalent Fractions4/58/1048/601/62/1212/725/610/1260/721/72/1412/8423 more rows ## What is the fraction 6/8 equivalent to? 3/4Thus, 6/8 is an equivalent fraction of 3/4. We can find some other equivalent fractions by multiplying the numerator and the denominator of the given fraction by the same number. Thus, the equivalent fractions of 3/4 are 6/8, 9/12, 12/16, and 15/20. ## How to make 3/6 equal to 1/2? Multiply the numerator and the denominator by the same number. Divide the numerator and the denominator by the same number. Thus, 3/6, 6/12, and 4/8 are equal to 1/2, when simplified. Hence they are all equivalent to 1/2. ## How to make equivalent fractions? Let us understand the two ways in which we can make equivalent fractions: 1 Multiply the numerator and the denominator by the same number. 2 Divide the numerator and the denominator by the same number. ## Do equivalent fractions have the same value in reduced form? Equivalent fractions have same value in reduced form. ## What is the equivalent of 2 6? The fraction 2 6 is equal to 1 3 when reduced to lowest terms. To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 1 3) by the same integer number, ie, multiply by 2, 3, 4, 5, 6 … and so on … ## How to find equivalent fractions? To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 13) by the same natural number, ie, multiply by 2, 3, 4, 5, 6 ## Is 1 3 a fraction? Important: 1 3 looks like a fraction, but it is actually an improper fraction. ## Can you convert fractions to decimals? This Equivalent Fractions Table/Chart contains common practical fractions. You can easily convert from fraction to decimal, as well as, from fractions of inches to millimeters. ## How to find equivalent ratios? As we previously mentioned, Equivalent Ratios are two ratios that express the same relationship between numbers. The Equivalent Ratio Calculator provides a table of equivalent ratios that have the same relationship between each other and directly with the ratio you enter into the calculator. We will look at how to calculate equivalent ratios shortly, first lets look at how to use the free online equivalent ratio calculator: 1 Enter a Ratio into the equivalent ratio calculator, for example, you could enter 7:25 2 Select the number of equivalent ratios that you would like to see in the table of results 3 The equivalent ratio calculator will calculate as you type and produce a lis of equivalent ratios in a table below the calculator 4 [Optional] Print or email the Table of Equivalent Ratios for later use ## What is a ratio? A ratio is a direct comparison of one number against another. A ratio calculator looks to define the relationship that compares between those two numbers ## Is there a formula for equivalent ratios? As equivalent ratios have the same value there is technically no equivalent ratio formula but the following equivalent ratio formula will help you with the manual math calculations. ## How to Calculate Ratios When calculating equivalent ratios you must multiply or divide both numbers in the ratio. This keeps both numbers in direct relation to each other. So, a ratio of 2/3 has an equivalent ratio of 4/6: in this ratio calculation we simply multiplied both 2 and 3 by 2. ## Mathematical facts about the ratio 1:2 If you wish to express the ratio 1:2 as n to 1 then the ratio would be: ## Equivalent ratio tables for decimal ratios ranging 1 : 2 to 2 : 3 The table below contains links to equivalent ratio examples with ratios in increments of 0.1 in the range 1:2 to 2:3 ## What is the equivalent of (3+7)+2? The expression equivalent to (3+7)+2 is 12. ## What is the equivalent fraction of 2/3? For example, if we multiply the numerator and denominator of 2/3 by 4 we get. 2/3 = 2×4 / 3×4 = 8/12 which is an equivalent fraction of 2/3. ## What is equivalent expression calculator? Equivalent Expression Calculator is a free online tool that displays the equivalent expressions for the given algebraic expression. BYJU’S online equivalent expression calculator tool makes the calculations and simplification faster and it displays the equivalent expression in a fraction of seconds. ## Is 3y+3 true? True, because when the numbers are in the parethasis that means you multiply what is outside of the parenthasis like the three by the numbers inside of the parenthasis so 3 x y and 3 x 1, therefore 3y+3 and 3 (y+1) is true. ## Is 3y+3 a simplified expression? The expressions 3y+3 and 3 (y+1) are equivalent expressions. Because 3 (y+1) can be simplified as 3y+3. ## What is the equivalent of 2:2 in Cameroon? Cameroon 2:2 (Hons) equivalent 2:1 (Hons) equivalent – Diplome dIngenieur: 13 out of 20 or GPA 3.2 – Licence/Bachelors degree: 15 out of 20 or GPA 3.5 – Diplome dIngenieur: 15 out of 20 or GPA 3.5 – Licence/Bachelors degree: 16 out of 20 or GPA 3.7 ## What degree is equivalent to 2:2 in Angola? Angola 2:2 (Hons) equivalent 2:1 (Hons) equivalent – Licenciado degree 13 out of 20 – Licenciado degree At least 15 out of 20 ## What is Belize 2:2? Belize 2:2 (Hons) equivalent 2:1 (Hons) equivalent – Bachelors degree from the Uni of West Indies • 2:2 – Bachelors degree from the Uni of West Indies • 2:1 English Required: No – UKVI rule
## Arrays and Area Models to The Standard Algorithm Did you know that the words “array” and “area model” appear in the Grade 1-8 Math Curriculum a combined 22 times? Not only do arrays and area models help to support the development of proportional reasoning when we formally introduce multiplication in primary, but they also help us understand how to develop strategies that lead to building number flexibility and the automaticity of math facts. Arrays and area models should be used as a tool and representation for many big ideas in mathematics including, but not limited to: • Multiplication • Distributive Property with Whole Numbers • Finding Area with Whole Number Dimensions • Perfect Squares & Square Roots • Multiplying a Binomial by Monomial • Multiplying a Binomial by Binomial (aka FOIL) • Factoring (Common, Simple/Complex Trinomials) • Completing the Square For many, the term “array” is not a familiar one. Luckily, the definition is fairly straightforward: In mathematics, an array is a group of objects ordered in rows and columns. Seems pretty simple, but they are extremely powerful in building a deep conceptual understanding as students learn multiplication and begin applying that knowledge to more abstract ideas requiring fluency with procedures. ## Where Multiplication Begins In The Ontario Curriculum • relate multiplication of one-digit numbers and division by one-digit divisors to real life situations, using a variety of tools and strategies (e.g., place objects in equal groups, use arrays, write repeated addition or subtraction sentences); • multiply to 7 x 7 and divide to 49 ÷ 7, using a variety of mental strategies (e.g., doubles, doubles plus another set, skip counting); • identify, through investigation, the properties of zero and one in multiplication (i.e., any number multiplied by zero equals zero; any number multiplied by 1 equals the original number) (Sample problem: Use tiles to create arrays that represent 3 x 3, 3 x 2, 3 x 1, and 3 x 0. Explain what you think will happen when you multiply any number by 1, and when you multiply any number by 0.); Consider 3 x 2, or “3 groups of 2”: This may seem like a simple and maybe even unnecessary representation, but having a visual that multiplying two numbers will always yield a rectangular array is an important concept that not only shows the interconnectedness of the Number Sense and Numeration strand to Measurement, but also implicitly provides students with insight into why more abstract mathematics works later on in grade 9 and 10. In grade 3, it is reasonable to believe that students are still working on counting and quantity including unitizing in order to skip count more fluently. By working with arrays, we can allow students to continue developing their ability to unitize and work with composing and decomposing numbers. ## Arrays and Perfect Squares Not only is it beneficial for students to understand that multiplying two quantities will yield an array covering a rectangular area, but it is also useful for students to discover without explicitly stating that when we make an array where the number of groups and number of items in each group are equal, the array is now a special rectangle; a square! While we don’t specifically discuss perfect squares until intermediate when students represent perfect squares and square roots using a variety of tools in grade 7 and estimate/verify the positive square roots of whole numbers in grade 8, I think it would be much easier for students to identify a perfect square and estimate the square root of a number if they have four years of concrete and visual work with perfect squares. ## How Multiplication With Arrays Lead to Area As we have witnessed in the previous examples, with every array a student builds, we are implicitly providing them with a window into the measurement strand with opportunities to think about perimeter and area. Not only can we easily make connections between an array and the area of a rectangle, but we can also better serve specific expectations involving estimation such as this grade 3 Measurement expectation: • estimate, measure (i.e., using centimetre grid paper, arrays), and record area. With the use of arrays in the Number Sense and Numeration strand, I can better serve my Measurement strand with problems that force students to estimate using visuals and then improve their predictions using concrete manipulatives. Something worth noting is the wording of the specific expectation in the grade 3 curriculum related to multiplication and division which states students are to multiply to 7 x 7 and divide to 49 ÷ 7, using a variety of mental strategies without any reference to memorization or automaticity. While I agree that knowing multiplication tables for intermediate and senior level math courses is a huge asset, I think working with multiplication early and often with concrete manipulatives is a great way to get there over other more traditional and/or rote strategies. ## Multiplying Larger Numbers – Creating a Need To Chunk When students enter grade 4, we extend our multiplication and division through the use of a variety of mental strategies to multiplying to 9 x 9 and dividing to 81 ÷ 9. As the factors and divisors get larger, we begin creating a need for some new strategies that will help us as we approach two-digit multiplication. In the grade 4 Patterning and Algebra strand, students are expected to: • identify, through investigation (e.g., by using sets of objects in arrays, by drawing area models), and use the distributive property of multiplication over addition to facilitate computation with whole numbers (e.g.,“I know that 9 x 52 equals 9 x 50 + 9 x 2. This is easier to calculate in my head because I get 450 + 18 = 468.”). Sounds kind of complicated. Ultimately, what the distributive property is trying to offer students is a way to “chunk” their factors into friendlier numbers to make multiplication easier. Let’s have a look at a basic example: If a student does not know how many objects there are in six groups of seven, they can use distributive property to “chunk” this multiplication problem into two or more smaller multiplication problems and add the products. In the example above, the student may have a comfort with multiples of 5. In the example below, the student might like working with 5 groups of 5 because the product, 25, is a friendly number to add up afterwards. Although not explicitly stated in the grade 4 curriculum, I think it is worth noting that the introduction of the distributive property is the first situation where a bracket could be used symbolically in a mathematical expression: 6 x 7 = 6(5 + 2) or 6 x 7 = 5(5 + 2) + 1(5 + 2) While a bracket could be used symbolically, I don’t think it would be developmentally appropriate or useful. However, when brackets do show up on the scene later on, how awesome would it be to connect them to our prior knowledge of arrays and distributive property? ## Base Ten Blocks – Making Two-Digit Multiplication Easier In grade 5, we continue to promote the use of mental strategies for addition, subtraction, and multiplication: • solve problems involving the addition, subtraction, and multiplication of whole numbers, using a variety of mental strategies (e.g., use the commutative property: 5 x 18 x 2 = 5 x 2 x 18, which gives 10 x 18 = 180); However, we also move towards finding more efficient ways to multiply factors greater than 9: • multiply two-digit whole numbers by two-digit whole numbers, using estimation, student-generated algorithms, and standard algorithms; Unfortunately, I think it is more common for educators to miss the first expectation about mental strategies as well as a huge portion of the second expectation, and rush straight for the standard algorithm. By continuing to use our knowledge of the distributive property with two-digit by two-digit multiplication, we can begin making things super friendly by splitting our arrays into chunks of 10. However, it can take a whole lot of square tiles and a whole lot of time: That’s where base ten blocks come in! What a genius idea to respect the theory of concreteness fading and allow students to begin this new layer of abstraction with the aide of concrete manipulatives to build a deep conceptual understanding. Now, we can do the same problem with much less concrete “pieces”, but still gain all the benefits of manipulatives. Also, how cool is it that base ten blocks are what I call “forced distribution” into chunks of friendly 10’s. As we introduce base ten blocks, we also begin to transition more explicitly from an array to an area model. Since we are no longer using single tiles to represent each single object or unit of an array, the “ten rods” and “hundred flats” are covering areas of 10 units-squared and 100 units-squared, respectively. The fun doesn’t stop here. Multiplying two-digit by two-digit numbers is extremely helpful using an area model with base ten blocks: ## Connecting Arrays & Area Models to the Standard Algorithm Speaking from experience as a math teacher in my own classroom in Ontario and having an opportunity to travel to observe in classrooms both near and far, I see a common trait where we rush to the algorithm. This race to the procedure might be motivated by teacher beliefs, the anxiety we feel due to limited time and a thick curriculum, or simply a lack of awareness to other approaches and strategies. In any case, I think using arrays and area models can serve as a great precursor to promoting students to begin creating their own algorithms and eventually, connect to the standard algorithm. Let’s make some connections between the standard algorithm and using arrays and area models. Consider 22 x 26 using the standard algorithm for multiplication: While there are many educators who do a great job of breaking the standard algorithm down into its working parts including doing their best to explain the impact place value has on each step to build a conceptual understanding, I think students who can use arrays and area models to multiply have an advantage to building a deeper conceptual understanding. If using arrays, area models and base ten blocks for multiplication is a new idea for you, then it might be worth making some connections between these representations and the standard algorithm. Have a look at the area model representation and the standard algorithm method for 22 x 26. Can you make any connections between the two representations/methods? What did you come up with? It might not be super obvious, but if we consider the two products created using the algorithm (132 and 440), we can see these products by looking vertically down the area model: Some might assume that because we end up with two products that the standard algorithm uses distribution to chunk the multiplication of 22 x 26 into two sets of factors. However, if we dig deeper, we will notice that both products are the result of two smaller chunks. When multiplying two-digit by two-digit factors with the standard algorithm, our base ten place value system allows for chunking into smaller “chunks”: • x units of 1 times x units of 1 • x units of 1 times x units of 10 • x units of 10 times x units of 1 • x units of 10 times x units of 10 In this case, the standard algorithm breaks up 22 x 26 using the distributive property into the following smaller factors: • 6 units of 1 times 2 units of 1; (6 x 2) • 6 units of 1 times 2 units of 10; (6 x 20) • 2 units of 10 times 2 units of 1; (20 x 2) • 2 units of 10 times 2 units of 10; (20 x 20) ## A “Conceptual” Algorithm Before the “Standard” Algorithm If we consider all of the deep thinking required to understand how the standard algorithm works conceptually, it might make sense to make the connections more explicit. Let’s come back to the grade 5 expectation multiply two-digit whole numbers by two-digit whole numbers, using estimation, student-generated algorithms, and standard algorithms. Note that student-generated algorithms sits in there and often times, we don’t provide enough opportunities for students to truly create their own strategies and procedures. Rather than using arrays and area models and then suddenly flying into the standard algorithm, what if we tried to guide students to develop their own “conceptual” algorithm by having them organize the smaller products they are creating when using base ten blocks? Consider this “conceptual” algorithm that might assist students in making the leap to the standard algorithm when developmentally appropriate: The connections don’t stop here, either. By explicitly addressing the conceptual understandings behind why the standard algorithm works, students can then apply that same thinking to create their own friendly chunks outside of those limited to 10s created when using base ten blocks. ## The Standard Algorithm is the Same As Multiplying Two Binomials You may recall the acronym “FOIL” which is commonly used for students to remember how to multiply two binomials. While I am guilty for teaching this memorization tool in my math class until only a few years ago, I now understand that using tricks like “FOIL” to teach important math concepts is not helpful (and maybe even harmful). What if instead of simply teaching students “FOIL” or “double-distribution”, which is a skill limited to the very specific case of multiplying two polynomials with two terms, we actually helped students to visualize what multiplying binomials really looks like? What we see in the previous example is: 9 x 12 = (5 + 4)(10 + 2) = 5 x 10 + 5 x 2 + 4 x 10 + 4 x 2 = 50 + 10 + 40 + 8 = 108 It might not be obvious to those who have never worked to make a connection, but what the standard algorithm we teach in grade 5 is actually the same procedure we teach students when multiplying binomials in grade 10. As we head into grade 9 and 10, the thinking becomes more abstract due to the use of variables. • multiply a polynomial by a monomial involving the same variable [e.g., 2x(x + 4), 2x^2(3x^2 – 2x + 1)], using a variety of tools (e.g., algebra tiles, diagrams, computer algebra systems, paper and pencil); • expand and simplify polynomial expressions involving one variable [e.g., 2x(4x + 1) – 3x(x + 2)], using a variety of tools (e.g., algebra tiles, computer algebra systems, paper and pencil); And here’s a couple examples of what these might look like if we use arrays and area models from grade 3 onwards:
## Virtual Help • Chat with library staff now The Learning Portal - College Libraries Ontario # Math: Basic Tutorials : Multiple-Step Linear Equations ## Multiple-Step Linear Equations Multiple-step linear equations are equations that involve multiple operations (e.g. subtraction and multiplication). This section explains how to solve these equations using the order of operations. ## Solving Equations Involving Multiple Operations Solving Equations Involving Multiple Operations Video transcript - RTF ## Examples Click on the titles below to view each example. Solve 6n minus 2 equals negative 3n plus 7, and verify the solution. Line 1: We want to remove the variables from the right side of the equation, so we add 3n to both sides to leave only constants on the right. The equation becomes 6n plus 3n minus 2 equals negative 3n plus 3n plus 7. Line 2: Combine the like terms to simplify so the equation becomes 9n minus 2 equals 7. Line 3: We want to remove the constants from the left side of the equation, so we add 2 to both side and the equation becomes 9n minus 2 plus 2 equals 7 plus 2. Line 4: Simplify to get 9n equals 7. Line 5: To isolate for n we must undo the multiplication, so we divide both sides by 9 and the equation becomes 9n over 9 equals 9 over 9. Line 6: Simplify to get the solution n equals 1. Line 7: Check you answer in the given equation 6n minus 2 equals negative 3n plus 7. Line 8: Substitute x equals 1 into the equation, so it is 6 times 1 minus 2 equals negative 3 times 1 plus 7. Line 9: Complete the multiplication to simplify the equation to 6 minus 2 equals negative 3plus 7. Line 10: Simplify to get 4 equals 4. Since n equals 1 makes 6n minus 2 equals negative 3n plus 7 a true statement, 1 is the solution to the equation. Solve 3 times open bracket x minus 2 close bracket minus 5 equals 4 times open bracket 2x plus 1 close bracket plus 5, and verify the solution. Line 1: Distribute to remove the brackets so the equation becomes 3x minus 6 minus 5 equals 8x plus 4 plus 5. Line 2: Combine like terms to simplify so the equation becomes 3x minus 11 equals 8x plus 9. Line 3: Since the numerical coefficient 8 is greater than 3 we will subtract 3x from both sides of the equation, so it becomes 3x minus 3x minus 11 equals 8x minus 3x plus 9. Line 4: Simplify to get the equation negative 11 equals 5x plus 9. Line 5: Subtract 9 from both sides of the equation to get the constants on the left, so the equation is negative 11 minus 9 equals 5x plus 9 minus 9. Line 6: Simplify to get negative 20 equals 5x. Line 7: Divide both sides by 5 to isolate x so the solution to the equation is negative 4 equals x. Line 8: Check you answer in the given equation 3 times open bracket x minus 2 close bracket minus 5 equals 4 times open bracket 2x plus 1 close bracket plus 5. Line 9: Substitute x equals negative 4 into the equation, so it is 3 times open bracket negative 4 minus 2 close bracket minus 5 equals 4 times open bracket 2 time negative 4 plus 1 close bracket plus 5. Line 10: Simplify inside the brackets so the equation is 3 times negative 6 minus 5 equals 4 time negative 7 plus 5. Line 11: Multiply to simplify the equation to negative 18 minus 5 equals negative 28 plus 5. Line 12: Simplify to get negative 23 equals negative 23. Since x equals negative 4 makes 3 times open bracket x minus 2 close bracket minus 5 equals 4 times open bracket 2x plus 1 close bracket plus 5 a true statement, negative 4 is the solution to the equation. ### Activity Try this activity to test your skills. If you have trouble, check out the information in the module for help.
