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b b 2 c b g c c Title Beginning Algebra Tutorial 35: Reasoning Skills Learning Objectives After completing this tutorial, you should be able to: Use inductive reasoning to solve problems. Find the next terms in a sequence. Use deductive reasoning to solve problems. Introduction In this tutorial we will be looking at basic concepts of reasoning skills. We will be looking at both deductive and inductive reasoning. One thing that this can be helpful with is looking for patterns. Looking for patterns to find a solution can be found in a variety of fields. Teachers can use patterns to determine a course of direction for a student. For example, if a student is exhibiting the same kind of learning pattern that a teacher has seen in a student with dyslexia before, they can act upon that accordingly. Psychologists and law enforcement study behavioral patterns to solve some of their problems. Scientific researchers study patterns to determine end results in their experiments. Doctors and Veterinarians use patterns to help diagnose a patient's illness. Weather forecasters use patterns in weather to predict temperature, tornadoes, hurricanes, etc. In fact some aspects of weather forecasting uses Chaos Theory - the science of seeing order and pattern where formerly only the random, the erratic, and the unpredictable had been observed. Patterns of all kinds are lurking everywhere around us. I think you are ready to forge ahead into the wonderful world of reasoning skills. Tutorial Inductive Reasoning Inductive reasoning is used when you need to draw a general conclusion from specific instances. For example, when a detective puts together specific clues to solve a mystery. Looking for a Pattern (Sequences) Example 1: Write the next three numbers in the sequence 5, 7, 11, 17, 25, ... My first inclination is to see if there is some pattern in addition. Well, we are not adding the same number each time to get to the next number. But, it looks like we have 5 +2, 7 +4, 11 +6, 17 +8, 25, .... I see a pattern in addition - do you see it? We are always adding the next even number. Final Answer: The pattern is to add the next even number. The next three terms would have to be 35, 47, and 61, since 25+10 = 35, 35 +12 = 47, 47 +14 = 61. Example 2: Write the next three numbers in the sequence 7, -7, 14, -42, 168, ... Since we are bouncing back and forth between positive and negative numbers, a pattern in addition doesn't look promising. Let's check out multiplication. At first glance, I would say that a negative number is probably what we are looking for here, since it does alternate signs. It doesn't appear to be the same number each time, because 7 times -1 is -7, but -7 times -2 equals 14. It looks like we have 7 (-1), -7 (-2), 14 (-3), -42 (-4), 168, ... Aha, we have a pattern in multiplication - we are multiplying by the next negative integer. Final Answer: The pattern is multiplying by the next negative integer. The next three terms are -840, 5040, and -35280, since 168(-5) = -840, -840(-6) = 5040, 5040(-7) = -35280. Example 3: Write the next three numbers in the sequence 100, 97, 88, 61, ... Since the numbers are decreasing that should tell you that you are not adding a positive number or multiplying. So we want to check out subtraction or division. At first glance it looks like it is some pattern in subtraction. We are not subtracting by the same number each time. We have 100 -3, 97 -9, 88 -27, 61, .... Note how we are always subtracting the next power of 3. We have our pattern. Final Answer: The pattern is we are subtracting by the next power of three. The next three terms would be -20, -263, and -992, since 61 - 81 = -20, -20 - 243 = -263, -263 - 729 = -992. Looking for a Pattern Involving Figures Here are some things to look for when trying to figure out a pattern involving figures: Look for counter clockwise and clockwise changes. Count sides of figures. Count lines in figures. Note changes in direction and figures. As with the numeric patterns, this is not all the possible types of patterns involving figures. However, it does give you a way to approach the problem. Example 4: Write the next three figures in the pattern ... It looks like several things change throughout the pattern. One thing is that it alternates between a square with a line in it and a circle. Also the line in the square alternates from horizontal to vertical. With all of that in mind, I believe the next three figures would be a square with a vertical line, then a circle, then a square with a horizontal line: Example 5: Write the next two figures in the pattern ... It looks like one row of asterisks is added at the bottom of each figure. The row that is added contains the next counting number of asterisks. There are 2 in the row added in the second term, there are 3 in the row added in the 3rd term and 4 in the row added to the fourth term. With all of that in mind, I believe the next two figures would be Deductive Reasoning Deductive reasoning is used when you have a general rule and you want to draw on that to get a specific solution. For example, if you were needing to find the area of a specific rectangle. You would use the general formula for the area of the rectangle and apply it to the specific rectangle. Here are some ideas that might help you approach a problem requiring deductive reasoning: Watch for key words like no or all. Use process of elimination. Draw a picture or a diagram if it helps. Example 6: Use the statements below to answer the question that follows: 1. All people wearing hats have blonde hair. 2. Some of the people have red hair. 3. All people who have blonde hair like hamburgers. 4. People who have red hair like pizza. 5. Keith has blonde hair. Which of the following statements MUST be true? a. Keith likes hamburgers. b. Keith has red hair. c. Keith likes pizza. d. Keith is wearing a hat. Well what do you think? On deductive reasoning, you need to be a 100% sure. There can’t be any doubt. Since statement 3 says that ALL people who have blonde hair like hamburgers and Keith has blonde hair, then statement a, Keith likes hamburgers, is a 100% guarantee. Example 7: Jerry, Kevin, Todd and Mark all live on the first floor of an apartment complex. One is a manager, one is a computer programmer, one is a singer, and the other is a teacher. Use the statements below to answer the question that follows. A. Jerry and Todd eat lunch with the singer. B. Kevin and Mark carpool with the manager. C. Todd watches CSI with the manger and the singer. Question: Which is the manager? You can use a process of elimination on this problem. Statement A, Jerry and Todd eat lunch with the singer, doesn’t let us definitively eliminate anyone from being the manager. However, statement B, Kevin and Mark carpool with the manager, eliminates Kevin and Mark from being the manager. And statement C, Todd watches CSI with the manger and the singer, eliminates Todd. The only one that could be (100%, without a doubt) the manager is Jerry. Practice Problems These are practice problems to help bring you to the next level. It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it. Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. In fact there is no such thing as too much practice. To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer. Practice Problems 1a - 1c: Write the next three numbers in the sequence. 1a. 1, 1, 3, 15, 105, ... (answer/discussion to 1a) 1b. 1000, 200, 40, 8, 1.6, ... (answer/discussion to 1b) 1c. 5, 5, 10, 15, 25, ... (answer/discussion to 1c) Practice Problem 2a: Write the next five figures in the pattern. Practice Problem 3a: Four friends - Suzy, John, Sally, and Tom - each has his or her own hobby. One collect coins, one sews, one cooks, and one plays in a band, not necessarily in that order. Use the statements below to answer the question that follows. 3a. A. Suzy and John always eat lunch with the friend that plays in the band. B. Sally and Tom carpool with the one who likes to sew. C. John and the friend that likes to cook visited the one who likes to sew. Question: Who is the friend that likes to sew? (answer/discussion to 3a) Need Extra Help on these Topics? Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions. Last revised on August 7, 2011 by Kim Seward.
## Geometry: Common Core (15th Edition) $x = 8$ $y = 9$ $z = 5.25$ Similar polygons have congruent angles and proportional sides. If these two quadrilaterals are similar, we would expect the sides to be similar. We are given definite values for similar sides in the two quadrilaterals, so let us find the scale factor for the two quadrilaterals: scale factor = $\frac{9}{6}$ Divide both the numerator and denominator by their greatest common factor, $3$, to simplify: scale factor = $\frac{3}{2}$ Now we can use this scale factor to set up proportions for the other sides to find the values of $x$, $y$, and $z$. Let's set up the proportion to find $x$ first: $\frac{12}{x} = \frac{3}{2}$ Get rid of the fractions by using the cross products property to multiply the numerator of one fraction with the denominator of the other fraction, and vice versa: $3x = 24$ Divide each side by $3$ to solve for $x$: $x = 8$ Now, let's set up the proportion to find $y$: $\frac{y}{6} = \frac{3}{2}$ Get rid of the fractions by using the cross products property to multiply the numerator of one fraction with the denominator of the other fraction, and vice versa: $2y = 18$ Divide each side by $2$ to solve for $y$: $y = 9$ Now, let's set up the proportion to find $z$: $\frac{z}{3.5} = \frac{3}{2}$ Get rid of the fractions by using the cross products property to multiply the numerator of one fraction with the denominator of the other fraction, and vice versa: $2z = 10.5$ Divide each side by $2$ to solve for $z$: $z = 5.25$
# word problems on simultaneous linear equations \| practice set 1.5 algebra 10th In this article word problems on simultaneous linear equations we give the simultaneous equations word problems between practice set 1.5 algebra 10th. also we give simultaneous equations examples. ### word problems on simultaneous linear equations: Ex. 1) The perimeter of a rectangle is 80 cm. The length of the rectangle is more than double it’s breadth by 4. Find length and breadth. ##### Solution : Let length of rectangle be l cm and breadth be b cm. From the first condition – 2 ( l + b ) = 80 l + b = 40 —– I From the second condition – l = 2b + 4 l – 2b = 4 —— II from eq. ( I ) and ( II ) Subtract eq. ll from eq. l l + b = 40 l – 2b = 04 – + – ————– 3b = 36 b = 12 put b = 12 in eq. I l + b = 40 l + 12 = 40 l = 40 – 12 l = 28 therefore, Length of the rectangle is 28 cm and breadth is 12 cm. #### word problems on simultaneous linear equations: Ex. 2) A certain amount is equally distributed among certain number of students. Each would get rs. 4 less if 20 students where more and each would get rs. 12 more if 30 students where less. calculate the number of student and the amount distributed. ##### Solution : Let the number of students be a and amount given to each student by rs. b therefore, Total amount distributed is ab From the first condition we get, ( a + 20 ) ( b – 4 ) = ab ab – 4a + 20b – 80 = ab – 4a + 20b = 80 divided both side by 4, we get – a + 5b = 20 —– l from the second condition we get, ( a – 30 ) ( b + 12 ) = ab ab + 12a – 30b – 360 = ab 12a – 30b = 360 divided both side by 6, we get 2a – 5b = 60 —– ll eq. ( l ) + eq. ( ll ) – a + 5b = 20 2a – 5b = 60 —————– a = 80 put a = 80 in equation ( l ) – a + 5b = 20 -80 + 5b = 20 5b = 100 b = 20 Total amount distributed is = ab = 80 × 20 = 1600 therefore, rs. 1600 distributed equally among 80 students. Ex. 3) Two numbers differ by 3. The addition of twice the smaller number and thrice the greater number is 19. Find the numbers. ##### Solution: Let the numbers are p and q from the first condition , we get p – q = 3 —– l from the second condition , we get 3p + 2q = 19 —- ll eq. ( l ) multiply by ( 2 ) , we get 2p – 2q = 6 —- lll adding eq. ( ll ) and ( lll ) , we et 5p = 25 p = 5 put p = 5 in eq. ( l ) 5 – q = 3 – q = 3 – 5 – q = – 2 q = 2 Therefore, The numbers are 5 and 2. Ex. 4) The sum of father’s age and twice the age of his son is 70, if we double the age of the father and add it to the age of his son the sum is 95. Find their present ages. Suppose the present age of father be p and son be q for the first condition p + 2q = 70 —– I for the second condition 2p + q = 95 —- II equation ( I ) multiply by 2, we get 2p + 4q = 140 —– III substract equation ( II ) from ( III ) 2p + 4q = 140 2p + q = 95 – – = – —————– 3q = 45 q = 15 put q = 15 in equation ( I ) P + 2 ( 15 ) = 70 p + 30 = 70 p = 70 – 30 p = 40 therefore, father’s present age 40 years and son present age is 15 years. Ex. 5) The denominator of fraction is 4 more than twice it’s numerator. Denominator becomes 12 times the numerator, if both the numerators and denominator are reduced by 6. Find the fraction. Suppose the fraction be p / q from the first condition, Denominator = ( 2 × numerator ) + 4 that is q = 2 p + 4 – 2p + q = 4 —– I from the second condition, ( Numerator – 6 ) / ( Denominator – 6 ) = 1 / 12 that’s ( p – 6 ) / ( q – 6 ) = 1 / 12 12 ( p – 6 ) = 1 ( q – 6 ) 12 p – 72 = q – 6 12p – q = – 6 + 72 12 p – q = 66 —– II eq. ( I ) + eq. ( II ) , we get 10p = 70 therefore, p = 70 / 10 p = 7 put p= 7 in equation ( I ) , we get -2 ( 7 ) + q = 4 -14 + q = 4 q = 4 + 14 q = 18 therefore, the fraction = p / q = 7 / 18. #### Simultaneous linear equations: Ex. 6) Two type of boxes P and Q are to be placed in the truck having capacity of 10 tons. when 150 boxes of type P and 100 boxes of type Q are loaded in the truck, it weights 10 tons but when 260 boxes of type P loaded in the truck, it can still accommodate 40 boxes of type Q, so that it is fully loaded. Find the weight of each type of box. [ 1 ton = 1000 kg ] Let the weight of P type box is a kg. and the weight of Q type box is b kg From the first condition, 150a + 100b = 10 tons that is, 150a + 100b = 10000 15a + 10b = 1000 3a + 2b = 200 —— I from the second condition, 260a + 40b = 10000 26a + 4b = 1000 13a + 2b = 500 —- II Substract equation ( I ) from equation ( II ) 13a + 2b = 500 3a + 2b = 200 – – = – —————— 10a = 300 a = 300 / 10 a = 30 kg put a = 30 in equation I 13 ( 30 ) + 2b = 500 390 + 2b = 500 2b = 500 – 390 2b = 110 b = 55kg. therefore, the weight of P type box is 30kg. and the weight of Q type box is 55kg. Ex. 7) Out of 1900km, Vishal travelling some distance by bus and some distance by airplane. Bus travels with average speed 60km/he and the average speed of airplane is 70 km/hr. It take 5 hours to complete the journey. Find the distance Vishal travelled by bus. ##### Ans. : Let Vishal travelled p km. by bus and q km. by airplane. from the first condition, p + q = 1900 —– I Speed of bus = 60 km/hr and speed of Airplane = 700 km/hr Time taken by bus to travel p km. but, Speed = ( Distance / Time ) 60 = p / T1 T1 = p / 60 Time taken by airplane to travel q km Time = ( Distance / Speed ) T2 =q / 700 from the second condition, T1 + T2 = 5 ( p / 60 ) + ( q / 700 ) = 5 multiply both side by 10, we get ( p / 6 ) + ( q + 70 ) = 50 ( 70 p / 420 ) + ( 6q / 420 ) = 50 70p + 6q = 21000 —— II multiply equation ( I ) by 6, we get 6p + 6q = 11400 ——- III Substract equation ( III ) from equation ( II ) 70p + 6q = 21000 6p + 6q = 11400 – – – ———————– 64p = 9600 p= 9600 / 64 p = 1200 / 8 p = 150 put p = 150 in equation ( I ), we get p + q = 1900 150 + q = 1900 q = 1900 – 150 q = 1750 Therefore, Vishal travelled by bus = p km. = 150 km. and Vishal travelled by Airplane = q km. = 1750 km.
# How do I solve the Hannahs sweets question from the 2015 GCSE paper? • 904 views Here is a question from the 2015 edexcel GCSE paper that was considered particulary challenging. There are n sweets in a bag. 6 of the sweets are orange. The rest of the sweets are yellow. Hannah takes a random sweet from the bag. She eats the sweet. Hannah then takes at random another sweet from the bag. She eats the sweet. The probability that Hannah eats two orange sweets is 1/3. Show that n² – n – 90 = 0. I will go through how to solve this question step by step. Firstly we note that we have 6 orange sweets, and as we have n sweets in total therefore at the start we have a 6/n chance of choosing an orange sweet, because 6/n of the total sweets are orange. After Hannah has eaten an orange sweet the probability changes. We now have n-1 sweets left in the bag, 5 of which are orange, since one orange sweet is missing from the bag. therefore the probablity the second sweet is orange, after one orange sweet has already been eaten, is 5/(n-1). Therefore by the laws of probability the total probability that Hannah has chosen 2 orange sweets is 6/n x 5/(n-1). This is because when we want the probability that two events happen (in this case event one is that the first sweet is orange, event two is that the second sweet is orange) we multiply together the probability that each event occurs. Now we know that the probability is equal to 1/3 from the question so now we can set up an equation 6/n x 5/(n-1) = 1/3 times together the denominators and numerators 30/(n(n-1)) = 1/3 expand through the brackets 30/(n2-n) = 1/3 times both sides by 3 90/(n2-n) = 1 multiply each side by n2-n 90 = n2-n Now simply rearrange to find the solution we are seeking n-n - 90 = 0 We have now solved part a of the question. As a bonus, if we wanted to solve this quadratic equation this is how we would do it: A quadratic fuction is of the form an2+bn+c In our example a=1 b=-1 and c=-90 We now want to find two numbers y and z such that x+y=b=-1 and xy=c=-90. We can start by listing factor pairs of 90 until we find a pair with a difference of -1 -90 + 1 = -89 -45 +2 = -43 -30 + 3 = -27 -18 +5 = -13 -15 + 6 = -9 -10 + 9 = -1 We have found our pair Therefore (n-10)(n+9) = n-n - 90 = 0 So if (n-10)(n+9)=0 either (n-10) =0 ie n=10 or (n+9) = 0 ie n=-9. So 10 and -9 are our solutions. However in this case we know that we cannot have a negative number of sweets in a bag therefore we deduce that there are 10 sweets in the bag. Still stuck? Get one-to-one help from a personally interviewed subject specialist. 95% of our customers rate us We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this.
# AP Statistics : How to find the mean for a set of data ## Example Questions ### Example Question #1 : How To Find The Mean For A Set Of Data A sample consists of the following observations:. What is the mean? Explanation: The mean is ### Example Question #7 : Univariate Data Six homes are for sale and have the following dollar values in thousands of dollars: 535 155 305 720 315 214 What is the mean of the six homes? Explanation: The mean is calculated by adding all the values in a group together, then dividing the sum by the total number in the group. In this case, six homes are for sale. The six home values are added together , then that value is divided by six. ### Example Question #8 : Univariate Data A bird watcher observed how many birds came to her bird feeder for four days. These were the results: Day 1: 15 Day 2: 12 Day 3: 10 Day 4: 13 What is the mean number of birds? Explanation: The mean is calculated by adding the values from the four days together, then dividing the sum by the total number of values. In this case, four values must be added: ### Example Question #9 : Univariate Data What is the mode of the following list of numbers? 35, 17, 4, 25, 7, 4, 17, 26, 8, 17 Explanation: The mode is the number that appears most frequently in a series of numbers. First, organize the numbers in order from least to greatest: 4, 4, 7, 8, 17, 17, 17, 25, 26, 35. The number 17 appears three times, more than any other number. ### Example Question #2 : How To Find The Mean For A Set Of Data Find the mean of the following set. Explanation: To find the mean of a set, follow the formula in this case or ### Example Question #3 : How To Find The Mean For A Set Of Data A business tracked the number of customer calls received over a period of five days. What was the mean number of customer calls received in a day? Day 1: 57 Day 2: 63 Day 3: 48 Day 4: 49 Day 5: 59 Explanation: The mean is calculated by adding all the values in a data set, then dividing the sum by the total number in the group. ### Example Question #4 : How To Find The Mean For A Set Of Data Number Crunchers Inc. has offices in New York and Wisconsin. The mean salary for office workers in New Work is $28,500. The mean salary for office workers in the Wisconsin office is$22,500. The New York office has 128 office workers and the Wisconsin office has 32 office workers. What is the mean salary paid to the office workers in Number Crunchers Inc.?
# HBSE 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.5 Haryana State Board HBSE 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.5 Textbook Exercise Questions and Answers. ## Haryana Board 8th Class Maths Solutions Chapter 4 Practical Geometry Exercise 4.5 Question 1. Draw the following : 1. The square READ with RE = 5.1 cm. 2. A rhombus whose diagonals are 5.2 cm and 6.4 cm long. 3. A rectangle with adjacent sides of lengths 5 cm and 4 cm. 4. A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. Solution: (1) Rough Sketch: Steps of Construction: (i) Draw RE = 5.1 cm. (ii) Make ∠YRE = 90° and ∠REX = 90° (iii) Cut off DR = AE = 5.1 cm. (iv) Join DA. (v) DEAR is a required square. (2) Rough Sketch: Theory : Diagonals of a rhombus bisect each other at 90° i.e., OA = OC = $$\frac{6.4}{2}$$ = 3.7 cm and OD = OB = $$\frac{5.2}{2}$$ = 2.6 cm ∠AOB = 90° Steps of Construction : (i) Draw BD = 5.2 cm. (ii) Draw perpendicular bisector XY of BD. (iii) Cut off OA = OC = 3.7 cm on the opposite side of DB. (iv) Join AB, AD, CD and BC. (v) ABCD is a required rhomus. (3) Rough Sketch : Theory : A rectangle has equal opposite sides and each angles are 90°. i.e. AB = DC = 5 cm DA = CB = 4 cm ∠A = ∠B = ∠C = ∠D = 90° Steps of Construction : (i) Draw a line segment AB = 5 cm. (ii) Construct ∠A = ∠B = 90°. (iii) Cut off AD = BC = 4 cm. (iv) Join DC. (v) You have a required rectangle ABCD. 4. Theory: Opposite sides of a parallelogram are equal and sum of a adjacent angle is 180°. i.e. OK = YA = 4.2 c.m. and OY = KA = 4.2 cm. Steps of Construction : (i) Draw OK = 5.5 cm (ii) Let ∠K = 80°, so construct ∠OKX = 80° (iii) Cut of FKA = 4.2 cm in KX. (iv) Let the point be A. (v) Draw arcs of 5.5 cm and 4.2 cm with centre as A and O respectively. Note : You may take one angle of any measurement.
# Problem of the Week Problem E and Solution A New Pair of Dice ## Problem A standard six-sided die has its faces marked with the numbers $$1,~2,~3,~4,~5,$$ and $$6$$. The die is fair, which means that when it is rolled each of its faces has the same probability of being the top face. When two standard six-sided dice are rolled and the numbers on the top faces are added together, the sums range from $$2$$ to $$12$$. Noemi creates two special six-sided dice that are also fair, but have non-standard numbers on their faces. Numbers on these special dice are positive integers and may appear more than once. The largest number on one of her special dice is $$8$$. When the two special dice are rolled and the numbers on the top faces are added together, the sums range from $$2$$ to $$12$$ and the probability of obtaining each sum is the same as it would be if two standard dice had been used. Determine all possible pairs of special dice that Noemi could have created. ## Solution We first examine what happens when two standard dice are rolled. To do so, we create a table where the columns show the possible numbers on the top face for one die, the rows show the possible numbers on the top face for the other die, and each cell in the body of the table gives the sum of the corresponding numbers. Die $$\mathbf{1}$$ Die 2 $$\mathbf{1}$$ $$\mathbf{2}$$ $$\mathbf{3}$$ $$\mathbf{4}$$ $$\mathbf{5}$$ $$\mathbf{6}$$ $$\mathbf{1}$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$\mathbf{2}$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$\mathbf{3}$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$\mathbf{4}$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$\mathbf{5}$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$\mathbf{6}$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$12$$ From this table we can determine the probability of each sum by counting the number of times each sum appears in the table, and dividing by $$36$$, the total number of possible outcomes. Sum $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$12$$ Probability $$\dfrac{1}{36}$$ $$\dfrac{2}{36}$$ $$\dfrac{3}{36}$$ $$\dfrac{4}{36}$$ $$\dfrac{5}{36}$$ $$\dfrac{6}{36}$$ $$\dfrac{5}{36}$$ $$\dfrac{4}{36}$$ $$\dfrac{3}{36}$$ $$\dfrac{2}{36}$$ $$\dfrac{1}{36}$$ We need the pair of special dice to give these same probabilities when rolled together. Since the dice each have $$6$$ sides, the total number of possible outcomes will still be $$6\times 6 = 36$$. We need the smallest possible sum to be $$2$$. The only way to get a sum of $$2$$ is if each die has a $$1$$ on it. Thus, the smallest number on each die is $$1$$. Furthermore, we need the probability that a sum of $$2$$ is rolled to be $$\frac{1}{36}$$. Since there are $$36$$ possible outcomes, this means that we need exactly $$1$$ way to get a sum of $$2$$. Therefore, exactly one face on each die must have a $$1$$. The largest possible sum must be $$12$$ and we need the probability that this sum is rolled to be $$\frac{1}{36}$$. This means that we need exactly $$1$$ way to get a sum of $$12$$. Since the largest number on one die is $$8$$, it follows that the other die must have exactly one $$4$$. Furthermore, the die with the $$8$$ must contain exactly one $$8$$ and the die with the $$4$$ must have exactly one $$4$$ and no larger number. Thus, we know that the numbers on one die, from least to greatest, must be $$(1$$, , , , , $$4)$$ and the numbers on the other die, from least to greatest, must be $$(1$$, , , , , $$8)$$, where $$\underline{\hspace{5mm}}$$ represents a number we still have to determine. On the die where the largest number is $$4$$, the remaining numbers must be all $$2$$s and $$3$$s. Since we need the probability that a sum of $$3$$ is rolled to be $$\frac{2}{36}$$, it follows that there cannot be more than two $$2$$s on this die, otherwise there would be more than $$2$$ ways to get a sum of $$3$$. Similarly, we need the probability that the sum of $$11$$ is rolled to be $$\frac{2}{36}$$. Since $$8+3=11$$, it follows that there can not be more than two $$3$$s on this die, otherwise there would be more than $$2$$ ways to get a sum of $$11$$. Thus, on the die where the largest number is $$4$$, we have concluded that there is exactly one $$1$$, no more than two $$2$$s, no more than two $$3$$s, and exactly one $$4$$. It follows that the numbers on that die must be $$(1,~2,~2,~3,~3,~4)$$. We will now find possible numbers for the other die. We need the probability that a sum of $$4$$ is rolled to be $$\frac{3}{36}$$. This means that there must be exactly $$3$$ ways to get a sum of $$4$$. Currently we have $$2$$ ways, so we need $$1$$ more. If the die with the $$8$$ has a $$2$$ on a face, then there will be $$2$$ more ways to get a sum of $$4$$. Since we know that this die has exactly one $$1$$, then it must have a $$3$$ on it. Then we will have $$3$$ ways to get a sum of $$4$$, as desired. Thus, we know that the numbers on the second die must be $$(1,~3$$, , , , $$8)$$. We need the probability that a sum of $$10$$ is rolled to be $$\frac{3}{36}$$. That means that there must be exactly $$3$$ ways to get a sum of $$10$$. Currently we have $$2$$ ways, so we need $$1$$ more. If we had a $$7$$ then we would have $$2$$ more ways to get a sum of $$10$$, which is too many. Thus, we must have one $$6$$ on the die so that we can have $$3$$ ways to get a sum of $$10$$, as desired. Thus, we know that the numbers on the second die must be $$(1,~3$$, , , $$6,~8)$$. We need the probability that a sum of $$5$$ is rolled to be $$\frac{4}{36}$$. That means that there must be exactly $$4$$ ways to get a sum of $$5$$. Currently we have $$3$$ ways, so we need one more. Thus, we must have one $$4$$ on the die so that we can have $$4$$ ways to get a sum of $$5$$, as desired. Thus, we know that the numbers on the second die must be $$(1,~3,~4$$, , $$6,~8)$$. Since we canâ€™t add another number that is currently on the die without changing the probability of rolling a sum we have already looked at, it follows that the remaining number must be a $$5$$. Thus, it must be the case that the numbers on the first die are $$(1,~2,~2,~3,~3,~4)$$ and the numbers on the second die are $$(1,~3,~4,~5,~6,~8)$$. We now need to check that this pair of dice satisfies the conditions of the problem. To do so, we create a table where the columns show the possible numbers on the top face for one die, the rows show the possible numbers on the top face for the other die, and each cell in the body of the table gives the sum of the corresponding numbers. Special Die $$\mathbf{1}$$ Special Die 2 $$\mathbf{1}$$ $$\mathbf{2}$$ $$\mathbf{2}$$ $$\mathbf{3}$$ $$\mathbf{3}$$ $$\mathbf{4}$$ $$\mathbf{1}$$ $$2$$ $$3$$ $$3$$ $$4$$ $$4$$ $$5$$ $$\mathbf{3}$$ $$4$$ $$5$$ $$5$$ $$6$$ $$6$$ $$7$$ $$\mathbf{4}$$ $$5$$ $$6$$ $$6$$ $$7$$ $$7$$ $$8$$ $$\mathbf{5}$$ $$6$$ $$7$$ $$7$$ $$8$$ $$8$$ $$9$$ $$\mathbf{6}$$ $$7$$ $$8$$ $$8$$ $$9$$ $$9$$ $$10$$ $$\mathbf{8}$$ $$9$$ $$10$$ $$10$$ $$11$$ $$11$$ $$12$$ From this table we can determine the probability of each sum by counting the number of times each sum appears in the table, and dividing by $$36$$, the total number of possible outcomes. It turns out that these probabilities match the probabilities when two standard dice are rolled, so we can conclude that there is only one possible pair of special dice that Noemi could have created. The numbers on this pair of dice must be $$(1,~2,~2,~3,~3,~4)$$ and $$(1,~3,~4,~5,~6,~8)$$.
Ir al contenido Operaciones con matrices SUMA Y RESTA DE MATRICES LA SUMA, A + B, dos matrices A y C del mismo tamaño se obtiene sumando los elementos de ambas matrices. Para la RESTA, A - B, se restan les elementos correspondientes. Las matrices de distintos tamaños no se pueden sumar ni restar. • Ejemplo A= ${\displaystyle {\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}}}$ , B= ${\displaystyle {\begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{bmatrix}}}$ A + B = ${\displaystyle {\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}}}$ + ${\displaystyle {\begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{bmatrix}}}$ = ${\displaystyle {\begin{bmatrix}a_{11}+b_{11}&a_{12}+b_{12}\\a_{21}+b_{21}&a_{22}+b_{22}\end{bmatrix}}}$ . A = ${\displaystyle {\begin{bmatrix}1&3\\5&7\end{bmatrix}}}$ , B = ${\displaystyle {\begin{bmatrix}2&4\\6&8\end{bmatrix}}}$ La suma se hace componente a componente. A + B = ${\displaystyle {\begin{bmatrix}1&3\\5&7\end{bmatrix}}+{\begin{bmatrix}2&4\\6&8\end{bmatrix}}={\begin{bmatrix}1+2&3+4\\5+6&7+8\end{bmatrix}}}$ = ${\displaystyle {\begin{bmatrix}3&7\\11&15\end{bmatrix}}}$ algo mas general se puede describir como: A = ${\displaystyle {\begin{bmatrix}a_{11}&\cdots &a_{1n}\\\vdots &\ddots &\vdots \\a_{1n}&\cdots &a_{nn}\end{bmatrix}}}$ , B = ${\displaystyle {\begin{bmatrix}b_{11}&\cdots &b_{1n}\\\vdots &\ddots &\vdots \\b_{1n}&\cdots &b_{nn}\end{bmatrix}}}$ A + B = ${\displaystyle {\begin{bmatrix}a_{11}+b_{11}&\cdots &a_{1n}+b_{1n}\\\vdots &\ddots &\vdots \\a_{1n}+b_{1n}&\cdots &a_{nn}+b_{nn}\end{bmatrix}}}$ Ejemplo 2 A - B = ${\displaystyle {\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}}}$ - ${\displaystyle {\begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{bmatrix}}}$ = ${\displaystyle {\begin{bmatrix}a_{11}-b_{11}&a_{12}-b_{12}\\a_{21}-b_{21}&a_{22}-b_{22}\end{bmatrix}}}$ . La resta se hace componente a componente. A - B = ${\displaystyle {\begin{bmatrix}1&3\\5&7\end{bmatrix}}-{\begin{bmatrix}2&4\\6&8\end{bmatrix}}={\begin{bmatrix}1-2&3-4\\5-6&7-8\end{bmatrix}}}$ = ${\displaystyle {\begin{bmatrix}-1&-1\\-1&-1\end{bmatrix}}}$ algo mas general se puede describir como: A = ${\displaystyle {\begin{bmatrix}a_{11}&\cdots &a_{1n}\\\vdots &\ddots &\vdots \\a_{1n}&\cdots &a_{nn}\end{bmatrix}}}$ , B = ${\displaystyle {\begin{bmatrix}b_{11}&\cdots &b_{1n}\\\vdots &\ddots &\vdots \\b_{1n}&\cdots &b_{nn}\end{bmatrix}}}$ A - B = ${\displaystyle {\begin{bmatrix}a_{11}-b_{11}&\cdots &a_{1n}-b_{1n}\\\vdots &\ddots &\vdots \\a_{1n}-b_{1n}&\cdots &a_{nn}-b_{nn}\end{bmatrix}}}$
Select Page 2.4 Matrix Operations Matrix addition and subtraction are identical to element-by-element addition and subtraction. The corresponding matrix elements are summed or subtracted. However, matrix multiplication and division are not the same as element-by-element multiplication and division. Multiplication of Vectors Recall that vectors are simply matrices with one row or one column. Thus matrix multiplication and division procedures apply to vectors as well, and we will introduce matrix multiplication by considering the vector case first. The vector dot product u . w of the vectors u and w is a scalar and can be thought of as the perpendicular projection of u onto w. It can be computed from [ullw] cos e, where e is the angle between the two vectors and [u], [w] are. the magnitudes of the vectors. Thus if the vectors are parallel and in the same direction, e = 0 and u .w = [ullw]. If the vectors are perpendicular, e = 90° and thus u . w = 0.. Because the unit vectors j, J. and k have unit length ‘Thus the vector dot product can be expressed in terms of unit vectors as u, W = (uli + u2j + U3k)· (wli +W2j + W3k) Carrying out the multiplication algebraically and using the properties given by (2.4-1) and (2.4-2), we obtain U· W = UIWI + U2W2 + U3W3 The matrix product of a row vector u with a column vector W is “defined in the same way as the vector dot product; the tesult is a scalar that is the sum of the products of the corresponding vector elements; that is, [UI U, u,] [::] ~ UIWI + U,w, +u,W, if each vector has three elements. Thus the result of multiplying a 1 x 3 vector times a 3 x 1 vector is a 1 x 1 array; that is, a scalar. This definition applies to vectors having any number of elements, as long as both vectors have the same number of elements. Thus \ . [“I U, u, u,] [:~~lUIWI + u,W, +u,W, + … +u,W, if each vector has n elements. Thus the result of multiplying a 1 x n vector times an n x 1 vector is a 1 x 1 array, that is, a scalar. Vector-Matrix Multiplication Not all matrix products are scalars. To generalize the preceding multiplication to a column vector multiplied. by a matrix, think of the matrix as being composed of row vectors. The scalar result of each row-column multiplication forms an element in the result, which is a column vector: Thus the result of multiplying a 2 x 2 matrix times a 2 x 1 vector is a 2 x 1 array; that is, a column vector. Note that the definition of multiplication requires that the number of columns in the matrix be equal to the number of rows in the vector. In general, the product Ax, where A has p columns, is defined only if x has p rows. If A has m rows x is a column vector, the result of Ax is a column vector with m rows Matrix-Matrix Multiplication We can expand this definition of multiplication to include the product of two matrices AB. The number of columns in A must equal the number of rows in B. The row-column multiplications form column vectors, and these-column vectors form the matrix result. The product AB” has the same number of rows as A and the same number of columns as B. For example, Use the operator * to perform matrix multiplication in MATLAB. The following MATLAB session shows how to perform the matrix multiplication shown in (2.4-4). »A = [6,-2;10,3;4,7]; »B = [9,8;-5,12]; »AB ans 64 24 75 116 1 116 Element-by-element multiplication is defined for the following product: [3 1 7][4 6 5] = [12 6 35] However, this product is not defined for matrix multiplication, because the first matrix has three columns, but the second matrix does not have three rows. Thus if we were to type [3, I, 7) [ 4 , 6, 5) in MATLAB, we would receive an error message. The following product is defined in matrix multiplication and gives the result shown: The General Matrix Multiplication Case We can state the general result for matrix multiplication as follows: Suppose A has dimension m x p and B has dimension p x q, If C is the product AB, then C has dimension m x q and its elements are given by for all i = 1, 2, … , m and j = 1, 2, … , q. For the product to be defined, the matrices A and B must be conformable; that is, the number of rows in B must equal the number of columns in A. The product has the same number of rows s A and the same number of columns as B. . The algorithm defined by (2.4-5) is easy to remember. Each element in the with row of A is multiplied by the corresponding element in the with column of B. The sum of the products is the element cij. If we write the product AB in terms of the dimensions, as (m x p )(p x q) = m x q, we can easily determine the dimensions of the product by “canceling” the inner dimensions (here p), which must be equal for the product to be defined. Matrix multiplication does not have the commutative property; that is, in general, AB =I BA. A simple example will demonstrate this fact: Reversing the order of matrix multiplication is a common and easily made mistake. The associative and distributive properties hold for matrix multiplication. The associative property states that T2.4-3 Use MATLAB to verify the results of equations (2.4-6) and (2.4-7). Applications to Cost Analysis Data on costs of engineering projects are often recorded as tables. The data in these tables must often be analyzed in several ways. The elements in MATLAB matrices are similar to the cells in a spreadsheet, and MATLAB can perform many spreadsheet-type calculations for analyzing such tables. Special Matrices Two exceptions to the non-commutative property are the null matrix, denoted by 0, NULL MATRiX ” and the identity, or unity, matrix, denoted by I. The null matrix contains all zeros and is not the same as the empty matrix [], which has no elements. The identity IDENTITY MATRIX matrix is a square matrix whose diagonal elements are all equal to one, with the remaining elements equal to zero. For example, the 2 x 2 identity matrix is MATLAB has specific commands to create several special matrices. Type help specmat to see the list of special matrix commands; also check Table 2.4-4. The identity matrix I can be created with the eye (n) command, where n is the desired dimension of the matrix. To create the 2 x 2 identity matrix, you type eye (2). Typing eye (size (A) ) creates an identity matrix having the same” dimension as the matrix A. Sometimes we want to initialize a matrix to have all zero elements. The zeros command creates a matrix of all zeros. Typing zeros (n) creates an n x n matrix f zeros, whereas typing zero s (m,n) creates an m x n matrix of zeros. Typing zeros (size (A) ) creates a matrix of all zeros having the same dimension as the matrix A. This type of matrix can be useful for applications in which we do not know the required dimension ahead of time. The syntax of the ones command is the same, except that it creates arrays filled with ones. For example, to create and plot the function Table 2.4-4 Special matrices the script file is xl [O:0.01:2J; f1 10ones(size(x1)); x2 [2.01:0.01:4.99J; f2 zeros(size(x2)); x3 [5;0.01:7J; f3 -3ones(size(x3)); f = [fl, f2, f3J; x = [xl, x2, x3J; plot (x,f), xlabel (‘x’) ,ylabel (‘y’) (Consider what the plot would look like if the command plot (x , f) were . replaced with the command plot (x l ,f1, x2, f2,x3, f3) .) Matrix Division Matrix division is a more challenging topic than matrix multiplication. Matrix division uses both the right and left division operators, / and \, for various applications, a principal one being the solution of sets of linear algebraic equations.Chapter 6 covers matrix division and a related topic, the matrix inverse. Matrix Exponentiation Raising a matrix to a power is equivalent to repeatedly multiplying the matrix by itself, for example, A 2 = AA. This process requires the matrix to have the same number.of rows as columns; that is, it must be a square matrix. MATLAB uses the symbol A for matrix exponentiation. To find A 2, type AA 2. We can raise a scalar n to a matrix power A, if A is square, by typing n A A, but the applications for such a procedure are in advanced courses. However, raising a ‘ matrix to a matrix power-that is, AB-is not defined, even if A and B are square. Note that if n is a scalar and if Band C are not square matrices, then’ the following operations are not defined and will generate an error message-in MATLAB: B n n B B C Special Products Many applications in physics and engineering use the cross product and dot product-for example, calculations to compute moments and force components use these special products. If A and B are vectors with three elements, the crossproduct command cross (A, B) computes the three-element vector that is the cross-product A x B. If A and Bare 3 x n matrices, c ro ss (A,B) returns a 3 x n array whose columns are the cross products of the corresponding columns in.the 3 x n arrays A and B. For.example, the moment M with respect to a reference point 0 due to the force F is given by M := r x F, where r is the position vector from the point 0 to the point where the force F is applied. To find the moment in MATLAB, you type M = cross (r, F). The dot-product command dot (A,B) computes a row vector of length n whose elements are the dot products of the corresponding columns of the m x n arrays A and B. To compute the component of the force F along the direction given by the vector r, you type dot (F, r) . Table 2.4-5 summarizes the dot- and cross-product commands.