# Poisson Distribution 09.08.2018 Episode #5 of the course Theory of probability by Polina Durneva Good morning! Today, we will talk about the Poisson distribution that is mainly used to predict the probability of a given number of events happening over a period of time. This distribution is closely associated with other types of probability distributions, such as the negative exponential distribution, that we will discuss later in the course. So, let’s get started! What Is Poisson Distribution? In the previous distributions that we discussed, an event could be classified as either success or failure. This can also be applicable to the Poisson distribution. It means that we want to know the probability that a number of success occurs over a fixed period of time. For example, you own a restaurant where 100 people come to on average every day. You might want to know the probability that more than 100 people will come to your restaurant today. To calculate such probability, you will use the Poisson distribution! A hundred people will be 100 of success, and a day will be a fixed period of time. Before we proceed to actual calculations of the probability, let’s discuss the main assumptions held for the Poisson distribution: 1. The intervals over which the events occur do not overlap. 2. The events are independent. 3. The probability that more than one event happens in a very short time period is approximately 0. Formula for Poisson Distribution It takes a while to derive the formula for the Poisson distribution that takes the following form: Probability (x) = (λxe)/x!, where x is the Poisson random variable and λ is the expected number of events (or the average). Let’s use our previous example with the restaurant to calculate the probability that 125 customers will come to our restaurant today. As we know, on average, there are 100 customers in our restaurant every day (this is λ). Then, the probability will be: P(x = 125) = (λxe)/x! = (100125e-100)/125! ≈ 0.00198 = 0.198%. This shows that the probability that we will have 125 customers today is less than 1%. Expected Value of Poisson Distribution As stated previously, λ is the expected value for the Poisson distribution. It means that the average value would be the rate of occurrence of an event. In our previous example, the expected value was 100 customers per day. Variance of Poisson Distribution The variance of this type of distribution is also easy to remember because it equals to the expected value. In the example we used, the variance would also be 100. That’s it for today! In the next three days, we will discuss discrete probability topics such as the uniform distribution (by the way, the uniform distribution can also be continuous, but we will not talk about this case in this course), the geometric distribution, and the negative binomial distribution. The rest of the course will be devoted to the continuous distributions. Take care, Polina Recommended book Lady Luck: The Theory of Probability by Warren Weaver Share with friends
# PROBABILITY of Compound Events. ## Presentation on theme: "PROBABILITY of Compound Events."— Presentation transcript: PROBABILITY of Compound Events Compound Events are when two or more events occur at the same time. Slide 2 Probability of Compound Events will be classified as either: Independent Events or Dependent Events Independent Events Whatever happens in one event has absolutely nothing to do with what will happen next because: The two events are unrelated OR You repeat an event with an item whose numbers will not change (eg.: spinners or dice) You repeat the same activity, but you REPLACE the item that was removed. The probability of two independent events, A and B, is equal to the probability of event A times the probability of event B. Slide 4 Independent Events Example: Suppose you spin each of these two spinners. What is the probability of spinning an even number and a vowel? P(even) = (3 evens out of 6 outcomes) P(vowel) = (1 vowel out of 5 outcomes) P(even, vowel) = S T R O P 1 2 3 6 5 4 Slide 5 Dependent Event What happens during the second event depends upon what happened before. In other words, the result of the second event will change because of what happened first. The probability of two dependent events, A and B, is equal to the probability of event A times the probability of event B. However, the probability of event B now depends on event A. Slide 6 Dependent Event Example: There are 6 black pens and 8 blue pens in a jar. If you take a pen without looking and then take another pen without replacing the first, what is the probability that you will get 2 black pens? P(black first) = P(black second) = (There are 13 pens left and 5 are black) THEREFORE……………………………………………… P(black, black) = Slide 7 Are these dependent or independent events? TEST YOURSELF Are these dependent or independent events? Tossing two dice and getting a 6 on both of them. 2. You have a bag of marbles: 3 blue, 5 white, and 12 red. You choose one marble out of the bag, look at it then put it back. Then you choose another marble. 3. You have a basket of socks. You need to find the probability of pulling out a black sock and its matching black sock without putting the first sock back. 4. You pick the letter Q from a bag containing all the letters of the alphabet. You do not put the Q back in the bag before you pick another tile. Slide 8 Independent Events Find the probability 1 5 5 8 P(jack, factor of 12) 40 x = 1 8 Slide 9 Independent Events Find the probability P(6, not 5) 1 6 5 6 5 36 x = Slide 10 Dependent Events Find the probability P(Q, Q) All the letters of the alphabet are in the bag the 1st time Do not replace the letter 1 26 25 650 x = Slide 11
Proof 2: Triangle area Two brothers share a slice of beer cake. Its face has a rectangle-shape. They cut it from corner to corner to get two equal halves. Those two halves namely have the same height, width and sloped side. They are completely equal with the same area. Also, they are triangles. And half the size of the original rectangle with the same height $h$ and length $l$: $$A_{triangle}=\frac12 \underbrace{A_{rectangle}}_{hl}=\frac12hl\quad_\blacksquare$$ The area of a rectangle is inputted here. Proof 10: Parallelogram and trapezium area A trapezium consists of a rectangle and two weird triangles at the ends. \begin{align}A_\text{trapezium}&= A_\text{rectangle}+ A_{\text{triangle } a}+ A_{\text{triangle } b}\\~&=\underbrace{lw_1}_\text{rectangle}+\underbrace{\frac12 w_al}_{\text{rectangle } a}+\underbrace{\frac12 w_b… Proof 1: Rectangle, square, box and cube area and volume ‘A room is $4\,\mathrm m$ long and $2\,\mathrm m$ wide’; a rectangle-shape. For each of the 4 metres there are 2 square metres across. That… Proof 9: Rhombus area A rhombus actually consists of four right-angled triangles, because the diagonals $d_1$ and $d_2$ are perpendicular. Those triangles are pair-wise equal. The rhombus area is… Proof 4: Circle and circle sector area A pizza slice’s edge is the pizza’s radius $r$. The slice is almost a triangle if it wasn’t for the rounding. That rounding has a… Proof 21: Sine and cosine values We here derive the sine and cosine values of a few chosen angles. Half ($\pi$) and quarter turns ($\frac \pi 2$, $-\frac \pi 2$): Cosine…
Basic Geometric Shapes: Square, Circle, Rectangle, and Triangle # Basic Geometric Shapes: Square, Circle, Rectangle, and Triangle Before moving on to more advance mathematical concepts in algebra and geometry, it is first important to have a thorough understanding of geometric shapes. Everyone is familiar with the most common shapes: square, circle, rectangle, and triangle. Each of these shapes is created by combining specific amounts of lines and/or curves. A square, for example, is a four sided figure created by connecting four line segments. All of the line segments are of the same length and they come together to form four right angles. A circle, on the other hand, has no straight lines. It is a combination of curves all connected. There are no angles to be found in a circle. A rectangle is created by connecting four line segments, in a similar manner to a square. However, a rectangle will typically have two line segments which are longer than the other two line segments. It could be described as an elongated square. Again, in a rectangle the four corners are created to form four right angles. A triangle consists of three line segments connected. Unlike a square and a rectangle the angles in a triangle can be of various measurements and are not always right angles. Triangles often take their name from the type of angles which can be found within the triangle itself. A triangle which has one right angle can be referred to as a right triangle. An acute triangle is created when all of the angles within the triangle measure less than ninety degrees. An obtuse triangle is created when one angle within the triangle measures more than ninety degrees. Finally, an equiangular triangle is created when all of the angles within the triangle measure sixty degrees. Additionally, triangles can be labeled or identified by the type of sides they may have. A scalene triangle has no congruent sides. An isosceles triangle has two congruent sides. An equilateral triangle has three congruent sides. Equiangular and equilateral triangles are two different terms for the same triangle. A polygon is a shape that is closed and is made up of only lines (no curves). It can have no open parts. Polygon in this case is a broader term to encompass a great many shapes including the square, rectangle and triangle. The circle described above cannot be considered a polygon because it is created by using curves, something which is not allowed by the simple definition of a polygon. A parallelogram is named because the opposite sides of the shape are parallel. In order to determine if the sides can be considered parallel, one must examine them closely. Parallel lines will never intersect, or cross, no matter how long they are extended. Therefore in a shape, if one imagines the lines extended and extended through eternity and they never will meet or touch, then they can be considered parallel. However, if the lines meet or touch, they cannot be considered parallel and the shape could not then be considered a parallelogram. In this way, the shape of a triangle cannot be considered a parallelogram, as the two lines opposite of each other meet at the point of the triangle. Since the lines meet, they are not parallel. An affix is prefix or a suffix which when added to a word changes the meaning of the word. Prefixes are added at the beginning of a word to change the meaning, while suffixes are added to the end of a word to change the meaning. Words can have a prefix, a suffix or both a prefix and a suffix. The broad term affix refers to both prefixes and suffixes. Typically, in geometric terms prefixes will be used frequently to help better define shapes. Prefixes themselves have specific meanings. It is these definitions which change the meanings of words. For example, the prefix un- means not. Therefore, if you add that prefix to the word able, which means capable of doing, the meaning is changed to mean not capable or not able. Prefixes are an important part in helping to determine the number of sides a specific polygon may have. For example, a pentagon uses the prefix pent- which means five because it has five sides. In the table below you will find some common prefixes which will help to distinguish how many sides a polygon will have. Some other prefixes which you might find in geometry include: • Circum- meaning around • A or an – meaning not or without • Ante- meaning before • Anti – meaning against • Con- meaning with or together • Cycl- meaning circle or wheel • Dia- meaning through, across or between • Equi – meaning equal • Exter- or Extra- meaning outside of • Homo- meaning same • Hyper- meaning above or over • Hypo- meaning below or less than • Morph- meaning shape • Peri- meaning around • Sect – meaning to cut • Semi- meaning half • Sub- meaning under or below • Super- meaning above • Trans- meaning across • Vert- meaning to turn This list is in no way totally complete, but understanding the meaning of basic prefixes can provide a more complete understanding of the terms, descriptions and definitions of items used throughout any study of geometry. The following are examples of two-dimensional shapes you should be able to recognize: In addition to being able to recognize specific shapes, it is important to be able to classify shapes. When classifying shapes, one is looking for what common characteristics they share. Shapes can be classified and sorted in a variety of ways. Think about how you could sort the shapes listed above into two groups. The above shapes could be classified by the number of sides the have, those with curves and those with no curves, the number of angles and a variety of other ways. Some additional ways shapes can be classified include: • Parallel lines – lines which if extended into infinity would never intersect. An example: railroad tracks • Right angles – two lines which intersect and form a 90 degree angle. An example: extend your arm at the shoulder straight out and then bend your elbow so that your hand is raised to the ceiling the angle at your elbow is a 90 degree angle and thus a right angle. • Curved lines/surfaces – any line/surface which is not straight. An example: a ball • Two shapes that make another shape – An example: two squares put together make a new shape a rectangle • Concentric circles - circles inside of circles • Concave –shapes which have dents • Convex – shapes without dents • Shapes used over and over to make a pattern – An example: bricks • Lines of Symmetry – a division of a shape where both parts look exactly the same – An example: square The classification of shapes provides commonalities upon which further ideas and conclusions can be drawn. It is through this process of classification which becomes the basis for theorems or underlying principles. This allows for the better understanding of the world which surrounds man. For example, it is through the understanding of a variety of shapes that mosaic pictures or other tiled images can be created. Without a complete understanding of the similarities between shapes, a variety of uses would be too complicated to be completed. It is through the understanding of these more simplistic properties or classifications that a deeper understanding can be based. For example, being able to recognize the idea of parallel lines within a group of shapes will allow for broader conclusions to be drawn. In this case, if shapes with parallel lines are combined longer, broader, and bigger parallel lines then exist. Therefore, combining shapes with parallel lines themselves create even more parallel lines. This can lead to a broader conclusion, combined parallel lines remain parallel. It is the ability to reach these broader conclusions that has lead to the determination of such ideas as theorems, which are then considered universal truths in the field of geometry. Exercises: • Using just the shapes above, how many different categories can you use to categorize them? • Find an example of a shape in your home which matches each of the above identified categories.