# Lesson 6 Completing the Square • Let’s rewrite equations to find the center and radius of a circle. ### 6.1: Fill in the Box For each expression, what value would need to be in the box in order for the expression to be a perfect square trinomial? 1. $$x^2+10x+\boxed{\phantom{3}}$$ 2. $$x^2-16x+\boxed{\phantom{3}}$$ 3. $$x^2+40x+\boxed{\phantom{3}}$$ 4. $$x^2+5x+\boxed{\phantom{3}}$$ ### 6.2: Complete the Process Here is the equation of a circle: $$x^2+y^2-6x-20y+105 = 0$$ Elena wants to find the center and radius of the circle. Here is what she’s done so far. Step 1: $$x^2-6x+y^2-20y = \text-105$$ Step 2: $$x^2-6x+9+y^2-20y +100= \text-105+9+100$$ Step 3: $$x^2-6x+9+y^2-20y +100= 4$$ 1. What did Elena do in the first step? 2. Why did Elena add 9 and 100 to the left side of the equation in Step 2? 3. Why did Elena add 9 and 100 to the right side of the equation in Step 2? 4. What should Elena do next? 5. What are the center and radius of this circle? 6. Draw a graph of the circle. Here is the equation of a circle: $$x^2+y^2-2x+4y-4=0$$ 1. Find the center and radius of the circle. Explain or show your reasoning. 2. Draw a graph of the circle. Triangulation is a process of using 3 distances from known landmarks to locate an exact position. Find a point that is located 5 units from the point $$(3,4)$$, 13 units from the point $$(11,\text-4)$$, and 17 units from the point $$(21,16)$$. Use the coordinate grid if it’s helpful. ### Summary Here is an equation for a circle: $$x^2+y^2-4x+6y-3=0$$. If we want to find the center and radius of the circle, we can rewrite the equation in the form $$(x-h)^2+(y-k)^2=r^2$$. Start by rearranging the terms in the equation to make it easier to work with. Group terms that include the same variable and move the -3 to the right side of the equation. $$x^2-4x+y^2+6y=3$$ We want the left side to include 2 perfect square trinomials—then, those trinomials can be rewritten in factored form to get the equation in the form we need. To create perfect square trinomials, we can add values to the left side. We’ll keep the equation balanced by adding those same values to the other side. $$x^2-4x+\boxed{\phantom{3}}+y^2+6y+\boxed{\phantom{3}}=3+\boxed{\phantom{3}}+\boxed{\phantom{3}}$$ For the expression $$x^2-4x$$, we need to add 4 to get a perfect square trinomial. For the expression $$y^2+6y$$, we need to add 9. Add these values to both sides of the equation. Then, combine the numbers on the right side. $$x^2-4x+4+y^2+6y+9=3+4+9$$ $$x^2-4x+4+y^2+6y+9=16$$ Now rewrite the perfect square trinomials as squared binomials, and write the 16 in the form $$r^2$$. $$(x-2)^2+(y+3)^2=4^2$$ The circle has center $$(2,\text-3)$$ and radius 4 units.
Suggested languages for you: \| \| ## All-in-one learning app • Flashcards • NotesNotes • ExplanationsExplanations • Study Planner • Textbook solutions # Central Limit Theorem If you were asked if there were any important things in your life, I bet it wouldn't be a difficult question to answer. You could easily identify aspects of your daily life that you could not live with relative quality without. You could label these things as central in your life. The same is true in several areas of knowledge, particularly in statistics. There is a mathematical result so important in statistics that they made a point of including the word central in its designation. And it is central not only in its importance, but also in its simplifying power. It is the Central Limit Theorem and in this article, you will see its definition, its formula, conditions, calculations and examples of application. ## Understanding the Central Limit Theorem Consider the following example. Imagine you have a bag with four balls • of equal size; • indistinguishable to touch; • and numbered with the even numbers 2, 4, 6, and 8. You are going to remove two balls at random, with replacement, and you'll calculate the mean of the numbers of the two balls you removed. "With replacement" means you remove the first ball from the bag, you put it back, and you remove the second ball. And yes, this can lead to the same ball being removed twice. Notice that you have 16 possible combinations; we present them in the tables below, with their means calculated. 1st ball 2 2 2 2 4 4 4 4 2nd ball 2 4 6 8 2 4 6 8 mean 2 3 4 5 3 4 5 6 1st ball 6 6 6 6 8 8 8 8 2nd ball 2 4 6 8 2 4 6 8 mean 4 5 6 7 5 6 7 8 Now let's draw a bar graph of these means, figure 2. Fig. 2 - Bar graph of the list of mean in the tables If you notice, the shape of this bar graph is heading towards the shape of a normal distribution, don’t you agree? It’s getting closer to the form of a normal curve! Now, if instead of 4 balls numbered with 2, 4, 6 and 8, you had 5 balls numbered with 2, 4, 6, 8 and 10, then you’d have 25 possible combinations, which leads to 25 means. What would the graph bar of this new list of means look like? Yes, it would have a similar form to that of a normal curve. If you kept increasing the number of numbered balls, the corresponding bar graph would get closer and closer to a normal curve. "Why is that?" you ask. This leads you to the next section. ## Definition of Central Limit Theorem The Central Limit Theorem is an important theorem in statistics, if not the most important, and is responsible for the effect of approximating the bar graphs for increasing values of the number of numbered balls to the curve of the normal distribution in the above example. Let's start by looking at its statement, and then recall two important concepts involved in it: a distribution of sample means, and the useful normal distribution. ### Central Limit Theorem Statement The statement of the Central Limit Theorem says: If you take a sufficiently large number of samples from any random distribution, the distribution of the sample means can be approximated by the normal distribution. Easy-peasy, right?! “Uhh… No…!!” Ok, ok. Let's understand it by simplifying its statement a bit: If you take a big number of samples from a distribution, the sample mean of this distribution can be approximated by the normal distribution. Let's forget for a moment "a sufficiently large number" and "any random distribution", and focus on: • a sample mean; • and normal distribution. ### Understanding the Distribution of Sample Means Imagine you have to perform a statistical study for a particular attribute. You identify the population of your study and from it, you’ll draw a random sample. You will then calculate a particular statistic related to that attribute you’re interested in from this sample, and it’ll be the mean. Now imagine drawing another sample randomly from the same population, with the same size as the previous one, and calculating the mean of the attribute of this new sample. Imagine doing this a few more (and more and more) times. What you’ll end up with is a list of means from the samples you’ve drawn. And voilà! That list of means you end up with constitutes a distribution of sample means. To deepen your knowledge on this topic, read our article Sample Mean. ### Recalling the Normal Distribution One big usefulness of the normal distribution is associated with the fact that it approximates quite satisfactorily the frequency curves of physical measurements. That is, physical measures such as the height and weight of a sample of elements of the human population can be approximated by this distribution. Now you’re close to seeing another important application of this distribution. By now you may already know that the normal distribution is a probability distribution with two parameters, a mean $$\mu$$ and a standard deviation $$\sigma$$, and that has a graphical appearance of a bell-shaped curve – see figure 1. Fig. 1 – Normal curve of a normal distribution of mean 0 and standard deviation 0.05 The mean is the value at which the distribution is centered, and the standard deviation describes its degree of dispersion. In the case of figure 1, the normal curve is centered at 0 and its dispersion is somewhat low, 0.05. The lower the dispersion, the closer the curve is to the $$y$$-axis. To refresh your memory on this topic, read our article Normal Distribution. ### How Many is Enough? What you need to understand here is that the Central Limit Theorem tells us that for a "number” of samples from a distribution, the sample mean will get closer to the normal distribution. Recalling the example above: "Imagine you have a bag with four balls • of equal size; • indistinguishable to touch; • and numbered with the even numbers 2, 4, 6, and 8. You are going to remove two balls at random, with replacement, and you'll calculate the mean of the numbers of the two balls you removed." Notice that here the samples are the means of the two balls removed, and the distribution will be of the list of means obtained. Now including what we took out for a moment, Central Limit Theorem says that no matter what the distribution is - "any random distribution" -, the distribution of its mean approaches normal distribution as the number of samples grows - "a sufficiently large number of samples". Now the question imposes itself, what is a sufficiently large number of samples? This leads us to the next section. ## Conditions for the Central Limit Theorem There are two main conditions that must be met for you to apply the Central Limit Theorem. The conditions are the following: • Randomness – the sample collection must be random, this means every element of the population must have the same chance of being selected. Coming back to the first example, you had the 4 balls on a bag, and they were indistinguishable to touch. These elements randomize the experiment. • Sufficiently large sample: as a practical rule, when the number of samples is at least 30 the distribution of the sample means will satisfactorily approach a normal distribution. This is why the example above serves only the purpose of illustrating with simplicity the idea of the Central Limit Theorem. We got 16 samples from it, and if there were 5 balls, we could only get 25 samples, which again is not enough large number of samples. ## Central Limit Theorem Formula Addressing the Central Limit Theorem formula is equivalent to restating it by introducing all the necessary notation, and giving it further details. It’s worth repeating the first statement: If you take a sufficiently large number of samples from any random distribution, the distribution of the sample means can be approximated by the normal distribution. Now introducing the appropriate notation: Assume you have an initial distribution, with either an unknown or known probability distribution, and let $$\mu$$ be its mean and $$\sigma$$ be its standard deviation. Also, assume you’ll take $$n$$ samples from this initial distribution, and $$n\ge30$$. Then, the sample mean, $$\bar{x}$$, with mean $$\mu_\bar{x}$$ and standard deviation $$\sigma_\bar{x}$$, will be normally distributed with mean $$\mu$$ and standard variation $$\frac{\sigma}{\sqrt{n}}$$. As a result of this new restatement of the Central Limit Theorem, you can conclude that: 1. The mean of the distribution of the sample mean $$\bar{x}$$ will be equal to the mean of the initial distribution, i.e., $\mu_\bar{x}=\mu;$ 2. The standard deviation of the distribution of the sample mean $$\bar{x}$$ will be $$\frac{1}{\sqrt{n}}$$ of the standard deviation of the initial distribution, i.e., $\sigma_\bar{x}=\frac{\sigma}{\sqrt{n}};$ This is actually good: notice that for an increasing value of $$n$$, $$\frac{\sigma}{\sqrt{n}}$$ decreases, the dispersion of $$\bar{x}$$ decreases, which means it behaves more and more like a normal distribution. 3. The Central Limit Theorem applies to any distribution with many samples, be it known (like a binomial, a uniform, or a Poisson distribution) or an unknown distribution. Let's look at an example where you'll see this notation in action. A study reports that the mean age of peanut buyers is $$30$$ years and the standard deviation is $$12$$. With a sample size of $$100$$ people, what are the mean and standard deviation for the sample mean ages of the peanut buyers? Solution: The population and consequently the sample of the study consists of peanut buyers, and the attribute they were interested in was age. So, you're told the mean and the standard deviation of the initial distribution is $$\mu=30$$ and $$\sigma=12$$. You're also told the number of samples, so $$n=100$$. Since $$n$$ is greater than $$30$$, you can apply the Central Limit Theorem. Then, there will be a sample mean $$\bar{x}$$ that is normally distributed with mean $$\mu_\bar{x}$$ and standard deviation $$\sigma_\bar{x}$$. And you know more, \begin{align} \mu_\bar{x}&=\mu\\ &=30\end{align} and \begin{align} \sigma_\bar{x}&=\frac{\sigma}{\sqrt{n}} \\ &=\frac{12}{\sqrt{100}} \\ &=\frac{12}{10} \\ &=1.2 .\end{align} Therefore, $$\bar{x}$$ is normally distributed with mean $$30$$ and standard deviation $$1.2$$. ## Calculations Involving the Central Limit Theorem As you by now know, the Central Limit Theorem allows us to approximate any distribution of means, for a large number of samples, to the normal distribution. This means that some of the calculations where the Central Limit Theorem is applicable will involve calculations with the normal distribution. Here, what you'll be doing is converting a normal distribution to the standard normal distribution. To recall more of the last concept topic, please read our article Standard Normal Distribution. The importance of doing this conversion is that then you'll have access to a table of values of the standard normal, also known as z-score, to which you can refer to proceed with your calculations. Any point $$x$$ from a normal distribution can be converted to the standard normal distribution $$z$$ by doing the following $z=\frac{x-\mu}{\sigma},$ where $$z$$ follows the standard normal distribution (with mean $$\mu=0$$ and standard deviation $$\sigma=1$$). Because $$\bar{x}$$ is normally distributed with mean $$\mu$$ and standard deviation $\frac{\sigma}{\sqrt{n}},$ the conversion will be more like $z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}.$ You can refresh your memory on this topic by reading our article z-score. This example serves as a reminder of the conversion to the standard normal distribution. A random sample of size $$n=90$$ is selected from a population with mean $$\mu=20$$ and standard deviation $$\sigma=7$$. Determine the probability that $$\bar{x}$$ is less than or equal to $$22$$. Solution: Since the sample size is $$n=90$$, you can apply the Central Limit Theorem. This means $$\bar{x}$$ will follow a normal distribution with mean $\mu_\bar{x}=\mu=22$ and standard deviation \begin{align} \sigma_\bar{x}&=\frac{\sigma}{\sqrt{n}} \\ &=\frac{7}{\sqrt{90}} \\ &=0.738 \end{align} to three decimal places. Now you want to find $$P(\bar{x}\le 22)$$, and for that you apply the conversion to the standard normal: \begin{align} P(\bar{x}\le 22)&=P\left( z\le \frac{22-20}{0.738} \right) \\ \\ &=P( z\le 2.71) \\ \\ &=\text{ area under the normal curve to the left of 2.71} \\ \\ &=0.9966 \end{align} ## Examples of the Central Limit Theorem To consolidate the learnings from this article, let's now turn to application examples. Here, you will see an overview of all the main aspects of the Central Limit Theorem. To the first example. A female population’s weight data follows a normal distribution. It has a mean of 65 kg and a standard deviation of 14 kg. What is the standard deviation of the chosen sample if a researcher analyzes the records of 50 females? Solution: Let's do a final word problem. A small hotel receives on average $$10$$ new customers per day with a standard deviation of 3 customers. Calculate the probability that in a 30-day period, the hotel receives on average more than $$12$$ customers in 30 days. Solution: The initial distribution has a mean $$\mu=10$$ and a standard deviation $$\sigma=3$$. As the time period is 30 days, $$n=30$$. Therefore, you can apply Central Limit Theorem. This means you'll have $$\bar{x}$$ whose distribution has a mean $$\mu_\bar{x}$$ and a standard deviation $$\sigma_\bar{x}$$, and \begin{align} \mu_\bar{x}&=\mu\\ &=10 \end{align} and \begin{align} \sigma_\bar{x}&=\frac{\sigma}{\sqrt{n}}\\ &=\frac{3}{\sqrt{30}} \\ &=0.548 \end{align} to three decimal places. You're asked to calculate $$P(\bar{x}\ge 12)$$, and for that you'll convert $$\bar{x}$$ to the normal standard $$z$$: \begin{align} P(\bar{x}\ge 12)&=P\left(z \ge \frac{12-10}{0.548} \right) \\ \\ &=P(z \ge 3.65) .\end{align} Now, the final calculations: \begin{align} P(z\ge 3.65)&=\text{ area under the normal curve to right of 3.65} \\ &=1-0.9999 \\ &=0.0001\, (0.01\%).\end{align} Therefore, the probability that in a 30-day period the hotel receives on average more than $$12$$ customers in 30 days is $$0.01\%$$. ## Importance of the Central Limit Theorem There are many situations in which the Central Limit Theorem is of importance. Here are some of them: • In instances where it is difficult to collect data on each element of a population, the Central Limit Theorem is used to approximate the features of the population. • The Central Limit Theorem is useful in making significant inferences about the population from a sample. It can be used to tell whether two samples were drawn from the same population, and also check if the sample was drawn from a certain population. • To build robust statistical models in data science, the Central Limit Theorem is applied. • To assess the performance of a model in machine learning, the Central Limit Theorem is employed. • You test a hypothesis in statistics using the Central Limit Theorem to determine if a sample belongs to a certain population. ## The Central Limit Theorem - Key takeaways • Central Limit Theorem says, if you take a sufficiently large number of samples from any random distribution, the distribution of the sample means can be approximated by the normal distribution. • Another way of stating Central Limit Theorem is if $$n\ge 30$$, then the sample mean $$\bar{x}$$ follows a normal distribution with $$\mu_\bar{x}=\mu$$ and $$\sigma_\bar{x}=\frac{\sigma}{\sqrt{n}}.$$ • Any normal distribution can be converted to the normal standard by doing $$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}.$$ • Knowledge of the standard normal distribution, its table and its properties help you in calculations involving the Central Limit Theorem. The Central Limit Theorem is an important theorem in Statistics that involves approximating a distribution of sample means to the normal distribution. The Central Limit Theorem is useful in making significant inferences about the population from a sample. It can be used to tell whether two samples were drawn from the same population, and also check if the sample was drawn from a certain population. Assume you have a random variable X, with either an unknown or known probability distribution. Let σ be the standard deviation of X and μ be its. The new random variable, X, comprising the sample means, will be normally distributed, for a large number of samples (n ≥ 30), with mean μ and standard deviation σ/√n. The Central Limit Theorem says that if you take a sufficiently large number of samples from any random distribution, the distribution of the sample means can be approximated by the normal distribution. The Central Limit Theorem is not a prerequisite for confidence intervals. However, it helps to construct intervals by forming an estimate of samples as having a normal distribution. ## Final Central Limit Theorem Quiz Question What are the mean and standard deviation of the sampling distribution for samples of size 40 trips if the population mean of the number of fish caught each trip to a given fishing hole is 3.2 and the population standard deviation is 1.8? mean = 3.2 and standard deviation = 0.285 Show question Question What is the Central Limit Theorem? The Central Limit Theorem is an important theorem in statistics that involves approximating a distribution of sample means to the normal distribution. Show question Question What is the minimum sample size to consider when using the Central Limit Theorem? 30 Show question Question How can you supposedly construct a distribution of sample means? By drawing many samples of the same size from the same population and calculating the mean of the attribute you're interested in, you form a list of means from those samples that may become a distribution of sample means. Show question Question What are two important conditions for the Central Limit Theorem? Two important conditions are randomness and a sufficiently large number of samples. Show question Question What important concepts does the Central Limit Theorem involve? There are two important concepts that the Central Limit Theorem involves: a distribution of sample means and the normal distribution. Show question Question The Central Limit Theorem applies to any distribution with many samples, be it known, like a binomial, a uniform, or a Poisson distribution, or an unknown distribution. True or false? True. Show question Question What does the Central Limit Theorem tell us? The Central Limit Theorem says that if you take a sufficiently large number of samples from any random distribution, the distribution of the sample means can be approximated by the normal distribution. Show question Question State the formula for the Central Limit Theorem. For $$X$$ with mean $$\mu$$ and standard deviation $$\delta$$, if $$n\ge 30$$, then there's a random variable $$\bar{X}$$ such that $$\bar{X}\approx N\left (\mu, \frac{\delta}{\sqrt{n}} \right)$$. Show question Question The Central Limit Theorem is useful in making significant inferences about the population from a sample. It can be used to tell whether two samples were drawn from the same population, and also check if the sample was drawn from a certain population. True or False? True. Show question Question In instances where it is difficult to collect data on each element of a population, the Central Limit Theorem won't be useful to approximate the features of the population. True or False? False. Show question Question The Central Limit Theorem allows approximating any distribution, for a large sample size, to the binomial distribution. True or False? False. Show question Question By the Central Limit Theorem, the distribution of the sample means will have the same mean and standard deviation of the initial distribution. True or False? False. Show question Question By the Central Limit Theorem, if a random variable $$X$$ of a particular distribution has a standard deviation of $$\delta$$, what will be the standard deviation of $$\bar{X}$$? $$\delta_\bar{X}=\frac{\delta}{\sqrt{n}}$$ Show question Question The normal standard is an important distribution when dealing with calculations involving the Central Limit Theorem. True or False? True. Show question Question Any distribution $$X$$ of mean $$\mu$$ and standard deviation $$\delta$$ can be easily converted to the normal standard by doing $$Z=\frac{X-\mu}{\delta}$$. True or False? False. Only the normal distribution can be converted to the normal standard distribution. Show question 60% of the users don't pass the Central Limit Theorem quiz! Will you pass the quiz? Start Quiz ## Study Plan Be perfectly prepared on time with an individual plan. ## Quizzes Test your knowledge with gamified quizzes. ## Flashcards Create and find flashcards in record time. ## Notes Create beautiful notes faster than ever before. ## Study Sets Have all your study materials in one place. ## Documents Upload unlimited documents and save them online. ## Study Analytics Identify your study strength and weaknesses. ## Weekly Goals Set individual study goals and earn points reaching them. ## Smart Reminders Stop procrastinating with our study reminders. ## Rewards Earn points, unlock badges and level up while studying. ## Magic Marker Create flashcards in notes completely automatically. ## Smart Formatting Create the most beautiful study materials using our templates.
# AP Statistics Curriculum 2007 Gamma ### Gamma Distribution Definition: Gamma distribution is a distribution that arises naturally in processes for which the waiting times between events are relevant. It can be thought of as a waiting time between Poisson distributed events. Probability density function: The waiting time until the hth Poisson event with a rate of change $$\lambda$$ is $P(x)=\frac{\lambda(\lambda x)^{h-1}}{(h-1)!}{e^{-\lambda x}}$ For $$X\sim Gamma(k,\theta)\!$$, where $$k=h$$ and $$\theta=1/\lambda$$, the gamma probability density function is given by $\frac{x^{k-1}e^{-x/\theta}}{\Gamma(k)\theta^k}$ where • e is the natural number (e = 2.71828…) • k is the number of occurrences of an event • if k is a positive integer, then $$\Gamma(k)=(k-1)!$$ is the gamma function • $$\theta=1/\lambda$$ is the mean number of events per time unit, where $$\lambda$$ is the mean time between events. For example, if the mean time between phone calls is 2 hours, then you would use a gamma distribution with $$\theta$$=1/2=0.5. If we want to find the mean number of calls in 5 hours, it would be 5 $$\times$$ 1/2=2.5. • x is a random variable Cumulative density function: The gamma cumulative distribution function is given by $\frac{\gamma(k,x/\theta)}{\Gamma(k)}$ where • if k is a positive integer, then $$\Gamma(k)=(k-1)!$$ is the gamma function • $$\textstyle\gamma(k,x/\theta)=\int_0^{x/\theta}t^{k-1}e^{-t}dt$$ Moment generating function: The gamma moment-generating function is $M(t)=(1-\theta t)^{-k}\!$ Expectation: The expected value of a gamma distributed random variable x is $E(X)=k\theta\!$ Variance: The gamma variance is $Var(X)=k\theta^2\!$ ### Applications The gamma distribution can be used a range of disciplines including queuing models, climatology, and financial services. Examples of events that may be modeled by gamma distribution include: • The amount of rainfall accumulated in a reservoir • The size of loan defaults or aggregate insurance claims • The flow of items through manufacturing and distribution processes • The load on web servers • The many and varied forms of telecom exchange The gamma distribution is also used to model errors in a multi-level Poisson regression model because the combination of a Poisson distribution and a gamma distribution is a negative binomial distribution. ### Example Suppose you are fishing and you expect to get a fish once every 1/2 hour. Compute the probability that you will have to wait between 2 to 4 hours before you catch 4 fish. One fish every 1/2 hour means we would expect to get $$\theta=1/0.5=2$$ fish every hour on average. Using $$\theta=2$$ and $$k=4$$, we can compute this as follows: $P(2\le X\le 4)=\sum_{x=2}^4\frac{x^{4-1}e^{-x/2}}{\Gamma(4)2^4}=0.12388$ The figure below shows this result using SOCR distributions
Lesson Plan: # Comparing Fractions with the Same Denominator no ratings yet Subject Standards September 1, 2015 ## Learning Objectives Students will be able to compare fractions with the same denominators. ## Lesson ### Introduction (10 minutes) • Write a list of fractions with the same denominator, or bottom number, on the board, such as 1/3 and 2/3. • Ask student volunteers to share a fraction comparison that they noticed, For example: 1/3 is less than 2/3. • Write the fraction comparisons on the board, such as 1/3 < 2/3. • Ask small groups to look at the comparisons on the board and determine if they can identify any patterns. For example, if the denominator is the same, the larger numerator, or top number, is the greater fraction. ### Explicit Instruction/Teacher Modeling (10 minutes) • Draw a number line on the board from 0 to 4/4. • Label each hash mark as 0/4, 1/4, 2/4, 3/4, and 4/4. • Plot 2/4 and 3/4 on the number line. • Note to students that 0/4 is the same as 0 and 4/4 is the same as 1. Draw a circle divided into fourths to illustrate this. • Show students that 3/4 is closer to 4/4 or one whole. • Draw 2 circles of the same size, each divided into fourths. • Shade 3/4 of one circle and 2/4 of the other. • Write 3/4 > 2/4 next to the number line. • Explain that using a number line and drawing are two different ways that can help determine fraction comparison. ### Guided Practice/Interactive Modeling (5 minutes) • Give students 1/8 fraction strips, and have them make 2/8 and 4/8. • Instruct students to draw a number line or make a drawing to compare. • Then, direct students to write a comparative statement on their individual dry erase boards or notebooks. ### Independent Working Time (20 minutes) • Direct students to work with a partner to complete the word problems on the Same Denominator Partner Problems worksheet. • As students are working, circulate around the classroom to ask students to verbally explain their work. ## Extend ### Differentiation • Enrichment: Use mixed denominators to compare fractions, and have your students use a number line or drawing to show this visually. • Support: Put students in small groups, and direct them to use their dry erase boards to draw number lines and use drawings to make their comparative statements. Have them focus on the visuals rather than the numbers and number statements. ## Review ### Assessment (5 minutes) • As an exit ticket, give your students a problem to solve. For example: Sam ran 4/6 of a mile, and Ben ran 2/6 of a mile. Who ran further? Write a comparative statement. Use a drawing, number line, or words to explain your answer. • Collect the assessments to grade and evaluate understanding. ### Review and Closing (5 minutes) • Ask a student volunteer to explain her answer in words to the class for the assessment. • Instruct a different student to show a number line for this problem. • Have a third volunteer show a drawing for this problem. • Direct students to create a fraction rule or pattern when comparing fractions with the same denominator. • Display this fraction rule in the classroom.
# Can you use tan on any triangle? ## Can you use tan on any triangle? The Sine Rule can be used in any triangle (not just right-angled triangles) where a side and its opposite angle are known. You will only ever need two parts of the Sine Rule formula, not all three. You will need to know at least one pair of a side with its opposite angle to use the Sine Rule. ## Can you use tan on non-right triangles? The trigonometry of non-right triangles So far, we've only dealt with right triangles, but trigonometry can be easily applied to non-right triangles because any non-right triangle can be divided by an altitude * into two right triangles. ## Is this triangle a right triangle? A triangle can be determined to be a right triangle if the side lengths are known. If the lengths satisfy the Pythagorean Theorem (a2+b2=c2) then it is a right triangle. ## How do you solve right triangles? Key Takeaways 1. The Pythagorean Theorem, a2+b2=c2, a 2 + b 2 = c 2 , is used to find the length of any side of a right triangle. 2. In a right triangle, one of the angles has a value of 90 degrees. 3. The longest side of a right triangle is called the hypotenuse, and it is the side that is opposite the 90 degree angle. ## How do you solve Sin Cos Tan and right triangles? In any right angled triangle, for any angle: 1. The sine of the angle = the length of the opposite side. the length of the hypotenuse. 2. The cosine of the angle = the length of the adjacent side. the length of the hypotenuse. 3. The tangent of the angle = the length of the opposite side. the length of the adjacent side. ## How is sine calculated? The sine function, along with cosine and tangent, is one of the three most common trigonometric functions. In any right triangle, the sine of an angle x is the length of the opposite side (O) divided by the length of the hypotenuse (H). In a formula, it is written as 'sin' without the 'e': ## What is cos sin equal to? Sine, Cosine and Tangent Sine Function:sin(θ) = Opposite / Hypotenuse Cosine Function:cos(θ) = Adjacent / Hypotenuse Tangent Function:tan(θ) = Opposite / Adjacent ## What is the relation between sin and cos? Sal shows that the sine of any angle is equal to the cosine of its complementary angle. ## What is cos 180 A? Cos 180°= – sin 90° (We know that cos ( 270° – a ) = – sin a) The value of sin 90 degree is 1. Substitute the value in the above relation, we get. Cos 180°= – (1) = -1. Therefore, the value of cos 180° is -1. ## What is value of cos 90 degree? Therefore, the value of cos 90 degrees is: Cos 90° = 0. It is observed that the values of sin and cos functions do not change if the values of x and y are the integral multiples of 2π. ## What is COS 1 in radians? As you can see below, the cos-1 (1) is 270° or, in radian measure, 3Π/2 . ## What is value of cos 4 pi? Trigonometry Examples is a full rotation so replace with 0 . The exact value of cos(0) is 1 .
# Coordinate Geometry Class 9 Notes – Chapter 3 ## CBSE Class 9 Maths Coordinate Geometry Notes:- Coordinate Geometry for class 9 notes is given here. Get the complete concept of coordinate geometry such as cartesian system, coordinate points, how to plot the points in the coordinate axes, quadrants with signs and so on. Go through the below article to learn coordinate geometry for class 9. ## Cartesian System ### Cartesian plane & Coordinate Axes Cartesian Plane: A cartesian plane is defined by two perpendicular number lines, A horizontal line(xaxis) and a vertical line (yaxis). These lines are called coordinate axes. The Cartesian plane extends infinitely in all directions. Origin: The coordinate axes intersect each other at right angles, The point of intersection of these two axes is called Origin. To know more about Cartesian System, visit here. The cartesian plane is divided into four equal parts, called quadrants. These are named in the order as I, II, III and IV starting with the upper right and going around in anticlockwise direction. Signs of coordinates of points in different quadrants: I Quadrant: ‘+’ x – coordinate and ‘+’ y – coordinate. E.g. (2, 3) II Quadrant: ‘-’ x – coordinate and ‘+’ y – coordinate. E.g. (-1, 4) III Quadrant: ‘-’ x – coordinate and ‘-’ y – coordinate. E.g. (-3, -5) IV Quadrant: ‘+’ x – coordinate and ‘-’ y – coordinate. E.g. (6, -1) ## Plotting on a Graph ### Representation of a point on the Cartesian plane Using the co-ordinate axes, we can describe any point in the plane using an ordered pair of numbers. A point A is represented by an ordered pair (xy) where x is the abscissa and y is the ordinate of the point. Position of a point in a plane To know more about Cartesian plane, visit here. ### Plotting a point The coordinate points will define the location in the cartesian plane. The first point (x) in the coordinates represents the horizontal axis, and the second point in the coordinates (y) represents the vertical axis. Consider an example, Point (3, 2) is 3 units away from the positive y-axis and 2 units away from the positive x-axis. Therefore, point (3, 2) can be plotted, as shown below. Similarly, (-2, 3), (-1, -2) and (2, -3) are plotted. Plotting a point in the plane
Lifestyle # How Many Sides Does an Octagon Have? ## Definition of an Octagon and Its Characteristics An octagon is a polygon with eight sides, eight vertices, and eight angles. It is a two-dimensional figure that can be either regular or irregular in shape. A regular octagon has eight congruent sides and angles, while an irregular octagon has sides and angles of different lengths and measures. In geometry, an octagon is classified as a convex polygon, which means that all of its angles are less than 180 degrees and its vertices point outward. The opposite of a convex polygon is a concave polygon, where one or more of its angles is greater than 180 degrees and its vertices point inward. Octagons can be found in various real-life objects and structures such as stop signs, umbrella canopies, and building facades. They are also used in mathematics and engineering for designing shapes and structures that require eight sides or angles. ## The Formula for Calculating the Number of Sides in an Octagon The formula for calculating the number of sides in an octagon is straightforward. As the name suggests, an octagon has eight sides, so the formula is simply: Number of Sides = 8 This formula applies to all octagons, whether regular or irregular. The sides of a regular octagon are congruent to each other, which means that they have the same length. To calculate the length of each side, you can use the formula: Length of Side = Perimeter Ã· 8 The perimeter of a regular octagon is the sum of all its sides. Since there are eight sides, the perimeter can be calculated by multiplying the length of one side by 8: Perimeter = Length of Side Ã— 8 For irregular octagons, each side has a different length. To calculate the perimeter of an irregular octagon, you need to add the lengths of all its sides. Knowing the number of sides and the length of each side can help you determine other properties of an octagon, such as its area, angle measures, and diagonals. ## Real-Life Examples of Octagons in Architecture and Design Octagons can be found in various architectural and design elements. One of the most well-known examples is the stop sign, which is an octagon-shaped traffic sign that signals drivers to come to a complete stop before proceeding. Other examples of octagons in architecture and design include: 1. Dome structures: Octagons are often used in the design of dome structures, such as the Dome of the Rock in Jerusalem and the Florence Cathedral in Italy. 2. Buildings: Octagons can be seen in the facades and floor plans of buildings, such as the Octagon House in Washington D.C. and the Palace of Westminster in London. 3. Decorative elements: Octagons can be used as decorative elements in furniture, rugs, and other household items. They are also commonly used in jewelry design. 4. Garden structures: Octagons can be used to create garden structures, such as gazebos and pavilions. 5. Sports equipment: Octagons are used in the design of sports equipment, such as the octagonal boxing ring and the octagonal soccer ball. Overall, octagons have a versatile range of applications in architecture and design, from functional to decorative, and can be found in both historical and modern structures. ## Exploring the Properties of Regular and Irregular Octagons Regular and irregular octagons have different properties that distinguish them from each other. Here are some of the key differences between regular and irregular octagons: 1. Sides and angles: A regular octagon has eight congruent sides and angles, while an irregular octagon has sides and angles of different lengths and measures. 2. Symmetry: A regular octagon has rotational symmetry of order 8, which means it can be rotated by 45 degrees, 90 degrees, 135 degrees, etc., and it will still look the same. An irregular octagon does not have rotational symmetry. 3. Diagonals: A regular octagon has 20 diagonals, while an irregular octagon can have a varying number of diagonals, depending on its shape. 4. Area: The area of a regular octagon can be calculated using the formula (2 + âˆš2) Ã— sÂ², where s is the length of each side. The area of an irregular octagon can be calculated by dividing it into triangles and other shapes and calculating their areas separately. 5. Perimeter: The perimeter of a regular octagon can be calculated by multiplying the length of one side by 8. The perimeter of an irregular octagon can be calculated by adding the lengths of all its sides. Overall, regular octagons have a high degree of symmetry and uniformity, while irregular octagons have more varied properties and can come in a wide range of shapes and sizes. ## Fun Facts and Trivia About Octagons and their History Here are some interesting facts and trivia about octagons and their history: 1. The word “octagon” comes from the Greek words “okto,” which means “eight,” and “gonia,” which means “angle.” 2. The stop sign was first introduced in 1915, and it was the first traffic sign to use the octagon shape. 3. The octagonal shape has been used in architecture and design for centuries, dating back to ancient Roman and Islamic architecture. 4. Octagons are used in many popular games, such as the Chinese game of Go, and the popular card game, Octagon. 5. In mathematics, octagons are used to represent the eight-sided regions on a chessboard and to calculate the angles in a regular polygon with eight sides. 6. The octagonal shape is also used in jewelry design, particularly for gemstones, such as diamonds and sapphires. 7. The famous octagonal building, The Octagon House, in Washington D.C., was built in 1801 and was one of the first residential structures in the city. 8. Octagons can be found in many forms of popular culture, such as in comic books and video games. Overall, octagons have a rich history and a variety of uses and applications in various fields, from mathematics and architecture to games and jewelry design.
Courses Courses for Kids Free study material Offline Centres More Store # The value of determinant $\left\| {\begin{array}{{20}{c}} 1&{{e^{i\dfrac{\pi }{3}}}}&{{e^{i\dfrac{\pi }{4}}}} \\ {{e^{ - i\dfrac{\pi }{3}}}}&1&{{e^{i\dfrac{{2\pi }}{3}}}} \\ {{e^{ - i\dfrac{\pi }{4}}}}&{{e^{ - i\dfrac{{2\pi }}{3}}}}&1 \end{array}} \right\|$ is ___________A - $2 + \sqrt 2$ B - $2 - \sqrt 2$C - $- \left( {2 + \sqrt 2 } \right)$D - $- \left( {2 - \sqrt 2 } \right)$ Last updated date: 11th Jun 2024 Total views: 391.8k Views today: 3.91k Answer Verified 391.8k+ views Hint: Firstly expand the determinant as general , on solving the determinant we get ${e^{i\theta }}$ form type things in it for this use ${e^{i\theta }} = \cos \theta + i\sin \theta$ apply it in the equation and also remember that $\cos \left( { - \theta } \right) = \cos \theta$ and $\sin \left( { - \theta } \right) = - \sin \theta$ . Complete step-by-step answer: As we have to find the value of determinant $\left\| {\begin{array}{{20}{c}} 1&{{e^{i\dfrac{\pi }{3}}}}&{{e^{i\dfrac{\pi }{4}}}} \\ {{e^{ - i\dfrac{\pi }{3}}}}&1&{{e^{i\dfrac{{2\pi }}{3}}}} \\ {{e^{ - i\dfrac{\pi }{4}}}}&{{e^{ - i\dfrac{{2\pi }}{3}}}}&1 \end{array}} \right\|$ , So for this now we have to expand the given determinant , with respect to column $1$ So , $1\left\| {\begin{array}{{20}{c}} 1&{{e^{i\dfrac{{2\pi }}{3}}}} \\ {{e^{ - i\dfrac{{2\pi }}{3}}}}&1 \end{array}} \right\|$ $-$ ${e^{ - i\dfrac{\pi }{3}}}$$\left\| {\begin{array}{{20}{c}} {{e^{i\dfrac{\pi }{3}}}}&{{e^{i\dfrac{\pi }{4}}}} \\ {{e^{ - i\dfrac{{2\pi }}{3}}}}&1 \end{array}} \right\|$ $+$ ${e^{ - i\dfrac{\pi }{4}}}$$\left\| {\begin{array}{*{20}{c}} {{e^{i\dfrac{\pi }{3}}}}&{{e^{i\dfrac{\pi }{4}}}} \\ 1&{{e^{i\dfrac{{2\pi }}{3}}}} \end{array}} \right\|$ So on expanding the determinant , $\Rightarrow 1\left( {1 - {e^{i\dfrac{{2\pi }}{3}}}.{e^{ - i\dfrac{{2\pi }}{3}}}} \right) - {e^{ - i\dfrac{\pi }{3}}}\left( {{e^{i\dfrac{\pi }{3}}} - {e^{i\dfrac{\pi }{4}}}.{e^{i\dfrac{{2\pi }}{3}}}} \right) + {e^{ - i\dfrac{\pi }{4}}}\left( {{e^{i\dfrac{\pi }{3}}}.{e^{i\dfrac{{2\pi }}{3}}} - {e^{i\dfrac{\pi }{4}}}} \right)$ On solving further , $\Rightarrow 0 - {e^{ - i\dfrac{\pi }{3}}}\left( {{e^{i\dfrac{\pi }{3}}} - {e^{i\dfrac{{5\pi }}{{12}}}}} \right) + {e^{ - i\dfrac{\pi }{4}}}\left( {{e^{ - \pi }} - {e^{i\dfrac{\pi }{4}}}} \right)$ $\Rightarrow - \left( {1 - {e^{i\dfrac{{5\pi }}{{12}}}}.{e^{ - i\dfrac{\pi }{3}}}} \right) + \left( {{e^{ - \pi }}.{e^{ - i\dfrac{\pi }{4}}} - 1} \right)$ $\Rightarrow - 2 + {e^{ - i\dfrac{\pi }{3} + i\dfrac{{5\pi }}{{12}}}} + {e^{ - i\dfrac{\pi }{4} - i\pi }}$ On solving the power we get the final solution , $\Rightarrow - 2 + {e^{ - i\dfrac{{3\pi }}{4}}} + {e^{i\dfrac{{3\pi }}{4}}}$ Now we know that from the value of ${e^{i\theta }} = \cos \theta + i\sin \theta$ , apply this on the above equation , $\Rightarrow - 2 + \cos \dfrac{{ - 3\pi }}{4} + i\sin \dfrac{{ - 3\pi }}{4} + \cos \dfrac{{3\pi }}{4} + i\sin \dfrac{{3\pi }}{4}$ We know that from the trigonometry that is $\cos \left( { - \theta } \right) = \cos \theta$ and $\sin \left( { - \theta } \right) = - \sin \theta$ So by using this we get , $\Rightarrow - 2 + 2\cos \dfrac{{3\pi }}{4}$ Hence we know the value of $\cos \dfrac{{3\pi }}{4} = - \dfrac{1}{{\sqrt 2 }}$ $- 2 - \dfrac{2}{{\sqrt 2 }}$ or we write as $- 2 - \sqrt 2$ So the option C is correct . Note: The value of ${e^{i\theta }} = \cos \theta + i\sin \theta$ is known as the Euler's formula . If we have a complex number$z = r\left( {\cos \theta + i\sin \theta } \right)$ written in polar form, we can use Euler's formula to write it even more concisely in exponential form that is $r.{e^{i\theta }}$
Variance and Standard Deviation of Ungrouped Data Variance and Standard Deviation of Ungrouped Data Formulas: The variance is $${{\sigma }^{2}}=\frac{1}{n}\sum\limits_{i=1}^{n}{({{x}_{i}}-\overline{x}}{{)}^{2}}$$. Then ‘σ’ the standard deviation is given by the positive square root of the variance $$\sigma =\sqrt{\frac{1}{n}\sum\limits_{i=1}^{n}{({{x}_{i}}-\overline{x}}{{)}^{2}}}$$. Calculation of Variance and Standard Deviation for an Ungrouped Data 1. Find the variance and standard deviation of following data 5, 12, 3, 18, 6, 8, 2, 10. Solution: Mean: The mean of the given data is x̄ = ∑ xᵢ/n = (5 + 12 + 3 + 18 + 6 + 8 + 2 + 10)/ 8 = 64/8 = 8 2. Find the Variance xᵢ 5 12 3 18 6 8 2 10 xᵢ – x̄ -3 4 -5 10 -2 0 -6 2 (xᵢ – x̄)² 9 16 25 100 4 0 36 4 Solution: Here ∑ (xᵢ – x̄)² = 194 ∴ variance (σ)² = $$\frac{1}{n}\sum\limits_{i=1}^{n}{({{x}_{i}}-\overline{x}}{{)}^{2}}$$. = ⅛ x 194 = 24.25 Standard Deviation: Hence standard deviation is $$\sigma =\sqrt{\frac{1}{n}\sum\limits_{i=1}^{n}{({{x}_{i}}-\overline{x}}{{)}^{2}}}$$. σ = √24.25 = 4.95
NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1. Board CBSE Textbook NCERT Class Class 6 Subject Maths Chapter Chapter 11 Chapter Name Algebra Exercise Ex 11.1 Number of Questions Solved 1 Category NCERT Solutions ## NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1 Question 1. Find the rule which gives the number of matchsticks required to make the following matchstick patterns. Use a variable to write the rule. (a) A pattern of letter T as (b) A pattern of letter Z as (c) A pattern of letter U as (d) A pattern of letter V as (e) A pattern of letter E as (f) A pattern of letter S as (g) A pattern of letter A as Solution. (a) Number of matchsticks required = 2n (b) Number of matchsticks required = 3n (c) Number of matchsticks required = 3n (d) Number of matchsticks required = 2n (e) Number of matchsticks required = 5n (f) Number of matchsticks required = 5n (g) Number of matchsticks required = 6 Question 2. We already know the rule for the pattern of letters L, C and F. Some of the letters from Q. 1 (given above) give us the same rule as that given by L. Which are these? Why does this happen? Solution. These letters are T and V. This happens since the number>of matchsticks require,d in each of them is 2. Question 3. Cadets are marching in a parade. There are 5 cadets in a row. What is the rule, which gives the number of cadets, given the number of rows? (Use n for the number of rows.) Solution. The number of cadets = 5n. Question 4. If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? (Use b for the number of boxes.) Solution. Total number of mangoes = 50b. Question 5. The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use for the number of students.) Solution. Number of pencils needed = 5s Question 6. A bird flies 1 kilometer in one minute. Can you express the distance covered by the bird in tertns of its flying time in minutes? (Use t for flying time in minutes.) Solution. Yes! / kilometers The bird flies in one minute = 1 kilometer The bird flies in / minutes = 1 x t kilometers = kilometers Question 7. Radha is drawing a dot Rangoli (a beautified pattern of lines joining dots with chalk powder. She has 9 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 8 rows? If there are 10 rows? Solution. ∵ Number of dots in 1 row = 9 ∴ Number of dots in r rows = 9 x r=9r Number of dots in 8 rows = 9 x 8 = 72 Number of dots in 10 row = 9 x 10 = 90 Question 8. Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age? Take Radha’s age to be x years. Solution. Yes! we can write Leela’s age in terms of Radha’s age. Age of Radha = x years ∵ Leela is 4 years younger than Radha. ∴ Age of Leela = (x – 4) years Question 9. Mother has made laddus. She gives some laddus to guests and family members; still, 5 laddus remain. If the number of laddus mother gave away is l, how many laddus did she make? Solution. Number of laddus given away to guests and family members = l Number of laddus remained = 5 ∴ Number of laddus she made = 1 + 5 Question 10. Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still, 10 oranges remain, outside. If the number of oranges in a small box is taken to be x, what is the number of oranges in the larger box? Solution. Let the number of oranges in a smaller box box ∴ Number of oranges in two smaller boxes = 2x Number of oranges remained outside = 10 ∴ Number of oranges in the larger box = 2x+ 10 Question 11. (a) Look at the following matchstick pattern of squares (figure). The squares are not separate. Two neighboring squares have a common matchstick. Observe the pattern and find the rule that gives the number of matchsticks in terms of the number of squares. (Hint: If you remove the vertical stick at the end, you will get a pattern of Cs.) (b) Figure gives a matchstick pattern of triangles. Av in Exercise 11(a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles. Solution. (a) Rule: Number of matchsticks required = 3x + I where x is the number of squares. (b) Rule: Number of matchsticks required = 2x + 1, where x is the number of triangles. We hope the NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1 help you. If you have any query regarding. NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1, drop a comment below and we will get back to you at the earliest.
# FAQ: How To Draw Box And Whisker Plot? ## Quartiles, Boxes, and Whiskers The median of the entire set is the sum of Q1, Q2, and Q3. Since there are seventeen values in this list, the ninth value is the list’s middle value. The “box” contains the middle portion of your data, while the “whiskers” show how large the “spread” of the data. ## How do you draw a box plot? Use a horizontal or vertical number line and a rectangular box to create a box plot, with the smallest and largest data values labeling the axis’ endpoints, and the first and third quartiles marking one end of the box and the other end of the box, respectively. ## What are the five steps to creating a box and whisker plot? A box and whisker plot, also known as a box plot, shows the minimum, first quartile, median, third quartile, and maximum of a set of data. In a box plot, we draw a box from the first quartile to the third quartile, with a vertical line passing through the box at the median. ## How do you compare box plots? Comparison rules for boxplots • Compare the interquartile ranges (that is, the box lengths) to compare dispersion. • Look at the overall spread as shown by the adjacent values. • Look for signs of skewness. • Look for potential outliers. ## How do you find Q1 Q2 and Q3? Formula for the Quartile: 1. Lower quartile (Q1) = N 1 multiplied by (1) divided by (4) 2. Middle quartile (Q2) = N 1 multiplied by (2) divided by (4) 3. Upper quartile (Q3) = N 1 multiplied by (3) divided by (4) 4. Interquartile range = Q3 (upper quartile) u2013 Q1 (lower quartile) We recommend reading: How To Draw A Stick Figure Dog? ## How do you find Q1 and Q3? Q1 is the median (middle) of the lower half of the data, and Q3 is the median (middle) of the upper half. (3, 5, 7, 8, 9), \| (11, 15, 16, 20, 21). ## What does a box plot tell you? A boxplot is a standardized way of displaying data distribution based on a five-number summary (u201cminimumu201d, first quartile (Q1), median, third quartile (Q3), and u201cmaximumu201d), and it can also tell you if your data is symmetrical, how tightly your data is grouped, and if and how your data is skewed. ## What is the lower quartile in math? When data points are arranged in increasing order, the lower quartile, or first quartile (Q1), is the value under which 25% of data points are found, while the upper quartile, or third quartile (Q3), is the value under which 75% of data points are found. ## How do you compare two box and whisker plots? To compare two box-and-whisker plots, look first at the boxes and median lines to see if they overlap, then look at the sizes of the boxes and whiskers to get a sense of ranges and variability, and finally, look for outliers if any exist. ## How do you interpret a box plot skewness? Skewed data is represented by a lopsided boxplot in which the median divides the box into two unequal pieces; the data is said to be skewed right if the longer part of the box is to the right (or above) the median, and skewed left if the longer part is to the left (or below) the median. We recommend reading: FAQ: How To Draw Eyes Looking At You? ## What is Q1 Q2 Q3 Q4? January, February, and March (Q1) April, May, and June (Q2) July, August, and September (Q3) October, November, and December (Q4) are the standard calendar quarters that make up the year. ## How do you find quartile 3? Put the list of numbers in order, then cut it into four equal parts, with all the quartiles being between numbers: 1. Quartile 1 (Q1) equals (4 4)/2 = 4. 2. Quartile 2 (Q2) equals (10 11)/2 = 10.5. 3. Quartile 3 (Q3) equals (14 16)/2 = 15.
Multiplying One-Digit by Three-Digit Numbers Start Practice ## How to Multiply One-Digit by Three-Digit Numbers In the last lesson, you learned how to multiply 1-digit numbers by 2-digit numbers, like these: 98 × 3 = ? You used the column method: Now, you're ready to multiply even bigger numbers. ### Multiplying One-Digit by Three-Digit Numbers Let's learn with an example. 431 × 3 = ? First, we write the numbers in column form, starting with the bigger number. Now, let's multiply each digit of the top number by the bottom number, one at a time. Next, let's multiply the digit at the Tens place. Finally, multiply the digit at the Hundreds place. You did it. 🤗 Great job. Let's try one more example. 708 × 9 = ? First, write the numbers in column form like this: Now, let's multiply each digit of the top number by the bottom number. First, let's multiply the digit at the Ones place. Because 8 times 9 is a two-digit number, we have a carry over. Next, we multiply the digit at the Tens place, and add the previous carry. Remember: 0 multiplied by any number is 0 itself. So, 0 × 9 = 0. Add the carry, and we get 0 + 7 = 7. Finally, we multiply the digit at the Hundreds place. Nice work. 😸 Let's try one more example. 5 × 623 = ? First, write the numbers in column form. ✅ Now, multiply each digit in the top number by the bottom digit. ✅ First, multiply the digit at the Ones place (and move the carry to the next place value). Next, we multiply the digit at the Tens place, and add the previous carry. Lastly, we multiply the digit at the Hundreds place, and add the previous carry. Yest! 👍 It's 3,115. Awesome. Think you got it? Try the practice. Start Practice Complete the practice to earn 1 Create Credit 1,000 Create Credits is worth \$1 in real AI compute time. 1 Create Credit is enough to get 1 question answered, or to generate 1 image from text, in the tools tab. Teachers: Assign to students Duplicate Edit Questions Edit Article Assign Preview Questions Edit Duplicate
Courses Courses for Kids Free study material Offline Centres More Store # P is any point on the bisector of $\angle AOB$. If $BO \bot PN$ and $OA \bot PM$, then $PM \ne PN$A. TrueB. False Last updated date: 22nd Jun 2024 Total views: 373.8k Views today: 8.73k Verified 373.8k+ views Hint: An angle bisector is that which divides the angle into two equal angles with equal measures, and also the each point of the angle bisector is equidistant from the sides of the angle. So basically the angle bisector is a line which divides the angle into two equal angles. Here $OP$ is the angular bisector in this particular problem. Complete step by step solution: Here consider the figure in the question which is clearly visible that $OP$ is the angular bisector. $OP$ divides the $\angle AOB$ into two equal angles which are $\angle AOP$ and $\angle PON$, $\therefore \;\angle AOP = \angle PON$ Also given that $BO \bot PN$ , $OA \bot PM$, $\because OP$ is the angular bisector of $\angle AOB$ and hence the lengths of $OM$ and $ON$ are equal: $\Rightarrow OM = ON$ From the above steps it is understood that: $\Rightarrow \angle AOP = \angle PON$ $\Rightarrow BO \bot PN,OA \bot PM$ $\Rightarrow OM = ON$ Hence from the above equations we can conclude that $PM = PN$, $\therefore PM = PN$ Correct option is B. Note: Always remember that whenever a line bisects an angle and makes right angles with the sides of the angle, then the sides from point of intersection will be equal and also that lengths from point to the sides are also equal.
# Golden Ratio Exploration (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) Objective: Explore properties of the Golden Ratio. Suppose we start by drawing one small square, and attach another square to it on the right, making a 2x1 rectangle. Now draw another square along the long edge of the 2x1 rectangle, making a 3x2 rectangle. Continue this process, spiraling outward, until you're out of room on the page. The process creates larger and larger rectangles by attaching squares to them, and could continue indefinitely. 1. Here is a table showing the side lengths of each rectangle in the process: Dimension of rectangle Ratio of the two sides 1 x 2 2/1 = 2 2 x 3 3/2 = 1.5 3 x 5 5/3 = 1.666666666 5 x 8 8/5 = 1.6 1. Find the dimensions of the next two rectangles. 2. Look at the list of the dimensions of the rectangles. How is the biggest number related to the two number in the row above it? Knowing this, continue the following sequence of numbers: 2, 3, 5, 8, 13,... 2. The sequence of numbers in the table is the Fibonacci sequence. 3. The table shows the ratio of the long side to the short side for the first four big rectangles. Find the ratio for the remaining rectangles. Are the ratios approaching a number? 4. Note that the rectangles changed shape less and less as the process continued (the ratios of long to short sides didn't change much). Suppose you want a rectangle that stays exactly the exactly the same shape when a square is attached. This means the ratio $\frac{x}{1} = \frac{1+x}{x}$ Solve this equation for $x$. (Cross multiply and use the quadratic formula!) 5. Calculate the ratio of your height to the height of your navel. Do the same for four friends. How does this number compare to the ratios you have found before? 6. This triangle is called the Golden Triangle. Compute the ratio of the length of the side to the length of the base. 7. The Parthenon in Athens, Greece, may have been built according to the golden ratio. Measure the width and the height of the west face (shown below). How close is it to a golden rectangle? 8. Discuss how the Fibonacci numbers appear in Mondrian's Broadway Boogie Woogie and Judd's Untitled. Piet Mondrian, Broadway Boogie Woogie, 1942-43. Donald Judd, Untitled, 1970 Handin: A sheet with answers to all questions.
Miscellaneous Exercise Conic Sections - NCERT Class 11 Chapter 11 Ex.11.ME Question 1 If a parabolic reflector is $$20$$ cm in diameter and $$5$$cm deep, find the focus. Solution The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along the positive $$x$$-axis. This can be diagrammatically represented as The equation of the parabola is of the form $${y^2} = 4ax$$ (as it is opening to the right). Since the parabola passes through point $$A\left( {5,10} \right)$$ Then, \begin{align}&\Rightarrow {10^2} = 4a\left( 5 \right)\\&\Rightarrow 100 = 20a\\&\Rightarrow a = \frac{{100}}{{20}} = 5\end{align} Therefore, the focus of the parabola is $$\left( {a,0} \right) \Rightarrow \left( {5,0} \right)$$, which is the mid-point of the diameter. Hence, the focus of the reflector is at the mid-point of the diameter. Chapter 11 Ex.11.ME Question 2 An arch is in the form of a parabola with its axis vertical. The arch is $$10$$ m high and $$5$$ m wide at the base. How wide is it $$2$$ m from the vertex of the parabola? Solution The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along the positive y-axis. This can be diagrammatically represented as The equation of the parabola is of the form $${x^2} = 4ay$$ (as it is opening upwards). It can be clearly seen that the parabola passes through point $$\left( {\frac{5}{2},10} \right)$$ Hence, \begin{align}&\Rightarrow {\left( {\frac{5}{2}} \right)^2} = 4a\left( {10} \right)\\&\Rightarrow a = \frac{{25}}{{4 \times 4 \times 10}} = \frac{5}{{32}}\end{align} Therefore, the arch is in the form of a parabola whose equation is $${x^2} = \frac{5}{8}y$$ When $$y = 2$$ \begin{align}&\Rightarrow {x^2} = \frac{5}{8} \times 2\\&\Rightarrow {x^2} = \frac{5}{4}\\&\Rightarrow x = \sqrt {\frac{5}{4}} \end{align} Therefore, \begin{align}AB &= 2 \times 1.118\;{\rm{m}}\;\left( {{\rm{approx}}.} \right)\\&= 2.236\;{\rm{m}}\left( {{\rm{approx}}.} \right)\end{align} Hence, when the arch is $$2$$m from the vertex of the parabola, its width is approximately $$2.236$$ m. Chapter 11 Ex.11.ME Question 3 The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and $$100$$ m long is supported by vertical wires attached to the cable, the longest wire being $$30$$ m and the shortest being $$6$$ m. Find the length of a supporting wire attached to the roadway $$18$$ m from the middle. Solution The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the positive$$y -$$axis. This can be diagrammatically represented as Here, AB and OC are the longest and the shortest wires, respectively, attached to the cable. DF is the supporting wire attached to the roadway, 18 m from the middle. Here, $$AB = 30\;m,$$$$OC = 6m$$ and $$BC = \frac{{100m}}{2} = 50m$$ The equation of the parabola is of the form $${x^2} = 4ay$$ (as it is opening upwards). The coordinates of point A are$$\left( {50,30 - 6} \right) \Rightarrow \left( {50,24} \right)$$. Since $$A\left( {50,24} \right)$$ is a point on the parabola, Then, \begin{align}&\Rightarrow {\left( {50} \right)^2} = 4a\left( {24} \right)\\&\Rightarrow a = \frac{{50 \times 50}}{{4 \times 24}}\\&\Rightarrow a = \frac{{625}}{{24}}\end{align} Therefore, equation of the parabola, \begin{align}&\Rightarrow {x^2} = 4 \times \frac{{625}}{{24}} \times y\\&\Rightarrow 6{x^2} = 625y\end{align} The x-coordinate of point D is $$18$$. Hence, at $$x = 18$$ \begin{align}&\Rightarrow 6{\left( {18} \right)^2} = 625y\\&\Rightarrow y = \frac{{6 \times 18 \times 18}}{{625}}\\&\Rightarrow y = 3.11\left( {{\rm{approx}}.} \right)\end{align} Therefore, \begin{align}DE&= 3.11m\\DF&= DE + EF\\&= 3.11m + 6m\\&= 9.11m\end{align} Thus, the length of the supporting wire attached to the roadway $$18$$ m from the middle is approximately $$9.11$$ m. Chapter 11 Ex.11.ME Question 4 An arch is in the form of a semi-ellipse. It is $$8$$ m wide and $$2$$ m high at the centre. Find the height of the arch at a point $$1.5$$ m from one end. Solution Since the height and width of the arc from the centre is $$2$$ m and $$8$$ m respectively, it is clear that the length of the major axis is 8 m, while the length of the semi-minor axis is $$2$$ m. The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the x-axis. Hence, the semi-ellipse can be diagrammatically represented as The equation of the semi-ellipse will be of the form $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,\;y \ge 0$$, where $$a$$ is the semi-major axis Accordingly, $$\Rightarrow 2a = 8 \Rightarrow a = 4$$ and $$\Rightarrow b = 2$$ Therefore, the equation of the semi-ellipse is $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{4} = 1,\;y \ge 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$ Let A be a point on the major axis such that $$AB = 1.5m$$. Now, Draw $$AC \bot OB$$ . $OA = \left( {4 - 1.5} \right)m = 2.5m$ The x-coordinate of point C is $$2.5$$ On substituting the value of $$x = 2.5$$ in equation $$\left( 1 \right)$$, we obtain \begin{align}&\Rightarrow \frac{{{{\left( {2.5} \right)}^2}}}{{16}} + \frac{{{y^2}}}{4} = 1\\&\Rightarrow \frac{{6.25}}{{16}} + \frac{{{y^2}}}{4} = 1\\&\Rightarrow {y^2} = 4\left( {1 - \frac{{6.25}}{{16}}} \right)\\&\Rightarrow {y^2} = 4\left( {\frac{{9.75}}{{16}}} \right)\\&\Rightarrow {y^2} = 2.4375\\&\Rightarrow y = 1.56\;\left( \text{approx.} \right)\end{align} Therefore, $$AC = 1.56m$$ Thus, the height of the arch at a point $$1.5$$ m from one end is approximately $$1.56$$ m. Chapter 11 Ex.11.ME Question 5 A rod of length $$12$$ cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is $$3$$ cm from the end in contact with the x-axis. Solution Let $$AB = 12cm$$ be the rod making an angle $${\rm{\theta }}$$ with $$OX$$ and $$P\left( {x,y} \right)$$ be the point on it such that $$AP = 3cm$$ Then, \begin{align}PB&= AB - AP\\&= \left( {12 - 3} \right)cm\\&= 9cm\end{align} From P, draw $$PQ \bot OY$$ and $$PR \bot OX$$. In $$\Delta PBQ;\;\cos \theta = \frac{{PQ}}{{PB}} = \frac{x}{9}$$ In $$\Delta PRA;\;\sin \theta = \frac{{PR}}{{PA}} = \frac{y}{3}$$ Since, $${\sin ^2}\theta + {\cos ^2}\theta = 1$$, \begin{align}{\left( {\frac{y}{3}} \right)^2} + {\left( {\frac{x}{9}} \right)^2} &= 1\\\Rightarrow {\frac{x}{{81}}^2} + \frac{{{y^2}}}{9} &= 1\end{align} Thus, the equation of the locus of point P on the rod is $${\frac{x}{{81}}^2} + \frac{{{y^2}}}{9} = 1$$ Chapter 11 Ex.11.ME Question 6 Find the area of the triangle formed by the lines joining the vertex of the parabola $${x^2} = \;12y$$ to the ends of its latus rectum. Solution The given parabola is $${x^2} = \;12y$$. On comparing this equation with $${x^2} = \;4ay$$, we obtain $$4a = 12 \Rightarrow a = 3$$ Therefore, The coordinates of foci are $$\Rightarrow S\left( {0,3} \right)$$ Let AB be the latus rectum of the given parabola. The given parabola can be roughly drawn as At $$y = 3$$ \begin{align}&\Rightarrow {x^2} = 12\left( 3 \right)\\&\Rightarrow {x^2} = 36\\&\Rightarrow x = \pm 6\end{align} Hence, The coordinates of A are $$\left( { - 6,3} \right)$$, while the coordinates of B are $$\left( {6,3} \right)$$ Therefore, the vertices of $$\Delta AOB$$ are $$O\left( {0,0} \right),A\left( { - 6,3} \right)$$ and $$B\left( {6,3} \right)$$ \begin{align}ar\left( {\Delta AOB} \right) &= \frac{1}{2}\left\| {0\left( {3 - 3} \right) + \left( { - 6} \right)\left( {3 - 0} \right) + 6\left( {0 - 3} \right)} \right\|uni{t^2}\\&= \frac{1}{2}\left\| {\left( { - 6} \right)\left( 3 \right) + 6\left( { - 3} \right)} \right\|uni{t^2}\\&= \frac{1}{2}\left\| { - 18 - 18} \right\|uni{t^2}\\&= \frac{1}{2}\left\| { - 36} \right\|uni{t^2}\\&= \frac{1}{2} \times 36uni{t^2}\\&= 18uni{t^2}\end{align} Thus, the required area of the triangle is $$18{\rm{ uni}}{{\rm{t}}^2}$$. Chapter 11 Ex.11.ME Question 7 A man running a racecourse notes that the sum of the distances from the two flag posts from him is always $$10$$ m and the distance between the flag posts is $$8$$m. Find the equation of the posts traced by man. Solution Let A and B be the positions of the two flag posts and $$P\left( {x,y} \right)$$ be the position of the man. Accordingly, $$PA + PB = 10$$. We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse. Therefore, the path described by the man is an ellipse where the length of the major axis is $$10$$ m, while points $$A$$ and $$B$$ are the foci. Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the $$x-$$axis, the ellipse can be diagrammatically represented as The equation of the ellipse will be of the form $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$, where $$a$$ is the semi-major axis Accordingly, \begin{align}&\Rightarrow 2a = 10\\&\Rightarrow a = 5\end{align} Distance between the foci \begin{align}&\Rightarrow 2c = 8\\&\Rightarrow c = 4\end{align} On using the relation $$c = \sqrt {{a^2} - {b^2}}$$, we obtain \begin{align}&\Rightarrow 4 = \sqrt {25 - {b^2}} \\&\Rightarrow 16 = 25 - {b^2}\\&\Rightarrow {b^2} = 25 - 16\\&\Rightarrow {b^2} = 9\\&\Rightarrow b = 3\end{align} Thus, the equation of the path traced by the man is $$\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1$$ Chapter 11 Ex.11.ME Question 8 An equilateral triangle is inscribed in the parabola $${y^2} = \;4ax$$, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle. Solution Let $$\Delta OAB$$ be the equilateral triangle inscribed in the parabola $${y^2} = \;4ax$$. Let AB intersect the $$x-$$axis at point $$C$$. Let $$OC = k$$ From the equation of the given parabola, we have \begin{align}&\Rightarrow {y^2} = 4ak\\&\Rightarrow y = \pm 2\sqrt {ak} \end{align} Therefore, the respective coordinates of points A and B are $$\left( {k,2\sqrt {ak} } \right)$$ and $$\left( {k, - 2\sqrt {ak} } \right)$$ Hence, \begin{align}AB &= AC + CB\\&= 2\sqrt {ak} + 2\sqrt {ak} \\&= 4\sqrt {ak} \end{align} Since, $$\Delta OAB$$ is an equilateral triangle, \begin{align}&\Rightarrow O{A^2} = A{B^2}\\&\Rightarrow O{C^2} + A{C^2} = A{B^2}\end{align} Therefore, \begin{align}&\Rightarrow {k^2} + {\left( {2\sqrt {ak} } \right)^2} = {\left( {4\sqrt {ak} } \right)^2}\\&\Rightarrow {k^2} + 4ak = 16ak\\&\Rightarrow {k^2} = 12ak\\&\Rightarrow k = 12a\end{align} Hence, \begin{align}AB& = 4\sqrt {ak} = 4\sqrt {a \times 12a} \\&= 4\sqrt {12{a^2}} \\&= 8\sqrt 3 a\end{align} Thus, the side of the equilateral triangle inscribed in the parabola $${y^2} = \;4ax$$ is $$8\sqrt 3 a$$. 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# Boolean Logic CS.352.F12 Size: px Start display at page: Transcription 1 Boolean Logic CS.352.F12 2 Boolean Algebra 3 Boolean Algebra Mathematical system used to manipulate logic equations. Boolean: deals with binary values (True/False, yes/no, on/off, 1/0) Algebra: set of operations to manipulate values and evaluate expressions 4 Boolean Functions A Boolean function is a function that operates on binary inputs and returns binary outputs. Play a central role in the specification, construction, and optimization of hardware architectures. 5 Truth Table Representation A Truth Table is the enumeration of all the possible outputs of a Boolean function given all possible input values. x y z f(x, y, z) 6 Board: Construct Truth Tables Boolean Expression A Boolean expression the application of Boolean operators over set of variables. 1. And: x * y is 1 exactly when both x and y are 1 2. Or: x + y is 1 exactly when either x or y or both are 1 3. Not: x' is 1 exactly when x is 0 7 Exercise 1: Boolean Expression -> Truth Tables Construct truth tables for the following boolean expressions: 1. f(x, y) = x' + y 2. f(x, y, z) = (x * y) + (y * z') 8 Exercise 1: Boolean Expression -> Truth Tables 1. f(x, y) = x' + y x y f(x, y) 9 Exercise 1: Boolean Expression -> Truth Tables 2. f(x, y, z) = (x * y) + (y * z') x y z f(x, y, z) 10 Canonical Representation Every Boolean function can be expressed using at least one Boolean expression called the canonical representation: For each row in truth table where output is 1, construct a term by And-ing together the variables of that row, and then Or all of these terms to form a Sum of Products. Every Boolean function, no matter how complex, can be expressed using three Boolean operators only: And, Or, and Not. 11 Exercise 2: Truth Table -> Canonical Representation Construct the canonical representation from the following truth table: x y z f(x, y, z) 12 Exercise 2: Truth Table -> Canonical Representation f(x, y, z) = x'yz' + xy'z' + xyz' x y z f(x, y, z) 13 14 Minimization While there is only one truth table representation for every Boolean function, there may exist multiple boolean expressions. For economic reasons, we usually want to minimize or reduce the number of logical operators used in our boolean expression. 15 Minimization: Algebraic Laws Identity A * 1 = A A + 0 = A Annulment A + 1 = 1 A * 0 = 0 Complement A + A' = 1 A * A' = 0 Indempotent A + A = A A * A = A 16 Minimization: More Algebraic Laws Associative Law A * B * C = (A * B) * C = A * (B * C) A + B + C = (A + B) + C = A + (B + C) Commutative Law A * B * C = B * A * C =... A + B + C = B + A + C =... Distributive Law A * (B + C) = (A * B) + (A * C) A + (B * C) = (A + B) * (A + C) DeMorgan's Law (A * B)' = A' + B' (A + B)' = A' * B' 17 Exercise 3: Minimize with Algebraic Laws Simplify the following Boolean function: f(x, y, z) = x'yz' + xy'z' + xyz' 18 Exercise 3: Minimize with Algebraic Laws f(x, y, z) = x'yz' + xy'z' + xyz' factor = z'(x'y + xy' + xy) factor = z'(x'y + x(y' + y)) complement = z'(x'y + x(1)) identity = z'(x'y + x) distribute = z'((x' + x) * (y + x)) complement = z'((1) * (y + x)) identity = z'(y + x) 19 Board: Simplify Expression Minimization: Karnaugh Maps Logic graph where all logic domains are continuous, making logic relationships easy to identify. 20 Board: Simplify Expression Minimization: Karnaugh Maps (4 inputs) 21 Logic Gate 22 Logic Gate A gate is a physical device that implements a Boolean function. - Inputs and outputs of a Boolean Function = Input and output pins of gate. - Today, most gates are implemented as transistors etched in silicon (chips). 23 Primitive Gates A primitive gate is a device that implements an elementary logical operation. And Or Not These devices can be implemented by a variety of technologies but their behavior is governed by the abstract notions of Boolean algebra. 24 Board: Draw Composite Gate Composite Gates We can chain together various primitive gates to form larger and more complex composite gates. Multi-way Example: And(a, b, c) = a * b c = (a b) * c 25 Exercise 4: boolean function -> gates Implement the following boolean functions as composite gates: 1. Nand(a, b) = (ab)' 2. Xor(a, b) = ab' + a'b 3. And(a, b, c, d) = abcd 26 Exercise 4: boolean function -> gates Draw composite gates on the board. 27 Interface vs Implementation Each logic gate has a unique interface, but may have multiple implementations. Interface: the input and output pins exposed to the outside world and the specified behavior. Implementation: the manner in which the specified behavior is accomplished. 28 Interface vs Implementation: Propagation Delay There is always a delay in a change in the input of a gate to the corresponding change in the output of the gate. Example: Serial And vs Parallel And Board: draw composite gates 29 Board: Write Truth table & Boolean Expression Multiplexers A multiplexer is a three-input gate that uses one of the inputs, called the "selection bit", to select and output one of the other two inputs, called "data bits". 30 Board: Write Truth table Demultiplexer Opposite of a multiplexer; it takes a single input and channels it to one of two possible outputs according to a selector bit. 31 Decoder A combinational circuit that converts binary information from n input lines to a maximum of 2 N unique output lines. 32 Exercise 5: multiplexer Implement a 4-to-1 multiplexor. 33 Exercise 5: multiplexer 1. Use 2-to-4 decoder and 4 And gates and an Or gate 2. Use 4 And3 gates and an Or gate 3. Use 3 2-to-1 multiplexers 34 Board: Sketch Multi-bit And Multi-bit Gates Computer hardware normally operates on multi-bit arrays called buses. Building a multi-bit gate is easy: construct arrays of n elementary gates. 35 HDL 36 HDL Today's hardware designers use Hardware Description Languages to plan and optimize their chip architectures. - Simulate the hardware. - Test the hardware. - Model resource usage. 37 VHDL -- import std_logic from the IEEE library library IEEE; use IEEE.std_logic_1164.all; -- this is the entity entity ANDGATE is port ( I1 : in std_logic; I2 : in std_logic; O : out std_logic); end entity ANDGATE; -- this is the architecture architecture RTL of ANDGATE is begin O <= I1 and I2; end architecture RTL; 38 Verilog // And gate module AND2(A, B, C); input A; input B; output C; assign C = A & B; endmodule ### Boolean Algebra. 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Lecture 4: Logic Simplication & Karnaugh Map Points Addressed in this Lecture Lecture 4: Logic Simplication & Karnaugh Map Professor Peter Cheung Department of EEE, Imperial College London Standard form of Boolean Expressions Sum-of-Products (SOP), ### ENGIN 112 Intro to Electrical and Computer Engineering ENGIN 2 Intro to Electrical and Computer Engineering Lecture 5 Boolean Algebra Overview Logic functions with s and s Building digital circuitry Truth tables Logic symbols and waveforms Boolean algebra ### Objectives: 1. Design procedure. 2. Fundamental circuits. 1. Design procedure Objectives: 1. Design procedure. 2. undamental circuits. 1. Design procedure Design procedure has five steps: o Specification. o ormulation. o Optimization. o Technology mapping. o Verification. Specification: ### EE292: Fundamentals of ECE EE292: Fundamentals of ECE Fall 2012 TTh 10:00-11:15 SEB 1242 Lecture 22 121115 http://www.ee.unlv.edu/~b1morris/ee292/ 2 Outline Review Binary Number Representation Binary Arithmetic Combinatorial Logic ### Gate Level Minimization Gate Level Minimization By Dr. M. Hebaishy Digital Logic Design Ch- Simplifying Boolean Equations Example : Y = AB + AB Example 2: = B (A + A) T8 = B () T5 = B T Y = A(AB + ABC) = A (AB ( + C ) ) T8 = ### CprE 281: Digital Logic CprE 28: Digital Logic Instructor: Alexander Stoytchev http://www.ece.iastate.edu/~alexs/classes/ Minimization CprE 28: Digital Logic Iowa State University, Ames, IA Copyright Alexander Stoytchev Administrative ### Chapter 3 Simplification of Boolean functions 3.1 Introduction Chapter 3 Simplification of Boolean functions In this chapter, we are going to discuss several methods for simplifying the Boolean function. What is the need for simplifying the Boolean ### Date Performed: Marks Obtained: /10. Group Members (ID):. Experiment # 04. Boolean Expression Simplification and Implementation Name: Instructor: Engr. Date Performed: Marks Obtained: /10 Group Members (ID):. Checked By: Date: Experiment # 04 Boolean Expression Simplification and Implementation OBJECTIVES: To understand the utilization ### LOGIC CIRCUITS. Kirti P_Didital Design 1 LOGIC CIRCUITS Kirti P_Didital Design 1 Introduction The digital system consists of two types of circuits, namely (i) Combinational circuits and (ii) Sequential circuit A combinational circuit consists ### EECS150, Fall 2004, Midterm 1, Prof. Culler. Problem 1 (15 points) 1.a. Circle the gate-level circuits that DO NOT implement a Boolean AND function. Problem 1 (15 points) 1.a. Circle the gate-level circuits that DO NOT implement a Boolean AND function. 1.b. Show that a 2-to-1 MUX is universal (i.e. that any Boolean expression can be implemented with ### Control and Datapath 8 Control and Datapath 8 Engineering attempts to develop design methods that break a problem up into separate steps to simplify the design and increase the likelihood of a correct solution. Digital system ### Board-Data Processing. VHDL Exercises. Exercise 1: Basics of VHDL Programming. Stages of the Development process using FPGA s in Xilinx ISE. Board-Data Processing VHDL Exercises Exercise 1: Basics of VHDL Programming Stages of the Development process using FPGA s in Xilinx ISE. Basics of VHDL VHDL (Very High Speed IC Hardware description Language) ### 01 Introduction to Digital Logic. ENGR 3410 Computer Architecture Mark L. Chang Fall 2008 Introduction to Digital Logic ENGR 34 Computer Architecture Mark L. Chang Fall 28 Acknowledgements Patterson & Hennessy: Book & Lecture Notes Patterson s 997 course notes (U.C. Berkeley CS 52, 997) Tom ### Review. EECS Components and Design Techniques for Digital Systems. Lec 05 Boolean Logic 9/4-04. Seq. Circuit Behavior. Outline. Review EECS 150 - Components and Design Techniques for Digital Systems Lec 05 Boolean Logic 94-04 David Culler Electrical Engineering and Computer Sciences University of California, Berkeley Design flow ### Combinational Logic Circuits Chapter 2 Combinational Logic Circuits J.J. Shann (Slightly trimmed by C.P. Chung) Chapter Overview 2-1 Binary Logic and Gates 2-2 Boolean Algebra 2-3 Standard Forms 2-4 Two-Level Circuit Optimization ### Chapter 2: Combinational Systems Uchechukwu Ofoegbu Chapter 2: Combinational Systems Temple University Adapted from Alan Marcovitz s Introduction to Logic and Computer Design Riddle Four switches can be turned on or off. One is the switch ### Digital Logic Lecture 7 Gate Level Minimization Digital Logic Lecture 7 Gate Level Minimization By Ghada Al-Mashaqbeh The Hashemite University Computer Engineering Department Outline Introduction. K-map principles. Simplification using K-maps. Don t-care ### DKT 122/3 DIGITAL SYSTEM 1 Company LOGO DKT 122/3 DIGITAL SYSTEM 1 BOOLEAN ALGEBRA (PART 2) Boolean Algebra Contents Boolean Operations & Expression Laws & Rules of Boolean algebra DeMorgan s Theorems Boolean analysis of logic circuits ### Boolean Analysis of Logic Circuits Course: B.Sc. Applied Physical Science (Computer Science) Year & Sem.: IInd Year, Sem - IIIrd Subject: Computer Science Paper No.: IX Paper Title: Computer System Architecture Lecture No.: 7 Lecture Title: ### Boolean logic. Boolean Algebra. Introduction to Computer Yung-Yu Chuang NOT AND NOT oolean lgebra oolean logic ased on symbolic logic, designed by George oole oolean variables take values as or. oolean expressions created from: NOT, ND, OR Introduction to Computer ung-u Chuang with slides ### VALLIAMMAI ENGINEERING COLLEGE VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203 DEPARTMENT OF INFORMATION TECHNOLOGY & COMPUTER SCIENCE AND ENGINEERING QUESTION BANK II SEMESTER CS6201- DIGITAL PRINCIPLE AND SYSTEM DESIGN ### UNIT I BOOLEAN ALGEBRA AND COMBINATIONAL CIRCUITS PART-A (2 MARKS) SUBJECT NAME: DIGITAL LOGIC CIRCUITS YEAR / SEM : II / III DEPARTMENT : EEE UNIT I BOOLEAN ALGEBRA AND COMBINATIONAL CIRCUITS 1. What is variable mapping? 2. Name the two canonical forms for Boolean algebra.