# Math 5 - Act. 30: Spinner Experiment Individual Utah LessonPlans ### Summary This activity provides students experience collecting and analyzing data. ### Materials #### Attachments • Spinner Face A (one per student) • 5-by-8-inch index cards, one each student • Spinner recording sheet, one to two for each student • Paper clips • Plastic straws, one 1/2- inch length per student • Scissors • Tape Math by All Means: Probability Grades 3-4 by Marilyn Burns About Teaching Mathematics: A K-8 Resource, 2nd Edition by Marilyn Burns ### Background for Teachers Part 1 of this lesson introduces children to probability through an experiment in which one outcome is more likely than the others. The experiment provides experience for children to collect and analyze data. The probability of spinning each number provides a context for talking about fractions and percents and engages students in comparing the areas of the regions of a circle. Making their own spinner gives children practice in following directions and helps develop their fine motor skills. ### Intended Learning Outcomes 2. Become mathematical problem solvers. 3. Reason mathematically. 4. Communicate mathematically. ### Instructional Procedures Invitation to Learn Hold up the sample spinner you made. Tell students they are going to make a spinner like yours. Ask what they notice about its face. Spin the spinner and point out how the indicator line tells what number the spinner lands on. Demonstrate for the children how to make a spinner. Instructional Procedures Directions for making a spinner: 1. Cut out the spinner face. 2. Cut the 5-by-8-inch index card in half. Mark a dot near the center of one of the halves. Using a straightedge, draw a line from the dot to one corner of the card. 3. Glue the spinner face to the side of the index card you did not draw on. Cut out the face of the spinner with its new, heavy backing. 4. Bend up the outside part of a paper clip. This part should point straight up when the paper clip is lying flat on your desk. 5. Use the paper clip to poke a hole in the center of the spinner face and through the dot near the center of the index card. 6. Push the bent end of the paper clip through the hole in the index card and use tape to secure the rest of the paper clip to the bottom of the card. Make sure the side of the card with the line is facing up. 7. Put the 1/2-inch length of plastic straw and then the spinner face on the paper clip. 8. Cover the tip of the paper clip with a piece of tape. Demonstrate for the students how to do the spinner experiment: 1. Write the numerals 1, 2, and 3 at the bottom of the first three columns on the spinner recording sheet. 2. Spin the spinner and record the number it lands of in the lowest square of its column. Point out to the students that they should start writing at the bottom of the columns. Do five or six spins, recording the number each time. 3. Tell students that they will continue spinning and recording until one number reaches the top of its column. 4. Children then cut out the three-column strip and post it on the chalkboard under 1, 2, or 3 heading. Pose a part of the problem: Ask, When you spin the spinner, is any number more likely to come up than any other number? Why do you think so? Explain: You can keep track of spins on a graph recording sheet that is 3 squares by 12 squares. Demonstrate for the class how to spin and record on the graph. After three or four spins, ask, What do you think the entire graph will look like when one number reaches the top of the paper? Ask, Which number do you think will reach the top of the recording sheet first? Write your prediction on paper. Present the problem to be solved: Explain: Each of you will use your spinner and graph recording sheet to conduct an experiment. When one of the numbers reaches the top, youve completed the experiment. Cut the 3 x 12 recording sheet apart from the graph paper. Then post your 3 x 12 recording sheet on the board under the winning number. (Have students tape their graphs to the board under the heading 1, 2 or 3). Students should conduct the experiment three times (using their entire graph recording sheet). Discuss the class results when all students are finished comparing the results to their predictions. Most likely, 3 was the winning number, but there will be instances of 1 or 2 as winner. Ask: how might we find out if 3 actually comes up in half of all the spins? ### Extensions You could have students take their graph recording sheets and cut apart each column. Then cut off any blank squares. Have small groups of students then tape their 1s end to end. Continue doing this with the 2s and 3s. Next combine each group’s strips of numbers together to get one long strip of 1s, 2s and 3s. Tape one end to the chalkboard. This will provide a visual for students to see which number actually won. See “Homework & Family Connections” to add to this activity. Homework & Family Connections Each student can take their spinner and one graph recording sheet home to conduct the experiment again. Explain to students that they are going to gather more data to add to their first experiment. Ask students why this would be helpful. If students do not suggest it, tell them the mathematical theory of probability says that the more times you spin a spinner, the closer the results will match the theoretical distribution. The next day, have students again cut apart their numbers, discard blank squares, and add to the class strip of 1s, 2s and 3s. Discuss the results. If there is time, have students actually count the number of 1s, 2s and 3s. This could be accomplished by cutting the strips into groups of ten, then compiling the 10s to form hundreds and so on. Some students may devise their own strategies for counting. Divide the strips among small groups for counting, then gather together for a class total. Discuss the total numbers finding out if there were any surprises with the additional data. Ask, “Does the mathematical theory of probability (gathering more data) seem to be evident in our experiment? How can you tell? Can you find a way to prove that 3 came up in half of the spins?” Add applicable questions and discussion. ### Assessment Plan Have children write about what they did, what they had predicted and what the results were. Pose these questions to students who might not know how to begin: How do the class results compare with your prediction? How did your individual experiments compare with the class results? Why do you think mathematicians say that a large sample of data is better for analyzing information than a small sample of data? Created: 09/03/2003 Updated: 02/03/2018
# ACT Math: Five Solutions (3/27/15) Hopefully you were able to find some time and a quiet place this weekend to try our Five Problems for Your Weekend (3/27/15). Below you’ll find the fully worked-out solutions to those problems. 7. D Carefully “FOIL” out the problem paying close attention to signs (3x – 4)(x + 5) 3x² + 15x – 4x – 20 3x² + 11x – 20 22. F Begin by getting zero on one side of the equation. Then factor and use the Zero Product Property. This is another problem where you need to be very careful with signs. x² + 10x = 24 x² + 10x – 24 = 0 (x + 12)(x – 2) = 0 x = -12 and x = 2 35. C There is a formula for finding arc length: s = θr. In order to use this formula, you need to convert the angle in degrees to an angle in radians by multiplying by π/180. For the angle in this problem that calculation would look like this: 30(π/180) = π/6. To find the arc length you can multiply that angle by the radius of the circle. For our arc, that would be (π/6)(6) = π. But what if you don’t remember the arc length formula? You could try more of a logical approach using what you know about the circumference of the circle. For this circle, the circumference would be C = 2πr = 2π(6) = 12π. The arc length we want is some fraction of that total circumference. But what fraction? Recall that a full circle is 360 degrees so our arc is 30/360 or 1/12 of the entire circumference. Multiplying 12π (our circumference) by 1/12 gets you an arc length of π, the same answer we got using the formula. 44. J Recognize that squaring a complex number is very much like squaring a binomial; it’s really a “FOIL problem in disguise.” (6 – i (6 – i)(6 – i) 36 – 6i – 6i + i² 36 – 12i + (-1) 35 – 12i You need to use the fact that i² = -1, but that fact is almost always (I think I really want to say always) given to you on the test. 55. B For every value of x, it is true that f(-x) = – f(x). This is another way of saying “opposite x‘s give you opposite y‘s. If you look at π, for instance, you’ll see that f(π) = 3. Now go look at -π and you’ll see that f(-π) = -3. This function is odd. If you have questions about these problems or anything else to do with the ACT, leave a comment below or send me an email at info@cardinalec.com.
• Home • / • Blog • / • Synthetic Division (With Steps & Examples) # Synthetic Division (With Steps & Examples) June 11, 2022 This post is also available in: हिन्दी (Hindi) There are several ways to divide a polynomial by another. The most basic way to do it is by using long division. But long division is very lengthy and time consuming. Especially for higher degree polynomials, long division take pretty long time to get a quotient and remainder. Synthetic Division is easy, fast and very straightforward. It helps students save a lot of time when compared with other traditional methods. ## What is a Polynomial? A polynomial is an algebraic expression made up of two or more terms subtracted, added, or multiplied. A polynomial can contain coefficients, variables, exponents, constants, and operators such as addition and subtraction. It is also important to note that a polynomial can’t have fractional or negative exponents. Examples of polynomials are: 4y2 – 2y + 7, -2x3 + 2x2 − 7x + 8, (¾)x2 – 7x + 3 etc. Like numbers, polynomials can undergo addition, subtraction, multiplication, and division. Examples of algebraic expressions that are not polynomials are: 5x4/5 + 7x + 3, 2x-2 + 7x – 3. (Having fractional or negative numbers as indices of a variable). ## Division of Polynomials Dividing polynomials is an algorithm to solve a rational number that represents a polynomial divided by a monomial or another polynomial. The divisor and the dividend are placed exactly the same way as we do for regular division. For example, if we need to divide 6x2 – 5x + 18 by 3x + 7, we write it in this way: The polynomial written on top of the bar is the numerator ( 6x2 – 5x + 18), while the polynomial written below the bar is the denominator (3x + 7). This can be understood by the following figure which shows that the numerator becomes the dividend and the denominator becomes the divisor. ## What is the Synthetic Division? Synthetic division is a shortcut method of dividing a polynomial by a linear polynomial (polynomial of degree 1). It is a simplified way of finding the zeroes of polynomials. One of the advantages of using this method over the traditional long method is that synthetic division is performed manually with less effort. Since synthetic division allows one to calculate without writing the variables while performing the polynomial division, it reduces the chances of making mistakes (the ones that students make while solving a problem using a long division method). In synthetic division, a linear binomial of the form (x – a) is used as a divisor. When we divide a polynomial P(x) of degree n by a linear polynomial (x – a), we get a polynomial quotient Q(x) of degree (n – 1) and a constant polynomial R as the remainder. Mathematically it can be represented as: P(x)/(x – a) = Q(x) + R/(x – a) Where, P(x) is a polynomial of degree n (x – a) is a binomial of degree 1 Q(x) is a polynomial of degree (n – 1). R is a constant polynomial, i.e., a number without variable Hence, we can use the synthetic division method to find the remainder quickly. ## Steps to Perform Synthetic Division Following steps are carried out to perform synthetic division on a polynomial P(x) of degree n and a binomial (x – a): Step 1: Write a for the divisor. Step 2: Write the coefficients of the dividend. Step 3: Bring the leading coefficient down. Step 4: Multiply the leading coefficient by a. Write the product in the next column. Step 5: Add the terms of the second column. Step 6: Multiply the result by a. Write the product in the next column. Step 7: Repeat steps 5 and 6 for the remaining columns. Step 8: Use the bottom numbers to write the quotient. The number in the last column is the remainder and has degree 0, the next number from the right has degree 1, the next number from the right has degree 2, and so on. ## Examples Let’s take a few examples to understand the process. ### Example 1 Divide (-3x3 + 19x2 – 30x + 15) by (x – 4) Note: (x – 4) is a linear polynomial (degree 1), so synthetic division is a fit case. Step 1: P(x) = (-3x3 + 19x2 – 30x + 15) (x – a) = (x – 4) => a = 4 Step 2: Coefficients of -3x3 + 19x2 – 30x + 15 are -3, 19, -30 and 15 Write a in the next line Now, draw a horizontal line Step 3: Leading coefficient is -3 and is brought down Step 4: Multiply the leading coefficient -3 by a(= 4). Write the product in the next column Step 5: Add the terms of the second column. Step 6: Multiply the result by a(4). Write the product in the next column. Step 7: Repeat steps 5 and 6 for the remaining columns, i.e., add and multiply progressively. Rightmost number, i.e., 7 is remainder Now, for remaining numbers (-3, 7 and -2) start from right: Power of variable x for -2 is 0, i.e., x0 Power of variable x for -7 is 1, i.e., x1 Power of variable x for 3 is 2, i.e., x2 Therefore, quotient Q(x) = -3x2 + 7x – 2 and remainder R = 7 Let’s check the result. Recall the division algorithm: Dividend = (Divisor × Quotient) + Remainder. Divisor = (x – 4) Quotient = (-3x2 + 7x – 2) Remainder = 7 (Divisor × Quotient) + Remainder = (x – 4) × (-3x2 + 7x – 2) + 7 = x × (-3x2 + 7x – 2) + (-4) × (-3x2 + 7x – 2) + 7 = (x × (-3x2)+ x × 7x + x × (- 2)) + (-4 × (-3x2) + (-4) × 7x + (-4) × (- 2)) + 7 = (-3x3 + 7x2 – 2x + 12x2 – 28x + 8) + 7 = -3x3 + 19x2 – 30x + 8 + 7= -3x3 + 19x2 – 30x + 15 is the same polynomial that we divide by (x – 4) ### Example 2 Let’s now divide a bit higher degree polynomial Divide (x6 + x5 + x4 + 22x3 + 8x2 + 14x + 2) by (x + 3) Step 1: P(x) = (x6 + x5 + x4 + 22x3 + 8x2 + 14x + 2) (x – a) = (x + 3) => a = -3 Step 2: Coefficients of x6 + x5 + x4 + 22x3 + 8x2 + 14x + 2 are 1, 1, 1, 22, 8, 14 and 2 Write a in the next line Now, draw a horizontal line Step 3: Leading coefficient is 1 and is brought down Step 4: Multiply the leading coefficient 1 by a(= -3). Write the product in the next column Step 5: Add the terms of the second column. Step 6: Multiply the result by a(-3). Write the product in the next column. Step 7: Repeat steps 5 and 6 for the remaining columns, i.e., add and multiply progressively. That finishes the process and the quotient is x5 – 2x4 + 7x3 + x2 + 5x – 1 and remainder is 5. Check: (x + 3) × (x5 – 2x4 + 7x3 + x2 + 5x – 1) + 5 = x6 + x5 + x4 + 22x3 + 8x2 + 14x + 2 ### Example 3 Divide (x4 – 6) by (x + 1) Here you can see that the terms for x3, x2 and x are not there. In all such cases, we add such terms with coefficient 0. Step 1: (x4 – 6) is written as (x4 + 0x3 + 0x2 + x – 6). And, (x + a) = (x + 1) => a = -1 Step 2: Coefficients of x4 + 0x3 + 0x2 + 0x – 6 are 1, 0, 0, 0 and -6 Write a in the next line Now, draw a horizontal line Step 3: Leading coefficient is 1 and is brought down Step 4: Multiply the leading coefficient 1 by a(= -1). Write the product in the next column Step 5: Add the terms of the second column. Step 6: Multiply the result by a(-1). Write the product in the next column. Step 7: Repeat steps 5 and 6 for the remaining columns, i.e., add and multiply progressively. That finishes the process and the quotient is x3 – x2 + x – 1 and remainder is -5. Check: (x + 1) × (x3 – x2 + x – 1) – 5 = x4 – x3 + x2 – x + x3 – x2 + x – 1 – 5 = x4 – 6 ### Example 4 Divide (4x6 – 9x4 + 2x3 + 3x2 – 8x – 12) by (2x + 3) Step 1: (4x6 – 9x4 + 2x3 + 3x2 – 8x – 12) is written as (4x6 + 0x5 – 9x4 + 2x3 + 3x2 – 8x – 12). (2x + 3) = 2(x + 3/2) Now, in order to get the form (x + a), we ignore 2 in 2(x + 3/2) And, (x + a) = (x + 3/2) => a = -3/2 Step 2: Coefficients of 4x6 + 0x5 – 9x4 + 2x3 + 3x2 – 8x – 12 are 4, 0, -9, 2, 3, -8 and -12 Write a in the next line Now, draw a horizontal line Step 3: Leading coefficient is 4 and is brought down Step 4: Multiply the leading coefficient 1 by a(= -3/2). Write the product in the next column Step 5: Add the terms of the second column. Step 6: Multiply the result by a(-3/2). Write the product in the next column. Step 7: Repeat steps 5 and 6 for the remaining columns, i.e., add and multiply progressively. That finishes the process and the quotient is 4x5 – 6x4 + 2x2 – 8 and remainder is 0. Since, in the beginning we ignored 2 in 2(x + 3/2), so here also, ignore 2 in the result 4x5 – 6x4 + 2x2 – 8 = 2(2x5 – 3x4 + x2 – 4) So, quotient is 2x5 – 3x4 + x2 – 4 and remainder is 0 Check: (2x + 3) × (2x5 – 3x4 + x2 – 4) = 4x6 – 6x5 + 2x3 – 8x + 6x5 – 9x4 + 3x2 – 12 = 4x6 – 9x4 + 2x3 + 3x2 – 8x – 12 ## Advantages and Limitations of Synthetic Division The advantages of using the synthetic division method are: • The calculation can be performed without variables • It requires only a few calculation steps • Unlike the polynomial long division method, there is no step involving subtraction and hence this method is less error-prone. The only limitation of the synthetic division method is that it is only applicable if the divisor is a linear polynomial (polynomial of degree 1). ## Conclusion We hope you enjoyed our post on how to do synthetic division of polynomials. There are a variety of ways to divide polynomials, but synthetic division is the most quick and easiest to remember.
##### Introduction to Math Proof Mathematics Tutor: None Selected Time limit: 1 Day Question No. 1: Let R be the relation on the real numbers given by xRy iff \|x-y\|≤2. Prove that this relation is Reflexive, Symmetric but not transitive. Question No. 2: Define a relation ~ on R by, x~y if x-y is rational. Prove that ~ is an equivalence relation. Question No. 3: Define a relation P that is reflexive but not symmetric and not transitive Apr 13th, 2015 1. For any real x, xRx because \|x - x\| = 0 <= 2. The relation is reflexive. If xRy, then \|x - y\| <=2. Since, \|y - x\| = \|x - y\|, we conclude that yRx and the relation is symmetric. Set x = 0, y = 2, and z = 4. Then \|x - y\| = 2 = \|y - z\|, however, \|x - z\| = 4 > 2. It means that xRy, yRz but x not R z. The relation is not transitive. 2. If x~x, then x - x = 0 is a rational number. The relation is reflexive. If x~y, then x - y is a rational number and y - x = -(x - y) is a rational number, too. The relation is symmetric. If x~y and y~z, then both x - y and y - z are rational numbers, x - z = (x - y) + (y - z) is a rational number and x~z. The relation is transitive. 3. Define the following relation R on the set of real numbers. xRy if and only if either x = y or x = 2, y = 1, or x = 1, y = 3. Then the relation R is reflexive. On the other hand, 2R1 is true but 1R2 is not true. Thus, the relation is not symmetric. Finally, 2R1 and 1R3 is true but 2R3 is not true, so the relation is not transitive. Apr 13th, 2015 ... Apr 13th, 2015 ... Apr 13th, 2015 Dec 3rd, 2016 check_circle
Views 3 months ago # CLASS_11_MATHS_SOLUTIONS_NCERT ## Class XI Chapter 12 – Class XI Chapter 12 – Introduction to Three Dimensional Geometry Maths ______________________________________________________________________________ Point A divides PQ in the ratio 1:2. Therefore, by section formula, the coordinates of point A are given by 1(10) 2 4 1( 16) 2 2 1(6) 2 4 , , 6, 4, 2 1 2 1 2 1 2 Point B divides PQ in the ratio 2:1. Therefore, by section formula, the coordinates of point B are given by 2(10) 14 2( 16) 12 2(6) 14 , , 8, 10,2 21 21 21 Thus, (6, â€“4, â€“2) and (8, â€“10, 2) are the points that trisect the line segment joining points P (4, 2, â€“6) and Q (10, â€“16, 6). Exercise Miscellaneous Question 1: Three vertices of a parallelogram ABCD are A (3, -1, 2), B (1, 2, -4) and C (-1, 1, 2). Find the coordinates of the fourth vertex. Solution 1: The three vertices of a parallelogram ABCD are given as A (3, â€“1, 2), B (1, 2, â€“4), and C (â€“1, 1, 2). Let the coordinates of the fourth vertex be D (x, y, z). We know that the diagonals of a parallelogram bisect each other. Therefore, in parallelogram ABCD, AC and BD bisect each other. ∴Mid-point of AC = Mid-point of BD 31 11 22 x1 y 2 z 4 , , , , 2 2 2 2 2 2 x 1 y 2 z 4 10,2 , , 2 2 2 Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor. Class XI Chapter 12 – Introduction to Three Dimensional Geometry Maths ______________________________________________________________________________ x1 y 2 z 4 1, 0, and 2 2 2 2 ⇒ x = 1, y = â€“2, and z = 8 Thus, the coordinates of the fourth vertex are (1, â€“2, 8). Question 2: Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and (6, 0, 0). Solution 2: Let AD, BE, and CF be the medians of the given triangle ABC. Since AD is the median, D is the mid-point of BC. 06 40 00 ∴Coordinates of point D = , , 2 2 2 = (3, 2, 0) 2 2 2 AD= 03 02 60 9 436 49 7 Since BE is the median, E is the mid-point of AC. 06 00 00 Coordinates of point E = , , 3,0,3 2 2 2 2 2 2 BE= 30 04 30 916 9 34 Since CF is the median, F is the mid-point of AB. 00 0 4 60 Coordinates of point F = , , 0,2,3 2 2 2 2 2 2 Length of CF = 60 02 03 36 49 49 7 Thus, the lengths of the medians of ΔABC are 7, 34 , and 7. Printed from Vedantu.com. 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A primer of infinitesimal analysis (2ed., CUP, 2008)(ISBN 0521887186)(O)(138s)_MCat_ Maths on track! - STEPS \| Engineers Ireland 103 Trigonometry Problems Stability Of Solitary Waves Of A Generalized Two-Component ... The Philosophical magazine; a journal of theoretical ... - Index of Teaching & Learning Plans - Project Maths Original - University of Toronto Libraries IIT-JEE 2010 - Career Point THE MATHS TEACHER'S HANDBOOK - Arvind Gupta March 2011 - Career Point View report - eResearch SA PDF file The Cubic Spline - STEM2 XT â MATHS Grade 11 â Equations SOME ELEMENTARY INEQUALITIES IN GAS DYNAMICS EQUATION Elliptic variational problems with critical exponent Convergence to Stochastic Integrals involving Non-linear Functions A Nonlinear Heat Equation with Temperature-Dependent ... - UFRJ Rudin's Principles of Mathematical Analysis: Solutions to ... - MIT Optimal Bounds on the Kuramoto-Sivashinsky Equation Felix Otto ...