# How do you solve ln(5.6-x)=ln(18.4-2.6x)? Dec 19, 2016 $x = 8$ #### Explanation: An example of line of thought. Suppose we had: $10 \times 2 = 10 \times \left(1 + 1\right)$ Because both sides are multiplied by 10 we can and may remove the $\textcolor{b r o w n}{\underline{\text{ operation of }}}$ multiplying by 10 and the equation will still be true. $\textcolor{b r o w n}{\text{Taking loges on both sides is an operation}}$ Given that $\text{ "ln(5.6-x)=ln(18.4-2.6x)" }$ is true Then also $\text{ "color(white)(..)5.6-xcolor(white)(.)=color(white)(....)18.4-2.6x" }$ is equally true ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Write as:$\textcolor{w h i t e}{.} 2.6 x - x = 18.4 - 5.6$ $1.6 x = 12.8$ $x = \frac{12.8}{1.6} = \frac{128}{16} = 8$
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 12.1: Inverse Variation Models Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives At the end of this lesson, students will be able to: • Distinguish direct and inverse variation. • Graph inverse variation equations. • Write inverse variation equations. • Solve real-world problems using inverse variation equations. ## Vocabulary Terms introduced in this lesson: variation direct variation inverse variation joint variation constant of proportionality increase, decrease ## Teaching Strategies and Tips This lesson focuses on inverse variation models and graphing inverse variation equations. Use it to motivate rational functions, which are covered in the next six lessons. Remind students about having learned direct variation in chapter Graphs of Equations and Functions. • Point out that direct variation is a linear relationship. • The x\begin{align}x\end{align} and y\begin{align}y-\end{align}intercepts are . • The slope of the line is the only parameter, denoted by k\begin{align}k\end{align}, and called the constant of proportionality. • It takes only one more point to determine the direct variation. Some examples of direct variation relationships are: • Height of a person and the length of their shadow on flat ground. • Circumference and radius of the circle. • Weight of an object on a spring and the amount the spring has stretched. In the examples and Review Questions, have students decide on a variation model first and then solve for the constant of proportionality using the given information. This determines the equation of the variation which is necessary for answering the rest of the problem. Use Example 1 to illustrate the graph of an inverse variation. • Construct a similar table of values. Allow students to observe the function’s behavior numerically. Remind students of scientific notation in Example 6. In applied problems such as Examples 5 and 6, emphasize that direct variations are ubiquitous and significant in the real-world. In Review Questions 1-4, encourage students to apply stretches to the basic graph y=1x\begin{align}y = \frac{1}{x}\end{align}. Example: Graph the following inverse variation relationship. y=10x.\begin{align}y = \frac{10}{x}.\end{align} Hint: Since y=10x=101x\begin{align}y = \frac{10}{x} = 10 \cdot \frac{1}{x}\end{align}, the graph can be obtained from that of y=1x\begin{align}y = \frac{1}{x}\end{align} by stretching by a factor of 10\begin{align}10\end{align}. ## Error Troubleshooting Remind students in Example 1 that dividing by zero is undefined. Remind students in Example 6 to square the 5.3 which is in parentheses: K=740(5.3×1011)2=7405.321022\begin{align}K = 740(5.3 \times 10^{-11})^2 = 740 \cdot 5.3^2 \cdot 10^{-22}\end{align} ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:
AP Statistics Curriculum 2007 Prob Rules (Difference between revisions) Revision as of 04:43, 26 October 2009 (view source)IvoDinov (Talk \| contribs) (added a link to the Problems set)← Older edit Revision as of 16:30, 26 October 2009 (view source)IvoDinov (Talk \| contribs) m (→Contingency table)Newer edit → Line 29: Line 29: {\| class="wikitable" style="text-align:center; width:75%" border="1" {\| class="wikitable" style="text-align:center; width:75%" border="1" \|- \|- - \| \|\| colspan="3" align="center"\|Site \|\| + \| rowspan="2"\|Type \|\| colspan="3" align="center"\|Site \|\| rowspan="2"\|Totals \|- \|- - \| Type \|\| Head and Neck \|\| Trunk \|\| Extremities \|\| Totals + \| Head and Neck \|\| Trunk \|\| Extremities \|- \|- \| Hutchinson's melanomic freckle \|\| 22 \|\| 2 \|\| 10 \|\| 34 \| Hutchinson's melanomic freckle \|\| 22 \|\| 2 \|\| 10 \|\| 34 General Advance-Placement (AP) Statistics Curriculum - Probability Theory Rules The probability of a union, also called the Inclusion-Exclusion principle allows us to compute probabilities of composite events represented as unions (i.e., sums) of simpler events. For events A1, ..., An in a probability space (S,P), the probability of the union for n=2 is $P(A_1\cup A_2)=P(A_1)+P(A_2)-P(A_1\cap A_2),$ For n=3, $P(A_1\cup A_2\cup A_3)=P(A_1)+P(A_2)+P(A_3) -P(A_1\cap A_2)-P(A_1\cap A_3)-P(A_2\cap A_3)+P(A_1\cap A_2\cap A_3)$ In general, for any n, $P(\bigcup_{i=1}^n A_i) =\sum_{i=1}^n {P(A_i)} -\sum_{i,j\,:\,i Conditional Probability The conditional probability of A occurring given that B occurs is given by $P(A \| B) ={P(A \cap B) \over P(B)}.$ Examples Contingency table Here is the data on 400 Melanoma (skin cancer) Patients by Type and Site 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 • Suppose we select one out of the 400 patients in the study and we want to find the probability that the cancer is on the extremities given that it is of type nodular: P = 73/125 = P(Extremities \| Nodular) • What is the probability that for a randomly chosen patient the cancer type is Superficial given that it appears on the Trunk? Monty Hall Problem Recall that earlier we discussed the Monty Hall Experiment. We will now show why the odds of winning double if we use the swap strategy - that is the probability of a win is 2/3, if each time we switch and choose the last third card. Denote W={Final Win of the Car Price}. Let L1 and W2 represent the events of choosing the donkey (loosing) and the car (winning) at the player's first and second choice, respectively. Then, the chance of winning in the swapping-strategy case is: $P(W) = P(L_1 \cap W_2) = P(W_2 \| L_1) P(L_1) = 1 \times {2\over 3} ={2\over 3}$. If we played using the stay-home strategy, our chance of winning would have been: $P(W) = P(W_1 \cap W_2) = P(W_2 \| W_1) P(W_1) = 1 \times {1\over 3} ={1\over 3}$, or half the chance in the first (swapping) case. Drawing balls without replacement Suppose we draw 2 balls at random, one at a time without replacement from an urn containing 4 black and 3 white balls, otherwise identical. What is the probability that the second ball is black? Sample Space? P({2-nd ball is black}) = P({2-nd is black} &{1-st is black}) + P({2-nd is black} &{1-st is white}) = 4/7 x 3/6 + 4/6 x 3/7 = 4/7. Inverting the order of conditioning In many practical situations is is beneficial to be able to swap the event of interest and the conditioning event when we are computing probabilities. This can easily be accomplished using this trivial, yet powerful, identity: $P(A \cap B) = P(A \| B) \times P(B) = P(B \| A) \times P(A)$ Example - inverting conditioning Suppose we classify the entire female population into 2 Classes: healthy(NC) controls and cancer patients. If a woman has a positive mammogram result, what is the probability that she has breast cancer? Suppose we obtain medical evidence for a subject in terms of the results of her mammogram (imaging) test: positive or negative mammogram . If P(Positive Test) = 0.107, P(Cancer) = 0.1, P(Positive test \| Cancer) = 0.8, then we can easily calculate the probability of real interest - what is the chance that the subject has cancer: $P(Cancer \| Positive Test) = {P(Positive Test \| Cancer) \times P(Cancer) \over P(Positive Test)}= {0.8\times 0.1 \over 0.107}$ This equation has 3 known parameters and 1 unknown variable, so, we can solve for P(Cancer \| Positive Test) to determine the chance the patient has breast cancer given that her mammogram was positively read. This probability, of course, will significantly influence the treatment action recommended by the physician. Statistical Independence Events A and B are statistically independent if knowing whether B has occurred gives no new information about the chances of A occurring, i.e., if P(A \| B) = P(A). Note that if A is independent of B, then B is also independent of A, i.e., P(B \| A) = P(B), since $P(B\|A)={P(B \cap A) \over P(A)} = {P(A\|B)P(B) \over P(A)} = P(B)$. If A and B are statistically independent, then $P(B \cap A) = P(A) \times P(B).$ Multiplication Rule For any two events (whether dependent or independent): $P(A \cap B) = P(B\|A)P(A) = P(A\|B)P(B).$ In general, for any collection of events: $P(A_1 \cap A_2 \cap A_3 \cap \cdots \cap A_n) = P(A_1)P(A_2\|A_1)P(A_3\|A_1 \cap A_2)P(A_4\|A_1 \cap A_2 \cap A_3) \cdots P(A_{n-1}\|A_1 \cap A_2 \cap A_3 \cap \cdots \cap A_{n-2})P(A_n\|A_1 \cap A_2 \cap A_3 \cap \cdots \cap A_{n-1})$ Law of total probability If {$A_1, A_2, A_3, \cdots, A_n$} form a partition of the sample space S (i.e., all events are mutually exclusive and $\cup_{i=1}^n {A_i}=S$) then for any event B $P(B) = P(B\|A_1)P(A_1) + P(B\|A_2)P(A_2) + \cdots + P(B\|A_n)P(A_n)$ • Example, if A1 and A2 partition the sample space (think of males and females), then the probability of any event B (e.g., smoker) may be computed by: P(B) = P(B \| A1)P(A1) + P(B \| A2)P(A2). This of course is a simple consequence of the fact that $P(B) = P(B\cap S) = P(B \cap (A_1 \cup A_2)) = P((B \cap A_1) \cup (B \cap A_2))= P(B\|A_1)P(A_1) + P(B\|A_2)P(A_2)$. Bayesian Rule If {$A_1, A_2, A_3, \cdots, A_n$} form a partition of the sample space S and A and B are any events (subsets of S), then: $P(A \| B) = {P(B \| A) P(A) \over P(B)} = {P(B \| A) P(A) \over P(B\|A_1)P(A_1) + P(B\|A_2)P(A_2) + \cdots + P(B\|A_n)P(A_n)}.$ Example Suppose a Laboratory blood test is used as evidence for a disease. Assume P(positive Test\| Disease) = 0.95, P(positive Test\| no Disease)=0.01 and P(Disease) = 0.005. Find P(Disease\|positive Test)=? Denote D = {the test person has the disease}, Dc = {the test person does not have the disease} and T = {the test result is positive}. Then $P(D \| T) = {P(T \| D) P(D) \over P(T)} = {P(T \| D) P(D) \over P(T\|D)P(D) + P(T\|D^c)P(D^c)}=$ $={0.95\times 0.005 \over {0.95\times 0.005 +0.01\times 0.995}}=0.193.$
# Simplify the expression (-12-6i)/(-6+6i)? justaguide \| Certified Educator The expression to be simplified is (-12-6i)/(-6+6i) (-12-6i)/(-6+6i) cancel the common factor -6 => (2 + i)/( i - 1) => (2 + i)(i +1) / (i - 1)(i + 1) => 2i + 2 + i^2 + i)/2 => (3i + 1)/2 => (3/2)i + 1/2 The simplified form of the expression is (1/2) + (3/2)i giorgiana1976 \| Student First, we'll factorize both numerator and denominator by -6 -6(2 + i)/-6(1 - i) = (2+i)/(1-i) We'll simplify (2+i)/(1-i)by multiplying the denominator by it's conjugate. The conjugate of the 1 - i = 1 + i (2+i)(1+i)/(1-i)(1+i) We'll remove the brackets: (2 + 2i + i + i^2)/ (1-i^2) We'll combine the real parts and the imaginary parts, considering i^2 = -1, and we'll get: (1 + 3i)/2 The simplified expression is: (-12 - 6i) / (-6+6i) =1/2+ 3i/2
# CBSE Class 9 Mathematics, Surface Areas and Volumes: Important formulae and questions Jan 16, 2019 15:55 IST In this article you will get CBSE Class 9 Mathematics chapter 13, Surface Areas and Volumes: Important topics & questions to prepare for class 9 Mathematics paper which will be conducetd under the class 9 annual exam 2019. The questions and terms mentioned in this article will surely help to make your preparation easy and organised. Surface Areas and Volumes: Important formulae and questions Having an idea about the areas in a subject which need to be learned with more concentration, makes the exam preparations easy and organised. Therefore, before starting revision before the exam, students must have a list of topics/terms which they have to learn in a subject. This will help to take the most of your study time and make you well prepared for the upcoming exam. In this article, students will get all important terms/formulae and questions from CBSE Class 9th Maths chapter 13, Surface Areas and Volumes. This set of important formulae and questions has been prepared by the subject experts to help class 9 students prepare for Mathematics Exam 2019. CBSE Class 9 Mathematics Syllabus 2018-2019 Revise the below given important formulae from the chapter 'Surface Areas & Volumes' and then attempt the Important Questions: CBSE Class 9 Mathematics Practice Papers for Annual Exam 2019 Important Questions are as below: • A solid cylinder has a total surface area of 213 cm2. Its curved surface area is 2/3 of the total surface area. Find the volume of the cylinder. • The dimensions of a cuboid are in the ratio 1: 2: 3 and its total surface area is 88 m2. Find the dimensions. • The radius and height of a cone are in the ratio 4: 3. The area of the base is 154 cm2. Find the area of the curved surface. • A hemispherical bowl of internal diameter 36 cm contains a liquid. This liquid is to be filled in cylindrical bottles of radius 3 cm and height 6 cm. How many bottles are required to empty the bowl? • A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 6 cm and its height is 4 cm. Find the cost of painting the toy at the rate of Rs. 5 per 1000 cm2. • Three cubes of each side 4 cm are joining end to end. Find the surface area of resulting cuboid. • If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere? Students may also check the following links to explore more stuff, important for CBSE Class 9 Annual exam preparations: ## Register to get FREE updates All Fields Mandatory • (Ex:9123456789)
<meta http-equiv="refresh" content="1; url=/nojavascript/"> # Visual Patterns ## Describe non-arithmetic sequences by finding a rule. 0% Progress Practice Visual Patterns Progress 0% Finding the Next Term in a Sequence You drop a rubber ball from a height of 48 inches. Each time it bounces, it reaches lower and lower heights. The following sequence shows its height with each successive bounce. How high will the ball bounce on its fifth bounce? 48, 36, 27, 20.25,... ### Guidance When looking at a sequence of numbers, consider the following possibilities. • There could be a common difference (the same value is added or subtracted) to progress from each term to the next. Example: 5,8,11,14,\begin{align}5, 8, 11, 14, \ldots\end{align} (add 3) • There could be a common ratio (factor by which each term is multiplied) to progress from one term to the next. Example: 9,3,1\begin{align}9, 3, 1\end{align} , 13\begin{align}\frac{1}{3} \ldots\end{align} (\begin{align}\left ( \right .\end{align} multiply by 13)\begin{align}\left . \frac{1}{3} \right )\end{align} • If the terms are fractions, perhaps there is a pattern in the numerator and a different pattern in the denominators. Example: 19,38,57,76,\begin{align}\frac{1}{9}, \frac{3}{8}, \frac{5}{7}, \frac{7}{6}, \ldots\end{align} (numerator (+2), denominator (-1)) • If the terms are growing rapidly, perhaps the difference between the term values is increasing by some constant factor. Example: 2,5,9,14,\begin{align}2, 5, 9, 14, \ldots\end{align} (add 3, add 4, add 5, ...) • The terms may represent a particular type of number such as prime numbers, perfect squares, cubes, etc. Example: 2,3,5,7,\begin{align}2, 3, 5, 7, \ldots\end{align} (prime numbers) • Consider whether each term is the result of performing an operation on the two prior terms. Example: 2,5,7,12,19,\begin{align}2, 5, 7, 12, 19, \ldots\end{align} (add the previous two terms) • Consider the possibility that the value is connected to the term number: Example: 0,2,6,12,\begin{align}0, 2, 6, 12, \ldots\end{align} In this example (0×1)=0,(1×2)=2,(2×3)=6,(3×4)=12,\begin{align}(0 \times 1) = 0, (1 \times 2) = 2, (2 \times 3) = 6, (3 \times 4) = 12, \ldots\end{align} This list is not intended to be a comprehensive list of all possible patterns that may be present in a sequence but they are a good place to start when looking for a pattern. #### Example A Find the next two terms in the sequence: 160,80,40,20,,\begin{align}160, 80, 40, 20, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align} Solution: Each term is the result of multiplying the previous term by 12\begin{align}\frac{1}{2}\end{align} . Therefore, the next terms are: 12(20)=10\begin{align}\frac{1}{2}(20)=10\end{align} and 12(10)=5\begin{align}\frac{1}{2}(10)=5\end{align} #### Example B Find the next two terms in the sequence: 0,3,7,12,18,,\begin{align}0, 3, 7, 12, 18, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align} Solution: The difference between the first two terms (30)\begin{align}(3-0)\end{align} is 3, the difference between the second and third terms (73)\begin{align}(7-3)\end{align} is 4, the difference between the third and fourth terms (127)\begin{align}(12-7)\end{align} is 5 and the difference between the fourth and fifth terms (1812)\begin{align}(18-12)\end{align} is 6. Each time we add one more to get the next term. The next difference will be 7, so 18+7=25\begin{align}18+7=25\end{align} for the sixth term. To get the seventh term, we add 8, so 25+8=33\begin{align}25+8=33\end{align} . #### Example C Find the next two terms in the sequence: 9,5,4,1,3,,\begin{align}9, 5, 4, 1, 3, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align} Solution: This sequence requires that we look at the previous two terms. To get the third term, the second term was subtracted from the first: 95=4\begin{align}9-5=4\end{align} . To get the fourth term, the third term is subtracted from the second: 54=1\begin{align}5-4=1\end{align} . Similarly: 41=3\begin{align}4-1=3\end{align} . Now, to get the next terms, continue the pattern: 13=2\begin{align}1-3=-2\end{align} and 3(2)=5\begin{align}3-(-2)=5\end{align} Intro Problem Revisit Each successive term in the sequence is the result of multiplying the previous term by 34\begin{align}\frac{3}{4}\end{align} . Therefore, the next term, the fifth, is: 34(20.25)=15.1875\begin{align}\frac{3}{4}(20.25)=15.1875\end{align} . Therefore, the ball reaches a height of 15.1875 inches on its fifth bounce. ### Guided Practice Find the next two terms in each of the following sequences: 1. 5,1,3,7,,\begin{align}-5, -1, 3, 7, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align} 2. 13,23,79,56,\begin{align}\frac{1}{3}, \frac{2}{3}, \frac{7}{9}, \frac{5}{6} \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align} 3. 1,4,9,16,,\begin{align}1, 4, 9, 16, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align} 1. Each term is the previous term plus 4. Therefore, the next two terms are 11 and 15. 2. The pattern here is somewhat hidden because some of the fractions have been reduced. If we “unreduced” the second and fourth terms we get the sequence: 13\begin{align}\frac{1}{3}\end{align} , 46\begin{align}\frac{4}{6}\end{align} , 79\begin{align}\frac{7}{9}\end{align} , 1012,,\begin{align}\frac{10}{12}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align} . Now the pattern can be observed to be that the numerator and denominator each increase by 3. So the next two terms are 1315\begin{align}\frac{13}{15}\end{align} and 1618\begin{align}\frac{16}{18}\end{align} . Reducing the last term gives us the final answer of 1315\begin{align}\frac{13}{15}\end{align} and 89\begin{align}\frac{8}{9}\end{align} . 3. This sequence is the set of perfect squares or the term number squared. Therefore the 5th\begin{align}5^{th}\end{align} and 6th\begin{align}6^{th}\end{align} terms will be 52=25\begin{align}5^2=25\end{align} and 62=36\begin{align}6^2=36\end{align} . ### Explore More Find the next three terms in each sequence. 1. 15,21,27,33,,,\begin{align}15, 21, 27, 33, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align} 2. 4,12,36,108,,,\begin{align}-4, 12, -36, 108, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align} 3. 51,47,43,39,,,\begin{align}51, 47, 43, 39, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align} 4. 100,10,1,0.1,,,\begin{align}100, 10, 1, 0.1, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align} 5. 1,2,4,8,,,\begin{align}1, 2, 4, 8, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align} 6. 72,53,34,15,,,\begin{align}\frac{7}{2}, \frac{5}{3}, \frac{3}{4}, \frac{1}{5}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align} Find the missing terms in the sequences. 1. 1,4,,16,25,\begin{align}1, 4, \underline{\;\;\;\;\;\;\;}, 16, 25, \underline{\;\;\;\;\;\;\;}\end{align} 2. 23,34,,56,\begin{align}\frac{2}{3}, \frac{3}{4}, \underline{\;\;\;\;\;\;\;}, \frac{5}{6}, \underline{\;\;\;\;\;\;\;}\end{align} 3. 0,2,,9,14,\begin{align}0, 2, \underline{\;\;\;\;\;\;\;}, 9, 14,\underline{\;\;\;\;\;\;\;}\end{align} 4. 1,,27,64,125,\begin{align}1, \underline{\;\;\;\;\;\;\;}, 27, 64, 125, \underline{\;\;\;\;\;\;\;}\end{align} 5. 5,,11,17,28,,73\begin{align}5, \underline{\;\;\;\;\;\;\;}, 11, 17, 28, \underline{\;\;\;\;\;\;\;}, 73\end{align} 6. 3,8,,24,,48\begin{align}3, 8, \underline{\;\;\;\;\;\;\;}, 24, \underline{\;\;\;\;\;\;\;}, 48\end{align} 7. 1,1,2,,5,,13\begin{align}1, 1, 2, \underline{\;\;\;\;\;\;\;}, 5, \underline{\;\;\;\;\;\;\;}, 13\end{align} 8. Do any of the problems above have a constant difference? If so, which ones and what is the constant? 9. Do any of the problems above have a common ratio? If so, which ones and what is the ratio?
<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. Inverse Trigonometric Ratios Solving for an angle given a trigonometric ratio. Estimated6 minsto complete % Progress Practice Inverse Trigonometric Ratios Progress Estimated6 minsto complete % Inverse Trigonometric Ratios What if you were told that the longest escalator in North America is at the Wheaton Metro Station in Maryland and is 230 feet long and is 115 ft high? What is the angle of elevation, x\begin{align}x^\circ\end{align}, of this escalator? After completing this Concept, you'll be able use inverse trigonometry to answer this question. Guidance The word inverse is probably familiar to you. In mathematics, once you learn how to do an operation, you also learn how to “undo” it. For example, you may remember that addition and subtraction are considered inverse operations. Multiplication and division are also inverse operations. In algebra you used inverse operations to solve equations and inequalities. When we apply the word inverse to the trigonometric ratios, we can find the acute angle measures within a right triangle. Normally, if you are given an angle and a side of a right triangle, you can find the other two sides, using sine, cosine or tangent. With the inverse trig ratios, you can find the angle measure, given two sides. Inverse Tangent: If you know the opposite side and adjacent side of an angle in a right triangle, you can use inverse tangent to find the measure of the angle. Inverse tangent is also called arctangent and is labeled tan1\begin{align}\tan^{-1}\end{align} or arctan. The “-1” indicates inverse. Inverse Sine: If you know the opposite side of an angle and the hypotenuse in a right triangle, you can use inverse sine to find the measure of the angle. Inverse sine is also called arcsine and is labeled sin1\begin{align}\sin^{-1}\end{align} or arcsin. Inverse Cosine: If you know the adjacent side of an angle and the hypotenuse in a right triangle, you can use inverse cosine to find the measure of the angle. Inverse cosine is also called arccosine and is labeled cos1\begin{align}\cos^{-1}\end{align} or arccos. Using the triangle below, the inverse trigonometric ratios look like this: tan1(ba)sin1(bc)cos1(ac)=mB=mB=mBtan1(ab)=mAsin1(ac)=mAcos1(bc)=mA\begin{align}\tan^{-1} \left ( \frac{b}{a} \right ) & = m \angle B && \tan^{-1} \left ( \frac{a}{b} \right ) = m \angle A\\ \sin^{-1} \left ( \frac{b}{c} \right ) & = m \angle B && \sin^{-1} \left ( \frac{a}{c} \right ) = m \angle A\\ \cos^{-1} \left ( \frac{a}{c} \right ) & = m \angle B && \cos^{-1} \left ( \frac{b}{c} \right ) = m \angle A\end{align} In order to actually find the measure of the angles, you will need you use your calculator. On most scientific and graphing calculators, the buttons look like [SIN1],[COS1]\begin{align}[\text{SIN}^{-1}], [\text{COS}^{-1}]\end{align}, and [TAN1]\begin{align}[\text{TAN}^{-1}]\end{align}. Typically, you might have to hit a shift or 2nd\begin{align}2^{nd}\end{align} button to access these functions. For example, on the TI-83 and 84, [2nd][SIN]\begin{align}[2^{nd}][\text{SIN}]\end{align} is [SIN1]\begin{align}[\text{SIN}^{-1}]\end{align}. Again, make sure the mode is in degrees. Now that we know how to use inverse trigonometric ratios to find the measure of the acute angles in a right triangle, we can solve right triangles. To solve a right triangle, you would need to find all sides and angles in a right triangle, using any method. When solving a right triangle, you could use sine, cosine or tangent, inverse sine, inverse cosine, or inverse tangent, or the Pythagorean Theorem. Remember when solving right triangles to only use the values that you are given. Example A Use the sides of the triangle and your calculator to find the value of A\begin{align}\angle A\end{align}. Round your answer to the nearest tenth of a degree. In reference to A\begin{align}\angle A\end{align}, we are given the opposite leg and the adjacent leg. This means we should use the tangent ratio. tanA=2025=45\begin{align}\tan A = \frac{20}{25} = \frac{4}{5}\end{align}, therefore tan1(45)=mA\begin{align}\tan^{-1} \left ( \frac{4}{5} \right ) = m \angle A\end{align}. Use your calculator. If you are using a TI-83 or 84, the keystrokes would be: [2nd][TAN](45)\begin{align}[2^{nd}][\text{TAN}]\left ( \frac{4}{5} \right )\end{align}[ENTER] and the screen looks like: So, mA=38.7\begin{align}m \angle A = 38.7^\circ\end{align} Example B A\begin{align}\angle A\end{align} is an acute angle in a right triangle. Use your calculator to find mA\begin{align}m \angle A\end{align} to the nearest tenth of a degree. a) sinA=0.68\begin{align}\sin A = 0.68\end{align} b) cosA=0.85\begin{align}\cos A = 0.85\end{align} c) tanA=0.34\begin{align}\tan A = 0.34\end{align} Solutions: a) mA=sin10.68=42.8\begin{align}m \angle A = \sin^{-1} 0.68 = 42.8^\circ\end{align} b) mA=cos10.85=31.8\begin{align}m \angle A = \cos^{-1} 0.85 = 31.8^\circ\end{align} c) mA=tan10.34=18.8\begin{align}m \angle A = \tan^{-1} 0.34 = 18.8^\circ\end{align} Example C Solve the right triangle. To solve this right triangle, we need to find AB,mC\begin{align}AB, m \angle C\end{align} and mB\begin{align}m \angle B\end{align}. Use AC\begin{align}AC\end{align} and CB\begin{align}CB\end{align} to give the most accurate answers. AB\begin{align}\underline{AB}\end{align}: Use the Pythagorean Theorem. 242+AB2576+AB2AB2AB=302=900=324=324=18\begin{align}24^2 + AB^2 & = 30^2\\ 576 + AB^2 & = 900\\ AB^2 & = 324\\ AB & = \sqrt{324} = 18\end{align} mB\begin{align}\underline{m \angle B}\end{align}: Use the inverse sine ratio. sinBsin1(45)=2430=45=53.1=mB\begin{align}\sin B & = \frac{24}{30} = \frac{4}{5}\\ \sin^{-1} \left ( \frac{4}{5} \right ) & = 53.1^\circ = m \angle B\end{align} mC\begin{align}\underline{m \angle C}\end{align}: Use the inverse cosine ratio. cosCcos1(45)=2430=45=36.9=mC\begin{align}\cos C & = \frac{24}{30} = \frac{4}{5}\\ \cos^{-1} \left ( \frac{4}{5} \right ) & = 36.9^\circ = m \angle C\end{align} Watch this video for help with the Examples above. Concept Problem Revisited To find the escalator’s angle of elevation, we need to use the inverse sine ratio. sin1(115230)=30The angle of elevation is 30.\begin{align}\sin^{-1} \left ( \frac{115}{230} \right ) = 30^\circ \qquad \text{The angle of elevation is}\ 30^\circ.\end{align} Vocabulary Trigonometry is the study of the relationships between the sides and angles of right triangles. The legs are called adjacent or opposite depending on which acute angle is being used. The three trigonometric (or trig) ratios are sine, cosine, and tangent. The inverse trig ratios, sin1, cos1\begin{align}\sin^{-1}, \ \cos^{-1}\end{align}, and tan1\begin{align}\tan^{-1}\end{align}, allow us to find missing angles when we are given sides. Guided Practice 1. Solve the right triangle. 2. Solve the right triangle. 1. To solve this right triangle, we need to find AB,BC\begin{align}AB, BC\end{align} and mA\begin{align}m \angle A\end{align}. AB\begin{align}\underline{AB}\end{align}: Use sine ratio. sin62ABAB=25AB=25sin6228.31\begin{align}\sin 62^\circ & = \frac{25}{AB}\\ AB & = \frac{25}{\sin 62^\circ}\\ AB & \approx 28.31\end{align} BC\begin{align}\underline{BC}\end{align}: Use tangent ratio. tan62BCBC=25BC=25tan6213.30\begin{align}\tan 62^\circ & = \frac{25}{BC}\\ BC & = \frac{25}{\tan 62^\circ}\\ BC & \approx 13.30\end{align} mA\begin{align}\underline{m \angle A}\end{align}: Use Triangle Sum Theorem 62+90+mAmA=180=28\begin{align}62^\circ + 90^\circ + m \angle A & = 180^\circ\\ m \angle A & = 28^\circ\end{align} 2. Even though, there are no angle measures given, we know that the two acute angles are congruent, making them both 45\begin{align}45^\circ\end{align}. Therefore, this is a 45-45-90 triangle. You can use the trigonometric ratios or the special right triangle ratios. Trigonometric Ratios tan45BC=15BC=15tan45=15sin45=15AC AC=15sin4521.21\begin{align}\tan 45^\circ & = \frac{15}{BC} && \sin 45^\circ = \frac{15}{AC}\\ BC & = \frac{15}{\tan 45^\circ} = 15 && \quad \ AC = \frac{15}{\sin 45^\circ} \approx 21.21\end{align} 45-45-90 Triangle Ratios \begin{align}BC = AB = 15, AC = 15 \sqrt{2} \approx 21.21\end{align} Practice Use your calculator to find \begin{align}m \angle A\end{align} to the nearest tenth of a degree. Let \begin{align}\angle A\end{align} be an acute angle in a right triangle. Find \begin{align}m \angle A\end{align} to the nearest tenth of a degree. 1. \begin{align}\sin A = 0.5684\end{align} 2. \begin{align}\cos A =0.1234\end{align} 3. \begin{align}\tan A = 2.78\end{align} Solve the following right triangles. Find all missing sides and angles. 1. Writing Explain when to use a trigonometric ratio to find a side length of a right triangle and when to use the Pythagorean Theorem. My Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Vocabulary Language: English Two angles are adjacent if they share a side and vertex. The word 'adjacent' means 'beside' or 'next-to'. Conic Conic sections are those curves that can be created by the intersection of a double cone and a plane. They include circles, ellipses, parabolas, and hyperbolas. cosine The cosine of an angle in a right triangle is a value found by dividing the length of the side adjacent the given angle by the length of the hypotenuse. Hypotenuse The hypotenuse of a right triangle is the longest side of the right triangle. It is across from the right angle. Pythagorean Theorem The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $a^2 + b^2 = c^2$, where $a$ and $b$ are legs of the triangle and $c$ is the hypotenuse of the triangle. sine The sine of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the hypotenuse. Slope Slope is a measure of the steepness of a line. A line can have positive, negative, zero (horizontal), or undefined (vertical) slope. The slope of a line can be found by calculating “rise over run” or “the change in the $y$ over the change in the $x$.” The symbol for slope is $m$ Tangent The tangent of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the side adjacent to the given angle.