# Surface Area and Volume ## Presentation on theme: "Surface Area and Volume"— Presentation transcript: Surface Area and Volume Surface Area of Prisms Surface Area = The total area of the surface of a three-dimensional object (Or think of it as the amount of paper you’ll need to wrap the shape.) Prism = A solid object that has two identical ends and all flat sides. We will start with 2 prisms – a rectangular prism and a triangular prism. Triangular Prism Rectangular Prism Surface Area (SA) of a Rectangular Prism Like dice, there are six sides (or 3 pairs of sides) Prism net - unfolded Add the area of all 6 sides to find the Surface Area. 6 - height 5 - width 10 - length SA = 2lw + 2lh + 2wh 6 - height 5 - width 10 - length SA = 2 (10 x 5) + 2 (10 x 6) + 2 (5 x 6) = 2 (50) + 2(60) + 2(30) = = 280 units squared Practice = 1208 ft squared 12 ft 10 ft 22 ft SA = 2lw + 2lh + 2wh = 2(22 x 10) + 2(22 x 12) + 2(10 x 12) = 2(220) + 2(264) + 2(120) = = 1208 ft squared Surface Area of a Triangular Prism 2 bases (triangular) 3 sides (rectangular) Unfolded net of a triangular prism 2(area of triangle) + Area of rectangles Area Triangles = ½ (b x h) = ½ (12 x 15) = ½ (180) = 90 Area Rect. 1 = b x h = 12 x 25 = 300 Area Rect = 25 x 20 = 500 15ft SA = SA = 1480 ft squared Practice Triangles = ½ (b x h) = ½ (8 x 7) = ½ (56) = 28 cm Rectangle 1 = 10 x 8 = 80 cm Rectangle 2 = 9 x 10 = 90 cm Add them all up SA = SA = 316 cm squared 9 cm 7 cm 8 cm 10 cm Surface Area of Pyramids Pyramids A pyramid has 2 shapes: One (1) square & Four (4) triangles Since you know how to find the areas of those shapes and add them. Or… Where S is the Slant Height and p is the perimeter of the base and you can use a formula… SA = ½ Sp + B Where S is the Slant Height and p is the perimeter of the base and B is the area of the Base SA = ½ Sp + B = ½ (8 x 26) + (7 x 6) area of the base SA = ½ Sp + B 5 Perimeter = (2 x 7) + (2 x 6) = 26 Slant height S = 8 ; SA = ½ Sp + B = ½ (8 x 26) + (7 x 6) area of the base = ½ (208) + (42) = = 146 units 2 SA = ½ Sp + B = ½ (18 x 24) + (6 x 6) Practice = ½ (432) + (36) 10 SA = ½ Sp + B = ½ (18 x 24) + (6 x 6) = ½ (432) + (36) = = 252 units2 Slant height = 18 Perimeter = 6x4 = 24 What is the extra information in the diagram? #3 Finding the Surface Area of a Cone The radius of the base of a cone is 5 m. Its slant height is 13 m. Find the surface area. Formula for Prisms VOLUME OF A PRISM The volume V of a prism is the area of its base B times its height h. V = Bh Note – the capital letter stands for the AREA of the BASE not the linear measurement. Try It V = Bh = (8 x 4) x 3 = (32) x 3 = 96 ft3 Find area of the base Multiply it by the height = 96 ft3 3 ft - height 4 ft - width 8 ft - length Practice V = Bh = (22 x 10) x 12 = (220) x 12 = 2640 cm3 12 cm 10 cm Notice that r2 is the formula for area of a circle. Cylinders VOLUME OF A CYLINDER The volume V of a cylinder is the area of its base, r2, times its height h. V = r2h Notice that r2 is the formula for area of a circle. Try It V = r2h The radius of the cylinder is 5 m, and the height is 4.2 m V = 3.14 · 52 · 4.2 Substitute the values you know. V = 329.7 Practice 13 cm - radius 7 cm - height V = r2h Start with the formula V = 3.14 x 132 x 7 substitute what you know = 3.14 x 169 x 7 Solve using order of Ops. = cm3 Lesson Quiz Find the volume of each solid to the nearest tenth. Use 3.14 for . 1. 2. 4,069.4 m3 861.8 cm3 3. triangular prism: base area = 24 ft2, height = 13 ft 312 ft3 Remember that Volume of a Prism is B x h where b is the area of the base. You can see that Volume of a pyramid will be less than that of a prism. How much less? Any guesses? If you said 2/3 less, you win! Volume of a Pyramid: V = (1/3) Area of the Base x height V = (1/3) Bh Volume of a Pyramid = 1/3 x Volume of a Prism + = + Find the volume of the square pyramid with base edge length 9 cm and height 14 cm. The base is a square with a side length of 9 cm, and the height is 14 cm. V = 1/3 Bh = 1/3 (9 x 9)(14) = 1/3 (81)(14) = 1/3 (1134) = 378 cm3 14 cm Practice V = 1/3 Bh = 1/3 (5 x 5) (10) = 1/3 (25)(10) = 1/3 250 = units3 The Rectangle This has 2 steps. To find the area we need base and height. Height is given (6) but the base is not as easy. Notice that the base is the same as the distance around the circle (or the Circumference). Find Circumference Formula is C =  x d = 3.14 x 6 (radius doubled) = 18.84 Now use that as your base. A = b x h = x 6 (the height given) = units squared Add them together Now add the area of the circles and the area of the rectangle together. = units squared The total Surface Area! Quiz Find the volume of each figure. a rectangular pyramid with length 25 cm, width 17 cm, and height 21 cm 2975 cm3 2. a triangular pyramid with base edge length 12 in. a base altitude of 9 in. and height 10 in. 360 in3 Surface Area of a Cylinder Review Surface area is like the amount of paper you’ll need to wrap the shape. You have to “take apart” the shape and figure the area of the parts. Then add them together for the Surface Area (SA) Parts of a cylinder A cylinder has 2 main parts. A rectangle and A circle – well, 2 circles really. Put together they make a cylinder. The Soup Can Think of the Cylinder as a soup can. You have the top and bottom lid (circles) and you have the label (a rectangle – wrapped around the can). The lids and the label are related. The circumference of the lid is the same as the length of the label. Area of the Circles Formula for Area of Circle A=  r2 = 3.14 x 32 = 28.26 But there are 2 of them so 28.26 x 2 = units squared Area of Rectangle Area of Circles Formula SA = ( d x h) + 2 ( r2) Label Lids (2) Area of Rectangle Area of Circles Practice Be sure you know the difference between a radius and a diameter! SA = ( d x h) + 2 ( r2) = (3.14 x 22 x 14) + 2 (3.14 x 112) = (367.12) + 2 (3.14 x 121) = (367.12) + 2 (379.94) = (367.12) + (759.88) = 1127 cm2 More Practice! SA = ( d x h) + 2 ( r2) = (3.14 x 11 x 7) + 2 ( 3.14 x 5.52) = (241.78) + 2 (3.14 x 30.25) = (241.78) + 2 (3.14 x 94.99) = (241.78) + 2 (298.27) = (241.78) + (596.54) = cm2 11 cm 7 cm Similar presentations
# Multiplying integers using models ## Multiplying Integers ``` ``` #### Objectives 1. Understand how to use models to multiply integers 2. Evaluate integer multiplication problems using mental math 3. Develop methods for using integer tiles 4. Apply rules to multiplying two negative numbers #### Warm up • 1. Evaluate: • a. - 15 + - 8 • b. 9 - 10 • c. 7 - (-2) • d. 6 + (-4) • 2. Writing: How is multiplication and addition related? • 3. Writing: What is the commutative property? #### Multiplying integers Now that you recalled the relationship between multiplication and addition we are ready to use our integer tiles to multiply integers. Evaluate: -3 x 4 If I wanted to change this into an addition problem this would mean -3 added together four times or visually: So we have 12 red tiles or -12. That’s it! So -3 x 4 = -12 Evaluate: 5 x -2 We can use the commutative property to rearrange this expression so that we can add 5 groups of -2. Our new expression looks like : -2 x 5 Using our tiles... Therefore our final answer is -10 So, 5 x -2 = -10 RULE: When multiplying integers, a positive times a negative will always be a negative. Then just multiply the two numbers to get a final answer. Unfortunately, tiles can not be used when we multiply a negative times a negative. However, we do have a rule. When multiplying a negative times a negative the result is always a positive number. However, there is a reason, as to why a negative times a negative is a positive. Explanation: Each number has an "additive inverse" associated to it (a sort of "opposite" number), which when added to the original number gives zero. This is in fact the reason why the negative numbers were introduced: so that each positive number would have an additive inverse. For example, the inverse of 3 is -3, and the inverse of -3 is 3. Note that when you take the inverse of an inverse you get the same number back again: "-(-3)" means "the inverse of -3", which is 3 (because 3 is the number which, when added to -3, gives zero). To put it another way, if you change sign twice, you get back to the original sign. Now, any time you change the sign of one of the factors in a product, you change the sign of the product: (-something) × (something else) is the inverse of (something) × (something else), because when you add them (and use the fact that multiplication needs to distribute over addition), you get zero. For example, (-3) x (-4) is the inverse of (-3) x (4), because when you add them and use the distributive law, you get (-3) x (-4) + (3) x (-4) = (-3 + 3) x (-4) = 0 x (-4)=0 So (-3) x (-4) is the inverse of (3) x (-4), which is itself (by similar reasoning) the inverse of (3) x (4). Therefore, (-3) x (-4) is the inverse of the inverse of 12; in other words, the inverse of -12; in other words, 12. The fact that the product of two negatives is a positive is therefore related to the fact that the inverse of the inverse of a positive number is that positive number back again. For example: Evaluate: -5 x -2= 10 Evaluate: -4 x -8= 32 Evaluate: -9 x -6= 45 • Here is a visual of the rules...don't memorize...think about why! • + means any positive number, - means any negative number) #### Assessment • Visit this website for more practice • Use mental math to perform the following operations 1. -9 x 2 2. -3 x -5 3. 6 x -8 4. 10 x 11
Edit Article # wikiHow to Divide Fractions by Fractions Dividing a fraction by a fraction might seem confusing at first, but it is really very simple. All you need to do is flip, multiply and reduce! This article will guide you through the process and show you that dividing fractions by fractions is really a breeze. ### Part 1 Understanding How to Divide Fractions by Fractions 1. 1 Think about what dividing by a fraction means. The problem 2 ÷ 1/2 is asking you: ”How many halves are in 2?” The answer is 4, because each unit (1) is made up of two halves, and there are 2 units total: 2 halves/1 unit * 2 units = 4 halves. • Try thinking about this same equation in terms of cups of water: How many half cups of water are in 2 cups of water? You could pour 2 half cups of water into each cup of water which means you are basically adding them, and you have two cups: 2 halves/1 cup * 2 cups = 4 halves. • All of this means that when the fraction you are dividing by is between 0 and 1, the answer will always be larger than the original number! This is true whether you are dividing whole numbers or fractions by a fraction. 2. 2 Understand that dividing is the opposite of multiplying. Therefore, dividing by a fraction can be accomplished by multiplying by its reciprocal. The reciprocal of a fraction (also called its “multiplicative inverse”) is just the fraction turned upside down, so that the numerator and denominator have switched places.[1] In a moment, we are going to divide fractions by fractions by finding the reciprocal of the second fraction and multiplying them together, but let’s look at some reciprocals first: • The reciprocal of 3/4 is 4/3. • The reciprocal of 7/5 is 5/7. • The reciprocal of 1/2 is 2/1, or 2. 3. 3 Memorize the following steps for dividing a fraction by a fraction. In order, the steps are: • Leave the first fraction in the equation alone. • Flip the second fraction over (find its reciprocal). • Multiply the numerators (top numbers) of the two fractions together. This result will be the numerator (top portion) of your answer. [2] • Multiply the denominators (bottom numbers) of the two fractions together. The result will be the denominator of your answer. • Simplify your fraction by reducing it to the simplest terms. 4. 4 Work through these steps on the example 1/3 ÷ 2/5. We will begin by leaving the first fraction alone, and changing the division sign to a multiplication sign: • 1/3 ÷ 2/5 = becomes: • 1/3 * __ = • Now we flip the second fraction (2/5) over to find its reciprocal, 5/2: • 1/3 * 5/2 = • Now multiply the numerators (top numbers) of the two fractions, 15 = 5. • 1/3 5/2 = 5/ • Now multiply the denominators (bottom numbers) of the two fractions, 32 = 6. • We now have: 1/3 5/2 = 5/6 • This particular fraction cannot be simplified further, so we have our answer. 5. 5 Try remembering the following rhyme to help you remember: "Dividing fractions, as easy as pie, Flip the second fraction, then multiply. And don't forget to simplify, Before it's time to say goodbye." [3] • Another helpful saying that tells you what to do with each part of the equation is: “Leave Me (the first fraction), Change Me (the division symbol), Turn Me Over (the second fraction).” ### Part 2 Dividing Fractions by Fractions in Action 1. 1 Begin with an example problem. Let’s use 2/3 ÷ 3/7. This question is asking us how many parts equal to 3/7 of a whole can be found in the value 2/3. Don’t worry; it’s not as hard as it sounds! 2. 2 Change the division sign to a multiplication sign. Your new equation should read: 2/3 * __ (we’ll fill in the blank in a moment.) 3. 3 Now find the reciprocal of the second fraction. This means flipping 3/7 over so that the numerator (3) is now on the bottom, and the denominator (7) is now on the top. The reciprocal of 3/7 is 7/3. Now write out your new equation: • 2/3 * 7/3 = __ 4. 4 Multiply your fractions. First multiply the numerators of the two fractions together: 2 * 7 = 14. 14 is the numerator (top value) of your answer. Then multiply the denominators of the two fractions together: 3 * 3 = 9. 9 is the denominator (bottom value) of your answer. You now know that 2/3 * 7/3 = 14/9. 5. 5 Simplify your fraction. In this case, because the numerator of the fraction is larger than the denominator, we know that our fraction is larger than 1, and we should convert it to a mixed fraction. (A mixed fraction is a whole number and a fraction combined, like 1 2/3.[4]) • First divide the numerator 14 by 9. 9 goes into 14 one time, with a remainder of 5, so you should write out your reduced fraction out as: 1 5/9 (“one and five ninths”). • Stop there, you have found your answer! You can determine that you cannot reduce the fraction further because the denominator is not evenly divisible by the numerator (9 cannot be divided evenly by 5) and the numerator is a prime number, or an integer that can only be divisible by one and itself.[5] 6. 6 Try another example! Let’s try the problem 4/5 ÷ 2/6 =. First change the division sign to a multiplication sign (4/5 * __ = ), then find the reciprocal of 2/6, which is 6/2. You know have the equation: 4/5 * 6/2 =__. Now multiply the numerators, 4 * 6 = 24, and the denominators 5* 2 = 10. You now have4/5 * 6/2 = 24/10. Now simplify the fraction. Because the numerator is larger than the denominator, we will need to convert this to a mixed fraction. • First divide the numerator by the denominator, (24/10 = 2 remainder 4). • Write the answer out as 2 4/10. We could still reduce this fraction further! • Note that 4 and 10 are both even numbers, so the first step in reducing them is to divide them each by 2. We end up with 2/5. • Because the denominator (5) cannot be divided evenly by the numerator (2), and it is a prime number, we know that it cannot be reduced further. Our answer is thus: 2 2/5. 7. 7 Get additional help with reducing fractions. You probably spent a lot of time learning to reduce fractions before trying to divide them by each other, but if you need a refresher or some more help, there are some great articles online that can help you a lot. [6] ## Community Q&A Search • How do I divide a whole number by a fraction? First, invert the fraction. Then multiply the whole number by the new numerator. Keep the new denominator. • What about 7/8 divided by 1/4? I keep getting 4, but my calculator says that's incorrect. Invert ¼ to 4/1, then multiply by 7/8, which gives you 28/8, or 3½. • How do I divide 3/4 by 3/5? Invert 3/5 and multiply: (3/4) x (5/3) = 15/12 = 5/4 or 1¼. • What is 5 3/4 shared among 4 people? wikiHow Contributor 5 3/4 shared with 4 people means 5 3/4 divided by 4. To do this, you need to turn the 4 into a fraction (4/1). Then you will need to flip the 2nd fraction(reciprocal) and turn the division sign into a multiplication sign (1/4). Next, you will need to turn the 5 3/4 into an improper fraction (23/4), that is, 5 x 4 + 3. Then, you multiply fractions as usual; the numerators together and the denominators together to get 23/16. You will need to simplify your answer and turn it into a mixed number fraction. The answer is equal to 1 7/16 meaning each person receives that much. = 5 3/4 divided by 4/1 = 5 3/4 x 1/4 = 23/4 x 1/4 = 23/16 = 1 7/16 • How do I find the reciprocal of a whole number? The reciprocal is a fraction whose numerator is 1 and whose denominator is the whole number. For example, if the whole number is 5, think of it as 5/1, then invert that fraction. • What is 2 1/2 divided by 1/2? Change the mixed number to an improper fraction. Invert ½ to become 2/1. Multiply 2/1 by the improper fraction, and reduce or simplify if possible. • How do I do this if one of the fractions is negative? That doesn't change anything but the answer, which will be negative. • How do I divide a fractions by a whole number? Multiply the denominator by the whole number, and reduce the resulting fraction if possible. • What is 32 divided by 26 in fractions? 32 ÷ 26 = 16/13. • What is 11/12 divided by 24/5? 11/12 ÷ 24/5 = (11/12) x (5/24) = 55/288. • How do I determine which fraction to put in the second and first in a word problem 200 characters left
# Cheryl climbed a set of stairs and stopped at the middle step. She then walked down 2 steps, up 4 steps, down 3 steps, and up 5 steps, and she was at the top of the stairs. How many steps are in the set of stairs? Jun 9, 2017 $9$ steps #### Explanation: We know that Cheryl was at the middle of the set of stairs. Let the middle step be $x$. Since she walked down 2 steps, that can be modelled as $x - 2$ We continue modelling it, until we get $x - 2 + 4 - 3 + 5$ $= x + 4$ Therefore, from the middle of the set of stairs to the top, there are 4 steps. That means that since $x$ is the median of the set of steps, it follows that there are also $4$ steps from the middle to the lowest step. Therefore, including the middle step, we have $4 + 4 + 1 = 9$ steps in total.