# Inverses, converses, and contrapositives ### Inverses, converses, and contrapositives #### Lessons Let $p$ be the hypothesis and $q$ be the conclusion. Then: An inverse statement is formed by negating both the hypothesis and conclusion of the conditional. In symbolic form it would be: ~$p$ $\to$ ~$q$ A converse statement is formed by switching the hypothesis and the conclusion of the conditional. In symbolic form it would be: $q\ \to p$ A contrapositive statement is formed by negating both the hypothesis and conclusion, AND switching them. In symbolic form it would be: ~$q$ $\to$ ~$p$ Statements which always have the same truth values are logical equivalents. Conditionals and contrapositives are logical equivalents, and inverses and converses are logical equivalents. • Introduction The inverse, converse, and contrapositive Overview: a) Inverse Statements b) Converse Statements c) Contrapositive Statements d) Logical equivalents • 1. Finding the inverse, converse and contrapositive Given the statements, write the inverse, converse, and contrapositive: a) Two intersecting lines create an angle. b) If today is Monday, then Kevin will play soccer. c) If $1+2=3$, then $1^2+2^2=3^2$. d) If the polygon is a triangle, then it has 3 sides. • 2. For each statement, write the inverse, converse, and contrapositive in symbolic form: a) $r \to s$ b) $p$ $to$ ~$q$ c) ~$m$ $\to$ $n$ d) ~$u$ $\to$ ~$v$ • 3. Truth value of inverse, converse and contrapositive Write the converse, and find the truth value of the converse: a) If $x+7=13$, then $x=6$. b) If $3$ is odd, then $3+1$ is even. c) If $2$ is an integer, then $2$ is a whole number. • 4. Assume that the conditional statement is true. Write the inverse and state whether the inverse is always true, sometimes true, or never true: a) $p \to q$ b) $q\ to p$ c) ~$q$ $\to$ ~$p$ d) $p$ $\to$ ~$q$ • 5. Logical Equivalents Find the truth value of the following conditionals. Then write the contrapositive and find the truth value of the contrapositive. Are the truth values the same? a) If (2+3)×4=20, then $2+3=5$. b) If $4$ is even, then $4+2$ is even. c) If $3$ is an integer, then $3$ is not a whole number. • 6. Finding truth values of original statements Find possible truth values for $p$ and $q$ where: a) $p \to q$ and ~$q$ $\to$ ~$p$ is both false? b) $p$ $\to$ $q$ and $q$ $\to$ $p$ is both true? c) $p \to q$ and $q \to p$ is both false? d) $q \to p$ and ~$q$ $\to$ ~$p$ is both true?
# Heron’s formula ## Formula $A = \sqrt{s(s-a)(s-b)(s-c)}$ ### Introduction A mathematician Heron (or Hero) of Alexandria derived a geometrical proof to express area of a triangle in algebraic form in terms of lengths of three sides and half-perimeter of the triangle. Hence, this formula is called as Heron’s formula or Hero’s formula. $a$, $b$ and $c$ are lengths of three sides of a triangle and its perimeter is denoted by $2s$. $2s \,=\, a+b+c$ $\implies$ $s \,=\, \dfrac{a+b+c}{2}$ The area of the triangle is denoted by either $A$ or $\Delta$. $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ This formula in algebraic form is called as Hero’s (or Heron’s) formula in geometry. #### Example Find area of a triangle, if $a = 5cm$, $b = 6cm$ and $c = 7cm$. Firstly, find the semi perimeter of the triangle. $s = \dfrac{5+6+7}{2}$ $\implies$ $s = \dfrac{18}{2}$ $\implies$ $\require{cancel} s = \dfrac{\cancel{18}}{{2}}$ $\implies$ $s = 9cm$ Now, find the area of the triangle. $A = \sqrt{9(9-5)(9-6)(9-7)}$ $\implies$ $A = \sqrt{9(4)(3)(2)}$ $\implies$ $A = \sqrt{9 \times 4 \times 3 \times 2}$ $\implies$ $A = \sqrt{36 \times 6}$ $\,\,\, \therefore \,\,\,\,\,\,$ $A = 6\sqrt{6} \, cm^2$ ### Proof Learn how to derive the hero’s formula in geometrical approach to find the area of a triangle. ###### Math Questions The math problems with solutions to learn how to solve a problem. Learn solutions Practice now ###### Math Videos The math videos tutorials with visual graphics to learn every concept. Watch now ###### Subscribe us Get the latest math updates from the Math Doubts by subscribing us.
• :00Days • :00Hours • :00Mins • 00Seconds A new era for learning is coming soon Suggested languages for you: Americas Europe Q44SE Expert-verified Found in: Page 805 ### Discrete Mathematics and its Applications Book edition 7th Author(s) Kenneth H. Rosen Pages 808 pages ISBN 9780073383095 # Find a minimum spanning tree of each of these graphs where the degree of each vertex in the spanning tree does not exceed 2. 1. Therefore, the resulting minimum spanning tree is given in the graph shown below. 2. Therefore, the resulting minimum spanning tree is given in the graph shown below. See the step by step solution ## Step 1: General form Definition of tree: A tree is a connected undirected graph with no simple circuits. Definition of Circuit: It is a path that begins and ends in the same vertex. Minimum spanning tree: A spanning tree with smallest possible sum of weights of its edges. ## Step 2: (a) Evaluate the minimum spanning tree of the given graph Given that, a graph is shown below. Noticed that the edges with the smallest weight 1 are $$\left\{ {b,c} \right\}$$ and $$\left\{ {b,d} \right\}$$. Since, both edges cannot be included, because we will need a degree of more than 2 to connect to a, f or e. So, we can only include one of the edges. For example, let us use $$\left\{ {b,c} \right\}$$, then we will also have to use $$\left\{ {c,d} \right\}$$ to connected to d. Then we need to connect b to another vertex (a, e, or f). we shouldn’t connect b to f as the weight on $$\left\{ {b,c} \right\}$$ is the largest weight of an edge incident to b. we can choose among the other two vertices. For example, let us choose $$\left\{ {a,b} \right\}$$. At last, noticed that $$\left\{ {a,f} \right\}$$ and $$\left\{ {e,f} \right\}$$ have to be in the minimum spanning tree. Because the graph either connected or contains a vertex with degree more than 2. The resulting minimum spanning tree is given in the graph shown below. Conclusion: The above graph represents the possibility of spanning tree. ## Step 3: (b) Evaluate the minimum spanning tree of the given graph Given that, a graph is shown below. Noticed that a is only connected t b, so $$\left\{ {a,b} \right\}$$ has to be included in the graph. Then, use at most one more edge incident to b. We cannot use the edges with weight 1,4 and 5. Because we will require the use of three edges incident to g, to d, to e or to g. We should use the edge of weight 2, which is $$\left\{ {b,c} \right\}$$. We cannot use any more edges incident to b. so, we have to include the edges $$\left\{ {c,d} \right\}$$ and $$\left\{ {d,g} \right\}$$. Since, $$\left\{ {e,g} \right\}$$ has a weight that is the sum of the weight of $$\left\{ {e,f} \right\}$$ and $$\left\{ {f,g} \right\}$$, we should include the two edges. So, the resulting minimum spanning tree is given in the graph shown below. Hence, the above graph represents the possibility of spanning tree.
Given (sqrt(x+1) - sqrt(2x+1)) / (sqrt(3x+4) - sqrt(2x+4) ] how do you find the limit as x approaches 0? Nov 22, 2016 Change the way it is written to avoid $\frac{0}{0}$. Explanation: By the time this problem is assigned, I assume students have seen things like ${\lim}_{x \rightarrow 0} \frac{\sqrt{9 + x} - 3}{x}$ which is found by 'rationalizing' the numerator: ${\lim}_{x \rightarrow 0} \frac{\left(\sqrt{9 + x} - 3\right)}{x} \cdot \frac{\left(\sqrt{9 + x} + 3\right)}{\left(\sqrt{9 + x} + 3\right)} = {\lim}_{x \rightarrow 0} \frac{1}{\left(\sqrt{9 + x} - 3\right)} = \frac{1}{6}$ In this problem, we have to try something, so late's use the same trick on both the numerator and denominator (sqrt(x+1)-sqrt(2x+1))/(sqrt(3x+4)-sqrt(2x+4)) = ((sqrt(x+1)-sqrt(2x+1)))/((sqrt(3x+4)-sqrt(2x+4))) * ((sqrt(x+1)+sqrt(2x+1))(sqrt(3x+4)+sqrt(2x+4)))/((sqrt(x+1)+sqrt(2x+1))(sqrt(3x+4)+sqrt(2x+4)) $= \frac{\left(x + 1 - \left(2 x + 1\right)\right) \left(\sqrt{3 x + 4} + \sqrt{2 x + 4}\right)}{\left(x + 4 - 4\right) \left(\sqrt{x + 1} + \sqrt{2 x + 1}\right)}$ $= \frac{- x \left(\sqrt{3 x + 4} + \sqrt{2 x + 4}\right)}{x \left(\sqrt{x + 1} + \sqrt{2 x + 1}\right)}$ $= \frac{- \left(\sqrt{3 x + 4} + \sqrt{2 x + 4}\right)}{\sqrt{x + 1} + \sqrt{2 x + 1}}$ $\text{ }$ (for $x \ne 0$) ${\lim}_{x \rightarrow 0} \frac{\sqrt{x + 1} - \sqrt{2 x + 1}}{\sqrt{3 x + 4} - \sqrt{2 x + 4}} = {\lim}_{x \rightarrow 0} \frac{- \left(\sqrt{3 x + 4} + \sqrt{2 x + 4}\right)}{\sqrt{x + 1} + \sqrt{2 x + 1}}$ $= - \frac{\sqrt{4} + \sqrt{4}}{\sqrt{1} + \sqrt{1}} = \frac{- 4}{2} = - 2$ The same algebra with simplified notation gets us $\frac{\sqrt{a} - \sqrt{b}}{\sqrt{c} - \sqrt{d}} = \frac{\left(\sqrt{a} - \sqrt{b}\right)}{\left(\sqrt{c} - \sqrt{d}\right)} \cdot \frac{\left(\sqrt{a} + \sqrt{b}\right) \left(\sqrt{c} + \sqrt{d}\right)}{\left(\sqrt{a} + \sqrt{b}\right) \left(\sqrt{c} + \sqrt{d}\right)}$ $= \frac{\left(a - b\right) \left(\sqrt{c} + \sqrt{d}\right)}{\left(c - d\right) \left(\sqrt{a} + \sqrt{b}\right)}$
# Question: What Are The Factors Of 4 And 18? ## What are the common factors of 12 and 18? In terms of numbers, the greatest common factor (gcf) is the largest natural number that exactly divides two or more given natural numbers. Example 1: 6 is the greatest common factor of 12 and 18.. ## Which is the biggest factor of 18? The factors of 18 are 1, 2, 3, 6, 9, 18. The factors of 27 are 1, 3, 9, 27. The common factors of 18 and 27 are 1, 3 and 9. The greatest common factor of 18 and 27 is 9. ## What are the common factors of 18 and 30? Factors for 18: 1, 2, 3, 6, 9, and 18. Factors for 30: 1, 2, 3, 5, 6, 10, 15, and 30….What is the Greatest Common Factor?Greatest Common Denominator (GCD)Highest Common Factor (HCF)Greatest Common Divisor (GCD) ## What are the common factors of 25? The factors of 25 are 1, 5, and 25. The common factors of 20 and 25 are 1 and 5. Example: What are the common factors of 15 and 30? ## Is 17 a perfect square? 17 is not a perfect square. ## How do you find the common factors? To find the GCF of two numbers:List the prime factors of each number.Multiply those factors both numbers have in common. If there are no common prime factors, the GCF is 1. ## What is the highest common factor of 12 15 and 18? We found the factors 12,15,18 . The biggest common factor number is the GCF number. So the greatest common factor 12,15,18 is 3. ## What are the common factors of 4 and 10? If there is more than one, the greatest number will be the greatest common factor. In this case, the only common number is 2, and so it is automatically the greatest common factor. The GCF = 2. Both 4 and 10 are even, so they must have a common factor of 2. ## What is the HCF of 18 24 and 30? As you can see when you list out the factors of each number, 6 is the greatest number that 18, 24, and 30 divides into. ## What are multiples of 18? The list of multiples of 18 are: 18,36,54,72,90,108,126,144,162,180,198,216,234,252,270,…. Sometimes multiples are misunderstood as factors also, which is not correct. Factors of 18 consists of only those numbers which are multiplied together to get the original number. ## What are the factors of 18? Table of Factors and MultiplesFactorsMultiples1, 3, 5, 1515301, 2, 4, 8, 1616321, 1717341, 2, 3, 6, 9, 18183641 more rows ## What are the factors of 4? 4 = 1 x 4 or 2 x 2. Factors of 4: 1, 2, 4. Prime factorization: 4 = 2 x 2, which can also be written 4 = 2². ## Is 24 a perfect square? 24 is NOT a perfect square. 24 is a natural number, but since there is no other natural number that can be squared to result in the number 24, 24 is NOT a perfect square. ## Is 18 a perfect square? 18 is not a perfect square. ## What is the highest common factor of 12 18 and 30? As you can see when you list out the factors of each number, 6 is the greatest number that 12, 18, and 30 divides into. ## What is the greatest common factor of 18 and 24? The greatest common factor is the greatest factor that divides both numbers. To find the greatest common factor, first list the prime factors of each number. 18 and 24 share one 2 and one 3 in common. We multiply them to get the GCF, so 2 * 3 = 6 is the GCF of 18 and 24. ## What is the GCF of 18 and 36? Greatest common factor (GCF) of 18 and 36 is 18. We will now calculate the prime factors of 18 and 36, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 18 and 36. ## Why is 18 not a perfect square? A number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. … Thus, the square root of 18 is not an integer, and therefore 18 is not a square number.
Home \| Learning \| History \| Fun \|Real-Life \| Help/Contact \| Index/Site-Map Lessons 1. Basics 2. Deductive Reasoning • 2.1 If-Then Statements • 2.2 How To Prove Theorems • 2.3 Pairs Of Angles • 2.4 Perpendicular Lines 3 - Parallel lines • 3.1 Parallel Lines • 3.2 Angles • 3.3 Inductive Reasoning 4 - Congruent Triangles • 4.1 Congruent Triangles • 4.2 Ways to prove it • 4.3 Bisectors 5 - Quadrilaterals • 5.1 Parallelograms • 5.2 Parallel lines Theorem • 5.3 Special Parallelograms • 5.4 Trapezoids 6 - Inequalities • 6.1 Inequalities • 6.2 Inequalities In A Triangle • 6.3 Inequalities In 2 Triangles 7 - Similar Polygon • 7.1 Ratios and Proportion • 7.2 Similar Polygons • 7.3 Similar Triangles 8 - Rt. Triangles • 8.1 Right Triangle • 8.2 Trig. In Geometry 9 - Circles • 9.1 Tangents, Arcs, and Chords • 9.2 Angles and Segment 10 - Constructions • 10.1 Construction • 10.2 Perpendicular Lines • 10.3 Parallel Lines • 10.4 Concurrent Lines • 10.5 Circles 11 - Areas of 2D objects • 11. 1 Areas Of Polygons • 11. 2 Circles and Similar Figures 12 - Areas and Volumes • 12.1 Prisms • 12.2 Pyramids • 12.3 Cylinders and cones • 12.4 Spheres • 12.5 Similar solids 13 - Coordinates • 13.1 Geometry and Algebra Trigonometry in Geometry Objective: • Learning about tangent, sine, and cosine rations for an acute angle • Using tangent, sine, and cosine rations to solve right triangle problems Lesson 8-2 Trigonometry in Geometry: Tangent Ratio: The measure of the leg opposite of the angle over the measure of the leg adjacent of the angle. Tangent of / A = (Leg opposite / A )/ (Adjacent leg / A) Tangent of / A = 3/4 Sine: The measure of the leg opposite of the angle over the measure of hypotenuse Sine of / A = (Leg opposite / A )/ Hypotenuse Sine of / A = 4/5 Cosine: The measure of the leg adjacent to the angle over the measure of the hypotenuse cosine of / A = (Leg adjacent to / A )/ Hypotenuse cosine of / A = 3/5 Easy way to remember trigonometric ratios: Soh, Cah, Toa or Oh Heck, Another Hour of Algebra Sin A= Opp/ hyp Cos A= Adj/ hyp Tan A= Opp/ adj Quickie Math Copyright (c) 2000 Team C006354
# Limit using L'Hopital's Rule Find $$\lim_{x \to 1} \sqrt{x-1}^{\,\sin(πx)}$$ using L'Hopital's Rule. Initially I get $0^0$ so I know I need to use the rule, but I don't know where to begin. Could you help me out with some steps on how to solve this limit? Let $u(x)=\sqrt{x-1}^{\sin(\pi x)}$. We have $$\begin{array}{lll} \lim_{x\rightarrow 1^+}\ln u(x)&=&\lim_{x\rightarrow 1^+}\sin(\pi x)\ln(\sqrt{x-1})\\ &=&\frac{1}{2}\lim_{x\rightarrow 1^+}\frac{\ln(x-1)}{\frac{1}{\sin(\pi x)}}\\ &\overset{H^\prime}{=}&\frac{1}{2}\lim_{x\rightarrow 1^+}\frac{\frac{1}{x-1}}{\frac{-\pi\cos(\pi x)}{\sin^2(\pi x)}}\\ &=&\frac{1}{2}\lim_{x\rightarrow 1^+}\frac{-\sin^2(\pi x)}{\pi(x-1)\cos(\pi x)}\\ &=&\frac{1}{2}\lim_{x\rightarrow 1^+}\frac{-\sin^2[\pi (x-1)]}{\pi(x-1)\cos(\pi x)}\\ &=&\frac{1}{2}\lim_{x\rightarrow 1^+}\frac{-\pi^2 (x-1)^2}{\pi(x-1)\cos(\pi x)}\\ &=&0. \end{array}$$ Hence $\displaystyle\lim_{x\rightarrow 1^+}\sqrt{x-1}^{\sin(\pi x)}=\lim_{x\rightarrow 1^+}u(x)=\lim_{x\rightarrow 1^+}e^{\ln u(x)}=e^{\displaystyle\lim_{x\rightarrow 1^+}\ln u(x)}=e^0=1.$ Let $L =\lim_{x \to 1} \sqrt{x-1}^{\sin(πx)}$. Now take Log on both sides and try now using L'hopitals rule. At the you have to solve for L. I assume you meant $$\lim_{x \to 1^+} \sqrt{x-1}^{\sin(\pi x)}$$ You could get away without L'hopitals rule. $$\underbrace{\lim_{x \to 1^+} \sqrt{x-1}^{\sin(\pi x)} = \lim_{t \to 0} t^{\sin(\pi(1+t^2))}}_{x = 1 + t^2} = \lim_{t \to 0} t^{\sin(\pi t^2)}$$ $$f(t) = t^{\sin(\pi t^2)} \implies \log(f(t)) = \sin \left(\pi t^2 \right) \log(t)$$ $$\left \vert \sin \left(\pi t^2 \right) \log(t) \right \vert \leq \left \vert \pi t^2 \log(t) \right \vert$$ Now $$\displaystyle \lim_{t \to 0} t^2 \log(t) = 0 \text{ (Why?)}$$ Hence, $$\lim_{t \to 0} \log(f(t)) = 0 \implies \lim_{t \to 0} f(t) = 1 \text{ (Why?)}$$
# The Square Root Of 69 Is 8 Something The Square Root of 69 is 8 Something â€“ Understanding the Maths Do you ever find yourself wondering what the square root of 69 is? It can be difficult to remember all the numbers, especially when it comes to maths. But with a little bit of knowledge, you can figure out the answer. The square root of 69 is 8 something. ## How to Calculate the Square Root of 69 Contents Calculating the square root of 69 is fairly simple. All you have to do is divide the number by itself. In this case, you would divide 69 by 69. This equation would be written as 69/69=1. This means that the square root of 69 is 1. ## Understanding the Result If you look at the equation, you can see that the square root of 69 is 8 something. This means that the number 8 is the root of the number 69. So, when you divide 69 by 8, you get the same result as when you divide 69 by itself. ## Why is the Square Root of 69 8 Something? The square root of 69 is 8 something because 8 is the root of the number 69. When you divide 69 by 8, you get the same result as when you divide 69 by itself. This is because the square root of a number is equal to the number divided by itself. ## Using the Square Root of 69 in Everyday Life The square root of 69 is 8 something, and this number can be used in everyday life. For example, if you are trying to determine the area of a rectangle, you can use the square root of 69 to help you. By multiplying the length of the rectangle by the width, you can get the area. ## Other Uses for the Square Root of 69 The square root of 69 is 8 something, and this number can also be used in other ways. For instance, if you want to calculate the circumference of a circle, you can use the square root of 69 to help you. By multiplying the diameter of the circle by the square root of 69, you can get the circumference. ## Understanding the Square Root of 69 The square root of 69 is 8 something, and understanding this number can help you with many everyday calculations. By dividing the number 69 by 8, you can get the same result as when you divide it by itself. This means that the square root of 69 is 8 something.
# Difference between revisions of "2015 UNCO Math Contest II Problems/Problem 10" ## Problem $\begin{tabular}[t]{\|c\|c\|c\|c\|}\hline & & & \\\hline & & & \\\hline \end{tabular}$ (a) You want to arrange $8$ biologists of $8$ different heights in two rows for a photograph. Each row must have $4$ biologists. Height must increase from left to right in each row. Each person in back must be taller than the person directly in front of him. How many different arrangements are possible? $\begin{tabular}[t]{\|c\|c\|c\|c\|c\|c\|}\hline & & & & & \\\hline & & & & & \\\hline \end{tabular}$ (b) You arrange $12$ biologists of $12$ different heights in two rows of $6$, with the same conditions on height as in part (a). How many different arrangements are possible? Remember to justify your answers. (c) You arrange $2n$ biologists of $2n$ different heights in two rows of $n$, with the same conditions on height as in part (a). Give a formula in terms of $n$ for the number of possible arrangements. ## Solution (a) $14$ (b) $132$ (c) $\frac{1}{n + 1} \binom{2n}{n}$ $= \binom{2n}{n}- \binom{2n}{n-1}= \binom{2n}{n}- \binom{2n}{n+1}=\frac{1}{2n+1} \binom{2n+1}{n}=\frac{(2n)!}{(n+1)!n!}$ ## See also 2015 UNCO Math Contest II (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byBONUS 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 All UNCO Math Contest Problems and Solutions Invalid username Login to AoPS
David Beckham Under The Microscope (11/10/2019) How will David Beckham get by on 11/10/2019 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is of questionable accuracy – do not take this too seriously. I will first find the destiny number for David Beckham, and then something similar to the life path number, which we will calculate for today (11/10/2019). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology experts. PATH NUMBER FOR 11/10/2019: We will take the month (11), the day (10) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. Here’s how it works. First, for the month, we take the current month of 11 and add the digits together: 1 + 1 = 2 (super simple). Then do the day: from 10 we do 1 + 0 = 1. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 2 + 1 + 12 = 15. This still isn’t a single-digit number, so we will add its digits together again: 1 + 5 = 6. Now we have a single-digit number: 6 is the path number for 11/10/2019. DESTINY NUMBER FOR David Beckham: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart: So for David Beckham we have the letters D (4), a (1), v (4), i (9), d (4), B (2), e (5), c (3), k (2), h (8), a (1) and m (4). Adding all of that up (yes, this can get tiring) gives 47. This still isn’t a single-digit number, so we will add its digits together again: 4 + 7 = 11. This still isn’t a single-digit number, so we will add its digits together again: 1 + 1 = 2. Now we have a single-digit number: 2 is the destiny number for David Beckham. CONCLUSION: The difference between the path number for today (6) and destiny number for David Beckham (2) is 4. That is greater than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A BAD RESULT. But don’t get too concerned! As mentioned earlier, this is of questionable accuracy. If you want to see something that we really strongly recommend, check out your cosmic energy profile here. Go see what it says for you now – you may be absolutely amazed. It only takes 1 minute. Abigale Lormen Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. Latest posts by Abigale Lormen (see all) Abigale Lormen Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
Session 7, Part C: Figurate Numbers In This Part: Square Numbers \| Triangular Numbers "Triangular numbers" describe the number of dots needed to make triangles like the ones below. The first triangular number is 1, the second is 3, and so on. Problem C5 Draw the next two triangles in this pattern. Problem C6 Fill in the table below: Number of dots on side of triangle Total number of dots (triangular number) 1 1 2 3 3 4 5 6 45 190 Number of dots on side of triangle Total number of dots (triangular number) 1 1 2 3 3 6 4 10 5 15 6 21 9 45 19 190 Problem C7 Graph the data in your table using graph paper or a spreadsheet, then describe your graph. How is it different from the linear and exponential graphs you've seen? Problem C8 Describe a rule relating the number of dots on the side of a triangle (the independent variable) and the total number of dots (the dependent variable). Problem C9 Describe any similarities and differences between your rules and graphs for the square and triangular numbers. In a way, they both have the same kind of rule. How would you describe it? Note 11 Think about how you would go from one output to the next. What changes? Can you describe these changes with a rule? Close Tip Think about how you would go from one output to the next. What changes? Can you describe these changes with a rule? You can create "figurate numbers" for any polygon shape. Below are pictures of the first few pentagonal and hexagonal numbers. The growth of figurate numbers is an example of a quadratic function. A quadratic function's formula will always involve squaring the input number: y = 3x2 + 5 is a quadratic function, and y = 3x + 5 is not. In Part D, we will explore the formulas and properties of quadratic functions in more detail. Session 7: Index \| Notes \| Solutions \| Video
# Surface Area of Triangular Pyramid Home » Math Vocabulary » Surface Area of Triangular Pyramid ## Surface Area of Triangular Pyramid: Introduction The surface area, or total surface area (TSA), of a triangular pyramid, is the total area of its 4 faces. Even though we have formulas to find the surface area of a pyramid with a triangular base, the basic idea of finding the surface is to add the areas of all the faces. It is measured in square units such as $in^{2}$, $ft^{2}$, etc. There are many types of pyramids, and all pyramids are named by the shape of their bases. For example, a pyramid, which has a rectangular base is a rectangular pyramid; the one that has a pentagon as a base is a pentagonal pyramid, and so on. So, a triangular pyramid has a base that is triangular in shape. Let’s first learn about triangular pyramids. ## What is a Triangular Pyramid? A triangular pyramid is a pyramid with a triangle as a base and 3 triangular faces. These 3 lateral triangular faces meet at one vertex. All triangular-based pyramids, either regular or irregular, have 4 vertices. They have 6 edges, out of which 3 are along the base and 3 are extending up from the base. Take a look at the image given below to understand its components. ## How to Find the Surface Area of a Triangular Pyramid A triangular pyramid has two types of surface areas: The lateral surface area and the total surface area. You can find the surface area of a triangular pyramid by adding the lateral area of the pyramid (three triangles) to the area of the pyramid’s base (a triangle again). Let us study both in detail. ### Lateral Surface Area The lateral surface area is the area of the side faces, or we can say lateral surface area (LSA) does not include the base. It can be calculated by subtracting the base from the total surface area. ### Lateral Surface Area of a Triangular Pyramid Formula The lateral surface area of a triangular pyramid is calculated using the following formula: The lateral surface area of a triangular pyramid $= \frac{1}{2}$ (Perimeter of the base $\times$ Slant height) We know that the base is a triangle. So, the perimeter of the base would be, $3 \times$ side for equilateral triangles. The lateral surface area of a triangular pyramid $= \frac{3}{2}$ (side slant $\times$ height) $= \frac{3}{2}(b \times l)$ Where, $b =$ side of the base triangular base $s =$ slant height ### Total Surface Area The total surface area of an object is the total area of the object’s surface. Thus, the total surface area is the sum of the base area and lateral surface area of the pyramid. ### Total Surface Area of a Triangular Pyramid Formula Here, h is the height of the pyramid. l is the slant height of the pyramid. The total surface area of a triangular pyramid is calculated using the following formula: The total surface area of a triangular pyramid $=$ Base Area $+$ Lateral Surface Area As we know, Area of a triangle $= 1\frac{2} \times$ base $\times$ height Total surface area of triangular pyramid $= (\frac{1}{2} \times$ Side of base $\times$ Height of base) $+ \frac{3}{2}$ (b $\times$ l) $= \frac{1}{2} \times$ (b $\times$ h) $+ \frac{3}{2}$ (b $\times$ l) where, $h =$ the height of the base triangle $b =$ the side of the triangular base $l =$ slant height ## Fun Facts! • A triangle-based pyramid is unique in comparison to other pyramids, as it only consists of one kind of 2D shape, a triangle. • A triangular pyramid has 4 triangular faces, 6 edges, and 4 vertices. • Triangular pyramids can be regular, irregular, and right-angled. • The base of a triangle pyramid can be any type of triangle, but is usually an equilateral triangle. This means that all sides and faces are the same. • The net of a triangle-based pyramid folds out to a large triangle made up of 4 smaller triangles. • The base of a right triangular pyramid is a right triangle. ## Conclusion In this article, we learned in detail about the triangular pyramid definition, and formulas and facts. Let’s solve some examples and practice problems to understand it better. ## Solved Examples 1. Determine the total surface area of a triangular pyramid whose base area is 36 sq. in, the perimeter of the triangle is 24 in, and the slant height of the pyramid is 28 in. Solution: Given data: Area of the triangular base $= 36\; in^{2}$ The slant height (l) $= 28$ in Perimeter (P) $= 24$ in We know that, Total surface area (TSA) of a triangular pyramid $= \frac{1}{2}$ perimeter of the base $\times$ slant height $+$ base area Thus, TSA $= (\frac{1}{2} \times 24 \times 28 ) + 36$ $= 12 \times 28 + 36$ $= 336 + 36$ $= 372\; in^{2}$ Hence, the total surface area of the given pyramid is $372\; in^{2}$. 2. Find the total surface area of a triangular pyramid with base lengths of 10 and base height of 8.7 and slant height of 14. (Note: Base is an equilateral triangle.) Solution: Given data: Base length $= 10$ in Base height $= 7$ in Slant height $= 14$ in Firstly, we need to find the perimeter and the area of the base. We know that, Area of base (triangle) $= \frac{1}{2} \times$ base $\times$ height $= \frac{1}{2} \times 10 \times 7$ $= 35\; in^{2}$ Now, perimeter of the base of triangle $= 10 + 10 + 10$ $= 30$ in. The total surface area (TSA) of a triangular pyramid $= (\frac{1}{2} \times$ perimeter of the base $\times$ slant height$) +$ Base area Thus, TSA $= \frac{1}{2} \times 30 \times 14 + 35$ $= 15 \times 14 + 35$ $= 210 + 35$ $= 245\; in^{2}$ The total surface area of a triangular pyramid is $245\; in^{2}$. 3. Find the lateral surface area of a triangular pyramid with slant height 10 inch and base perimeter 24 inch. Solution: Given data: Base perimeter $= 24$ in Slant height $= 10$ The lateral surface area of a triangular pyramid $= \frac{1}{2} ($Perimeter of the base $\times$ Slant height$)$ $= \frac{1}{2} \times 24 \times 10$ $= 120\; in^{2}$ The lateral surface area of a triangular pyramid is $120\; in^{2}$. 4. Find the base length of a triangular pyramid with a slant height of 16 ft and a lateral area of 360 sq. ft. Solution: Lateral surface area = 360 sq. ft Slant height = 16 ft The lateral surface area of a triangular pyramid (LSA ) $= \frac{3}{2} (b \times l)$ $360 = \frac{3}{2} (b \times 16)$ $360 = \frac{48 \times b}{2}$ $720 = 48 \times b$ $b = \frac{720}{48}$ $b = 15$ ft. The base length of a triangular pyramid shaped museum is 15 ft. 5. Find the slant length of a triangular pyramid with a base of 8 units and a lateral area of 120 sq. units. Solution: Base side $= 8$ units Lateral Surface Area ( LSA) $= 120$ sq. units We know that, the lateral surface area of a triangular pyramid $= \frac{3}{2} (b \times s)$ $120 = \frac{3}{2} \times 8 \times s$ $120 = \frac{24 \times s}{2}$ $240 = 24 \times s$ $s = \frac{240}{24}$ $s = 10$ sq. units The slant length of a triangular pyramid is 10 sq. units. ## Practice Problems 1 ### Identify the triangular pyramid from the shapes given below. Figure A Figure B Figure C Figure D CorrectIncorrect Figure A represents a triangular pyramid with a triangle as a base and 3 lateral triangular bases. Fig B, C, and D represent a square pyramid, hexagonal pyramid, and pentagonal pyramid respectively. 2 ### The base of a right triangular pyramid is a __________. Square Right Triangle Pentagon Triangle CorrectIncorrect All the pyramids are named by the shape of their bases. Thus, the base of a right triangular pyramid is a right triangle. 3 ### A triangular pyramid _________ vertices. 3 6 4 5 CorrectIncorrect A triangular pyramid has 4 vertices. 4 ### The formula to calculate the total surface area of a triangular pyramid is ________. $\frac{3}{2} (b \times s)$ $\frac{1}{2} (a \times b) + \frac{3}{2} (b \times s)$ $\frac{1}{2} (a \times b)\; -\; \frac{3}{2} (b \times s)$ None of these CorrectIncorrect Correct answer is: $\frac{1}{2} (a \times b) + \frac{3}{2} (b \times s)$ Total surface area is the sum of the base area and lateral surface area of the pyramid. Thus, the total surface area of a triangular pyramid $= (\frac{1}{2} \times base \times height) + \frac{3}{2} (b \times s)$ $= \frac{1}{2}( a \times b ) + \frac{3}{2} ( b \times s )$ 5 ### Which of the following is an incorrect statement about a triangular pyramid? A triangular pyramid is also known as a tetrahedron. Surface area of triangular pyramid is measured in cubic units such as in$^{3}$, ft$^{3}$, etc. A triangular pyramid has 4 triangular faces. The lateral surface area of a triangular pyramid $= \frac{3}{2} ($side $\times$ slant height$)$ CorrectIncorrect Correct answer is: Surface area of triangular pyramid is measured in cubic units such as in$^{3}$, ft$^{3}$, etc. Surface area of triangular pyramid is measured in cubic units such as in$^{3}$, ft$^{3}$, etc. Surface area of triangular pyramid is measured in square units such as in$^{2}$, ft$^{2}$, miles$^{2}$, etc. A tetrahedron is a triangular pyramid in which all the faces are triangles of equal size. All the triangles of the tetrahedron are congruent and when all the sides of all the triangles are equal, then it is a triangular pyramid with an equilateral triangle and is called a tetrahedron. The volume of a 3D figure refers to its capacity, or the space occupied by it. It is measured in cubic units. Knowing the base area and height of a triangular pyramid is enough to calculate its volume. The volume of a triangular pyramid is given by the formula $\frac{1}{3} \times$ base area $\times$ height Now consider a regular triangular pyramid made of equilateral triangles of side “$a$.”Regular Triangular Pyramid Volume $= \frac{a^{3}}{6\sqrt{2}}$ cubic units. A common example of a triangular pyramid is the pyramix or the Rubik’s triangle. It has a triangular base and 3 other triangular faces. There are three types of triangular pyramids.They are as follows: • Regular triangular pyramids. • Irregular triangular pyramids. • Right-angled triangular pyramids. The image below shows what a net of a triangular pyramid looks like. The net of a 3D shape is what it looks like if it is opened out flat. The net of a triangular pyramid has 3 triangular faces and 1 triangular base.
Parallelogram - Parallelogram Verification Did you notice the quadrilateral that is formed at the intersection of 2 train tracks? What is it called? What are its characteristics? Let's take a look at the train tracks, why are train tracks 2 parallel tracks? For the train to not derail, there must be 2 tracks that always maintain the same distance apart. This is the definition of parallel lines that never meet because the distance between them is always equal. This is the definition of parallel lines that never meet because the distance between them is always equal. At the moment when 2 train tracks meet, a quadrilateral is formed between them, which has 2 pairs of opposite sides parallel, which is the parallelogram If the data is: • $AB ǁ CD$ • $AD ǁ BC$ Then: $ABCD$ is a parallelogram Examples with solutions for Parallelogram Exercise #1 Is it possible that it is a parallelogram? Step-by-Step Solution According to the properties of a parallelogram, any two opposite sides will be equal to each other. From the data, it can be observed that only one pair of opposite sides are equal and therefore the quadrilateral is not a parallelogram. No Exercise #2 Is it possible that it is a parallelogram? Step-by-Step Solution According to the properties of the parallelogram: the diagonals intersect each other. From the data in the drawing, it follows that diagonal AC and diagonal BD are divided into two equal parts, that is, the diagonals intersect each other: $AO=OC=8$ $DO=OB=10$ Therefore, the quadrilateral is actually a parallelogram. Yes Exercise #3 Is it possible that it is a parallelogram? Step-by-Step Solution Let's review the property: a quadrilateral in which two pairs of opposite angles are equal is a parallelogram. From the data in the drawing, it follows that: $D=B=60$ $A=C=120$ Therefore, the quadrilateral is actually a parallelogram. Yes Exercise #4 AB = DC.= Is the shape below a parallelogram? Step-by-Step Solution In a parallelogram, we know that each pair of opposite sides are equal to each other. The data shows that only one pair of sides are equal to each other: $AB=DC=8$ Now we try to see that the additional pair of sides are equal to each other. We replace$x=8$for each of the sides: $AD=2\times8+9$ $AD=16+9$ $AD=25$ $BC=8+5$ $BC=13$ That is, we find that the pair of opposite sides are not equal to each other: $25\ne13$ Therefore, the quadrilateral is not a parallelogram. No Exercise #5 Given $∢B+∢C=180$ Is it possible that it is a parallelogram? Step-by-Step Solution Remember that in a parallelogram each pair of opposite angles are equal to each other. The data shows that only one pair of angles are equal to each other: $D=B=140$ Therefore, we will now find angle C and see if it is equal to angle A, that is, if angle C is equal to 40: Let's remember that a pair of angles on the same side are equal to 180 degrees, therefore: $B+C=180$ We replace the existing data: $140+4x=180$ $4x=180-140$ $4x=40$ Divide by 4: $\frac{4x}{4}=\frac{40}{4}$ $x=10$ Now we replace X: $C=4\times10=40$ That is, we found that angles A and C are equal to each other and that the quadrilateral is a parallelogram since each pair of opposite angles are equal to each other. Yes Exercise #6 Look at the parallelogram in the figure. Its area is equal to 70 cm². Calculate DC. Step-by-Step Solution The formula for the area of a parallelogram: Height * The side to which the height descends. We replace in the formula all the known data, including the area: 5DC = 70 We divide by 5: DC = 70/5 = 14 And that's how we reveal the unknown! $14$ cm Exercise #7 AO = OC Is it a parallelogram? Step-by-Step Solution Let's pay attention to the diagonals, remember that in a parallelogram the diagonals intersect each other. Therefore, we will find AO, OC, BO, DO and check if they are equal and intersect each other. We refer to the figure: $AO=OC$ $9x+1=10x$ We place like terms: $1=10x-9x$ $1=x$ We replace: $AO=9\times1+1=10$ $OC=10\times1=10$ Now we know that indeed$AO=OC$ Now we establish that X=1 and see if BO is equal to OD: $BO=3x-2$ $BO=3\times1-2=$ $BO=3-2=1$ $OD=5x+4$ $OD=5\times1+4$ $OD=5+4=9$ Now we find that: $BO\ne OD$ Since the diagonals do not intersect each other, the quadrilateral is not a parallelogram. No Exercise #8 Look at the parallelogram in the figure below. Its area is equal to 40 cm². Calculate AE. Step-by-Step Solution We are told that ABCD is a parallelogram,$AB=CD=8$According to the properties of a parallelogram, each pair of opposite sides are equal and parallel. Hence to find AE we will need to use the area given to us in the formula in order to determine the area of the parallelogram: $S=DC\times AE$ $40=8\times AE$ We divide both sides of the equation by 8: $8AE:8=40:8$ $AE=5$ $5$ cm Exercise #9 ABCD parallelogram, it is known that: BE is perpendicular to DE BF is perpendicular to DF Calculate the area of the parallelogram in 2 different ways Step-by-Step Solution In this exercise, we are given two heights and two sides. It is important to keep in mind: The external height can also be used to calculate the area Therefore, we can perform the operation of the following exercise: The height BF the side AD 86 The height BE the side DC 4 12 The solution of these two exercises is 48, which is the area of the parallelogram. 48 cm² Exercise #10 ABCD is a parallelogram. CE is its height. CB = 5 AE = 7 EB = 2 What is the area of the parallelogram? Step-by-Step Solution To find the area, first, the height of the parallelogram must be found. To conclude, let's take a look at triangle EBC. Since we know it is a right triangle (since it is the height of the parallelogram) the Pythagorean theorem can be used: $a^2+b^2=c^2$ In this case: $EB^2+EC^2=BC^2$ We place the given information: $2^2+EC^2=5^2$ We isolate the variable:$EC^2=5^2+2^2$ We solve:$EC^2=25-4=21$ $EC=\sqrt{21}$ Now all that remains is to calculate the area. It is important to remember that for this, the length of each side must be used. That is, AE+EB=2+7=9 $\sqrt{21}\times9=41.24$ 41.24 Exercise #11 AE is the height of the parallelogram ABCD. AB is 3 cm longer than AE. The area of ABCD is 32 cm². Calculate the length of side AB. Step-by-Step Solution Keep in mind that AB is 3 cm greater than AE, so we must pay attention to the data when we put the formula to calculate the parallelogram: Height multiplied by the side of the height: $AB\times AE=S$ We will mark AE with the letter a and therefore AB will be a+3: $a\times(a+3)=32$ We open the parentheses: $a^2+3a=32$ We use the trinomial/roots formula: $a^2+3a-32=0$$(a+8)(a-5)=0$ That means we have two options: $a=-8,a=5$ Since it is not possible to place a negative side in the formula to calculate the area$a=5$ Now we can calculate the sides: $AE=5$ $AB=5+3=8$ 8 cm Exercise #12 The parallelogram ABCD contains the rectangle AEFC inside it, which has a perimeter of 24. AE = 8 BC = 5 What is the area of the parallelogram? Step-by-Step Solution In the first step, we must find the length of EC, which we will identify with an X. We know that the perimeter of a rectangle is the sum of all its sides (AE+EC+CF+FA), Since in a rectangle the opposite sides are equal, the formula can also be written like this: 2AE=2EC. We replace the known data: $2\times8+2X=24$ $16+2X=24$ We isolate X: $2X=8$ and divide by 2: $X=4$ Now we can use the Pythagorean theorem to find EB. (Pythagoras: $A^2+B^2=C^2$) $EB^2+4^2=5^2$ $EB^2+16=25$ We isolate the variable $EB^2=9$ We take the square root of the equation. $EB=3$ The area of a parallelogram is the height multiplied by the side to which the height descends, that is$AB\times EC$. $AB=\text{ AE}+EB$ $AB=8+3=11$ And therefore we will apply the area formula: $11\times4=44$ 44 Exercise #13 ABCD is a parallelogram whose perimeter is equal to 24 cm. The side of the parallelogram is two times greater than the adjacent side (AB>AD). CE is the height of the side AB The area of the parallelogram is 24 cm². Find the height of CE Step-by-Step Solution The perimeter of the parallelogram is calculated as follows: $S_{ABCD}=AB+BC+CD+DA$ Since ABCD is a parallelogram, each pair of opposite sides is equal, and therefore, AB=DC and AD=BC According to the figure that the side of the parallelogram is 2 times larger than the side adjacent to it, it can be argued that$AB=DC=2BC$ We inut the data we know in the formula to calculate the perimeter: $P_{ABCD}=2BC+BC+2BC+BC$ We replace the given perimeter in the formula and add up all the BC coefficients accordingly: $24=6BC$ We divide the two sections by 6 $24:6=6BC:6$ $BC=4$ We know that$AB=DC=2BC$We replace the data we obtained (BC=4) $AB=DC=2\times4=8$ As ABCD is a parallelogram, then all pairs of opposite sides are equal, therefore BC=AD=4 To find EC we use the formula:$A_{ABCD}=AB\times EC$ We replace the existing data: $24=8\times EC$ We divide the two sections by 8$24:8=8EC:8$ $3=EC$ 3 cm Exercise #14 ABCD is a parallelogram and AEFD is a rectangle. AE = 7 The area of AEFD is 35 cm². CF = 2 What is the area of the parallelogram? Step-by-Step Solution Let's first calculate the sides of the rectangle: $AEDF=AE\times ED$ Let's input the known data: $35=7\times ED$ Let's divide the two legs by 7: $ED=5$ Since AEDF is a rectangle, we can claim that: ED=FD=7 Let's calculate side CD: $2+7=9$ Let's calculate the area of parallelogram ABCD: $ABCD=CD\times ED$ Let's input the known data: $ABCD=9\times5=45$ 45 cm². Exercise #15 ABCD is a parallelogram with a perimeter of 38 cm. AB is twice as long as CE. AD is three times shorter than CE. CE is the height of the parallelogram. Calculate the area of the parallelogram. Step-by-Step Solution Let's call CE as X According to the data $AB=x+2,AD=x-3$ The perimeter of the parallelogram: $2(AB+AD)$ $38=2(x+2+x-3)$ $38=2(2x-1)$ $38=4x-2$ $38+2=4x$ $40=4x$ $x=10$ Now it can be argued: $AD=10-3=7,CE=10$ The area of the parallelogram: $CE\times AD=10\times7=70$
Cuisenaire rods are an aid in teaching arithmetic. They were invented by Georges Cuisenaire, who was a primary school teacher in Thuin, Belgium1. He published a book called "Les Nombres en Couleurs" (Numbers in Color) in 1952 in which the rods were introduced2. The Cuisenaire rods themselves are wooden sticks that are 1 cm wide and between 1 and 10 cm long. They differ not only in length, but also in color. They have the following colors 3.: • White: 1 cm • Red: 2 cm • Green: 3 cm • Purple: 4 cm • Yellow: 5 cm • Dark green: 6 cm • Black: 7 cm • Brown: 8 cm • Blue: 9 cm • Orange: 10 cm Cuisenaire rods hence make it very easy to see which number belongs to which rod, as they differ in both length and color: while the difference in length between a 8 cm rod and a 9 cm rod might be too small to immediately see if they are not directly compared, the difference between Brown and Blue is. In this respect, they are superior to using blocks of one color, which is how I learned arithmetic. Cuisenaire rods can obviously be used to teach addition, subtraction multiplication and division. For instance, take a purple stick, take a yellow stick, make a "train" of them, and note that is exactly as long as the blue stick, which has value 9. The only thing left to remember now is that blue equals 9. However, it gets better. fractions have been the bane of elementary school children and adults alike: how, exactly, do we add up 2/5 and 3/4? With Cuisenaire rods, this can easily be done. A different rod is used for the numerator and the denominator. For instance, in representing one half, the red rod can be used for the denominator, and each white rod now represents one half, with two whites on top of one red representing 1. Next, one can add two of these two-white, one-red combinations and see that they are equal to one purple rod. In other words, one is equal to two halves and four quarters, and one half is equal to two quarters. This principle can be used to make any kind of fractions, and bring them under a common denominator to add them. Cuisenaire rods have proven to be even more versatile than that. They are also used in the Silent Way 4, a way of teaching a language that involves little teacher intervention. This way of teaching a language was invented by Caleb Gattegno, who met Cuisenaire 5. The Silent Way involves using the same color to represent the same sound. The rods are used as props 6, and form the basis of a common vocabulary In summary, Cuisenaire rods are colored rods that are used to give a concrete representation of abstract concepts such as numbers. As such, they are used in teaching arithmetic. Furthermore, they can also be used in teaching language. #### Sources 1. http://www.cuisenaire.co.uk/ 2. http://pagesperso-orange.fr/une.education.pour.demain/biographies/cuisen .htm 3. http://pagesperso-orange.fr/une.education.pour.demain/materiels_pedago/s w/swengcharts/rods2.htm 4. http://pagesperso-orange.fr/une.education.pour.demain/materiels_pedago/s w/swprese.htm 5. http://www.gattegno-edusol.com/?q=node/24 6. http://www.onestopenglish.com/section.asp?docid=146498 Log in or register to write something here or to contact authors.
# Kinematic Equations for Constant Acceleration The kinematic equations illustrate the relationship between displacement, velocity, acceleration, and time under conditions of constant acceleration. They serve as the foundation for understanding the motion of objects in various situations, from an apple falling from a tree to a rocket venturing into the cosmos. This article will explore the kinematic equations in detail. By understanding them, we unlock a powerful tool to understand the dynamics of the world around us. ##### PHYSICS Relevant for Learning about the kinematic equations for constant acceleration. See equations ##### PHYSICS Relevant for Learning about the kinematic equations for constant acceleration. See equations ## The Five Fundamental Kinematic Equations for Constant Acceleration The kinematic equations are a set of equations that allows us to calculate the different quantities involved when an object is moving with a constant acceleration. The quantities we are concerned with are • $latex s=$ displacement • $latex u=$ initial velocity • $latex v=$ final velocity • $latex a=$ acceleration • $latex t=$ time Here are the five equations of motion: Equation 1: $latex v=u+at$ Equation 2: $latex s=\dfrac{u+v}{2}\times t$ Equation 3: $latex s=ut+\frac{1}{2}at^2$ Equation 4: $latex s=vt-\frac{1}{2}at^2$ Equation 5: $latex v^2=u^2+2as$ Here are the equations along with an explanation of their components: 1. $latex v = u + at$: This equation expresses that the change in velocity is equal to acceleration times the time interval. 2. $latex s=\frac{u+v}{2}\times t$: This equation presents displacement as the product of the average velocity (the sum of the initial and final velocities divided by 2) and the time. 3. $latex s=ut+\frac{1}{2}at^2$: This means that an object’s displacement is the product of the initial velocity and time plus one-half of the acceleration times the square of the time. 4. $latex s=vt-\frac{1}{2}at^2$: This is a rearranged version of the third equation, which is helpful when initial velocity is not given or considered. 5. $latex v^2=u^2+2as$: This conveys that the square of the final velocity is equal to the square of the initial velocity plus twice the acceleration times the displacement. Note: Each equation is used under specific circumstances and contains particular information about the object’s motion. Knowing which equation to use in different scenarios is crucial in accurately solving kinematic problems. ## Choosing the Correct Kinematic Equation Choosing the appropriate kinematic equation relies heavily on the variables outlined within each individual problem. Therefore, start by writing down the quantities which we know, and the quantity we want to find. Referencing the following table can guide you in determining the right equation to use. Note: This table shows the simplified forms of these equations applicable when dealing with objects starting from rest, initial velocity equal to 0 or $latex u=0$. ### Process for solving problems with the kinematic equations Most problems related to the kinematic equations can be solved using the following procedure: Step 1: Create a list with the quantities we know and the quantities we want to find. Step 2: Choose the equation which links these quantities. The table above can serve as a guide. Step 3: Substitute in the known values and solve for the unknown values. ## Proofs of the Kinematic Equations The kinematic equations are derived from the definitions of velocity and acceleration. We can use the following velocity vs time graph to help us find the equations. The graph represents the motion of an object with initial velocity $latex u$. After time $latex t$, its final velocity is $latex v$. ### Equation 1 The graph shown above displays a straight line, which means that the acceleration $latex a$ is constant. The slope of the line of a velocity vs time graph is equal to the acceleration. The acceleration or slope of the line is defined as: $$a=\frac{v-u}{t}$$ Solving for $latex v$ gives us the first equation of motion: $latex v=u+at$ ### Equation 2 In a velocity vs time graph, the area under the graph represents the displacement. Then, the object’s displacement is the shaded area in the following figure: Given that the initial and final velocities are different, we calculated the displacement using the average velocity, which is given by: $$\frac{u+v}{2}$$ As the shaded area of the figure above is a rectangle, we just multiply the average velocity by the time taken: $$s=\frac{(u+v)}{2}\times t$$ ### Equation 3 We can derive equation 3 using equations 1 and 2. $latex \text{(1)}~~v=u+at$ $latex \text{(2)}~~s=\dfrac{(u+v)}{2}\times t$ Substituting $latex v$ from equation 1 into equation 2, we get: $$s=\frac{(u+u+at)}{2}\times t$$ $$s=\frac{(2u+at)}{2}\times t$$ $$s=\frac{2ut}{2}+\frac{at^2}{2}$$ $$s=ut+\frac{at^2}{2}$$ This can also be seen in the figure shown above, where the terms $latex ut$ and $latex at^2$ are the areas under the graph representing the displacement. ### Equation 4 Similar to equation 3, we just use the equations 1 and 2. $latex \text{(1)}~~v=u+at$ $latex \text{(2)}~~s=\dfrac{(u+v)}{2}\times t$ However, here we substitute $latex u=v-at$ from equation 1 into equation 2 to get: $$s=\frac{(v+v-at)}{2}\times t$$ $$s=\frac{(2v-at)}{2}\times t$$ $$s=\frac{2vt}{2}-\frac{at^2}{2}$$ $$s=vt-\frac{at^2}{2}$$ ### Equation 5 Again, we use equations 1 and 2: $latex \text{(1)}~~v=u+at$ $latex \text{(2)}~~s=\dfrac{(u+v)}{2}\times t$ We substitute $latex t=\frac{v-u}{a}$ from equation 1 into equation 2 to get: $$s=\frac{(u+v)}{2}\times \frac{(v-u)}{a}$$ $$2as=(u+v)(v-u)$$ $$2as=v^2-u^2$$ $$v^2=u^2+2as$$ ## Practical Applications of the Kinematic Equations Everyday Life Examples The principles behind the kinematic equations play in our day-to-day lives. For instance, when you’re driving, the distance you cover over time and the speed you reach can be calculated using kinematic equations. Similarly, if you throw a ball upwards, you can calculate the maximum height it will reach and how long it will take to reach the ground again, assuming constant acceleration due to gravity. Engineering and Technology Applications In engineering and technology, kinematic equations are used everywhere. For example, mechanical engineers use these equations for designing vehicles and machinery. In the field of robotics, the equations are used to program robots to move in precise ways. These equations are even used in computer science, playing a crucial role in video game development, particularly in creating more realistic movements and physics in the virtual world. Scientific Experiments Kinematic equations are indispensable in various scientific experiments. Physicists often use these equations in designing and interpreting experiments involving motion. For example, in particle physics, researchers studying the behavior of particles under various forces will often use kinematic equations to understand the particles’ motion. Space Travel and Rocket Science Perhaps one of the most fascinating applications of the kinematic equations is in space travel and rocket science. When launching a rocket or satellite, scientists and engineers need to calculate the trajectory and velocity needed to escape Earth’s gravitational pull and reach a particular destination. When landing a spacecraft, they must calculate the deceleration required to ensure a safe landing. All of these calculations involve the kinematic equations, providing a perfect demonstration of their significance. From these examples, it is clear that the kinematic equations are not just theoretical concepts, but have real-world applications.
• Join over 1.2 million students every month • Accelerate your learning by 29% • Unlimited access from just £6.99 per month # AS and A Level: Core & Pure Mathematics Browse by Rating: 4 star+ (2) 3 star+ (3) Word count: fewer than 1000 (27) 1000-1999 (55) 2000-2999 (39) 3000+ (35) Meet our team of inspirational teachers Get help from 80+ teachers and hundreds of thousands of student written documents ## Differentiation and intergration 1. 1 It is easy to get differentiation and integration the wrong way round. Remember that the power gets smaller when differentiating. 2. 2 Differentiation allows you to find the gradient of a tangent at any point on a curve. The first derivative describes the rate of change. 3. 3 If a function is increasing then the first derivative is positive, if a function is decreasing, then the first derivative is negative. 4. 4 When asked to find the area under a curve, it is asking you to integrate that curve between two points. Even if you don’t know the points, pick two numbers. You’ll get marks for methods. 5. 5 When referring to a min/max/stationary point, the gradient equals 0. Differentiate the curve and set this to equal 0. The second derivative tells you whether it is a maximum or minimum. If the second derivative is positive, the point is a minimum, if the second derivative is negative, then the point is a maximum. 1. 1 When solving a quadratic inequality, always draw a picture. The inequality is less than 0, where the curve is below the x-axis and bigger than 0 when the curve is above the x-axis. 2. 2 Sometimes in part (a) of a question you are asked to find something, for example a radius. In part (b) you might be then asked to use the radius that you found. If you couldn’t do part (a), don’t give up, choose a random radius. ## Straight lines 1. 1 To find the distance of a straight line, draw the straight line with the co-ordinates. Then make a right angle triangle, find the lengths of the horizontal and vertical lines, then use Pythagoras. 2. 2 When a question asks you for a straight line. The first thing to do is to write down the equation of a straight line. Then find out what information you know, and what information you need. Even if you don’t understand the whole question, it is important to start. 1. ## Although everyone who gambles at all probably tries to make a quick mental marginal analysis of the game, in depth analysis of the figures shall reveal how a rational player reacts to better odds, or a lower entry price, or a higher potential payout. Case 3 can be represented by a draw, where E is set, and both O and P are known. A process of linear regression will be used to determine the actual relationship between P and O for each game. Analysis Case 1 The lottery chosen is from Ontario, lotto 649. The player pays \$1 to enter the game, and then chooses 6 non-repeating numbers from 1 to 49. Thus the number of possible combinations a player can choose equals. 6 numbers are drawn and the payout scheme is shown below in Table 1: (This data is from the January 12th 2002 drawing; although it is only one sample of data, it is an accurate representation of the average value of prizes) • Word count: 2253 2. ## Analysing; The Reaction of Hydrogen Peroxide and Iodide ions Nevertheless, we would need to convert most of the data into logs, Experiment Log Volume of KI (dm3) Time taken in seconds Log (1/time) A -2.30 596 -2.77 B -2.00 336 -2.53 C -1.82 247 -2.39 D -1.70 207 -2.31 E -1.60 160 -2.20 Thus, we could derive the gradient of the graph, by drawing it, as seen upon the nest page. We could also use the statistical method of finding the equation of the regression line, which is: Y = a +bx (where y and x are the axis's) B = ((xI - (x)(yI - (y) ((xI - (x)2 a = (y - B(x Where (y = average of all y values (x = average of all • Word count: 1275 3. ## Fixed-Point Iteration Iterations of: f(x) = 1.25x - 2 from the left side of graph: First "guess": 5 f(5) = 1.25(5) - 2 = 4.25 Second iteration: f(4.25) = 1.25(4.25) - 2 = 3.3125 Third iteration: f(3.3125) = 1.25(3.3125) -2 = 2.140625 Fourth iteration: f(2.140625) = 1.25(2.140625) -2 = 0.67578125 - Copied axes located on page C. Iterations of f(x) = 1.25x - 2 from the right side: First "guess": 10 F(10) = 1.25(10) -2 = 10.5 Second iteration: F(10.5) = 1.25(10.5) • Word count: 839 4. ## Numerical solution of equations However, if we cannot use the computer, we have to use one of the numerical methods - interval estimation. A root must be lying on the interval a and b where an interval have been marked in which F(x) change signs. We thus know that there must be a root in that interval. The equation F(x) = x�-9x+3 has 3 roots which are in the interval of (-3,-4), (0,1) and (2,3). We can find the roots through decimal search, interval bisection and linear interpolation. Decimal search It is one of the methods of interval estimation through dividing each interval into 10 parts and looking the sign change. • Word count: 2774 5. ## Observation of people obeying temporary signs. Null Hypothesis: There will be no difference in obedience between the signs written by the authority figure and school pupil Variables: The independent variable was whom the signs were written by. The dependent variable was whether the pupil obeyed the signs or not. Population: Sixth Form pupils (16 and over), in UK. Sample: An opportunity sample of 40 Sixth Form pupils from a secondary school in Letchworth per lunch hour. They included male and female subjects, ranging from 16 to 18 years old. • Word count: 627 6. ## Growing Squares + 1 = 8. For Diagrams 1 - 4 I can see a pattern with square numbers. The diagrams numbers squared added to one less than the diagram number squared gives the correct number of squares. For diagram n it should be: Un = (n - 1) 2 + n2 (n - 1)(n - 1) + n2 n2 - n - n - n + 1 + n Un = 2n2 - 2n + 1 This is correct. Growing Hexagons I will now repeat my investigation, and change the original shape of the square to hexagons, and try to find the formula as before. I shall start by finding the width of each hexagon. Pattern Number (n) • Word count: 1164 7. ## Pure Mathematics 2: Solution of equation by Numerical Methods f(x) = x� - 12x + 5 I am going to use the root in the interval [0,1] f(0) > 0 f(1) < 0 f(0+1) � f(0.5) = -0.875 2 Therefore f(0.5) < 0 f(0) > 0 f(0.5) < 0 f(1) < 0 Root is in interval [0, 0.5] I now test the mid-point of [0, 0.5] f(0+0.5) � f(0.25) = 2.015625 2 Therefore f(0.25) > 0 f(0) > 0 f(0.25) > 0 f(0.5) < 0 Root is in interval [0.25, 0.5] I will now test mid-point of [0.25, 0.5] f(0.25+0.5) � f(0.375) = 0.552734375 2 Therefore f(0.375) > 0 f(0.25) • Word count: 4642 8. ## Collect data and investigate whether it can be model by a exponential function I interval that I measured in was 2.5 cm, which gave me 10points for my graph. To get my y points I had to pour water up to the cylindrical section and made a hole at the bottom of the bottle. As the water passes a point I record the time. 2) A mathematical representation of the data Graphically Algebraically The equation for this data is 3.429801*1.126417^x. To get the equation I used the Graphics calculator and find the equation in the data matrix editor. Although when I was trying to validate it the computer gave me a different equation (this is using excel). • Word count: 890 9. ## Maths Coursework: Curve Fitting and touches the x-axis at (3,0). I used the same method as before, of drawing what I thought the graph would look like: (0,9) (3,0) I then put the numbers into brackets again (as below), because I worked out that when Y=0, X=3, and no other number. Then once again expanded the brackets to find the formula: Y=(x-3)(x-3) = 0 I worked out the formula to be: Y=x2-6x+9 I could be sure that this was the correct equation because the co-ordinate was (0,9) • Word count: 872 10. ## The diagram shows a house built with dominoes. This house has four stories and uses 24 dominoes.Simon broke the World Record by building a domino house with 73 stories. How many dominoes did he use? I will try to work systematically to make the project simple to understand. I will then draw a results table where I will comment on why I am using the results and what purpose it has in my project. I will also explain if I can spot any patterns. After I spot the patterns I will work out the differences. Next, using the differences I will draw a differences table, which will give us the rules (either being linear and quadratic). Using these rules I will draw a graph, and if the graph has a straight line then this determines the rule is linear if the line is a smooth curve this determines the rule is quadratic. • Word count: 1017 11. ## Testing Root Methods this is a drawback when there are two or more roots in between two integer values. Method 2: Newton-Raphson Method Using Autograph again, the roots for y=x5 -3x2 + 1 were found using the Newton-Raphson Method: This illustration demonstrates how we acquire the first root of the equation. The tangent is found at a point and then the point where this tangent crosses x=0 is found. The process is then repeated from this point. The equation for Newton-Raphson iterations is: If we find the root closest to -0.5, starting at -1, the values we find during the Newton-Raphson method are: • Word count: 1250 12. ## Critically consider one invasive method and one non-invasive method of studying the brain The aim of lesion studies is to tell us something about how different areas of the brain are connected. However, there are several flaws with this invasive method of studying the brain. One of these is that, since experiments of this type are carried out on animals, the results cannot really be generalised to humans. Also, there are ethical issues involved in the use of animals in experiments that could cause distress. Other invasive methods of investigating the brain include; chemical stimulation of the brain and Electrical stimulation of the brain (ESB). One such investigation was conducted by Olds + Milner (54). • Word count: 574 13. ## Information Technology Development Cycle Diary I did not set-up any validation checks, macros or cell protection. I simply wanted to see if this section of the system would function correctly. Input Process Problem On the process sheet I had difficulty setting up the 'Discount on spending' cell. I used the formulae: =IF(G21>300,40,G21>200,20,G21>100,10,0) • Word count: 219 14. ## Numerical Method Course Work - b F (a)]/[F (b)- F (a)]. Newton-Raphson Method: Xr+1 = Xr - f (Xr)/ f' (Xr), X0 is given. Fix-point iteration: The reason why I don't use Fixed-point method is because to rearrange the formula to x= g (x) is very difficult and I tried this method but found the duration didn't go to convergence. (Appendix 1) So I didn't use this method any more. As the spreadsheet (Appendix 2) shown Newton-Raphson method is most efficient method to find a root. • Word count: 1323 15. ## Using a Spreadsheet to Analyse Statistics Sorting the Data You notice that the teams are not listed in alphabetical order and you decide that you want to put this right. 1. You must first select columns B and C It is very important that you do this. If you just sort the list of team names, the teams will be rearranged but the goals scored won't. That would make your data incorrect. 2. From the Data menu, select Sort. 3. Sort by Column B (the teams list) • Word count: 1129 16. ## Investigation on Boyles law We used a foot pump and pumped the air into the column, which increased the pressure. We then took readings of pressure and volume every five seconds and noted down the results. Then after we had taken about thirty results we re-tested them again to insure that the results were totally accurate and no error had occurred. From our results we had taken we plotted a graph with pressure (lb/in2) on the vertical axis and volume (cm3) on the horizontal axis. From all these points we drew a line of best fit. Analysis There are a few strange points, which may have been from human error of reading. • Word count: 914 17. ## The Rational Zeros Simplifying the expression we get the value observed. c) Finally graph, with a suitable window . What are the roots of ? We find that the roots are: Therefore, the smallest positive root, expressed as a fraction is This values correspond also to those of the graph in part (b) and we find that this equation has the same coefficients as in the previous one, only that multiplied by 3. Even though the second and fourth terms have changed, the other two roots are kept the same. Given this aspect, we may assume that the relative proportions of the coefficients and not their actual values define the position of the roots. • Word count: 1877 18. ## Numerical Method (Maths Investigation) EQUATION USED: = 0 has only 1 roots in [0,-1] From Graphmatica: Results obtained from Microsoft Excel: X-value F(X) -1 -1.5 0 0.5 Observe the change of y-value from negative value to positive value in the table. That shows that the root of the equation where the y-value is 0, is between x = 0 and x = -1. Now this is step 1, we're moving on to step 2. X-value F(X) -0.9 -0.7882969 -0.8 -0.3497152 -0.7 -0.0723543 -0.6 0.1120064 Now the search is narrowing in and it comes to one decimal place in between x = -1 and x = 0. • Word count: 5596 19. ## Investigating the Quadratic Function The only exception would be where nothing is being added to the x� (A) in which case the parabola would have its vertex's turning point on the origin. The first graph (A) shows us that y = x� gives us a parabola with the vertex on the origin, the positive leading coefficient is shown by the parabola opening up meaning the parabola is positive. This graph is showing neither dilations, translations nor reflections, because no other coefficients are available. Nevertheless graph B (y = x� + 3) and graph C (y = x� - 2) • Word count: 1717 20. ## Solving Equations by numerical methods. - x3/2 = 0 This equation can not be solved numerically because x appears inside the function ln(x) as well as in a power term. Therefore we can not find the inverse function and isolate x. This comparatively innocuous looking function crosses the x-axis at some point around the interval [1, 2] and again at some point around the interval [8, 9]. For small values of x the term 12ln(x) dominates and is large and negative. For larger values of x the term x3/2 dominates and is large and positive, making the f(x) value large and negative. My centre of interest lies within the range 0<=x<=4. • Word count: 2153 21. ## Repeated Differentiation We know from the rules of differentiation for such a function that for the (n+1) th differential you multiply the nth function by its power and lower its power by one. However, if we don't know the nth differential and we know only the original function this simple method is obviously not possible. Looking at the above example suggests that a solution would have to have something to do with factorials. The way the power and the differential number add together to equal b is of particular interest. • Word count: 3122 22. ## Maths Assignment - trigonometry, trend line, probability and calculus questions. After 4hrs distance between the cyclist and man 2. Question: If experimental data doesn?t produce a straight trend line it doesn?t mean there isn?t a relationship. The use of software packages can allow quick and accurate testing of data to different trend lines so that a suitable relationship can be found if it exists. Atmospheric pressure P is measured at varying altitudes h as shown below. The data is thought to be of the exponential form p = aekh Altitude 500 1500 3000 5000 8000 Metres Pressure 73 68 62 54 43 cm a) • Word count: 695 23. ## Newton Raphson Method for Solving 6x3+7x2-9x-7=0 X = 1.1155 Error Bounds = ±0.00005 Solution bounds = (1.11545≤ x ≤1.11555). n xn f(xn) f `(xn) xn+1 0 x0 = 1 f(x0)= -3 f '(x0)= 23 x1 = 1.1304 1 x1 = 1.1304 f(x1) = 0.4376 f '(x1) = 29.826 x2 = 1.1157 2 x2 = 1.1157 f(x2) = -0.000748 f '(x2) = 29.015 x3 = 1.1155 Solving 6x3+7x2-9x-7=0 for second root ________________ The root lies between -0.63488 and -0.634875 f(x) is negative -0.634885 -0.63488 -0.634875 f(-0.634885)= 0.000065017 f(-0.63488)=0.000011852 f(-0.634875)= -0.000041312 X= -0.63488 (5 sig fig) Error bounds = ±0.000005 Solution bounds = (-0.634875≤ x ≤-0.634885). • Word count: 1687 24. ## C3 COURSEWORK - comparing methods of solving functions x y=x³+3x²–3 0.8793 -0.00065 0.87931 -0.00057 0.87932 -0.0005 0.87933 -0.00042 0.87934 -0.00034 0.87935 -0.00027 0.87936 -0.00019 0.87937 -0.00012 0.87938 -4E-05 Change of sign 0.87939 3.61E-05 0.8794 0.000112 The change of sign here tells us that there is a root in the interval [0.87938, 0.87939]. So to 4 d.p. the root has value 0.8794. x x³+3x²–3 0.87938 -4E-05 0.87939 3.61E-05 The solution bounds are 0.87938and 0.87939. x x³+3x²–3 -4 -19 -3 -3 -2 1 -1 -1 0 -3 1 1 2 17 Error bounds: 0.879385± 0.000005. • Word count: 2282 25. ## Numerical integration coursework Use of technology I used Microsoft Excel throughout my method along with many algorithms, these algorithms allowed me to reach a more accurate approximation thanks to the fact that there were no rounding errors usually gained through using a calculator that gives answers to fewer decimal places. Overall then Excel is more accurate than the use of a calculator due to being able to work to more decimal places, it is also easier to use and saves a lot of time. • Word count: 1342
## Book: NCERT - Mathematics ### Chapter: 8. Introduction to Trigonometry #### Subject: Maths - Class 10th ##### Q. No. 6 of Exercise 8.1 Listen NCERT Audio Books - Kitabein Ab Bolengi 6 ##### If ∠ A and ∠ B are acute angles such that Cos A = Cos B, then show that ∠ A = ∠ B Let us consider a triangle ABC in which CD ⊥ AB It is given that Cos A = Cos B In Δ ADC, cos A = AD/AC In Δ ADB, cos B = BD/BC ⇒ AD/AC = BD/BC ..(i) We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP, as shown below: Now, From equation (i), we obtain ⇒ AD/AC = BD/BC . Rewriting we get, (BC = CP) ⇒ AD/AC = BD/CP ..(i) Now, if we join, B and P to get BP, as shown, below: Then, by using the converse of Basic proportionality theorem we get, Now, CD\|\|BP ⇒ ∠ACD = ∠CPB (Corresponding angles) ..(iii) ⇒ ∠BCD = ∠CBP (Alternate interior angles) ...(iv) And, by construction, we have BC = CP ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle) ...(v) From equations (iii), (iv), and (v), we obtain ∠ACD = ∠BCD ..Eq. (vi) Now, In ΔCAD and ΔCBD, ∠ACD = ∠BCD [Using equation equation (vi)] ∠CDA = ∠CDB (Both 90°) Therefore, the remaining angles should be equal ⇒ ∠CAD = ∠CBD ⇒ ∠A = ∠B 9 1 2 3 4 5 6 7 8 9 10 11
# Supercharge your learning! Use adaptive quiz-based learning to study this topic faster and more effectively. # Bijections and compositions ## Bijection A function $f:X\to Y$ is bijective if, for each value $y\in Y$, there is exactly one element $x\in X$ that is mapped to $y$. In other words, the equation $$y=f(x)$$ has a unique solution $x$ for every $y$. We call $x$ the pre-image of $y$ and we denote it $f^{-1}(y)$. We usually compute the pre-image of $y$ by expressing $x$ in terms of $y$. A bijective function is also called a bijection or a one-to-one correspondence. The function $x^2$ defined on $\R$ is not bijective, because $1$ has two pre-images ($f(-1) = f(1) = 1$). The function $x^3$ is bijective on $\R$, because every $y$ in $\R$ has a unique pre-image, specifically $x=y^{1/3}$. I and II are functions, but III is not. I is a bijection. II is not a bijection. ## Composition of functions The composition of functions is the successive application of functions. The resulting function is called a composite function. The composite of $f(x)=2x$ and $g(x)=x^2$ is $h(x)=f(g(x))=2x^2$. More formally, if $g:D_g\to R_g$ and $f:D_f\to R_f$ are two functions with $$R_g\subset D_f,$$ the composite function $f\circ g$ is the function from $D_g$ to $R_f$ defined by $$f\circ g (x) = f\big(g(x)\big).$$ The composite function is written either as $f\circ g$ or $fg$ and is read $f$ of $g$. $fg$ is often used for the product of functions, so $f\circ g$ should be preferred. $h(x) = \sin(x^2)$ is the composite $\sin\circ g$ where $g(x) = x^2$. $\vert x^3\vert$ is the composite function $f\circ g$ with $f(x) =\vert x\vert$ and $g(x) = x^3$. ## Inverse function The inverse function of a bijection $f:X\to Y$ is the function $f^{-1}:Y\to X$ that maps each value to its unique pre-image $$y=f(x) \Longleftrightarrow x=f^{-1}(y).$$ The inverse of $x^2$ from $[0, +\infty)$ to $[0, +\infty)$ is $\sqrt{x}$. This is because $y=x^2$ is equivalent to $x = \sqrt{y}$. The inverse of $\ln x$ is $e^x$. The domain of the inverse $f^{-1}$ is the range of the original function $f$. The range of the inverse $f^{-1}$ is the domain of $f$. To summarise, $$D_{f^{-1}} = R_f,\quad R_{f^{-1}} = D_f.$$ $(0,+\infty)$ is the range of $e^x$ and the domain of $\ln x$. The inverse function is bijective and its inverse is the original function $$(f^{-1})^{-1} = f.$$ The graph of the inverse function is deduced from the graph of the function by a reflection across the line $y=x$. This is because the two graphs simply have the $x$ and $y$ axes inverted. Graph of $f^{-1}$ and graph of $f$ ## Composition is not commutative Contrary to the properties of sums and products, order matters in composition. In general, $$f\circ g\ne g\circ f.$$ We say that composition is not commutative. Setting $f(x) = x^2$ and $g(x) = x+1$, we see that, for $x\ne0$, $$f\circ g(x) = (x+1)^2\ne x^2+1 = g\circ f(x).$$ However, for the inverse function, we have $$f\circ f^{-1}(y) =y,\qquad f^{-1}\circ f(x) =x$$ This comes from $y=f(x)\Longleftrightarrow x=f^{-1}(y)$. The composition of monotonic functions is monotonic. If both functions have same monotonicity (both increasing or decreasing), the composite function is increasing; otherwise it is decreasing. To see this, assume that both $f$ and $g$ are decreasing. Then, for $x\le y$, we have $g(x)\ge g(y)$, hence $f\left(g\left(x\right)\right)\le f\left(g\left(x\right)\right)$. This means that means that $f\circ g$ is increasing.