# nth root (Redirected from N-th root algorithm) In mathematics, an nth root of a number x, where n is usually assumed to be a positive integer, is a number r which, when raised to the power n yields x: ${\displaystyle r^{n}=x,}$ where n is the degree of the root. A root of degree 2 is called a square root and a root of degree 3, a cube root. Roots of higher degree are referred by using ordinal numbers, as in fourth root, twentieth root, etc. The computation of an nth root is a root extraction. For example: • 3 is a square root of 9, since 32 = 9. • −3 is also a square root of 9, since (−3)2 = 9. Any non-zero number considered as a complex number has n different complex nth roots, including the real ones (at most two). The nth root of 0 is zero for all positive integers n, since 0n = 0. In particular, if n is even and x is a positive real number, one of its nth roots is real and positive, one is negative, and the others (when n > 2) are non-real complex numbers; if n is even and x is a negative real number, none of the nth roots is real. If n is odd and x is real, one nth root is real and has the same sign as x, while the other (n – 1) roots are not real. Finally, if x is not real, then none of its nth roots are real. Roots of real numbers are usually written using the radical symbol or radix with ${\displaystyle {\sqrt {x}}}$ denoting the positive square root of x if x is positive, and ${\displaystyle {\sqrt[{n}]{x}}}$ denoting the real nth root, if n is odd, and the positive square root if n is even and x is nonnegative. In the other cases, the symbol is not commonly used as being ambiguous. In the expression ${\displaystyle {\sqrt[{n}]{x}}}$, the integer n is called the index, ${\displaystyle {\sqrt {{~^{~}}^{~}\!\!}}}$ is the radical sign or radix, and x is called the radicand. When complex nth roots are considered, it is often useful to choose one of the roots as a principal value. The common choice is the one that makes the nth root a continuous function that is real and positive for x real and positive. More precisely, the principal nth root of x is the nth root, with the greatest real part, and, when there are two (for x real and negative), the one with a positive imaginary part. A difficulty with this choice is that, for a negative real number and an odd index, the principal nth root is not the real one. For example, ${\displaystyle -8}$ has three cube roots, ${\displaystyle -2}$, ${\displaystyle 1+i{\sqrt {3}}}$ and ${\displaystyle 1-i{\sqrt {3}}.}$ The real cube root is ${\displaystyle -2}$ and the principal cube root is ${\displaystyle 1+i{\sqrt {3}}.}$ An unresolved root, especially one using the radical symbol, is sometimes referred to as a surd[1] or a radical.[2] Any expression containing a radical, whether it is a square root, a cube root, or a higher root, is called a radical expression, and if it contains no transcendental functions or transcendental numbers it is called an algebraic expression. Roots can also be defined as special cases of exponentiation, where the exponent is a fraction: ${\displaystyle {\sqrt[{n}]{x}}=x^{1/n}.}$ Roots are used for determining the radius of convergence of a power series with the root test. The nth roots of 1 are called roots of unity and play a fundamental role in various areas of mathematics, such as number theory, theory of equations, and Fourier transform. ## History An archaic term for the operation of taking nth roots is radication.[3][4] ## Definition and notation The four 4th roots of −1, none of which are real The three 3rd roots of −1, one of which is a negative real An nth root of a number x, where n is a positive integer, is any of the n real or complex numbers r whose nth power is x: ${\displaystyle r^{n}=x.}$ Every positive real number x has a single positive nth root, called the principal nth root, which is written ${\displaystyle {\sqrt[{n}]{x}}}$. For n equal to 2 this is called the principal square root and the n is omitted. The nth root can also be represented using exponentiation as x1/n. For even values of n, positive numbers also have a negative nth root, while negative numbers do not have a real nth root. For odd values of n, every negative number x has a real negative nth root. For example, −2 has a real 5th root, ${\displaystyle {\sqrt[{5}]{-2}}=-1.148698354\ldots }$ but −2 does not have any real 6th roots. Every non-zero number x, real or complex, has n different complex number nth roots. (In the case x is real, this count includes any real nth roots.) The only complex root of 0 is 0. The nth roots of almost all numbers (all integers except the nth powers, and all rationals except the quotients of two nth powers) are irrational. For example, ${\displaystyle {\sqrt {2}}=1.414213562\ldots }$ All nth roots of integers are algebraic numbers. The term surd traces back to al-Khwārizmī (c. 825), who referred to rational and irrational numbers as audible and inaudible, respectively. This later led to the Arabic word "أصم‎" (asamm, meaning "deaf" or "dumb") for irrational number being translated into Latin as "surdus" (meaning "deaf" or "mute"). Gerard of Cremona (c. 1150), Fibonacci (1202), and then Robert Recorde (1551) all used the term to refer to unresolved irrational roots.[5] ### Square roots The graph ${\displaystyle y=\pm {\sqrt {x}}}$. A square root of a number x is a number r which, when squared, becomes x: ${\displaystyle r^{2}=x.}$ Every positive real number has two square roots, one positive and one negative. For example, the two square roots of 25 are 5 and −5. The positive square root is also known as the principal square root, and is denoted with a radical sign: ${\displaystyle {\sqrt {25}}=5.}$ Since the square of every real number is a positive real number, negative numbers do not have real square roots. However, for every negative real number there are two imaginary square roots. For example, the square roots of −25 are 5i and −5i, where i represents a number whose square is −1. ### Cube roots The graph ${\displaystyle y={\sqrt[{3}]{x}}}$. A cube root of a number x is a number r whose cube is x: ${\displaystyle r^{3}=x.}$ Every real number x has exactly one real cube root, written ${\displaystyle {\sqrt[{3}]{x}}}$. For example, ${\displaystyle {\sqrt[{3}]{8}}=2}$ and ${\displaystyle {\sqrt[{3}]{-8}}=-2.}$ Every real number has two additional complex cube roots. ## Identities and properties Expressing the degree of an nth root in its exponent form, as in ${\displaystyle x^{1/n}}$, makes it easier to manipulate powers and roots. ${\displaystyle {\sqrt[{n}]{a^{m}}}\equiv (a^{m})^{1/n}\equiv a^{m/n}.}$ Every positive real number has exactly one positive real nth root, and so the rules for operations with surds involving positive radicands ${\displaystyle a,\;b}$ are straightforward within the real numbers: {\displaystyle {\begin{aligned}{\sqrt[{n}]{ab}}&\equiv {\sqrt[{n}]{a}}{\sqrt[{n}]{b}}\\{\sqrt[{n}]{\frac {a}{b}}}&\equiv {\frac {\sqrt[{n}]{a}}{\sqrt[{n}]{b}}}\end{aligned}}} Subtleties can occur when taking the nth roots of negative or complex numbers. For instance: ${\displaystyle {\sqrt {-1}}\times {\sqrt {-1}}\neq {\sqrt {-1\times -1}}=1,\quad }$ but rather ${\displaystyle \quad {\sqrt {-1}}\times {\sqrt {-1}}=i\times i=i^{2}=-1.}$ Since the rule ${\displaystyle {\sqrt[{n}]{a}}\times {\sqrt[{n}]{b}}={\sqrt[{n}]{ab}}}$ strictly holds for non-negative real radicands only, its application leads to the inequality in the first step above. ## Simplified form of a radical expression A non-nested radical expression is said to be in simplified form if[6] 1. There is no factor of the radicand that can be written as a power greater than or equal to the index. 2. There are no fractions under the radical sign. 3. There are no radicals in the denominator. For example, to write the radical expression ${\displaystyle {\sqrt {\tfrac {32}{5}}}}$ in simplified form, we can proceed as follows. First, look for a perfect square under the square root sign and remove it: ${\displaystyle {\sqrt {\tfrac {32}{5}}}={\sqrt {\tfrac {16\cdot 2}{5}}}=4{\sqrt {\tfrac {2}{5}}}}$ Next, there is a fraction under the radical sign, which we change as follows: ${\displaystyle 4{\sqrt {\tfrac {2}{5}}}={\frac {4{\sqrt {2}}}{\sqrt {5}}}}$ Finally, we remove the radical from the denominator as follows: ${\displaystyle {\frac {4{\sqrt {2}}}{\sqrt {5}}}={\frac {4{\sqrt {2}}}{\sqrt {5}}}\cdot {\frac {\sqrt {5}}{\sqrt {5}}}={\frac {4{\sqrt {10}}}{5}}={\frac {4}{5}}{\sqrt {10}}}$ When there is a denominator involving surds it is always possible to find a factor to multiply both numerator and denominator by to simplify the expression.[7][8] For instance using the factorization of the sum of two cubes: ${\displaystyle {\frac {1}{{\sqrt[{3}]{a}}+{\sqrt[{3}]{b}}}}={\frac {{\sqrt[{3}]{a^{2}}}-{\sqrt[{3}]{ab}}+{\sqrt[{3}]{b^{2}}}}{\left({\sqrt[{3}]{a}}+{\sqrt[{3}]{b}}\right)\left({\sqrt[{3}]{a^{2}}}-{\sqrt[{3}]{ab}}+{\sqrt[{3}]{b^{2}}}\right)}}={\frac {{\sqrt[{3}]{a^{2}}}-{\sqrt[{3}]{ab}}+{\sqrt[{3}]{b^{2}}}}{a+b}}\,.}$ Simplifying radical expressions involving nested radicals can be quite difficult. It is not obvious for instance that: ${\displaystyle {\sqrt {3+2{\sqrt {2}}}}=1+{\sqrt {2}}}$ The above can be derived through: ${\displaystyle {\sqrt {3+2{\sqrt {2}}}}={\sqrt {1+2{\sqrt {2}}+2}}={\sqrt {1^{2}+2{\sqrt {2}}+{\sqrt {2}}^{2}}}={\sqrt {\left(1+{\sqrt {2}}\right)^{2}}}=1+{\sqrt {2}}}$ ## Infinite series The radical or root may be represented by the infinite series: ${\displaystyle (1+x)^{\frac {s}{t}}=\sum _{n=0}^{\infty }{\frac {\prod _{k=0}^{n-1}(s-kt)}{n!t^{n}}}x^{n}}$ with ${\displaystyle \|x\|<1}$. This expression can be derived from the binomial series. ## Computing principal roots The nth root of an integer k is only an integer if k is the product of nth powers of integers. In all other cases the nth root of an integer is an irrational number. For instance, the fifth root of 248832 is ${\displaystyle {\sqrt[{5}]{248832}}={\sqrt[{5}]{3^{5}\cdot 2^{5}\cdot 2^{5}}}=12}$ and the fifth root of 34 is ${\displaystyle {\sqrt[{5}]{34}}={\sqrt[{5}]{2\cdot 17}}=2.024397458\ldots ,}$ where here the dots signify not only that the decimal expression does not end after a finite number of digits, but also that the digits never enter a repeating pattern, because the number is irrational. Since for positive real numbers a and b the equality ${\displaystyle \;{\sqrt[{n}]{a/b}}={\sqrt[{n}]{a}}/{\sqrt[{n}]{b}}\;}$ holds, the above property can be extended to positive rational numbers. Let ${\displaystyle r=p/q}$, with p and q coprime and positive integers, be a rational number, then r has a rational nth root, if both positive integers p and q have an integer nth root, i.e., ${\displaystyle \;r=p\cdot {\tfrac {1}{q}}\;}$ is the product of nth powers of rational numbers. If one or both nth roots of p or q are irrational, the quotient is irrational, too. ### Using Newton's method The nth root of a number A can be computed with Newton's method. Start with an initial guess x0 and then iterate using the recurrence relation ${\displaystyle x_{k+1}={\frac {1}{n}}\left({(n-1)x_{k}+{\frac {A}{x_{k}^{n-1}}}}\right)}$ until the desired precision is reached. Depending on the application, it may be enough to use only the first Newton approximant: ${\displaystyle {\sqrt[{n}]{x^{n}+y}}\approx x+{\frac {y}{nx^{n-1}}}.}$ For example, to find the fifth root of 34, note that 25 = 32 and thus take x = 2, n = 5 and y = 2 in the above formula. This yields ${\displaystyle {\sqrt[{5}]{34}}={\sqrt[{5}]{32+2}}\approx 2+{\frac {2}{5\cdot 16}}=2.025.}$ The error in the approximation is only about 0.03%. Newton's method can be modified to produce a generalized continued fraction for the nth root which can be modified in various ways as described in that article. For example: ${\displaystyle {\sqrt[{n}]{z}}={\sqrt[{n}]{x^{n}+y}}=x+{\cfrac {y}{nx^{n-1}+{\cfrac {(n-1)y}{2x+{\cfrac {(n+1)y}{3nx^{n-1}+{\cfrac {(2n-1)y}{2x+{\cfrac {(2n+1)y}{5nx^{n-1}+{\cfrac {(3n-1)y}{2x+\ddots }}}}}}}}}}}};}$ ${\displaystyle {\sqrt[{n}]{z}}=x+{\cfrac {2x\cdot y}{n(2z-y)-y-{\cfrac {(1^{2}n^{2}-1)y^{2}}{3n(2z-y)-{\cfrac {(2^{2}n^{2}-1)y^{2}}{5n(2z-y)-{\cfrac {(3^{2}n^{2}-1)y^{2}}{7n(2z-y)-\ddots }}}}}}}}.}$ In the case of the fifth root of 34 above (after dividing out selected common factors): ${\displaystyle {\sqrt[{5}]{34}}=2+{\cfrac {1}{40+{\cfrac {4}{4+{\cfrac {6}{120+{\cfrac {9}{4+{\cfrac {11}{200+{\cfrac {14}{4+\ddots }}}}}}}}}}}}=2+{\cfrac {4\cdot 1}{165-1-{\cfrac {4\cdot 6}{495-{\cfrac {9\cdot 11}{825-{\cfrac {14\cdot 16}{1155-\ddots }}}}}}}}.}$ ### Digit-by-digit calculation of principal roots of decimal (base 10) numbers Pascal's Triangle showing ${\displaystyle P(4,1)=4}$. Building on the digit-by-digit calculation of a square root, it can be seen that the formula used there, ${\displaystyle x(20p+x)\leq c}$, or ${\displaystyle x^{2}+20xp\leq c}$, follows a pattern involving Pascal's triangle. For the nth root of a number ${\displaystyle P(n,i)}$ is defined as the value of element ${\displaystyle i}$ in row ${\displaystyle n}$ of Pascal's Triangle such that ${\displaystyle P(4,1)=4}$, we can rewrite the expression as ${\displaystyle \sum _{i=0}^{n-1}10^{i}P(n,i)p^{i}x^{n-i}}$. For convenience, call the result of this expression ${\displaystyle y}$. Using this more general expression, any positive principal root can be computed, digit-by-digit, as follows. Write the original number in decimal form. The numbers are written similar to the long division algorithm, and, as in long division, the root will be written on the line above. Now separate the digits into groups of digits equating to the root being taken, starting from the decimal point and going both left and right. The decimal point of the root will be above the decimal point of the radicand. One digit of the root will appear above each group of digits of the original number. Beginning with the left-most group of digits, do the following procedure for each group: 1. Starting on the left, bring down the most significant (leftmost) group of digits not yet used (if all the digits have been used, write "0" the number of times required to make a group) and write them to the right of the remainder from the previous step (on the first step, there will be no remainder). In other words, multiply the remainder by ${\displaystyle 10^{n}}$ and add the digits from the next group. This will be the current value c. 2. Find p and x, as follows: • Let ${\displaystyle p}$ be the part of the root found so far, ignoring any decimal point. (For the first step, ${\displaystyle p=0}$). • Determine the greatest digit ${\displaystyle x}$ such that ${\displaystyle y\leq c}$. • Place the digit ${\displaystyle x}$ as the next digit of the root, i.e., above the group of digits you just brought down. Thus the next p will be the old p times 10 plus x. 3. Subtract ${\displaystyle y}$ from ${\displaystyle c}$ to form a new remainder. 4. If the remainder is zero and there are no more digits to bring down, then the algorithm has terminated. Otherwise go back to step 1 for another iteration. #### Examples Find the square root of 152.2756. 1 2. 3 4 / \/ 01 52.27 56 01 100·1·00·12 + 101·2·01·11 ≤ 1 < 100·1·00·22 + 101·2·01·21 x = 1 01 y = 100·1·00·12 + 101·2·01·12 = 1 + 0 = 1 00 52 100·1·10·22 + 101·2·11·21 ≤ 52 < 100·1·10·32 + 101·2·11·31 x = 2 00 44 y = 100·1·10·22 + 101·2·11·21 = 4 + 40 = 44 08 27 100·1·120·32 + 101·2·121·31 ≤ 827 < 100·1·120·42 + 101·2·121·41 x = 3 07 29 y = 100·1·120·32 + 101·2·121·31 = 9 + 720 = 729 98 56 100·1·1230·42 + 101·2·1231·41 ≤ 9856 < 100·1·1230·52 + 101·2·1231·51 x = 4 98 56 y = 100·1·1230·42 + 101·2·1231·41 = 16 + 9840 = 9856 00 00 Algorithm terminates: Answer is 12.34 Find the cube root of 4192 to the nearest hundredth. 1 6. 1 2 4 3 / \/ 004 192.000 000 000 004 100·1·00·13 + 101·3·01·12 + 102·3·02·11 ≤ 4 < 100·1·00·23 + 101·3·01·22 + 102·3·02·21 x = 1 001 y = 100·1·00·13 + 101·3·01·12 + 102·3·02·11 = 1 + 0 + 0 = 1 003 192 100·1·10·63 + 101·3·11·62 + 102·3·12·61 ≤ 3192 < 100·1·10·73 + 101·3·11·72 + 102·3·12·71 x = 6 003 096 y = 100·1·10·63 + 101·3·11·62 + 102·3·12·61 = 216 + 1,080 + 1,800 = 3,096 096 000 100·1·160·13 + 101·3·161·12 + 102·3·162·11 ≤ 96000 < 100·1·160·23 + 101·3·161·22 + 102·3·162·21 x = 1 077 281 y = 100·1·160·13 + 101·3·161·12 + 102·3·162·11 = 1 + 480 + 76,800 = 77,281 018 719 000 100·1·1610·23 + 101·3·1611·22 + 102·3·1612·21 ≤ 18719000 < 100·1·1610·33 + 101·3·1611·32 + 102·3·1612·31 x = 2 015 571 928 y = 100·1·1610·23 + 101·3·1611·22 + 102·3·1612·21 = 8 + 19,320 + 15,552,600 = 15,571,928 003 147 072 000 100·1·16120·43 + 101·3·16121·42 + 102·3·16122·41 ≤ 3147072000 < 100·1·16120·53 + 101·3·16121·52 + 102·3·16122·51 x = 4 The desired precision is achieved: The cube root of 4192 is about 16.12 ### Logarithmic calculation The principal nth root of a positive number can be computed using logarithms. Starting from the equation that defines r as an nth root of x, namely ${\displaystyle r^{n}=x,}$ with x positive and therefore its principal root r also positive, one takes logarithms of both sides (any base of the logarithm will do) to obtain ${\displaystyle n\log _{b}r=\log _{b}x\quad \quad {\text{hence}}\quad \quad \log _{b}r={\frac {\log _{b}x}{n}}.}$ The root r is recovered from this by taking the antilog: ${\displaystyle r=b^{{\frac {1}{n}}\log _{b}x}.}$ (Note: That formula shows b raised to the power of the result of the division, not b multiplied by the result of the division.) For the case in which x is negative and n is odd, there is one real root r which is also negative. This can be found by first multiplying both sides of the defining equation by −1 to obtain ${\displaystyle \|r\|^{n}=\|x\|,}$ then proceeding as before to find \|r\|, and using r = −\|r\|. ## Geometric constructibility The ancient Greek mathematicians knew how to use compass and straightedge to construct a length equal to the square root of a given length, when an auxiliary line of unit length is given. In 1837 Pierre Wantzel proved that an nth root of a given length cannot be constructed if n is not a power of 2.[9] ## Complex roots Every complex number other than 0 has n different nth roots. ### Square roots The square roots of i The two square roots of a complex number are always negatives of each other. For example, the square roots of −4 are 2i and −2i, and the square roots of i are ${\displaystyle {\tfrac {1}{\sqrt {2}}}(1+i)\quad {\text{and}}\quad -{\tfrac {1}{\sqrt {2}}}(1+i).}$ If we express a complex number in polar form, then the square root can be obtained by taking the square root of the radius and halving the angle: ${\displaystyle {\sqrt {re^{i\theta }}}=\pm {\sqrt {r}}\cdot e^{i\theta /2}.}$ A principal root of a complex number may be chosen in various ways, for example ${\displaystyle {\sqrt {re^{i\theta }}}={\sqrt {r}}\cdot e^{i\theta /2}}$ which introduces a branch cut in the complex plane along the positive real axis with the condition 0 ≤ θ < 2π, or along the negative real axis with π < θ ≤ π. Using the first(last) branch cut the principal square root ${\displaystyle \scriptstyle {\sqrt {z}}}$ maps ${\displaystyle \scriptstyle z}$ to the half plane with non-negative imaginary(real) part. The last branch cut is presupposed in mathematical software like Matlab or Scilab. ### Roots of unity The three 3rd roots of 1 The number 1 has n different nth roots in the complex plane, namely ${\displaystyle 1,\;\omega ,\;\omega ^{2},\;\ldots ,\;\omega ^{n-1},}$ where ${\displaystyle \omega =e^{\frac {2\pi i}{n}}=\cos \left({\frac {2\pi }{n}}\right)+i\sin \left({\frac {2\pi }{n}}\right)}$ These roots are evenly spaced around the unit circle in the complex plane, at angles which are multiples of ${\displaystyle 2\pi /n}$. For example, the square roots of unity are 1 and −1, and the fourth roots of unity are 1, ${\displaystyle i}$, −1, and ${\displaystyle -i}$. ### nth roots Geometric representation of the 2nd to 6th roots of a complex number z, in polar form re where r = \|z \| and φ = arg z. If z is real, φ = 0 or π. Principal roots are shown in black. Every complex number has n different nth roots in the complex plane. These are ${\displaystyle \eta ,\;\eta \omega ,\;\eta \omega ^{2},\;\ldots ,\;\eta \omega ^{n-1},}$ where η is a single nth root, and 1, ωω2, ... ωn−1 are the nth roots of unity. For example, the four different fourth roots of 2 are ${\displaystyle {\sqrt[{4}]{2}},\quad i{\sqrt[{4}]{2}},\quad -{\sqrt[{4}]{2}},\quad {\text{and}}\quad -i{\sqrt[{4}]{2}}.}$ In polar form, a single nth root may be found by the formula ${\displaystyle {\sqrt[{n}]{re^{i\theta }}}={\sqrt[{n}]{r}}\cdot e^{i\theta /n}.}$ Here r is the magnitude (the modulus, also called the absolute value) of the number whose root is to be taken; if the number can be written as a+bi then ${\displaystyle r={\sqrt {a^{2}+b^{2}}}}$. Also, ${\displaystyle \theta }$ is the angle formed as one pivots on the origin counterclockwise from the positive horizontal axis to a ray going from the origin to the number; it has the properties that ${\displaystyle \cos \theta =a/r,}$ ${\displaystyle \sin \theta =b/r,}$ and ${\displaystyle \tan \theta =b/a.}$ Thus finding nth roots in the complex plane can be segmented into two steps. First, the magnitude of all the nth roots is the nth root of the magnitude of the original number. Second, the angle between the positive horizontal axis and a ray from the origin to one of the nth roots is ${\displaystyle \theta /n}$, where ${\displaystyle \theta }$ is the angle defined in the same way for the number whose root is being taken. Furthermore, all n of the nth roots are at equally spaced angles from each other. If n is even, a complex number's nth roots, of which there are an even number, come in additive inverse pairs, so that if a number r1 is one of the nth roots then r2 = –r1 is another. This is because raising the latter's coefficient –1 to the nth power for even n yields 1: that is, (–r1)n = (–1)n × r1n = r1n. As with square roots, the formula above does not define a continuous function over the entire complex plane, but instead has a branch cut at points where θ / n is discontinuous. ## Solving polynomials It was once conjectured that all polynomial equations could be solved algebraically (that is, that all roots of a polynomial could be expressed in terms of a finite number of radicals and elementary operations). However, while this is true for third degree polynomials (cubics) and fourth degree polynomials (quartics), the Abel–Ruffini theorem (1824) shows that this is not true in general when the degree is 5 or greater. For example, the solutions of the equation ${\displaystyle x^{5}=x+1}$ cannot be expressed in terms of radicals. (cf. quintic equation) ## Proof of irrationality for non-perfect nth power x Assume that ${\displaystyle {\sqrt[{n}]{x}}}$ is rational. That is, it can be reduced to a fraction ${\displaystyle {\frac {a}{b}}}$, where a and b are integers without a common factor. This means that ${\displaystyle x={\frac {a^{n}}{b^{n}}}}$. Since x is an integer, ${\displaystyle a^{n}}$and ${\displaystyle b^{n}}$must share a common factor if ${\displaystyle b\neq 1}$. This means that if ${\displaystyle b\neq 1}$, ${\displaystyle {\frac {a^{n}}{b^{n}}}}$ is not in simplest form. Thus b should equal 1. Since ${\displaystyle 1^{n}=1}$ and ${\displaystyle {\frac {n}{1}}=n}$, ${\displaystyle {\frac {a^{n}}{b^{n}}}=a^{n}}$. This means that ${\displaystyle x=a^{n}}$ and thus, ${\displaystyle {\sqrt[{n}]{x}}=a}$. This implies that ${\displaystyle {\sqrt[{n}]{x}}}$ is an integer. Since x is not a perfect nth power, this is impossible. Thus ${\displaystyle {\sqrt[{n}]{x}}}$ is irrational.
Sie sind auf Seite 1von 33 # Differential Equations Chapter 01: Introduction Brannan Copyright 2010 by John Wiley & Sons, Inc. Chapter 1 Introduction In ## this introductory chapter we try to give perspective to your study of differential equations. We first formulate several problems to introduce terminology and to illustrate some of the basic ideas that we will book. We introduce three major methods geometrical, analytical, and numerical for investigating the solutions of these problems. We then discuss several ways of classifying differential equations in order to provide organizational structure for the book. Chapter 1 Introduction 1.1 Mathematical Models, Solutions, and Direction Fields 1.2 Linear Equations: Method of Integrating Factors 1.3 Numerical Approximations: Eulers Method 1.4 Classification of Differential Equations ## 1.1 Mathematical Models, Solutions, and Direction Fields Differential Equations Real world relations involving rates at which one variable, say y, may change with respect to another variable, t. Most often, these relations take the form of equations containing y and certain of the derivatives y', y'' , . . . , y(n) of y with respect to t. The resulting equations are then referred to as differential equations. Mathematical Models A ## differential equation that describes some physical process is referred to as a mathematical model of the process. ## Ex: Heat Transfer: Newtons Law of Cooling. du/dt= k(u T ), or u' = k(u T ). Terminology u' = k(u T ) - time t is an independent variable, - temperature u is a dependent variable because it depends on t, - k and T0 are parameters in the model. The equation is an ordinary differential equation because it has one, and only one, independent variable. A solution to above equation is a differentiable function u = (t) that satisfies the equation. Terminology (Ctd.) All ## possible solutions to diff. eq. is called the general solution to the equation. The geometrical representation of the general solution is an infinite family of curves in the tu-plane called integral curves. An initial condition to the differential equation is a value of u for a particular t, such as u(0)=70. Diff eq. together with the initial condition forms an initial value problem. ## Several integral curves of u' = 1.5(u 60). The general solution to This diff. eq. is u = 60 + ce1.5t . c=10 is the solution satisfying initial condition u(0)=70. y' = f(t,y) ## A solution of dy/dt = f (t, y) is a differentiable function y = (t) such that '(t) = f (t, (t)). Hence f (t, y) specifies the slope of a solution passing through the point y at time t. This information can be displayed geometrically by drawing a line segment with slope f (t, y) at the point (t, y). A direction field (or Slope filed) consists of a large number of such line segments, usually drawn on a rectangular grid. A direction field provides a geometric representation of the flow of solutions of the differential equation. ## Direction field and equilibrium solution u = 60 for u' = 1.5(u 60). A solution that does not change with time is called the equilibrium solution. ## Example - Heating and Cooling of a Building Equation: du/dt + ku = kT0 + kA sin t Solution: u = T0 + kA/(k2 + 2) (k sin t cos t) + cekt Example: Let k = 1.5 day1, T0 = 60F, A = 15F, and = 2 be given. Then, du/dt + 1.5u = 90 + 22.5 sin 2t (25) with corresponding general solution u = 60 + 22.5/(2.25 + 42) (1.5 sin 2t 2 cos 2t) + ce1.5t . ## Example - Heating and Cooling of a Building (Ctd) The steady-state solution is the solution to the equation As t . In this example, it is: ## U(t) = 60 + (22.5/2.25 + 42) (1.5 sin 2t 2 cos 2t) ## 1.2 Linear Equations: Method of Integrating Factors DEFINITIONS: A differential equation that can be written in the form (standard form) dy/dt+ p(t)y = g(t) is said to be a first order linear equation in the dependent variable y. If h(t)=0, it is said to be homogeneous; otherwise, the equation is nonhomogenous. If powers of y are there equation becomes nonlinear. ## The Method of Integrating Factors for Solving y' + p(t)y = 1. Put the equation in standard form: g(t). y' + p(t)y = g(t). 2. Calculate the integrating factor (t) = ep(t)dt 3. Multiply the equation by (t) and write it in the form [(t)y]' = (t)g(t). 4. Integrate this equation to obtain (t)y = (t)g(t) dt + c. 5. Solve for y. Example Solve the initial value problem ty' + 2y = 4t2, y(1) = 2. 1. Standard form: y' + (2/t)y = 4t. 2. Integrating factor (t) = e2/tdt =t2 3. Multiply the equation by (t) and write it in the form [t2y]' = 4t3 4. Integrate and obtain t2y = t4+ c. 5. Solve for y. y = t2 + 1/t2 , t > 0 ## Integral Curves for the initial value problem ty' + 2y = 4t2 ## 1.3 Numerical Approximations: Eulers Method When drawing direction field, many tangent line segments at successive values of t almost touch each other. By linking these an approximation to a solution of the differential equation can be obtained. 1. Can we carry out this linking of tangent lines in a simple and systematic manner? 2. If so, does the resulting piecewise linear function provide an approximation to an actual solution of the differential equation? 3. If so, can we say anything about the accuracy of the approximation? It turns out that the answer to each question is affirmative. Euler Formula For the initial value problem y = f (t, y), y(t0) = y0, the Euler method is the numerical approximation algorithm yn+1 = yn + f (tn, yn)(tn+1 tn), n = 0, 1, 2, . . . If uniform step size h is used. tn+1 = tn + h, yn+1 = yn + h f (tn, yn), To use Eulers method, you simply evaluate the equations repeatedly. ## Euler Formula (Ctd.) The result at each step is used to execute the next step. Thereby generate a sequence of values y1, y2, y3, . . . that approximate the values of the solution (t) at the points t1, t2, t3, . . . . Instead of a sequence of points, to get a function to approximate the solution (t), we use the piecewise linear function constructed from the collection of tangent line segments. That is, let y be given by y = y0 + f (t0, y0)(t t0) in [t0, t1], and in general y is given by y = yn + f (tn, yn)(t tn) in [tn, tn+1]. ## Example: Euler Method Problem: Use Eulers method with a step size h = 0.2 to approximate the solution of the initial value problem dy/dt + 1/2y = 3/2 t, y(0) = 1 on the interval 0 t 1. Answer: To use Eulers method, note that f (t, y) = 1/2y + 3/2 t, so using the initial values t0 =0, y0 = 1, we have f0 = f (t0, y0) = f (0, 1) = 0.5 + 1.5 0 = 1.0. (Ctd.) ## The tangent line approximation for t in [0, 0.2] is y = 1 + 1.0(t 0) = 1 + t. Setting t = 0.2, we find the approximate value y1 of the solution at t = 0.2, namely, y1 = 1.2. At the next step we have f1 = f (t1, y1) = f (0.2, 1.2) = 0.6 + 1.5 0.2 = 0.7, and y = 1.2 + 0.7(t 0.2) = 1.06 + 0.7t for t in [0.2, 0.4]. Evaluating at t = 0.4, we obtain y2 = 1.34. ## Example: Euler Method (Ctd.) 1.4 Classification of Differential Equations: only ordinary 1.- IfOrdinary andderivatives Partial appear in the differential Differential Equations ## equation then it is an ordinary differential equation. - If derivatives are partial derivatives then the equation is called a partial differential equation. 2.Systems of Differential Equations If ## there is a single unknown function, then one equation is sufficient. However, if there are two or more unknown functions, then a system of equations is required. For example, the LotkaVolterra, or predatorprey, equations are important in ecological modeling. They have the form dx/dt = ax xy dy/dt = cy + xy 3. Order The order of a differential equation is the order of the highest derivative that appears in the equation. The equations in the preceding sections are all first order equations. The equation ay'' + by' + cy = f (t), where a, b, and c are given constants, and f is a given function, is a second order equation. Equations ## The ordinary differential equation F(t, y, y', . . . , y(n)) = 0 is said to be linear if F is a linear function of the variables y, y', . . . , y(n).Thus the general linear ordinary differential equation of order n is a0(t)y(n)+ a1(t)y(n-1)+ +an(t)y = g(t). An equation that is not of this form is a nonlinear equation. Some Terminology The process of approximating a nonlinear equation by a linear one is called linearization. A solution to y(n) = f (t, y, y', y'', . . . , y(n-1)) on the interval < t < is a function such that ', '', . . . , (n) exist and satisfy (n)(t) = f [t, (t), '(t), . . . , (n-1)(t)] for every t in < t < . How ## can we tell whether some particular equation has a solution? This is the question of existence of a solution How many solutions it has, and specified to single out a particular solution? This is the question of uniqueness. Can we actually determine a solution, and if so, how? Equations A ## computer can be an extremely valuable tool in the study of differential equations. There exists extremely powerful and general software packages that can perform a wide variety of mathematical operations. Among these are Maple, Mathematica, and MATLAB. CHAPTER SUMMARY Differential equations are used to model systems that change continuously in time. The mathematical investigation of a differential equation usually involves one or more of three general approaches: geometrical, analytical, and numerical. ## 1.1 Geometrical Approach A solution of dy/dt = f (t, y) is a differentiable function y = (t) such that '(t) = f (t, (t)). Hence f (t, y) specifies the slope of a solution passing through the point y at time t. This information can be displayed geometrically by drawing a line segment with slope f (t, y) at the point (t, y). A direction field (or Slope filed) consists of a large number of such line segments, usually drawn on a rectangular grid. A direction field provides a geometric representation of the flow of solutions of the differential equation. ## 1.2 Linear Equations Analytical Approach dy/dt + p(t)y = g(t).Multiplying by the integrating Integrating then yields y = 1[g dt + c]. ## 1.3 Numerical Approach For the initial value problem y = f (t, y), y(t0) = y0, the Euler method is the numerical approximation algorithm tn+1 = tn + h, yn+1 = yn + h f (tn, yn), n = 0, 1, 2, . . . that geometrically consists of a finite number of connected line segments, each of whose slopes is determined by the slope at the initial point of each segment. ## 1.4 Classification of Differential Equations Differential equations are classified according to various criteria for a sensible treatment of theory, solution methods, and solution behavior: ordinary differential equations and partial differential equations single (or scalar) equations and systems of equations linear equations and nonlinear equations first order equations and higher order equations