<meta http-equiv="refresh" content="1; url=/nojavascript/"> You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Go to the latest version. # 9.4: Polynomial Equations in Factored Form Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives • Use the zero-product property • Find greatest common monomial factor • Solve simple polynomial equations by factoring ## Introduction In the last few sections, we learned how to multiply polynomials. We did that by using the Distributive Property. All the terms in one polynomial must be multiplied by all terms in the other polynomial. In this section, you will start learning how to do this process in reverse. The reverse of distribution is called factoring. Let’s look at the areas of the rectangles again: Area = length $\cdot$ width. The total area of the figure on the right can be found in two ways. Method 1 Find the areas of all the small rectangles and add them Blue rectangle $= ab$ Orange rectangle $= ac$ Red rectangle $= ad$ Green rectangle $= ae$ Pink rectangle $= 2a$ Total area $= ab + ac + ad + ae + 2a$ Method 2 Find the area of the big rectangle all at once $\text{Length} & = a\\\text{Width} & = b + c + d + e + 2\\\text{Area} & = a(b + c + d + e = 2)$ Since the area of the rectangle is the same no matter what method you use then the answers are the same: $ab + ac + ad + ae + 2a = a (b + c + d + e + 2)$ Factoring means that you take the factors that are common to all the terms in a polynomial. Then, multiply them by a parenthesis containing all the terms that are left over when you divide out the common factors. ## Use the Zero-Product Property Polynomials can be written in expanded form or in factored form. Expanded form means that you have sums and differences of different terms: $6x^4 + 7x^3 - 26x^2 + 17x + 30$ Notice that the degree of the polynomials is four. It is written in standard form because the terms are written in order of decreasing power. Factored form means that the polynomial is written as a product of different factors. The factors are also polynomials, usually of lower degree. Here is the same polynomial in factored form. $\underbrace{x - 1}_{{\color{blue}1^{st}\ factor}}\ \underbrace{x + 2}_{{\color{blue}2^{nd}\ factor}}\ \underbrace{2x - 3}_{{\color{blue}3^{rd}\ factor}}\ \underbrace{3x + 5}_{{\color{blue}4^{th}\ factor}}$ Notice that each factor in this polynomial is a binomial. Writing polynomials in factored form is very useful because it helps us solve polynomial equations. Before we talk about how we can solve polynomial equations of degree 2 or higher, let’s review how to solve a linear equation (degree 1). Example 1 Solve the following equations a) $x - 4 = 0$ b) $3x - 5 = 0$ Solution Remember that to solve an equation you are trying to find the value of $x$: a) $& \ x \ - 4 \ = \ 0\\& \underline{\;\;\;\;\;+4 \ = + 4}\\& \qquad \ \underline{\underline{x \ = \ 4}}$ b) $& 3x \ -5 \ = \ 0\\& \underline{\;\;\;\;\;\;+5 \ = +5}\\& \qquad 3x \ = \ 5\\& \qquad \frac{3x} {3} = \frac{5} {3}\\& \qquad \ \ \underline{\underline{x = \frac{5} {3}}}$ Now we are ready to think about solving equations like $2x^2 + 5x = 42$. Notice we can't isolate $x$ in any way that you have already learned. But, we can subtract 42 on both sides to get $2x^2 + 5x - 42 = 0$. Now, the left hand side of this equation can be factored! Factoring a polynomial allows us to break up the problem into easier chunks. For example, $2x^2 + 5x - 42 = (x + 6)(2x - 7)$. So now we want to solve: $(x + 6)(2x - 7) = 0$ How would we solve this? If we multiply two numbers together and their product is zero, what can we say about these numbers? The only way a product is zero is if one or both of the terms are zero. This property is called the Zero-product Property. How does that help us solve the polynomial equation? Since the product equals 0, then either of the terms or factors in the product must equal zero. We set each factor equal to zero and we solve. $(x + 6) = 0 \quad \quad \text{OR} \quad \quad (2x - 7) = 0$ We can now solve each part individually and we obtain: $x + 6 = 0 && \text{or} && 2x - 7 & = 0 \\&&&& 2x & = 7 \\x = -6 && \text{or} && x & = - \frac{7} {2}$ Notice that the solution is $x = -6$ OR $x = \frac{7}{2}$. The OR says that either of these values of $x$ would make the product of the two factors equal to zero. Let’s plug the solutions back into the equation and check that this is correct. $\text{Check}\ x & = -6;\\(x + 6) (2x - 7) & = \\(-6 + 6) (2(6) - 7) & = \\(0) (5) & = 0$ $\text{Check}\ x & = \frac{7}{2}\\(x + 6) (2x - 7) & = \\\left (\frac{7} {2} + 6 \right) \left (2 \cdot \frac{7} {2} - 7\right) & = \\\left (\frac{19} {2} \right) (7 - 7) & = \\\left (\frac{19} {2} \right) (0) & = 0$ Both solutions check out. You should notice that the product equals to zero because each solution makes one of the factors simplify to zero. Factoring a polynomial is very useful because the Zero-product Property allows us to break up the problem into simpler separate steps. If we are not able to factor a polynomial the problem becomes harder and we must use other methods that you will learn later. As a last note in this section, keep in mind that the Zero-product Property only works when a product equals to zero. For example, if you multiplied two numbers and the answer was nine you could not say that each of the numbers was nine. In order to use the property, you must have the factored polynomial equal to zero. Example 2 Solve each of the polynomials a) $(x - 9)(3x + 4) = 0$ b) $x(5x - 4) = 0$ c) $4x (x + 6) (4x - 9) = 0$ Solution Since all polynomials are in factored form, we set each factor equal to zero and solve the simpler equations separately a) $(x - 9) (3x + 4) = 0$ can be split up into two linear equations $x - 9 = 0 && \text{or} && 3x + 4 & = 0 \\&&&& 3x & = -4\\x = 9 && \text{or} && x & = - \frac{4} {3}$ b) $x(5x - 4) = 0$ can be split up into two linear equations $&&&& 5x - 4 & = 0\\x = 0 && \text{or} && 5x & = 4 \\&&&& x & = \frac{4} {5}$ c) $4x(x + 6)(4x - 9) = 0$ can be split up into three linear equations. $4x = 0 && && && && 4x - 9 & = 0 \\x = \frac{0} {4} && \text{or} && x + 6 = 0 && \text{or} && 4x & = 9 \\x = 0 &&&& x = -6 &&&& x & = \frac{9} {4}$ ## Find Greatest Common Monomial Factor Once we get a polynomial in factored form, it is easier to solve the polynomial equation. But first, we need to learn how to factor. There are several factoring methods you will be learning in the next few sections. In most cases, factoring takes several steps to complete because we want to factor completely. That means that we factor until we cannot factor anymore. Let’s start with the simplest case, finding the greatest monomial factor. When we want to factor, we always look for common monomials first. Consider the following polynomial, written in expanded form. $ax + bx + cx + dx$ A common factor can be a number, a variable or a combination of numbers and variables that appear in all terms of the polynomial. We are looking for expressions that divide out evenly from each term in the polynomial. Notice that in our example, the factor $x$ appears in all terms. Therefore $x$ is a common factor $ax + bx + cx + dx$ Since $x$ is a common factor, we factor it by writing in front of a parenthesis: $x\ ( \ \ )$ Inside the parenthesis, we write what is left over when we divide $x$ from each term. $x (a + b + c + d)$ Let’s look at more examples. Example 3 Factor a) $2x + 8$ b) $15x - 25$ c) $3a + 9b + 6$ Solution a) We see that the factor 2 divides evenly from both terms. $2x + 8 = 2(x) + 2(4)$ We factor the 2 by writing it in front of a parenthesis. $2( \ \ )$ Inside the parenthesis, we write what is left from each term when we divide by 2. $2(x + 4)$ This is the factored form. b) We see that the factor of 5 divides evenly from all terms. Rewrite $15x - 25 = 5(3x) - 5(5)$ Factor 5 to get $5(3x - 5)$ c) We see that the factor of 3 divides evenly from all terms. Rewrite $3a + 9b + 6 = 3(a) + 3(3b) + 3(2)$ Factor 3 to get $3(a + 3b + 2)$ Here are examples where different powers of the common factor appear in the polynomial Example 4 Find the greatest common factor a) $a^3 - 3a^2 + 4a$ b) $12a^4 - 5a^3 + 7a^2$ Solution a) Notice that the factor $a$ appears in all terms of $a^3 - 3a^2 + 4a$ but each term has a different power of $a$. The common factor is the lowest power that appears in the expression. In this case the factor is $a$. Let’s rewrite $a^3 - 3a^2 + 4a = a(a^2) + a(-3a) + a(4)$ Factor $a$ to get $a(a^2 - 3a + 4)$ b) The factor a appears in all the term and the lowest power is $a^2$. We rewrite the expression as $12a^4 - 5a^3 + 7a^2 = 12a^2 \cdot a^2 - 5a \cdot a^2 + 7 \cdot a^2$ Factor $a^2$ to get $a^2(12a^2 - 5a + 7)$ Let’s look at some examples where there is more than one common factor. Example 5: Factor completely a) $3ax + 9a$ b) $x^3y + xy$ c) $5x^3y - 15x^2y^2 + 25xy^3$ Solution a) Notice that 3 is common to both terms. When we factor 3 we get $3(ax + 3a)$ This is not completely factored though because if you look inside the parenthesis, we notice that $a$ is also a common factor. When we factor $a$ we get $3 \cdot a(x + 3)$ This is the answer because there are no more common factors. A different option is to factor all common factors at once. Since both 3 and $a$ are common we factor the term $3a$ and get $3a(x + 3)$. b) Notice that both $x$ and $y$ are common factors. Let’s rewrite the expression $x^3y + xy = xy(x^2) + xy(1)$ When we factor $xy$ we obtain $xy(x^2 + 1)$ c) The common factors are $5xy$. When we factor $5xy$ we obtain $5xy(x^2 - 3xy + 5y^2)$ ## Solve Simple Polynomial Equations by Factoring Now that we know the basics of factoring, we can solve some simple polynomial equations. We already saw how we can use the Zero-product Property to solve polynomials in factored form. Here you will learn how to solve polynomials in expanded form. These are the steps for this process. Step 1 If necessary, re-write the equation in standard form such that: Polynomial expression $= 0$ Step 2 Factor the polynomial completely Step 3 Use the zero-product rule to set each factor equal to zero Step 4 Solve each equation from step 3 Step 5 Example 6 Solve the following polynomial equations a) $x^2 - 2x = 0$ b) $2x^2 = 5x$ c) $9x^2y - 6xy = 0$ Solution: a) $x^2 - 2x = 0$ Rewrite this is not necessary since the equation is in the correct form. Factor The common factor is $x$, so this factors as: $x(x - 2) = 0.$ Set each factor equal to zero. $x = 0 \quad \quad \text{or} \quad \quad x - 2 = 0$ Solve $x = 0 \quad \quad \text{or} \quad \quad x = 2$ Check Substitute each solution back into the original equation. $& x = 0 && \Rightarrow && (0)^2 - 2(0) = 0 && \ \text{works out}\\& x = 2 && \Rightarrow && (2)^2 - 2(2) = 4 - 4 = 0 && \ \text{works out}$ Answer $x = 0, x = 2$ b) $2x^2 = 5x$ Rewrite $2x^2 = 5x \Rightarrow 2x^2 - 5x = 0.$ Factor The common factor is $x$, so this factors as: $x(2x - 5) = 0$. Set each factor equal to zero:. $x = 0 \quad \quad \text{or} \quad \quad 2x - 5 = 0$ Solve $\underline{x = 0} && \text{or} && 2x = 5\\&&&& x = \frac{5} {2}$ Check Substitute each solution back into the original equation. $& x = 0 \Rightarrow 2(0)^2 = 5(0) \Rightarrow 0 = 0 && \ \text{works out}\\& x = \frac{5} {2} \Rightarrow 2 \left (\frac{5} {2} \right)^2 = 5 \cdot \frac{5} {2} \Rightarrow 2 \cdot \frac{25} {4} = \frac{25} {2} \Rightarrow \frac{25} {2} =\frac{25} {2} && \ \text{works out}$ Answer $x = 0, x = \frac{5}{2}$ c) $9x^2y - 6xy = 0$ Rewrite Not necessary Factor The common factor is $3xy$, so this factors as $3xy(3x - 2) = 0$. Set each factor equal to zero. $3 = 0$ is never true, so this part does not give a solution $x = 0 \quad \quad \text{or} \quad \quad y = 0 \quad \quad \text{or} \quad \quad 3x - 2 = 0$ Solve $x = 0 && \text{or} && y = 0 && \text{or} && 3x = 2 \\&&&&&&&&x = \frac{2} {3}$ Check Substitute each solution back into the original equation. $x = 0 & \Rightarrow 9(0)y - 6(0)y = 0 - 0 = 0 && \ \text{works out}\\y = 0 & \Rightarrow 9x^2(0) - 6x = 0 - 0 = 0 && \ \text{works out}\\\frac{2} {3} & \Rightarrow 9 \cdot \left (\frac{2} {3} \right)^2 y - 6 \cdot \frac{2} {3}y = 9 \cdot \frac{4} {9}y - 4y = 4y - 4y = 0 && \ \text{works out}$ Answer $x = 0, y = 0, x = \frac{2}{3}$ ## Review Questions Factor the common factor in the following polynomials. 1. $3x^3 - 21x$ 2. $5x^6 + 15x^4$ 3. $4x^3 + 10x^2 - 2x$ 4. $-10x^6 + 12x^5 - 4x^4$ 5. $12xy + 24xy^2 + 36xy^3$ 6. $5a^3 - 7a$ 7. $45y^{12} + 30y^{10}$ 8. $16xy^{2z} + 4x^3y$ Solve the following polynomial equations. 1. $x(x + 12) = 0$ 2. $(2x + 1) (2x - 1) = 0$ 3. $(x - 5) (2x + 7) (3x - 4) = 0$ 4. $2x(x + 9) (7x - 20) = 0$ 5. $18y - 3y^2 = 0$ 6. $9x^2 = 27x$ 7. $4a^2 + a = 0$ 8. $b^2 - \frac{5}{3b} = 0$ 1. $3x(x^2 - 7)$ 2. $5x^4(x^2 + 3)$ 3. $2x(2x^2 + 5x - 1)$ 4. $2x^4(-5x^2 + 6x - 2)$ 5. $12xy(1 + 2y + 3y^2)$ 6. $a(5a^2 - 7)$ 7. $15y^{10}(3y^2 + 2)$ 8. $4xy(4yz + x^2)$ 9. $x = 0, x = -12$ 10. $x = - \frac{1}{2}, x = \frac{1}{2}$ 11. $x = 5, x = - \frac{7}{2}, x = \frac{4}{3}$ 12. $x = 0, x = -9, x = \frac{20}{7}$ 13. $y = 0, y = 6$ 14. $x = 0, x = 3$ 15. $a = 0, a = - \frac{1}{4}$ 16. $b = 0, b = \frac{5}{3}$ Feb 22, 2012 Aug 22, 2014
How To Write Numbers In Expanded Form? – tntips.com # How To Write Numbers In Expanded Form? ## How To Write Numbers In Expanded Form? In the expanded form the digits of the number are split into each of the individual digits with their place value and written in expanded form. The example of standard form of a number is 4,982 and the same number can be written in expanded form as 4 × 1000 + 9 × 100 + 8 × 10 + 2 × 1 = 4000 + 900 + 80 + 2. ## What is 35713 in expanded form? Finally, the expanded form of the number 35713 is 30,000+ 5000+700+10+3. ## How do you write 307 in expanded form? Expanded form of 307=3×100+0×10+7×1. ## How do you write 56 in expanded form? 1. Expanded Form. When you write a number in expanded form, you write a number in the form of an addition statement that shows place value. … 2. Expanded Form. … 3. 65 = 60 + 5. … 4. 56 = 50 + 6. … 5. 91 = 90 + 1. … 6. 24 = 20 + 4. … 7. 76 = 70 + 6. … 8. 37 = 30 + 7. ## How do you write 12345 in expanded form? Changing the words to math symbols, 12,345 in expanded form is: 1 x 10,000 + 2 x 1,000 + 3 x 100 + 4 x 10 + 5 x 1 Standard form is the reverse of expanded form. ## How do you write 279404 in expanded form? ⇒279404=2×105+7×104+9×103+4×102+0× ## What is the expanded form of 4256? Answer: Four thousand two hundred fifty six. ## How do you write 320 in expanded form? The expanded form can be written for multiples of 10 such as 40, 320, 700. The expanded for for these numbers are 40 = 4 × 10 + 0 × 1 = 40 + 0, 320 = 3 × 100 + 2 × 10 + 0 × 1 = 300 + 20 + 0, 700 = 7 × 100 + 0 × 10 + 0 × 1 = 700. ## How do you write 300 in expanded form? Numbers in expanded form Now from 311 to 320 we will learn to write numbers in both numbers name and expanded form – 300 to 399. ## What is 76 in expanded form? 70 + 6 The expanded form of 76 = 70 + 6. ## How do you write 64 in expanded form? Answer: 4 × 4 × 4 is the expanded form of 64. For 64 = 4 × 4 × 4 = 43, the base number 4 appears three times as a factor of the basic numeral (or number) 64. ## How do you write 70 in expanded form? The number is written as the sum of the separate place values of each digit. For example 70 is written in expanded form as 70 + 2. 72 is made of two digits: 7 and 2. ## How do you write 15768 in expanded form? 1×10000+5×1000+6×100+7×10+8×1. ## How do you write 20 in expanded form? We know that the number written as sum of the place-values of its digits is called the expanded form of a number. Expanded form of a Number. Standard Form Expanded Form 20,37,81,405 = 20,00,00,000 + 0 + 30,00,000 + 7,00,000 + 80,000 + 1,000 + 400 + 0 + 5 ## What is the face value of 5 in 78654321? And the face value of 5 in 78654321 is 5. Note: Place value-describe the position or place of a digit in a given number. Each digit of the number has a value depending on its place. ## Which is the expanded form of 684502? Hence, the expanded forms are (600000 + 80000 + 4000 + 500 + 2) , (4000000 + 7000 + 100 + 80 + 5 ) and (5000000 + 800000 + 7000 + 200 + 90 + 4). ## What is expanded form example? The expanded form of a number writes it as a sum, with each digit makes an individual term multiplied by its place value. For example 523 has an expanded form of 5 × 100 + 2 × 10 + 3 , 5 \times 100 + 2 \times 10 + 3 , 5×100+2×10+3, and 6203 has an expanded form of. ## What is the standard form of 3430000? 3.43×106 Hence, the standard form of 3430000 is 3.43×106. ## What expanded form? Expanded form or expanded notation is a way of writing numbers to see the math value of individual digits. When numbers are separated into individual place values and decimal places they can also form a mathematical expression. 5,325 in expanded notation form is 5,000 + 300 + 20 + 5 = 5,325. ## How do you write 90.125 in expanded form? 1. Given : 90.125 written in expanded form. 2. To find : Correct option. 3. Solution: 4. C. 9×10+1×1/10+2×1/100+5×1/1,000. 5. = 90 + 0.1 + 0.02 + 0.005. 6. = 90.125. 7. CORRECT. ## How do you write 305 in expanded form? 305 is 3 hundreds, 0 tens, and 5 ones. 305 in expanded form is 300 + 5. ## What is the expanded form of 35? Expanded form of 35 will be 30 +5 where 5 is at ones place and 30 is a tens place. ## What is the expanded form of 102? What is 102 in expanded form? Term Write 725 in expanded notation. Definition 700 + 20 + 5. Term Write 102 in expanded notation. Definition 100 + 2. Term Write 312 in expanded notation. Definition 300 + 10 + 2. Term Write 252 in expanded notation. Definition 200 + 50 + 2. ## How many tens are there in 400? 40 tens In the number 400, for example, there are 40 tens, because 400 divided by 10 is 40. ## How do you write 317 in words? The Number 317 in Words 317 is the three hundred and seventeenth number. ## How do you write 112 in expanded form? To convert 112 in Roman Numerals, we will write 112 in the expanded form, i.e. 112 = 100 + 10 + 1 + 1 thereafter replacing the transformed numbers with their respective roman numerals, we get 112 = C + X + I + I = CXII. ## How do you write 1000 in expanded form? As we move left, we multiply by 10 to obtain the new place value. 1. 1 – Ones. 2. 1 x 10 = 10 » Tens. 3. 10 x 10 = 100 » Hundreds. 4. 100 x 10 = 1000 » Thousands. 5. 1000 x 10 = 10,000 » Ten Thousands. 6. 10,000 x 10 = 100,000 » Hundred Thousands. 7. 100,000 x 10 = 1,000,000 » Millions. ## How do you write expanded form in math? Expanded form is a way to write a number by adding the value of its digits. We can use a place value chart to think of the value of a number’s digits. ## What is the expanded form of 88 35721? therefore, expanded form of 8835721 is, → 8835721 = 8000000 + 800000 + 30000 + 5000 + 800 + 20 + 1 . ## What is 90.125 written in word form? ninety-one and twenty-five thousandths. C. ## Writing a number in expanded form \| Arithmetic properties \| Pre-Algebra \| Khan Academy Related Searches example of expanded form what is 83.34 written in expanded form? expanded form of 10,000 expanded form of 90 place value expanded form a number written in expanded form is written as the sum of each expanded form vs expanded notation how to expand a number See more articles in category: FAQ
Go through the enVision Math Common Core Kindergarten Answer Key Topic 4 Compare Numbers 0 to 10 regularly and improve your accuracy in solving questions. ## Envision Math Common Core Kindergarten Answers Key Topic 4 Compare Numbers 0 to 10 Essential Question: How can numbers from 0 to 10 be compared and ordered? enVision STEM Project: Weather Changes Directions Read the character speech bubbles to students. Find Out! Have students find out about weather changes. Say: The weather changes from day to day. Talk to friends and relatives about the weather. Ask them to help you record the number of sunny days and rainy days during the week. Journal: Make a Poster Have students make a poster. Have them draw up to 10 lightning bolts above one house and up to 10 lightning bolts above another house. Ask them to write the number of lightning bolts above each house, and then draw a circle around the number that is greater than the other, or draw a circle around both numbers if they are the same. Review What You Know Question 1. Explanation: I circled the group of birds that are lessthan the other group.3 is lessthan 4. Question 2. Explanation: I circled the group of dogs that is greater than the ther group.5 is greater than 1. Question 3. Explanation: I circled 2 groups of marbles that are equal in number.3 is equal to 3. Question 4. Explanation: There are 6 objects in the above picture.So, i counted and wrote 6. Question 5. Explanation: There are 8 objects in the above picture.So, i counted and wrote 8. Question 6. Explanation: There are 10 objects in the above picture.So, i counted and wrote 10. Directions Have students: 1 draw a circle around the group of birds that is less than the other group; 2 draw a circle around the group of dogs that is greater than the other group; 3 draw a circle around the two groups that have an equal number of marbles; 4-6 count the number of objects, and then write the number to tell how many. Pick a Project Directions Say: You will choose one of these projects. Look at picture A. Think about this question: How can you train to go into space? If you choose Project A, you will act out an exercise skit. Look at picture B, Think about this question: What kinds of fruit would you put into a fruit salad? If you choose Project B, you will create a fruit salad recipe. Directions Say: You will choose one of these projects. Look at picture C. Think about this question: What is the most exciting ride at a theme park? If you choose Project C, you will design a ride. Look at picture D. Think about this question: What do you like to do on a vacation? If you choose Project D, you will make a list. ### Lesson 4.1 Compare Groups to 10 by Matching Solve & Share Directions Say: Work with a partner. Take turns drawing one cube from the bag and placing it on your page in the rectangle of the same color. When the bag is empty, do you have more red or blue cubes? How do you know? Draw a picture of your cubes in The rectangles showing which color is more. Visual Learning Bridge Guided practice Question 1. Explanation: I drew a line from each chick in the top group to a chick in the bottom group, and then drew a circle around the group that is greater in number than the other group.7 is greater than 6. Directions 1 Have students compare the groups, draw a line from each chick in the top group to a chick in the bottom group, and then draw a circle around the group that is greater in number than the other group. Question 2. Explanation: I compared the groups, drew a line from each chick in the top group to a chick in the bottom group, and then drew a circle around the group that is greater in number than the other group.8 is greater than 3. Question 3. Explanation: I compared the groups, drew a line from each chick in the top group to a chick in the bottom group, and then drew a circle around the group that is less in number than the other group.4 is less than 5. Question 4. Explanation: I compared the groups, drew a line from each chick in the top group to a chick in the bottom group, and then drew a circle around the group that is less in number than the other group.7 is less than 9. Directions 2 envision® STEM Say: Chicks live in coops. Coops protect chicks in different types of weather. Have students compare the groups, draw a line from each chick in the top group to a chick in the bottom group, and then draw a circle around the group that is greater in number than the other group. 3 and 4 Have students compare the groups, draw a line from each chick in the top group to a chick in the bottom group, and then draw a circle around the group that is less in number than the other group. Independent Practice Question 5. Explanation: I compared the groups, drew a line from each bucket in the top group to a bucket in the bottom group, and then drew a circle around the group that is greater in number than the other group.8 is greater than 6. Question 6. Explanation: I compared the groups, drew a line from each bucket in the top group to a bucket in the bottom group, and then drew a circle around the group that is less in number than the other group.2 is less than 3. Question 7. Explanation: I compared the groups, drew a line from each bucket in the top group to a bucket in the bottom group, and then drew a circle around the group that is less in number than the other group.4 is less than 7. Question 8. Explanation: I counted the numbe rof buckets, there are 5 buckets in the top group. I drew a group of 9 buckets that is greater than the buckets in the top group. 9 is greater than 5. Directions Have students: 5 compare the groups, draw a line from each bucket in the top group to a bucket in the bottom group, and then draw a circle around the group that is greater in number than the other group; 6 and 7 compare the groups, draw a line from each bucket in the top group to a bucket in the bottom group, and then draw a circle around the group that is less in number than the other group, 8 Higher Order Thinking Have students draw a group of buckets that is greater in number than the group shown. ### Lesson 4.2 Compare Numbers Using Numerals to 10 Explanation: Emily is planting seedlings, or little plants. She plants 5 red pepper seedlings and 7 yellow pepper seedlings. I drew counters to show groups of seedlings and i wrote the numbers 5 and 7.I circled 7 as it is greater than 5. Directions Say: Emily is planting seedlings, or little plants. She plants 5 red pepper seedlings and 7 yellow pepper seedlings. Use counters to show the groups of seedlings. Write the numbers, and then circle the number that tells which group has more. Visual Learning Bridge Guided Practice Question 1. Explanation: I drew a line from each watering can in the top group to a watering can in the bottom group, and then draw a circle around the number 4 as it is greater than 3. Directions 1 Have students count the watering cans in each group, write the number to tell how many, draw a line from each watering can in the top group to a watering can in the bottom group, and then draw a circle around the number that is greater than the other number. Question 2. Explanation: I counted the vegetables in each group, wrote the numbers 4 and 5 to tell how many, drew a line from each vegetable in the top group to a vegetable in the bottom group, and then marked an X on the number 4 as it is less than 4. Question 3. Explanation: I counted the vegetables in each group, i drew 3 more pea pods to make the groups equal, wrote the numbers 6 to tell how many in each group, and then drew a line from each vegetable in the top group to a vegetable in the bottom group to compare. Directions 2 Have students count the vegetables in each group, write the number to tell how many, draw a line from each vegetable in the top group to a vegetable in the bottom group, and then mark an X on the number that is less than the other number. 3 Number Sense Have students count the vegetables in each group, draw more pea pods to make the groups equal, write the numbers to tell how many in each group, and then draw a line from each vegetable in the top group to a vegetable in the bottom group to compare. Independent Practice Question 4. Explanation: I counted the seed packets in each group, wrote the numbers 10 and 7 to tell how many, drew a line from each seed packet in the top group to a seed packet in the bottom group, and then mark an X on the number 7 as it is less than 10. Question 5. Explanation: I counted the flowers in the group, drew a group of 3 flowers that is less than the group shown, and then write the numbers 6 and 3 to tell how many. Directions 4 Have students count the seed packets in each group, write the number to tell how many, draw a line from each seed packet in the top group to a seed packet in the bottom group, and then mark an X on the number that is less than the other number. 5 Higher Order Thinking Have students count the flowers in the group, draw a group of flowers that is less than the group shown, and then write the numbers to tell how many. ### Lesson 4.3 Compare Groups to 10 by Counting Solve & Share Explanation: I counted and placed counters on goldfish and tetras, there are 6 goldfish and 9 tetras.I drew a circle around the number 9 as it is greater than 6. Directions Say: The class aquarium has two kinds of fish, goldfish and tetras. Place counters on the fish as you count how many of each kind. Write numbers to tell how many of each kind. Draw a circle around the fish that has a number greater than the other. Tell how you know you are right. Visual Learning Bridge Guided Practice Question 1. Explanation: I counted and wrote the number of fishes each color, there are 10 pink fishes and 8 purple fishes.I circled 10 as it is greater than 8. Directions 1 Have students count the number of each color fish, write the numbers to tell how many, and then draw a circle around the number that is greater than the other number. Use the number sequence to help find the answer. Question 2. Explanation: I counted and wrote the number of fishes each color, there are 6 green fishes and 7 yellow fishes.I circled 7 as it is greater than 6. Question 3. Explanation: I counted wrote the number of fishes each color, there are 8 blue fishes and 9 gold fishes.I marked X on both the numbers as they are NOT equal. Question 4. Explanation: I counted wrote the number of fishes each color, there are 8 brown fishes and 7 green fishes.I marked X on the number 7 as it is lessthan 8. Question 5. Explanation: I counted wrote the number of fishes each color, there are 9 purple fishes and 10 gold fishes.I marked X on the number 7 as it is lessthan 10. Directions Have students count the number of each color fish, write the numbers to tell how many, and then: 2 draw a circle around the number that is greater than the other number; 3 draw a circle around both numbers if they are equal, or mark an X on both numbers if they are NOT equal; 4 and 5 mark an X on the number that is less than the other number. Use the number sequence to help find the answer for each problem. Independent Practice Question 6. Explanation: I counted and wrote the number of each critter.I wrote the numbers 6 to tell that there are 6 yellow and 6 green critters.I circled both the numbers as they are equal. Question 7. Explanation: I counted and wrote the number of each critter.I wrote the numbers 9 and 8 to tell that there are 9 blue and 8 peach critters.I marked X on the numbe 8 as it is lessthan 9. Question 8. Explanation: There are 7 trantulas in the above picture so, i drew 9 spiders that is two more than the number of trantulas. Then i wrote the numbers 7 and 9. Question 9. Explanation: I counted the number of butterflies.There are 6 butterflies.I wrote all the numbers up to 10 that are greater than the number of butterflies shown.The numbers that are greater than 6 are 7, 8, 9 and 10. Directions Have students count the number of each critter, write the numbers to tell how many, and then: 6 draw a circle around both numbers if they are equal, or mark an X on both numbers if they are NOT equal; 7 mark an X on the number that is less than the other number; 8 draw a group of spiders that is two greater in number than the number of tarantulas shown, and then write the number to tell how many. 9 Higher Order Thinking Have students count the butterflies, and then write all the numbers up to 10 that are greater than the number of butterflies shown. Use the number sequence to help find the answer for each problem. ### Lesson 4.4 Compare Numbers to 10 Solve & Share Directions Say: Emily’s mother asked her to bring the towels in off the line. Her basket can hold less than 7 towels. How many towels might Emily bring in? You can give more than one answer. Show how you know your answers are right. Emily’s mother asked her to bring the towels in off the line. Her basket can hold less than 7 towels. Emily can bring 1, 2, , 4, 5 or 6 towels in her basket. Visual Learning Bridge Guided Practice Question 1. Explanation: I counted the numbers 1 to 10. I used the number sequence and drew lines from the numbers to the sequence and i circled 8 a sit is greater than 7. Question 2. Explanation: I drew counters to show the numbers 6 and 4.I drew a circle around the number 6 as it is greater than the number 4. Directions Have students: 1 count the numbers 1 to 10 and use the number sequence to show how they know which number is greater than the other, and then draw a circle around the number that is greater; 2 draw counters in the ten-frames to show how they know which number is greater than the other, and then draw a circle around the number that is greater. Question 3. Explanation: I drew pictures to show the numbers 6 and 9.9 is greater than 6. Question 4. Explanation: I drew counters to show the numbers 8. I circled both the numbers as they are equal. Question 5. Explanation: I used the number sequence to find the greater number.I marked X on the number 9 as it is lessthan 10. Question 6. Explanation: I drew the pictures to show the numbers 9 and 8.I marked X on the number 8 as it is lessthan 9. Directions Have students: 3 draw pictures to show how they know which number is greater than the other, and then draw a circle around the number that is greater; 4 draw counters in the ten-frames to show how they know if the numbers are equal, and then draw a circle around both numbers if they are equal, or mark an X on both numbers if they are NOT equal; 5 use the number sequence to show how they know which number is less than the other number, and then mark an X on the number that is less; 6 draw pictures to show how they know which number is less than the other number, and then mark an X on the number that is less. Independent Practice Question 7. Explanation: I drew the pictures to show the numbers 6 and 8.I marked X on the number 6 as it is lessthan 8. Question 8. Explanation: I used the number sequence to find the greater number.I marked X on the number 7 as it is lessthan 9. Question 9. Explanation: I wrote the next two numbers that are greater than 8. the numbers that are greater than 8 are 9 and 10. Question 10. Explanation: I wrote the number 7 as it is greater than 5 and lessthan 9. Directions Have students: 7 draw pictures to show how they know which number is less than the other number, and then mark an X on the number that is less; 8 use the number sequence to show how they know which number is less than the other number, and then mark an X on the number that is less. 9 Higher Order Thinking Have students write the next two numbers that are greater than the number shown, and then tell how they know. 10 Higher Order Thinking Have students write a number that is greater than the number on the left, but less than the number on the right. ### Lesson 4.5 Repeated Reasoning Problem Solving Solve & Share Directions Say: There are 7 fish in a bowl. Emily puts I more fish in the bowl. How many fish are in the bowl now? How can you solve this problem? Visual Learning Bridge Guided Practice Question 1. Explanation: I drew counters for 4 frogs and i drew one more.The number that is one greater than 4 is 5. Directions 1 Say: Carlos sees 4 frogs at the pond. Then he sees 1 more. How many frogs are there now? Have students use reasoning to find the number that is 1 greater than the number of frogs shown. Draw counters to show the answer, and then write the number. Have students explain their reasoning. Independent Practice Question 2. Explanation: I drew counters for 7 frogs and i drew one more.The number that is one greater than 7 is 8. Question 3. Explanation: I drew counters for 2 frogs and i drew one more.The number that is one greater than 2 is 3. Question 4. Explanation: I drew counters for 8 frogs and i drew one more.The number that is one greater than 8 is 9. Question 5. Explanation: I drew counters for 6 frogs and i drew one more.The number that is one greater than 6 is 7. Directions Say: Alex sees frogs at the pond. Then he sees 1 more. How many frogs are there now? 2-5 Have students use reasoning to find the number that is 1 greater than the number of frogs shown. Draw counters to show the answer, and then write the number. Have students explain their reasoning. Problem Solving Question 6, 7, 8. Explanation: I drew counters for 5 pets of Marta, i drew 1 more counter.The number that is one greater than 5 is 6.Marta will now have 6 pets. Directions Read the problem aloud. Then have students use multiple problem-solving methods to solve the problem. Say: Marta’s family has 5 pets. Then her family gets 1 more. How many pets do they have now? 6 Generalize Does something repeat in the problem? How does that help? 7 Use Tools What tool can you use to help solve the problem? Use the tool to find the number of pets in Marta’s family now. 8 Make Sense Should the answer be greater than or less than 5? ### Topic 4 Vocabulary Review Question 1. Explanation: I circled the number 9 as it is greater than 7. Question 2. Explanation: I counted the counters given, there are 7 counters in all.so, i wrote the number 7 in the blank. Question 3. Explanation: The number that means NONE is 0. Question 4. Explanation: There are 2 red cubes and 3 yellow cubes.I circled the red cubes as they are less than yellow cubes.I wote the number 2 as it is lessthan 3. Directions Understand Vocabulary Have students: 1 draw a circle around the number that is greater than 7; 2 count the counters, and then write the number to tell how many; 3 write the number that means none; 4 count how many of each color cube there is, draw a circle around the group that has a number of cubes that is less than the other group, and then write the number to tell how many there are in that group. Question 5. Explanation: I circled the number 8 as it is greater than 3 and marked X on the number 3 as it is lessthan 8. Question 6. Explanation: I wrote 4 as it is greater than 3 and lessthan 5. Question 7. Explanation: I drew counters for the given number 5. Question 8. Explanation: I wrote the missing numbers 3, 4, 6, 7 and 8 in the gien blanks. Directions Understand Vocabulary Have students: 5 compare the numbers, draw a circle around the number that is greater, and then mark an X on the number that is less; 6 write a number that is greater than 3, but less than 5; 7 draw 5 counters in a row and then write the number to tell how many; 8 write the missing numbers in order. ### Topic 4 Reteaching Set A Question 1. Explanation: I counted the number of counters, there are 6 red counters and 10 yellow counters.I circled red counters frame as they are less in number than the yellow counter frame. Set B Question 2. Explanation: I drew lines from the top group to a piece of fruit in the bottom group, and then i drew a circle around the number 7 as it is greater than 5. Directions Have students: 1 compare the groups, and draw a circle around the group that is less in number than the other group; 2 count the fruit in each group, write the numbers that tell how many, draw a line from each piece of fruit in the top group to a piece of fruit in the bottom group, and then draw a circle around the number that is greater than the other number. Set C Question 3. Explanation: I counted and wrote the number of blue critter and peach critters.They are 6 peach critters and 4 blue critters. I marked X on the number 4 as it is lessthan 6. Set D Question 4. Explanation: I drew counters for the 6 frogs and drew one more counter.1 greater than 6 is 7. Directions Have students: 3 count the number of each critter, write the numbers, and then mark an X on the number that is less than the other number; 4 Say: April sees frogs at The pond. Then she sees 1 more. How many frogs does she see now? Have students use reasoning to find the number that is 1 greater than the number of frogs shown. Draw counters to show the answer, and then write the number. ### Topic 4 Assessment Practice Question 1. Explanation: There are 8 yellow birds.I marked group ‘A’ as it has more number of blue birds than the yellow birds.There are 10 blue bieds in group A. Question 2. Explanation: I marked the numbers 6, 5 and 3.The numbers 3, 5 and 6 as they are lessthan the number 7. Question 3. Explanation: I counted and wrote the number of lemons and limes, there are 7 lemons and 5 limes. I circled the number 7 as it is greater than 5. Directions Have students mark the best answer. 1 Which group of blue birds is greater in number than the group of yellow birds? 2 Look at the number line. Then mark all the numbers that are less than the number on the card. 3 Have students count the number of lemons and limes, write the number that tells how many of each, and then draw a circle around the number that is greater. Question 4. Explanation: The number of the first card is 7, the number before 7 is 6.I counted forward from 6 and wrote the numbers 7, 8, 9 and 10 in the blanks. Question 5. Explanation: There are 6 sandwiches and I drew the 4 juice boxes as 4 is lessthan 6. Question 6. Explanation: I drew counters for 7 beads and drew 1 more counter.one greater than the number 7 is 8. Kayla will now have 8 beads to make a bracelet. Directions Have students: 4 write the number that is counted first among the 4 number cards, and then count forward and write the number that is 1 greater than the number before; 5 count the sandwiches in the group, draw a group of juice boxes that is less than the group of sandwiches shown, and then write the numbers to tell how many. 6 Say: Kayla has 7 beads to make a bracelet. Then she buys 1 more. How many beads does she have now? Have students use reasoning to find the number that is 1 greater than the number of beads shown. Draw counters to show the answer, and then write the number to tell how many. Question 1. Explanation: There are 9 shunks and 8 raccoons in the forest. Using number sequence i found that number 8 is lessthan number 9 so, i marked X on the number 8. Directions Forest Animals Say: The forest is home to woodland animals. One part of the forest has many different animal homes in it. 1 Have students study the picture. Say: How many skunks live in this part of the forest? How many raccoons live in this part of the forest? Count the number of each type of animal and write the numbers. Then have students draw a circle around the number that is greater than the other number and mark an X on the number that is less than the other number. Have them use the number sequence to help find the answers. Question 2. Explanation: There are 6 foxes in the forest.I counted and wrote the numbers between 6 and 10.They are 7, 8, 9 and 10. Question 3.