Education.com Try Brainzy Try Plus # Distance Problems Help (page 2) based on 5 ratings By — McGraw-Hill Professional Updated on Sep 26, 2011 ## Traveling Towards Eachother When two bodies travel towards each other (from opposite directions) the rate at which the distance between them is shrinking is also the sum of their individual rates. #### Example 1 Dale left his high school at 3:45 and walked towards his brother’s school at 5 mph. His brother, Jason, left his elementary school at the same time and walked toward Dale’s high school at 3 mph. If their schools are 2 miles apart, when will they meet? The rate at which the brothers are moving towards each other is 3 + 5 = 8 mph. Let t represent the number of hours the boys walk. The boys will meet after an hour or 15 minutes; that is, at 4:00. #### Example 2 A jet airliner leaves Dallas going to Houston, flying at 400 mph. At the same time, another jet airliner leaves Houston, flying to Dallas, at the same rate. How long will it take for the two airliners to meet? (Dallas and Houston are 250 miles apart.) The distance between the jets is decreasing at the rate of 400 + 400 = 800 mph. Let t represent the number of hours they are flying. The planes will meet after hours or minutes or 18 minutes 45 seconds. Find practice problems and solutions at Distance Problems Practice Problems - Set 3. ## Traveling at Right Angles For distance problems in which the bodies are moving away from each other or toward each other at right angles (for example, one heading east, the other north), the Pythagorean Theorem is used. This topic will be covered in the last chapter. Some distance problems involve the complication of the two bodies starting at different times. For these, you need to compute the head start of the first one and let t represent the time they are both moving (which is the same as the amount of time the second is moving). Subtract the head start from the distance in question then proceed as if they started at the same time. #### Example 1 A car driving eastbound passes through an intersection at 6:00 at the rate of 30 mph. Another car driving westbound passes through the same intersection ten minutes later at the rate of 35 mph. When will the cars be 18 miles apart? The eastbound driver has a 10-minute head start. In 10 minutes ( hours), that driver has traveled miles. So when the westbound driver passes the intersection, there is already 5 miles between them, so the question is now “How long will it take for there to be 18 – 5 = 13 miles between two bodies moving away from each other at the rate of 30 + 35 = 65 mph?” Let t represent the number of hours after the second car has passed the intersection. In of an hour or minutes, an additional 13 miles is between them. So 12 minutes after the second car passes the intersection, there will be a total of 18 miles between the cars. That is, at 6:22 the cars will be 18 miles apart. #### Example 2 Two employees ride their bikes to work. At 10:00 one leaves work and rides southward home at 9 mph. At 10:05 the other leaves work and rides home northward at 8 mph. When will they be 5 miles apart? The first employee has ridden miles by the time the second employee has left. So we now need to see how long, after 10:05, it takes for an additional miles to be between them. Let t represent the number of hours after 10:05. When both employees are riding, the distance between them is increasing at the rate of 9 + 8 = 17 mph. After hour, or 15 minutes, they will be an additional miles apart. That is, at 10:20, the employees will be 5 miles apart. #### Example 3 Two boys are 1250 meters apart when one begins walking toward the other. If one walks at a rate of 2 meters per second and the other, who starts walking toward the first boy four minutes later, walks at the rate of 1.5 meters per second, how long will it take for them to meet? The boy with the head start has walked for 4(60) = 240 seconds. (Because the rate is given in meters per second, all times will be converted to seconds.) So, he has traveled 240(2) = 480 meters. At the time the other boy begins walking, there remains 1250 – 480 = 770 meters to cover. When the second boy begins to walk, they are moving toward one another at the rate of 2 + 1.5 = 3.5 meters per second. The question becomes “How long will it take a body moving 3.5 meters per second to travel 770 meters?” Let t represent the number of seconds the second boy walks. The boys will meet 220 seconds, or 3 minutes 40 seconds, after the second boy starts walking. #### Example 4 A plane leaves City A towards City B at 9:10, flying at 200 mph. Another plane leaves City B towards City A at 9:19, flying at 180 mph. If the cities are 790 miles apart, when will the planes pass each other? In 9 minutes the first plane has flown miles, so when the second plane takes off, there are 790 – 30 = 760 miles between them. The planes are traveling towards each other at 200 + 180 = 380 mph. Let t represent the number of hours the second plane flies. Two hours after the second plane has left the planes will pass each other; that is, at 11:19 the planes will pass each other. Find practice problems and solutions at Distance Problems Practice Problems - Set 4. 150 Characters allowed ### Related Questions #### Q: See More Questions Top Worksheet Slideshows
# How does a scale ruler work? Getty Images A scale ruler is the three-sided ruler used by architects and readers of blueprints to convert between scaled drawings and the actual dimensions without having to resort to any mathematical calculations. An architect uses the scale ruler to convert dimensions into a smaller drawing of a building plan. The reader of the blue print will then use a scale ruler to translate the drawing into the real sizes for construction. ## What is a scale ruler? A scale ruler is the three-sided ruler used by architects and readers of blueprints to convert between scaled drawings and the actual dimensions without having to resort to any mathematical calculations. An architect uses the scale ruler to convert dimensions into a smaller drawing of a building plan. The reader of the blueprint will then use a scale ruler to translate the drawing into the real sizes for construction. ## Choosing the correct scale On each side of a scale ruler at the far left before the zero mark is a number reflecting the scale of the rules on that particular side. Using the appropriate scale is crucial and should be the first step. If reading a blueprint, the appropriate scale will be written on the plans. If drafting a plan, choosing an appropriate scale will depend on the size of the drawing compared to the actual dimensions described. • On each side of a scale ruler at the far left before the zero mark is a number reflecting the scale of the rules on that particular side. • If drafting a plan, choosing an appropriate scale will depend on the size of the drawing compared to the actual dimensions described. ## Measuring with a scale ruler Common scales are 1:10, 1:50 and 1:100. The first number represents the distance as measured on the ruler and the larger number is how many times bigger the feature is in the real world. To read a blueprint with a scale ruler once the appropriate scale is determined, line up the zero mark with the beginning of the length to be measured. If the distance falls exactly on a line of the ruler, that is the measure in metres. Make sure the correct marks are being read, as most scale rulers have two scales per side for maximum efficiency. • Common scales are 1:10, 1:50 and 1:100. • To read a blueprint with a scale ruler once the appropriate scale is determined, line up the zero mark with the beginning of the length to be measured. If the distance being measured does not fall exactly on a line, the exact measure will be the number of metres corresponding to the nearest line passed plus a certain number of cm. Determining the cm will require a second measurement. Before removing the scale ruler, first mark the appropriate metre line on the drawing. The first cm of the scale ruler is subdivided into tenths, corresponding to cm. Line up the zero line with the mark representing the metre line of the first measurement, and use the detailed marking to find the number of cm. The full measurement will be the number of metres plus this number of cm. Even with this second step, using a scale ruler is much easier and faster than making complex conversions using maths. It also minimises the chance of error in the calculations. With a little practice, using a scale ruler becomes a simple but essential tool for the contractor and architect alike.
## MATHS:: Lecture 16 :: PERMUTATION AND COMBINATION PERMUTATION AND COMBINATION Fundamental Counting Principle If a first job can be done in m ways and a second job can be done in n ways then the total number of ways in which both the jobs can be done in succession is m x n. For example, consider 3 cities Coimbatore, Chennai and Hyderabad. Assume that there are 3 routes (by road) from Coimbatore to Chennai and 4 routes from Chennai to Hyderabad. Then the total number of routes from Coimbatore to Hyderabad via Chennai is 3 x 4 =12. This can be explained as follows. For every route from Coimbatore to Chennai there are 4 routes from Chennai to Hyderabad. Since there are 3 road routes from Coimbatore to Chennai, the total number of routes is 3 x 4 =12. The above principle can be extended as follows. If there are n jobs and if there are mi ways in which the ith job can be done, then the total number of ways in which all the n jobs can be done in succession ( 1st job, 2nd job, 3rd job… nth job) is given by m1 x m2 x m3 …x mn . Permutation Permutation means arrangement of things. The word arrangement is used, if the order of things is considered. Let us assume that there are 3 plants P1, P2, P3. These 3 plants can be planted in the following 6 ways namely P1 P2 P3 P1 P3 P2 P2 P1 P3 P2 P3 P1 P3 P1 P2 P3 P2 P1 Each arrangement is called a permutation. Thus there are 6 arrangements (permutations) of 3 plants taking all the 3 plants at a time. This we write as 3P3. Therefore 3P3 = 6. Suppose out of the 3 objects we choose only 2 objects and arrange them. How many arrangements are possible? For this consider 2 boxes as shown in figure. I Box II Box Permutation Since we want to arrange only two objects and we have totally 3 objects, the first box can be filled by any one of the 3 objects, (i.e.) the first box can be filled in 3 ways. After filling the first box we are left with only 2 objects and the second box can be filled by any one of these two objects. Therefore from Fundamental Counting Principle the total number of ways in which both the boxes can be filled is 3 x 2 =6. This we write as 3P2 = 6. In general the number of permutations of n objects taking r objects at a time is denoted by nPr. Its value is given by Note: 1 • nPn = n ! (b ) nP1= n. (c) nP0= 1. Examples: 1. Evaluate 8P3 Solution: 2. Evaluate 11P2 Solution: 3. There are 6 varieties on brinjal, in how many ways these can be arranged in 6 plots which are in a line? Solution Six varieties of brinjal can be arranged in 6 plots in 6P6 ways. 6P6 = = 6! [0! = 1] = 6 x 5 x 4 x 3 x 2x1 = 720. Therefore 6 varieties of brinjal can be arranged in 720 ways. 4. There are 5 varieties of roses and 2 varieties of jasmine to be arranged in a row, for a photograph. In how many ways can they be arranged, if (i) all varieties of jasmine together (ii)All varieties of jasmine are not together. Solution i) Since the 2 varieties of jasmine are inseparable, consider them as one single unit. This together with 5 varieties of roses make 6 units which can be arranged themselves in 6! ways. In every one of these permutations, 2 varieties of jasmine can be rearranged among themselves in 2! ways. Hence the total number of arrangements required = 6! x 2! = 720 x 2 = 1440. ii)The number of arrangements of all 7 varieties without any restrictions =7! = 5040 Number of arrangements in which all varieties of jasmine are together = 1440. Therefore number of arrangements required = 5040 -1440 = 3600. Combinatination Combination means selection of things. The word selection is used, when the order of thing is immaterial. Let us consider 3 plant varieties V1, V2 & V3. In how many ways 2 varieties can be selected? The possible selections are 1) V1 & V2 2) V2 & V3 3) V1 & V3 Each such selection is known as a combination. There are 3 selections possible from a total of 3 objects taking 2 objects at a time and we write 3C2= 3. In general the number of selections (Combinations) from a total of n objects taking r objects at a time is denoted by n Cr. Combination Probability using Combinations Relation between nPr and nCr We know that n Pr = nCr x r! (or) ------------(1) But we know ---------(2) Sub (2) in (1) we get Another formula for nCr We know that nPr = n. (n-1). (n-2)…(n-r+1) \ Example 1. Find the value of 10C3. Solution: Note -1 • nC0 = 1 • nC1 = n • nCn = 1 d) nCr = nCn-r Examples 1. Find the value of 20C18 Solution We have 20C18 = 20C20-18=20C2 = =190 2. How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all the prizes? Solution Any one prize can be given to any one of the 3 boys and hence there are 3 ways of distributing each prize. Hence, the 4 prizes can be distributed in 34= 81 ways. 3. A team of 8 students goes on an excursion, in two cars, of which one can accommodate 5 and the other only 4. In how many ways can they travel? Solution There are 8 students and the maximum number of students can accommodate in two cars together is 9. We may divide the 8 students as follows Case I: 5 students in the first car and 3 in the second Case II: 4 students in the first car and 4 in the second In Case I: 8 students are divided into groups of 5 and 3 in 8C3 ways. Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4 ways. Therefore, the total number of ways in which 8 students can travel is 8C3 + 8C4 = 56 + 70 = 126. 4. How many words of 4 consonants and 3 vowels can be made from 12 consonants and 4 vowels, if all the letters are different? Solution 4 consonants out of 12 can be selected in 12C4 ways. 3 vowels can be selected in 4C3 ways. Therefore, total number of groups each containing 4 consonants and 3 vowels = 12C4 * 4C3 Each group contains 7 letters, which can be arranging in 7! ways. Therefore required number of words = 124 * 4C3 * 7!
Area of Rectangle Formula When we talk about some plane figures, we think of their shape, region or boundary. We compare the objects on the basis of their size and area. We all know that we need some measure to compare them. And one such measure is its area. All the objects that lie in a plane acquire some region of a flat surface. The measure of the surface enclosed by a closed figure is known as its area. There are different geometrical closed shapes that exist namely square, rectangle, triangle, circle, etc. In this article, we will mainly be focusing on the understanding of the area of rectangle formula with some practical examples, its calculation, and units. Let us get begin! Area of Rectangle Formula What is a Rectangle? Let us first understand the shape and structure of a rectangle. A rectangle is a quadrilateral with four sides. It has straight lines on all 4 sides. The opposite sides of a rectangle are parallel and of equal length. Since a rectangle has four sides, it has four angles. All angles of a rectangle are equal. It is an equiangular rectangle with four right angles which is 90 degrees. Another property of a rectangle is that has two diagonals of equal length. Diagonal is a line that is drawn inside the rectangle connecting the opposite corners or vertices. Because of this property, we say that the two diagonals of a rectangle are congruent. An object in two-dimensional geometry must have measured for length and breadth. Here, in the case of a rectangle, its length and breadth are not equal. The area of a rectangle is the product of its two sides, multiplying the length and its breadth. The unit of area of the rectangle is square units. Let us now take a deep look into the formula. Formula A= l x b Where, A Area of rectangle l Length of the rectangle b Breadth of rectangle Solved Examples Now that we have some clarity about the concept and meaning of area of rectangle, let us try some examples to deepen our understanding of the subject. Q: The length and the breadth of a rectangular piece of land are 500 m and 300 m respectively find its area. Also, find the cost of painting the sheet, if a painting of an area of one square meter costs 50 paise. Ans: The Area of a rectangle (A) = l x b, where l is the length and b are the breadths of the rectangle. Given, l = 500 m and b = 300 m. Thus, Area of the rectangle A = 500 m x 300 m = 1, 50,000 square meters. Now, let us calculate the cost of painting the land. Cost of painting of an area of 1 square meter = Re 0.50 So, cost of painting the total area of the rectangular land = 0.50 x 150000 = Rs 75000. Q: The area of a rectangular sheet is 500 cmÂ². If 25 cm is the length of the sheet, find its width. Ans: It is given that the Area of the rectangular sheet = 500 cmÂ² and the Length (l) =25 cm We know that the Area of the rectangle = l Ã— b (where b = width of the sheet) Therefore, width b = $$\frac{Area}{l}$$ Width b = $$\frac{500}{25}$$ = 20 cm Share with friends Customize your course in 30 seconds Which class are you in? 5th 6th 7th 8th 9th 10th 11th 12th Get ready for all-new Live Classes! Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes. Ashhar Firdausi IIT Roorkee Biology Dr. Nazma Shaik VTU Chemistry Gaurav Tiwari APJAKTU Physics Get Started One response to “Equation Formula” 1. KUCKOO B says: I get a different answer for first example. I got Q1 as 20.5 median 23 and Q3 26
Dear Mockbankers Today in Banker’s Bounty, you will get Decimal and Fraction concept and previous year questions. Step 1: Watch the video prepared by MockBank’s expert. Step 2: Read the concepts that follow. Step 3: Put a timer before you solve the previous year questions on decimal and fraction in quiz and record how much time it takes you to solve it. Step 4: Send us questions in comments section below which took maximum time and we will tell you a faster way to solve it. Basic Concepts Fraction A fraction is an expression that indicates the quotient of two quantities. Examples of fractions: 1/2, 1/3, -8/5 A fraction has two parts, namely Numerator and Denominator. Numerator is the number at the top of the fraction and denominator is the number at the bottom of the fraction. For example, in the fraction 1/5, 1 is the numerator and 5 is denominator. The denominator of a fraction cannot be zero TYPES OF FRACTION: 1. Common Fraction: A common fraction (also known as Vulgar fraction and simple fraction) is a fraction in which both numerator and denominator are integers (As with other fractions, the denominator cannot be zero). Example 1/2, 1/5, 2/7 etc. 2. Decimal Fraction: A decimal fraction is a fraction in which denominator is an integer power of ten. (The term decimals are commonly used to refer decimal fractions). Generally, a decimal fraction is expressed using decimal notation and its denominator is not mentioned explicitly. Example1/10 = 0.11/100 = 0.011/1000 = 0.001 151/100 = 1.51 Decimal fractions can also be expressed using scientific notation with negative exponents Example: 5.192 × 10−6 is a decimal fraction which represents .000005192. 3. Conversion of decimal to fraction Put 1 in the denominator under the decimal point and annex with it as many zeros as is the number of digits after the decimal point. Now, remove the decimal point and reduce the fraction to its lowest terms. Example 0.5 = 5/10 = 1/2 0.75 = 75/100 = 3/4 1.25 = 125/100 = 5/4 Now that you have seen the video and taken the mock test , let us now start our timers and attempt the mock test below 1. 11/30 is closest to? Question 1 of 10 2. If the numerator of a certain fraction is increased by 200% and the denominator is increased by 150% the new fraction thus formed is 9/10, what is the original fraction ? Question 2 of 10 3. If the numerator of a fraction is increased by 400% and the denominator is increased by 300%,the resultant fraction is 15/14 What was the original fraction? Question 3 of 10 4. 295 + 395 when divided by 5 leaves remainder as – Question 4 of 10 5. A fraction becomes 1/2 on subtracting 1 from the numerator and adding 2 to the denominator and reduces to 1/3 on subtracting 7 from the numerator and 2 from the denominator. The fraction is- Question 5 of 10 6. The fourth proportional to the numbers 16, 9 and 8 is Question 6 of 10 7. If the fractions 5/6, 7/8, 6/7, 8/9 are arranged in descending order then the correct sequence is – Question 7 of 10 8. After measuring 120 metres of rope, it was discovered that the metre rod was 3cm longer. The true length of the rope measured is? Question 8 of 10 9. Find the greatest number of five digits which when divided by 3, 5, 8, 12 have 2 as remainder. Question 9 of 10 10. A fraction becomes 1/3 when 1 is subtracted from its numerator as well as from its denominator. If 1 is added to both its numerator and denominator it becomes 1/2 Find the fraction. Question 10 of 10 1. For every 100 cm, there is an extra 3cm. Now, there are 120 metres. so, 120*3=360 extra cms Which means 3 metre and 60cms extra. And=123m 60cms. 2. fter measuring 120 metres of rope, it was discovered that the metre rod was 3cm longer. The true length of the rope measured is?
Select Page Exercises of linear inequalities can be solved by following a process similar to that used to solve linear equations. In general, the techniques used to solve linear equations are also useful for solving inequalities. The most important difference when solving inequalities is that when we divide or multiply the entire expression by a negative number, the inequality sign has to be switched. The examples with answers that we will see will show the process of solving linear inequalities. ##### ALGEBRA Relevant for Learning to solve linear inequalities problems. See examples ##### ALGEBRA Relevant for Learning to solve linear inequalities problems. See examples ## Linear inequalities – Examples with answers The following linear inequalities examples have their respective solution. The solution details the step-by-step process that can be followed to find the answer. It is recommended that you try to solve the exercises yourself before looking at the answer. ### EXAMPLE 1 Solve and graph the inequality . • We start by writing the original problem: • To solve for the variable, we add 5 to both sides of the inequality: • After simplifying, the expression reduces to: • To solve, we divide both sides by 3: • We graph the inequality with an open point since 2 is not included in the solution. The solution is all the numbers to the right of 2: ### EXAMPLE 2 Solve and graph the inequality . In this exercise, we will look at what happens to the inequality when we divide by a negative number: • We have the inequality: • We subtract 6 from both sides to solve for the variable: • After simplifying, the expression reduces to: • Now, we have to divide by -4 to get the answer: • The 1 is not part of the solution, so we use an open point to indicate the solutions on the right side of the 1: ### EXAMPLE 3 Solve and graph the inequality . In this case, we have variables on both sides. We have to move the variables to one side and the constants to the other. It doesn’t matter which side contains the variables, but it is common to move the variables to the left: • We subtract 2 and 2x from both sides to solve for the variable: • Simplifying the inequality, we have: • We divide both sides by 2 and simplify to get the answer: • Here, the 4 is part of the solution, so we use a closed point to indicate this: ### EXAMPLE 4 Solve the inequality . We have to move the variables to one side of the inequality and the constants to the other: • We have: • We isolate the variable by subtracting 4 and 5x from both sides : • By simplifying, we get: • We have to divide both sides by -3 to get the answer: ### EXAMPLE 5 Solve the inequality . In this case, we have parentheses, so we use the distributive property to remove parentheses and simplify: • We write the original problem: • We apply the distributive property: • We add 6 from both sides and subtract 4x to solve for the variable: • After simplifying, the expression reduces to: • By dividing both sides by 2, we have: ### EXAMPLE 6 Solve the inequality . Here, we have to remove the parentheses from both sides and combine like terms to simplify: • We have: • We apply the distributive property to both sides and combine like terms: • We isolate the variable by subtracting 20 and adding 2x to both sides: • We simplify to obtain: • We divide both sides by 10 and simplify to get the answer: ### EXAMPLE 7 Solve the inequality . Similar to the previous problem, we simplify the parentheses on both sides and combine like terms: • We write the original problem: • We apply the distributive property and combine like terms: • We subtract 8x from both sides to solve for the variable: • Simplifying, we have: • We divide by -6 and simplify to get the answer: ## Linear inequalities – Practice problems Test your knowledge of linear inequalities with the following problems. Solve the inequalities and choose your answer. After clicking “Check”, you can check if you got the correct answer.
# Using Patterns of Integer Exponents Save this PDF as: Size: px Start display at page: ## Transcription 1 8.EE.1 Know and apply the properties of integer exponents to generate equivalent numerical expressions. How can you develop and use the properties of integer exponents? The table below shows powers of 5, 4, and 3. 8.EE.1 Using Patterns of Integer Exponents What pattern do you see in the powers of 5? What pattern do you see in the powers of 4? What pattern do you see in the powers of 3? Reflect Complete the table for the values of Complete the table for the values of Complete the table for the values of 1. Make a Conjecture Write a general rule for the value of 2. Make a Conjecture Write a general rule for the value of where and n is an integer. Lesson /8 2 A 8.EE.1 Exploring Properties of Integer Exponents What pattern do you see when multiplying two powers with the same base? Use your pattern to complete this equation: What pattern do you see when dividing two powers with the same base? Use your pattern to complete this equation: Complete the following equations: What pattern do you see when raising a power to a power? Mathematical Practices Do the patterns you found in parts A D apply if the exponents are negative? If so, give an example of each. Use your pattern to complete this equation: 34 Unit 1 2/8 3 Complete the following equation: What pattern do you see when raising a product to a power? Reflect Use your pattern to complete this equation: Let m and n be integers. Simplifying Expressions with Powers You can use the general rules you found in the Explore Activities to simplify expressions involving powers. EXAMPLE 1 8.EE.1 my.hrw.com Lesson /8 4 Math On the Spot my.hrw.com Applying Properties of Integer Exponents The general rules you found in the Explore Activities are summarized below. You can use them to simplify more complicated expressions. My Notes EXAMPLE 2 8.EE.1 Simplify within parentheses. Use properties of exponents. Simplify. Apply the rule for negative exponents and add. Simplify within parentheses. Use properties of exponents. Use properties of exponents. Simplify. 36 Unit 1 4/8 5 Guided Practice Find the value of each power. (Explore Activity 1) Use properties of exponents to write an equivalent expression. (Explore Activity 2) 22. Summarize the rules for multiplying powers with the same base, dividing powers with the same base, raising a product to a power, and raising a power to a power. Lesson /8 6 35. Vocabulary Identify the property that is being applied at each step to simplify the expression. 36. Communicate Mathematical Ideas Camille simplified an expression as shown. Discuss why this method is justified. 38 Unit 1 6/8 7 37. Explain why the exponents cannot be added in the product 38. List three ways to express as a product of powers. 39. Astronomy The distance from Earth to the moon is about miles. The distance from Earth to Neptune is about miles. Which distance is the greater distance, and about how many times greater is it? 40. Critique Reasoning A student claims that is greater than 1. Explain whether the student is correct or not. Find the missing exponent. 44. Communicate Mathematical Ideas Why do you subtract exponents when dividing powers with the same base? 46. Represent Real-World Problems In computer technology, a kilobyte is bytes in size. A gigabyte is bytes in size. The size of a terabyte is the product of the size of a kilobyte and the size of a gigabyte. What is the size of a terabyte? Lesson /8 8 A toy store is creating a large window display of different colored cubes stacked in a triangle shape. The table shows the number of cubes in each row of the triangle, starting with the top row. 48. Look for a Pattern Describe any pattern you see in the table. 49. Using exponents, how many cubes will be in Row 6? How many times as many cubes will be in Row 6 than in Row 3? 50. Justify Reasoning If there are 6 rows in the triangle, what is the total number of cubes in the triangle? Explain how you found your answer. Work Area 51. Critique Reasoning A student simplified the expression Do you agree with this student? Explain why or why not. 53. Persevere in Problem Solving A number to the 12th power divided by the same number to the 9th power equals 125. What is the number? 40 Unit 1 8/8 ### The wavelength of infrared light is meters. The digits 3 and 7 are important but all the zeros are just place holders. Section 6 2A: A common use of positive and negative exponents is writing numbers in scientific notation. In astronomy, the distance between 2 objects can be very large and the numbers often contain many ### Sometimes it is easier to leave a number written as an exponent. For example, it is much easier to write 4.0 Exponent Property Review First let s start with a review of what exponents are. 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EE.6.1 Write and evaluate numerical expressions involving wholenumber Unit 2 2 2 Unit : Operations and Properties Mathematics Common Core Domain Mathematics Common Core Cluster Mathematics Common Core Standard Resources NS..2 Fluently divide multi-digit numbers using the ### Please Excuse My Dear Aunt Sally Really Good Stuff Activity Guide Order of Operations Poster Congratulations on your purchase of the Really Good Stuff Order of Operations Poster a colorful reference poster to help your students remember ### Performance Level Descriptors Grade 6 Mathematics Performance Level Descriptors Grade 6 Mathematics Multiplying and Dividing with Fractions 6.NS.1-2 Grade 6 Math : Sub-Claim A The student solves problems involving the Major Content for grade/course with ### Rational Exponents. Given that extension, suppose that. Squaring both sides of the equation yields. a 2 (4 1/2 ) 2 a 2 4 (1/2)(2) a a 2 4 (2) SECTION 0. Rational Exponents 0. OBJECTIVES. Define rational exponents. 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Content Skills Learning Targets Assessment Resources & Technology St. Michael-Albertville Middle School East Teacher: Dawn Tveitbakk Grade 8 Math September 2014 UEQ: (new) CEQ: WHAT IS THE LANGUAGE OF ALGEBRA? HOW ARE FUNCTIONS USED? HOW CAN ALGEBRA BE USED TO SOLVE ### parent ROADMAP MATHEMATICS SUPPORTING YOUR CHILD IN GRADE FIVE TM parent ROADMAP MATHEMATICS SUPPORTING YOUR CHILD IN GRADE FIVE 5 America s schools are working to provide higher quality instruction than ever before. The way we taught students in the past simply does ### EQUATIONS. Main Overarching Questions: 1. What is a variable and what does it represent? EQUATIONS Introduction to Variables, Algebraic Expressions, and Equations (2 days) Overview of Objectives, students should be able to: Main Overarching Questions: 1. Evaluate algebraic expressions given ### 8-6 Radical Expressions and Rational Exponents Index 8-6 Radical Expressions and Rational Exponents Numbers and Types of Real Roots Case Roots Examples Finding Real Roots Find all real roots. 1a. fourth roots of 256 1b. sixth roots of 1 1c. cube roots ### Section 1.9 Algebraic Expressions: The Distributive Property Section 1.9 Algebraic Expressions: The Distributive Property Objectives In this section, you will learn to: To successfully complete this section, you need to understand: Apply the Distributive Property. ### New York State Mathematics Content Strands, Grade 6, Correlated to Glencoe MathScape, Course 1 and Quick Review Math Handbook Book 1 New York State Mathematics Content Strands, Grade 6, Correlated to Glencoe MathScape, Course 1 and The lessons that address each Performance Indicator are listed, and those in which the Performance Indicator ### SQUARES AND SQUARE ROOTS 1. 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For example, 5 6 = 6 5 = 30, since: 5 6=6+6+6+6+6=30 6 5=5+5+5+5+5+5=30 To develop a rule ### 7 th \$Grade\$ \$Summer\$Math\$Packet\$ 13 Write, read, and evaluate expressions in which letters stand for numbers. 13a Write expressions that record operations with numbers and with letters standing for numbers. Objective: Write an algebraic ### Reteaching. Properties of Operations - Properties of Operations The commutative properties state that changing the order of addends or factors in a multiplication or addition expression does not change the sum or the product. Examples: 5 ### parent ROADMAP MATHEMATICS SUPPORTING YOUR CHILD IN GRADE SIX TM parent ROADMAP MATHEMATICS SUPPORTING YOUR CHILD IN GRADE SIX 6 America s schools are working to provide higher quality instruction than ever before. The way we taught students in the past simply does
###### Norm Prokup Cornell University PhD. in Mathematics Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule. #### Next video playing in 10 Introduction to Polar Coordinates - Problem 3 # Introduction to Polar Coordinates - Problem 2 Norm Prokup ###### Norm Prokup Cornell University PhD. in Mathematics Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule. Share We're plotting points in a polar coordinate system, and changing from polar coordinates to rectangular. Here is my first example. Let's plot the point A which is 4, 2 pi over 3. Now the first thing I want to do is, face the direction 2 pi over 3. So imagine you're standing on the pole, and you want to turn into the direction 2 pi over 3, which is this direction. You want to walk forward four units 1, 2, 3, 4 and that's your point A; 4, 2 pi over 3. I'll also draw a little segment connecting it to the pole. That makes it easier to see the angle 2 pi over 3. Let's find the coordinates of this point. Now the rectangular coordinates, I can find them using these two formulas here; x equals r cosine theta, and y equals r sine theta. If I know r and theta, I can get x and y using these. So first the x coordinate. X equals r, which is 4, cosine theta, which is 2 pi over 3. Now the cosine of 2 pi over 3 is -1/2. So it's 4 times -1/2, which is -2. The y coordinate is r, which is 4, times the sine of theta which is 2 pi over 3. The sine of 2 pi over 3 is root 3 over 2. So this is 4 times root 3 over 2 which is 2 root 3. So I have -2, and 2 root 3. These are my rectangular coordinates. So polar coordinates, rectangular coordinates. Let's try point B. Point B has coordinates r is 4, and theta is negative pi over 3. So going to my polar graph, I face the direction negative pi over 3 which is negative pi over 6, 2 pi over 6, negative pi over 3 is this direction. I want to go 4 in that direction, so 1, 2, 3, 4. So this is my point B; 4, negative pi over 3. Now you may notice that this point is a reflection of this point around the origin. It's a 180 degree around the origin from point A. That means that its x and y coordinates, are going to be the opposites of this guy. If this guy has an x coordinate of -2, B is going to have an x coordinate of +2 by symmetry. This guy has a y coordinate of 2, root 3, so this guy has a y coordinate of -2, root 3 by symmetry. So our coordinates are 2, -2 root 3. Those are our rectangular coordinates. Now let's do point C. Point C is -4, pi over 3. So r is -4, theta is pi over 3. Pi over 3, that's pi over 6, that's pi over 3. So I have to go -4 in this direction which means walk backwards -1, -2, -3, -4. This is point C; -4, pi over 3. Again, you see that there is symmetry with this point, and the others. Suppose I use the symmetry with point B, how are the rectangular coordinates of these two points going to be related? It looks like they are on the same y value, but opposite x value. So let me use that. The x value for B was 2, so the x value for this guy since it's opposite will be -2. The y value was -2, root 3, and this guy should have the same y value. So the rectangular coordinates are -2, -2 root 3. That's it. You can often use symmetry to get rectangular coordinates from polar coordinates, as long as your points are symmetrical. In the future, we'll show you some tricks for finding the symmetries between points that are given polar coordinates.

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