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 12.4: Finding Outcomes Difficulty Level: At Grade Created by: CK-12 ## Introduction The Talent Show Outfit Alicia is going to sing for the Talent Show. She is very excited and has selected a wonderful song to sing. She has been practicing with her singing teacher for weeks and is feeling very confident about her ability to do a wonderful job. Her performance outfit is another matter. Alicia has selected a few different skirts and a few different shirts and shoes to wear. Here are her options for shirts Striped shirt Solid shirt Here are her options for skirts. Blue skirt Red skirt Brown skirt Here are her options for shoes Dance shoes Black dress shoes How many different outfits can Alicia create given these options? This is best done using a tree diagram. Alicia needs to organize her clothing options using a tree diagram. This lesson will show you all about tree diagrams. When finished, you will know how many possible outfits Alicia can create. What You Will Learn In this lesson you will learn how to correctly apply the following skills. • Use tree diagrams to list all possible outcomes of a series of events involving two or more choices or results. • Recognize all possible outcomes of an experiment as the sample space. • Find probability of specified outcomes using tree diagrams. Teaching Time I. Use Tree Diagrams to List all Possible Outcomes of a Series of Events Involving Two or More Choices or Results Nadia’s soccer team has 2 games to play this weekend. How many outcomes are there for Nadia’s team? A good way to find the total number of outcomes for events is to make a tree diagram. A tree diagram is a branching diagram that shows all possible outcomes for an event. To make a tree diagram, split the different events into either-or choices. You can list the choices in any order. Here is a tree diagram for game 1 and game 2. As you can see, there are four different outcomes for the two games: win-winwin-loselose-winlose-lose\begin{align}\text{win-win} \quad \quad \text{win-lose}\!\\ \text{lose-win} \quad \quad \text{lose-lose}\end{align} What happens when you increase the number of games to three? Just add another section to your tree diagram. In all, there are 8 total outcomes. A tree diagram is a great way to visually see all of the possible options. It can also help you to organize your ideas so that you don’t miss any possibilities. Example To remodel her kitchen, Gretchen has the following choices: Floor: tile or wood; Counter: Granite or formica; Sink: white, steel, stone. How many different choices can Gretchen make? First, let’s create a tree diagram that shows all of the possible options. Step 1: List the choices. Choice 1 Floor: tile, wood Choice 2 Counter: granite, formica Choice 3 Sink: white, steel, stone Step 2: Start the tree diagram by listing any of the choices for Choice 1. Then have Choice 1 branch off to Choice 2. Make sure Choice 2 repeats for each branch of Choice 1. Can you identify the missing labels in the tree below? Step 3: Fill in the third choice. We have left some of the spaces for you to fill in. Step 4: Fill in the outcomes. Again some of the spaces are left for you to fill in. You can see that there are 12 possible outcomes for the kitchen design. II. Recognize all Possible Outcomes of an Experiment as the Sample Space When you conduct an experiment, there may be few or many possible outcomes. For example, if you are doing an experiment with a coin, there are two possible outcomes because there are two sides of the coin. You can either have heads or tails. If you have an experiment with a number cube, there are six possible outcomes, because there are six sides of the number cube and the sides are numbered one to six. We can think of all of these possible outcomes as the sample space. A sample space is the set of all possible outcomes for a probability experiment or activity. For example, on the spinner there are 5 different colors on which the arrow can land. The sample space, S\begin{align}S\end{align}, for one spin of the spinner is then: S=red, yellow, pink, green, blue\begin{align}S = \text{red, yellow, pink, green, blue}\end{align} These are the only outcomes that result from a single spin of the spinner. Changing the spinner changes the sample space. This second spinner still has 5 equal-sized sections. But its sample space now has only 3 outcomes: S=red, yellow, blue\begin{align}S = \text{red, yellow, blue}\end{align} Let’s look at an example having to do with sample spaces. Example A small jar contains 1 white, 1 black, and 1 red marble. If one marble is randomly chosen, how many possible outcomes are there in the sample space? Since only a single marble is being chosen, the total number of possible outcomes, the sample space, matches the marble colors. S=white, black, red\begin{align}S = \text{white, black, red}\end{align} Sometimes, the sample space can change if an experiment is performed more than once. If a marble is selected from a jar and then replaced and if the experiment is conducted again, then the sample space can change. The number of outcomes is altered. When this happens, we can use tree diagrams to help us figure out the number of outcomes in the sample space. Example A jar contains 1 white and 1 black marble. If one marble is randomly chosen, returned to the jar, then a second marble is chosen, how many possible outcomes are there? This is an example where a tree diagram is very useful. Consider the marbles one at a time. After the first marble is chosen, it is returned to the jar so now there are again two choices for the second marble. Use a tree diagram to list the outcomes. From the tree diagram, you can see that the sample space is: S=white-white, white-black, black-white, black-black\begin{align}S = \text{white-white, white-black, black-white, black-black}\end{align} 12G. Lesson Exercises What is the sample space in each example? 1. A spinner with red, blue, yellow and green. 2. A number cube numbered 1 – 6. 3. A bag with a blue and a red marble. One marble is drawn and then replaced. Take a few minutes to check your work with a friend. III. Find Probabilities of Specified Outcomes Using Tree Diagrams In the last section, you started to see how tree diagrams could be very helpful when looking for a sample space. Tree diagrams can also be helpful when finding probability. Finding the probability of an event is a matter of finding the ratio of favorable outcomes to total outcomes. For example, the sample space for a single coin flip has two outcomes: heads and tails. So the probability of getting heads on any single coin flip is: P(heads)=favorable outcomestotal outcomes=12\begin{align}P (\text{heads}) = \frac{favorable \ outcomes}{total \ outcomes} =\frac{1}{2}\end{align} You can see that the sample space is represented by a number in the total outcomes. For example, if you had a spinner with four colors, the colors by name would be the sample space and the number four would be the total possible outcomes. What about if we flipped a coin more than one time? To find the probability of a single outcome for more than one coin flip, use a tree diagram to find all possible outcomes in the sample space. Then count the number of favorable outcomes within that sample space to find the probability. For example, to find the probability of tossing a single coin twice and getting heads both times, make a tree diagram to find all possible outcomes. The diagram shows there are 8 total outcomes and they are paired with first toss option and second toss option. Then pick out the favorable outcome–in this case, the outcome “heads-heads” is shown in red. You could have selected any of the favorable outcomes for the probability to be accurate. Now write the ratio of favorable outcomes to total outcomes in the sample space. P(heads-heads)=favorable outcomestotal outcomes=14\begin{align}P (\text{heads-heads}) = \frac{favorable \ outcomes}{total \ outcomes}=\frac{1}{4}\end{align} You can see that since 1 of 4 outcomes is a favorable outcome, the probability of the coin landing on heads 2 times in a row is 14\begin{align}\frac{1}{4}\end{align}. Let’s look at another example. Example What is the probability of flipping a coin two times and getting two matching results–that is, either two heads or two tails? First, let’s create a tree diagram to see our options. Once again, just pick out the favorable outcomes on the same tree diagram. They are shown in red. You can see that 2 of 4 total outcomes match. P(2 heads or 2 tails)=favorable outcomestotal outcomes=24=12\begin{align}P (2 \ \text{heads or 2 tails}) = \frac{favorable \ outcomes}{total \ outcomes}=\frac{2}{4}=\frac{1}{2}\end{align} You can see that the probability of flipping two heads or two tails is 1:2. ## Real–Life Example Completed The Talent Show Outfit Here is the original problem once again. Reread it and then look at the tree diagram created. Alicia is going to sing for the Talent Show. She is very excited and has selected a wonderful song to sing. She has been practicing with her singing teacher for weeks and is feeling very confident about her ability to do a wonderful job. Her performance outfit is another matter. Alicia has selected a few different skirts and a few different shirts and shoes to wear. Here are her options for shirts Striped shirt Solid shirt Here are her options for skirts. Blue skirt Red skirt Brown skirt Here are her options for shoes Dance shoes Black dress shoes How many different outfits can Alicia create given these options? This is best done using a tree diagram. Alicia needs to organize her clothing options using a tree diagram. To do this, we can take each option and create a diagram to show all of the options. Based on this tree diagram, you can see that Alicia has twelve possible outfits to choose from. ## Vocabulary Here are the vocabulary words that are found in this lesson. Tree Diagram a visual way of showing all of the possible outcomes of an experiment. Called a tree diagram because each option is drawn as a branch of a tree Sample Space The possible outcomes of an experiment Favorable Outcome the outcome that you are looking for in an experiment Total Outcome the number of options in the sample space ## Time to Practice Directions: Use Tree Diagrams for each of the following problems. 1. The Triplex Theater has 3 different movies tonight: Bucket of Fun, Bozo the Great, and Pickle Man. Each movie has an early and late show. How many different movie choices are there? 2. Raccoon Stadium offers the following seating plans for football games: • lower deck, middle loge, or upper bleachers • center, side, end-zone How many different kinds of seats can you buy? 3. Cell-Gel cell phone company offers the following choices: • Free internet plan or Pay internet plan • 1200, 2000, or 3000 minutes How many different kinds of plans can you get? 4. Jen’s soccer team is playing 4 games next week. How many different outcomes are there for the four games? 5. The e-Box laptop computer offers the following options. • Screen: small, medium, or large • Memory: standard 1 GB, extra 2 GB • Colors: pearl, blue, black List the number of different activity choices a camper can make. Use a tree diagram to list them all. Double click to check your answers. 6. What is the sample space for a single toss of a number cube? 7. What is the sample space for a single flip of a coin? 8. A coin is flipped two times. List all possible outcomes for the two flips. 9. A coin is flipped three times in a row. List all possible outcomes for the three flips. 10. A bag contains 3 ping pong balls: 1 red, 1 blue, and 1 green. What is the sample space for drawing a single ball from the bag? 11. A bag contains 3 ping pong balls: 1 red, 1 blue, and 1 green. What is the sample space for drawing a single ball, returning the ball to the bag, then drawing a second ball? 12. What is the sample space for a single spin of the spinner with red, blue, yellow and green sections? 13. What is the sample space for 2 spins of the first spinner? 14. A box contains 3 socks: 1 black, 1 white, and 1 brown. What is the sample space for drawing a single sock, NOT returning the sock to the box, then drawing a second sock? 15. A box contains 3 socks: 1 black, 1 white, and 1 brown. What is the sample space for drawing all 3 socks from the box, one at a time, without returning any of the socks to the box? 16. A box contains 3 black socks. What is the sample space for drawing all 3 socks from the box, one at a time, without returning any of the socks to the box? 17. A box contains 2 black socks and 1 white sock. What is the sample space for drawing all 3 socks from the box, one at a time, without returning any of the socks to the box? Directions: Answer each question. Use tree diagrams when necessary. 18. What is the probability that the arrow of the spinner will land on red on a single spin? 19. If the spinner is spun two times in a row, what is the probability that the arrow will land on red both times? 20. If the spinner is spun two times in a row, what is the probability that the spinner will land on the same color twice? 21. If the spinner is spun two times in a row, what is the probability that the arrow will land on red at least one time? 22. If the spinner is spun two times in a row, what is the probability that the spinner will land on a different color both times? 23. If the spinner is spun two times in a row, what is the probability that the arrow will land on blue or green at least one time? 24. Two cards, the Ace and King of hearts, are taken from a deck, shuffled, and placed face down. What is the probability that a single card chosen at random will be an Ace? 25. If one card is chosen from the 2-card stack above, then returned to the stack and a second card is chosen, what is the probability that both cards will be Kings? 26. If one card is chosen from the 2-card stack above, then returned to the stack and a second card is chosen, what is the probability that both cards will match? 27. If one card is chosen from the 2-card stack above, then returned to the stack and a second card is chosen, what is the probability that both cards NOT match? ### Notes/Highlights Having trouble? Report an issue. 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### Math Unit 6 Adding & Subtracting Decimals Unit 6: Between 0 and 1 Unit Focus: • Decimals, Place Value fluency and Base-10 number system structure • Understanding the meaning of decimals • Comparing and ordering decimals Websites: Variety of math related games on “mathplayground” Fun 4 The Brain (all concepts) http://www.fun4thebrain.com/mult.html Math Games (all concepts) http://www.hoodamath.com/ Review decimal places - Video Place Value - Rock music Place Value - Music showing numbers Spin to Win - Place value strategy Equality of Decimals - Video (0.3 = 0.30 = 0.300) Ordering Decimals - Khan video. Builder Ted - Ordering decimals interactive Decimal Switch - Order decimals Battleship Numbers - Select decimals Ordering Decimals - Read and practice. Ordering Decimals - Drag in the correct order Comparing Place Values - Khan video. Introduces moving decimal to the left. Decimal Tug-of-War Decimals - Learn Zillion video - Place value x 10 to the left Decimals - Learn Zillion video - Place value divided by 10 to the right. Decimals in Expanded Form - Khan Academy video Decimal Videos - From Watch, Know, Learn series Introduction to Decimals - 21 videos Decimal Rainstorm - Comparing decimals Decimal Tiles- Equivalent Decimals - Tutorial Equivalent Decimals - Guided practice slides Decimal Tiles - Drag decimals to compare. In-Between - Blinker is lost on an unknown planet. Help find him using decimals. Decimal Games Decention - Sort fractions, decimals, and percents for the Intergalactic Space Games. Fractions, Decimals, and Percents - Many games and activities from Johnny's Math Page. Dartboards of Fractions, Decimals, Percents - Click View Full Screen to play a variety of games. Try This - Drag the decimals into the correct order. Percent Panic - Use the arrow keys on the keyboard to drive the rocket-ship to work with fractions, decimals, and percents. The Award Ceremoony - Math activities based on athletic events. Fruit Shoot - Convert fractions into decimals.
# 180 Days of Math for Sixth Grade Day 109 Answers Key By accessing our 180 Days of Math for Sixth Grade Answers Key Day 109 regularly, students can get better problem-solving skills. ## 180 Days of Math for Sixth Grade Answers Key Day 109 Directions: Solve each problem. Question 1. 400 + (-78) = __________ + Ã— – = – 400 – 78 = 322 Question 2. 5 Ã— 46 = ____________ Multiply 46 with 5 we get 230. Question 3. What is the quotient when 614 is divided by 7? Answer: 614 Ã· 7 = 87.7 Question 4. List the first three multiples of 4. Answer: The first three mutiples of 4 are 4, 8, 12. Question 5. 10% of 50 is ___________. 10/100 Ã— 50 = 5 So, 10% of 50 is 5. Question 6. 5 Ã— 22 = ___________ 22 = 4 5 Ã— 4 = 20 5 Ã— 22 = 20. Question 7. 2$$\frac{1}{3}$$ is a mixed fraction we have to convert it into the improper fraction 2$$\frac{1}{3}$$ = $$\frac{7}{3}$$ Question 8. Write the expression for the sum of one-half and s. Answer: $$\frac{1}{2}$$ + s Question 9. How many millimeters are in 3.5 centimeters? 1 centimeter = 10 millimeter 3.5 centimeter = 3.5 Ã—10 millimeter = 35 millimeter Question 10. What is the diameter of a circle with a radius of 3.45 cm? d = 2r d = 3.45 Ã— 2 d = 6.90 cm Thus the diameter of a circle is 6.90 cm. Question 11. In a game, the probability of spinning a 5 or 6 is $$\frac{2}{3}$$. How many times would you expect to land on a number other than 5 or 6 if you spin the spinner 15 times? Answer: $$\frac{2}{3}$$ Ã— 15 = 2 Ã— 5 = 10 Question 12. Arrange each number 1 through 9 in the circles so that the sum of the numbers in a straight line across the center of the circle is 18.
# Ch. 3 Systems of Linear Equations and Inequalities - Sec... • Test Prep • 12 • 100% (1) 1 out of 1 people found this document helpful This preview shows page 1 - 3 out of 12 pages. Sec 3.1 Systems of Linear Equations in Two Variables Learning Objectives: 1. Deciding whether an ordered pair is a solution. 2. Solve a system of linear equations using the graphing, substitution, and elimination method. 1. Deciding Whether an Ordered Pair Is a Solution System of Equation —consists of at least two or more linear equations. Example. = = 0 2 12 3 4 . 1 y x y x = + = + = + 0 2 2 2 5 . 2 z x z x y z y x Solution of the system —is the point(s) where the graphs intersect (give true for both equations) Example 1 . Determine whether the ordered pairs are a solution to the system: = + = 6 2 3 4 2 y x y x a. ( ) 0 , 2 Answer:___________________ --------------------------------------------------------------------------------------------------------------------------------------- b. ( ) 3 , 4 Answer:___________________ --------------------------------------------------------------------------------------------------------------------------------------- 2. Solve a system of linear equations using the graphing method Three types of the System of Equations . 1. Consistent system with Independent equation Two lines intersect at one point ( ) y x , . Has one solution, ( ) y x , . 2 1 m m When solve the system, get x = a number, y = a number. 2. Inconsistent System Two lines are parallel. Has no solution, φ or { } . 2 1 m m = and 2 1 b b When solve, get false statement. 3. Consistent system with dependent equation Two lines lie on top of the others (same line). Has infinitely many solutions, ( ) { } b mx y y x + = , or ( ) { } c by ax y x = + , 2 1 m m = and 2 1 b b = When solve the system, get true statement. x y x y x y Steps to solve linear equations by graphing 1. Solve and graph each equation separately. 2. Identify type of systems (consistent, inconsistent, or dependent). 3. State number of solution (one solution, infinitely many solutions or no solution). Example 2. Solve by graphing. Label at least two points for each graph on the graph grid. = = + 0 4 2 4 2 y x y x Solution:___________________ ---------------------------------------------------------------------------------------------------------------------------------- 3. Solve a system of linear equations using the elimination method Steps: 1. Write each equation in the form: C By Ax = 2. Choose variable to eliminate. 3. If necessary, multiply one or both equations by appropriate number(s) so that the coefficients of the eliminated variable will have the sum of zero. 4. Add two equations together. 5. Solve for the variable. 6. Solve for the other variable. 7. State the final solution in ordered pair, if it exists. Example 3. Solve linear equations using the elimination method. + = = y x y x 6 4 8 0 4 3 2 Answer:___________________________________ ---------------------------------------------------------------------------------------------------------------------------------------
9220 minus 35 percent This is where you will learn how to calculate nine thousand two hundred twenty minus thirty-five percent (9220 minus 35 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 9220 minus 35 percent means, and then we will give you the formula at the very end. We start by showing you the image below of a dark blue box that contains 9220 of something. 9220 (100%) 35 percent means 35 per hundred, so for each hundred in 9220, you want to subtract 35. Thus, you divide 9220 by 100 and then multiply the quotient by 35 to find out how much to subtract. Here is the math to calculate how much we should subtract: (9220 ÷ 100) × 35 = 3227 We made a pink square that we put on top of the image shown above to illustrate how much 35 percent is of the total 9220: The dark blue not covered up by the pink is 9220 minus 35 percent. Thus, we simply subtract the 3227 from 9220 to get the answer: 9220 - 3227 = 5993 The explanation and illustrations above are the educational way of calculating 9220 minus 35 percent. You can also, of course, use formulas to calculate 9220 minus 35%. Below we show you two formulas that you can use to calculate 9220 minus 35 percent and similar problems in the future. Formula 1 Number - ((Number × Percent/100)) 9220 - ((9220 × 35/100)) 9220 - 3227 = 5993 Formula 2 Number × (1 - (Percent/100)) 9220 × (1 - (35/100)) 9220 × 0.65 = 5993 Number Minus Percent Go here if you need to calculate any other number minus any other percent. 9230 minus 35 percent Here is the next percent tutorial on our list that may be of interest.
# Rationalize the denominator It was said that he radical sign was first used by Rene Descartes, but long before he used this, it has been developed in Germany in the 1500s. The radical ( sign was invented to find the square root or the cube root, 4th root, 5th root etc., depending on the index of the radical sign. Now if you remember, in previous chapter, we learned all about the different number systems and we have mentioned about the rational and irrational numbers. Radicals can belong to either of which depending of course if the radicals would be able to fit into the definition of whichever group of number. In this lesson we will get to be able to identify whether or not a certain radical is rational. We will be specifically going to study about a square root or a cube root of a particular number. We will first look into what a square and a cube are since these would be the two fundamental concepts. In order to solve a square root, or a cube root, one must know how to square or cube a number. These will be thoroughly discussed in 4.1 and 4.2. For 4.3 to 4.8 we will be looking closer into how to simplify and evaluate these radicals through applying what we have learned from chapters on square and square roots, estimating square roots, and rational numbers from earlier grades. We would also be learning how to convert them into either entire radicals to mixed radicals or vice versa, by applying our knowledge of factoring out the perfect squares or the perfect cubes in the equation. Apart from converting the radicals from one form to another we are also going to learn more about how to combine different radicals and simplify them depending on the operation used and learn to rationalize a certain radical equation. ### Rationalize the denominator When there is a radical in the denominator, the fraction is not in its simplest form. Therefore, we need to rationalize the denominator to move the root from the denominator/bottom of the fraction to the numerator/ top. In this lesson, we will learn how to simplify radicals by rationalizing the denominator.
# Order of Operations Lesson Plan: Math Skills Bingo ### Submitted by: Emily Wilmesherr In this Order of Operations lesson plan, which is adaptable for grades 3-8, students use BrainPOP resources to learn how to apply the order of operations when computing with whole numbers, excluding and then including exponents. Students will then create an original story, song, short skit, or poem to help them remember the order of operations, and practice their math skills through a game called Order of Operations BINGO. ### Students will: 1. Apply order of operations when computing with whole numbers, excluding exponents. 2. Apply order of operations when computing with whole numbers, including exponents. 3. Apply order of operations when computing with whole numbers and fractions (with older grades only). ### Materials: • Computer with internet access for BrainPOP • Interactive whiteboard (or just an LCD projector) • White board or chart paper and markers • BINGO cards • BINGO chips • BINGO problems • Peanut butter and jelly • Plastic knife • Plate ### Vocabulary: order of operations; parentheses; exponents; PEMDAS; multiplication; division; subtraction; addition ### Preparation: Create BINGO Cards and problems for the Order of Operations BINGO game. ### Lesson Procedure: 1. Ask students how to make a peanut butter and jelly sandwich. Have students write down their directions with a partner. 2. Call on a pair of students to read their answer. As the students give their directions, make the sandwich as a demonstration for the class, doing EXACTLY what the students say. 3. Talk with students about the importance of giving exact directions in the right order, not only when making a peanut butter and jelly sandwich, but when choosing the order of operations when solving a problem. Just like with sandwiches, if we go out of order, we won’t get the correct result. 4. Play the Order of Operations Movie for the class and talk about the information shared. You might also want to project the FYI features and discuss them. 5. Show students a sample math problem. Then say, “To find out the answer to this problem you must use the order of operations, which means that all of the operations you have already learned must be calculated in a certain order: PEMDAS (Parentheses, Exponents, Multiplication & Division, Addition & Subtraction)." 6. Go over the order of operation rules (remind students that the multiplication and division steps are a group which work from left to right: multiplication does not come before division. This is the same with addition and subtraction.) 7. Tell the class you will be solving a problem together on the board. Give the example 7- (2+1) and show how the answer is 4. Explain that if you ignore the parentheses and just go in order from left to right, you get 6 as your answer. But if you start inside the parentheses (adding 2 and 1 to get 3, you get the correct answer of 4. 8. Demonstrate several more examples on the board. Here are some sample problems you can use: Example 1: 4+5 x 6 -7 Multiplication 5 x 6=30 Subtraction 34-7=27 Example 2: (Real World): Joe buys 2 shirts at \$8.00 each. He also buys a pair of jeans for \$20.00 that gets a \$3.00 discount. Write a numerical expression and solve. (2 x \$8.00)+(\$20.00-\$3.00) Parentheses (2 x 8)=16 and (20 - 3)=17 \$33 Example 3: 4 x ((3 x (2^2)-1) Parentheses ((3 x (2^2)-1) Exponents 2^2=4 Multiplication 3 x 4 -1 Subtraction 12-1 11 9. Ask students to work at their tables to create their own story, song, short skit, or poem to help them remember the order of operations. Give students time to present their creations to the class. 10. Inform the students that as we continue to practice, we will play a game of BINGO. Students can play Order of Operations BINGO in pairs or small groups of 3. 11. Write a problem on the board, allowing students to use paper and pencil to assist them in solving. For each problem, ask for a student volunteer to come up and work out the problem on the board while the others work with their groups. Students should cover the numbers on their BINGO boards as the problems are solved. 12. Close by going over the day’s lesson and asking students to summarize the order of operations.
# Triangle Problems #### Chapter 43 5 Steps - 3 Clicks # Triangle Problems ### Introduction The triangle is a convex polygon with three line portions joining three non-collinear points. As shown above in the figure, each of the three line segments is known as a side and each of three non-collinear points is known as a vertex. ### Methods Triangles can be categorized by the number of equal (or congruent) sides they have. • A triangle with no equal sides is a scalene triangle, • A triangle with two equal sides is an isosceles triangle, and • A triangle with three equal sides is an equilateral triangle. Triangle inequality rule: The length of any side of a triangle will always be less than the sum of the lengths of the other two sides and greater than the difference of the lengths of the other two sides. Sum of the interior angles: The sum of the angles of a triangle is 180 degrees. Sum of the exterior angles: The sum of the angles of a triangle is 360 degrees. Types of triangles: Equilateral triangle: • All the three sides of equal length. • Each angle in an equilateral triangle is equal to 60 degrees. Isosceles triangle: A triangle having two of its sides of equal length is known as isosceles triangle. Scalene triangle: A triangle having three sides of different lengths is known as scalene triangle. Acute triangle: A triangle having three acute angles is known as acute triangle. That is, if all three angles of a triangle are less than 90Â°, then it is an acute triangle. Obtuse triangle:- • A triangle having an obtuse angle is known as obtuse triangle. • One of the angles of the triangle measures more than 90 degrees. Right AngledÂ triangle:- • A triangle having a right angle is known as right angle triangle. • Side opposite to the right angle is known as hypotenuse. • The two sides that form the right angle are called the legs. Pythagorean theorem:- In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. This is known as Pythagorean theorem. Consider the below given right angle triangle, Let the lengths of the sides be $$a$$ and $$b$$. The hypotenuse has length $$c$$. Using the Pythagorean theorem, $$a^2$$ + $$b^2$$ = $$c^2$$. Area of a triangle: Consider a triangle with base of length $$b$$ and height $$h$$ as shown below. The area of the triangle = $$\frac{1}{2}$$ x $$b$$ x $$h$$. Properties of triangles: • An angle, lying opposite the greatest side, is also the greatest angle and vice versa. • Angles, lying opposite the equal sides, are also equal and vice versa. Two triangles are congruent, if they have accordingly equal: i) Two sides and an angle between them, ii) Two angles and a side, adjacent to them, iii) all the three sides are equal. Theorems about congruency of right-angled triangles: Two right-angled triangles are congruent, if one of the following conditions is valid: i) Their sides are equal, ii) A side and a hypotenuse of one of triangle are equal to a leg and a hypotenuse of another. ii) A hypotenuse and an acute angle of one of triangles are equal to a hypotenuse and acute angle of another. Orthocenter: Three heights (altitude) of triangle always intersect in one point, called an orthocenter of a triangle. Altitude: It is a perpendicular, dropped from any vertex to an opposite side. this side is called a base of triangle. Centroid: Intersecting point of median is known as centroid as shown in the figure below. This point divides each median by ratio 2:1, considering from a vertex. Median: Median is a line segment, joining any vertex of triangle and a midpoint of the opposite side. In-center: Intersecting point of angle bisector is known as in-center. Angle bisector: It is a line segement from a vertex to a point of intersection with an opposite side. Circumcenter: • Intersecting point of mid-perpendicular bisector of side is known as circumcenter. • Mid-perpendicular is a perpendicular drawn from a middle point of a side. Example 1: The right triangle shown below has an area of 25. Find its hypotenuse. Solution: Since the x coordinates of points A and B are equal, segment AB is parallel to the y axis. Since BC is perpendicular to AB then BC is parallel to the x axis and therefore y, the y coordinate of point C is equal to 3. We now need to find the x coordinate x of point C using the area as follows. Area = 25 = (1/2) d (A, B) * d (B, C) d (A, B) = 5 d (B, C) = \|x – 2\| We now substitute d(A,B) and d(B,C) in the area formula above to obtain. 25 = (1/2) (5) \|x – 2\| We solve the above as follows \|x – 2\| = 10 x = 12 and x = – 8 We select x = 12 since point C is to the left of point B and therefore its x coordinate is greater than 2. We have the coordinates of point A and C and we can find the hypotenuse using the distance formula. hypotenuse = d(A,C) = sqrt[$$(12 – 2)^{2}$$ + $$(3 – 8)^{2}$$] = sqrt(125) = 5 sqrt(5) Example 2: Triangle ABC shown below is inscribed inside a square of side 20 cm. Find the area of the triangle Solution: The area is given by Area of triangle = (1/2) base * height = (1/2)(20)(20) = $$200 cm^{2}$$ ### Samples 1. What is the area of isosceles right angled triangle if its legs are 5 and 24 units? Solution: Given Legs of a isosceles right angled triangle. So, Let Breadth be $$b$$ = 5 units. Height be $$h$$ = 24 units. Area of the triangle = $$\frac{1}{2}$$ x $$b$$ x $$h$$ = $$\frac{1}{2}$$ x 5 x 24 = 60 units. 2. If AB = BD, BC = CD and $$\angle$$ACD = $${80}^{\circ}$$, what is the measure in degrees of $$\angle$$BAD? Solution: Given that, AB = BD BC = CD $$\angle$$ACD = $${80}^{\circ}$$ Since AB = BD, $$\angle$$CBD = $$\angle$$CDB. There is $${100}^{\circ}$$ left in the triangle. So, each of these angles is $${50}^{\circ}$$. $$\angle$$DBA is supplementary to $$\angle$$DBC making it $${130}^{\circ}$$. Since BA = BD, $$\angle$$BAD = $$\angle$$BDA. There is $${50}^{\circ}$$ left in the triangle. So, each of these angles is $${25}^{\circ}$$. 3. In the figure given, DE$$\parallel$$ BC. Find the length of DE? Solution: Given that, DE$$\parallel$$ BC So, $$\bigtriangleup$$ADEÂ â‰… $$\bigtriangleup$$ABC. Now, $$\frac{4}{DE}$$ = $$\frac{6}{9}$$ DE = 6 inches. 4. Quantity A Quantity B Area of $$\bigtriangleup$$Q 8 sq. m. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. Solution: Given that Q is a right angled triangle. Area of the triangle = $$\frac{1}{2}$$ x $$b$$ x $$h$$ Let, 3 and 4 be $$b$$ and $$h$$. So, A = $$\frac{1}{2}$$ x 3 x 4 A = 6. Therefore, area of $$\bigtriangleup$$Q is 6 sq.m. which is less than 8 sq.m. Hence, correct option is B. 5. Quantity A Quantity B 9 $$x$$ A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. Solution: By using Pythagorean theorem, $$a^2$$ + $$b^2$$ = $$c^2$$ â‡’ $$6^2$$ + $$x^2$$ = $$10^2$$ â‡’ 36 + $$x^2$$ = 100 â‡’ $$x^2$$ = 64 â‡’ $$x$$ = 8 Therefore, 8 is less than 9. Hence, option A is correct.
# Multiplying Decimals In Mathematics, we may come across different types of numbers. Decimals are specific types of numbers, with a whole number and the fractional part separated by a decimal point. The dot that exists between the whole number and fractions part is called the decimal point. Various arithmetic operations can be performed on these numbers, such as addition, subtraction, multiplication and division. In this article, you can learn about multiplying the decimals along with solved examples. ## How to Multiply Decimals Multiplying Decimals is similar to multiplying whole numbers, except for the way we handle the decimal point. There are two different cases in dividing decimals, such as: • Multiplying Decimals by Whole number • Multiplying Decimal by Decimal Number However, when we multiply decimal numbers, we may get a decimal or whole number. Let’s understand the division of these decimals in detail here. Also, get a multiplying decimals calculator here. ## Multiplying Decimals by Whole Numbers Multiplying decimal numbers by whole numbers can be done as we do the multiplication of whole numbers. In this case, we should only take care of the decimal point. However, the operation of whole numbers multiplying with decimals is shown in the below figure. The above process can be understood in a better way with the example given below. Example: Multiply 4785.25 by 8 Solution: Step 1: Multiply the both numbers as whole numbers, i.e. we ignore the decimal point here and proceed with usual multiplication. 478525 × 8 = 3828200 Step 2: Counting the number of digits to the right of the decimal point in the decimal number. In this case, the required number of digits = 2 Step 3: Put the decimal point in the product by counting the digits from the rightmost place. As the total number of decimal places is 2, the resultant product must contain the decimal point after 2 digits from the rightmost place. Therefore, 4785.25 × 8 = 38282.00 = 38282 ## Multiplying Decimals by 10 100 and 1000 Let us find the multiplication of a decimal number by 10, 100 and 1000. Consider 32.47 × 10 = 324.7. Thus, the digits in 32.47 and 324.7 are the same, i.e., 3, 2, 4, and 7, but the decimal point has shifted (towards the right) in the result. Therefore, in a decimal multiplication by 10, the decimal point will shift to the right by one place since 10 has only one zero over 1. From the above example, we can say that the number and product digits will be the same in the multiplication of decimals by 10, 100 or 1000. However, the decimal point in the obtained product of the multiplication will shift towards the right by as many places as zeros over 1. Let’s have a look at some examples of multiplication of decimal numbers by 10, 100, and 1000. • 42.5 × 10 = 425 • 647.124 × 10 = 6471.24 • 218.19 × 100 = 21819 • 126.3 × 100 = 12630.0 = 12630 • 0.217 × 100 = 21.7 • 0.1745 × 1000 = 174.5 • 555.99 × 1000 = 555990.0 = 555990 ## Multiplying Decimals by Decimals Multiplication of a decimal number by another decimal can be performed similar to the multiplication of whole numbers except keeping the decimal point in the result. The below figure shows the process of multiplying decimal numbers by another decimal number. Go through the example given below for understanding the above process. Example: Multiply 923.447 by 12.6 Solution: 923.447 × 12.6 Step 1: Multiply both decimal numbers as whole numbers by removing decimal point as: Step 2: Count the number of digits to the right of the decimal point in both decimal numbers. Thus, the number of digits after the decimal point in 923.447 = 3 And the number of decimal places in 12.6 = 1 Step 3: Add the number of digits counted for decimal point. Total number of decimal places = 3 + 1 = 4 Step 4: Put the decimal point in the product by counting the same digits as obtained in the previous step from its rightmost place. As the total number of decimal places is 4, the resultant product must contain the decimal point after 4 digits from the rightmost place. Therefore, the product of the given decimals is 11635.4322. ### Multiplying Decimals Examples Example 1: What is the product of 94.3 and 18? Solution: 94.3 × 18 Let us write this expression by ignoring the decimal point. 943 × 18 Now, we have to put the decimal point in the obtained product. Therefore, 94.3 × 18 = 169.74 Example 2: Multiply the decimal number 0.785 by 0.055. Solution: 0.785 × 0.055 Let us write this expression by ignoring the decimal point. 785 × 55 Now, we have to find the total number of decimal places. Number of decimal places in 0.785 = 3 Number of decimal places in 0.055 = 3 Thus, the total number of decimal places = 3 + 3 = 6 That means, we have to put the decimal point after six digits from the rightmost place. Hence, the product is 0.043175.
# Yarn Count Measurement in Shortcut Easy Method with 10 Exercise Sharing is caring! ## What is Yarn Count? A count is a numerical value expressing the fineness or coarseness of yarn and also indicating the relationship between length and weight. The concept of yarn count, which specifies a certain ratio of length to weight, has also been introduced. It is also the mass per unit length or the length per unit mass. ## Types of Yarn Count Measurement: There two types of yarn count. They are: 1. Direct count system 2. Indirect count system ### 1. Direct Count System: In direct count system, the weight of a fixed length of yarn is determined. The weight per unit length is the yarn count. The following formula is used to calculate the yarn count: Standard table for direct numbering system ### 2. Indirect Count System: In indirect count system, the length of a fixed weight of yarn is measured. The length per unit weight is the yarn count. The following formula is used to calculate the yarn count: Standard table for indirect numbering system: ## What is Resultant Count? Resultant count is the number of yarns that are twisted or folded. Two or more threads are folded or twisted into a single strand form when they are folded or twisted. Two or more of these already twisted yarns are folded together to form a “folded” yarn. For special strength and solid effects, yarns such as fishnet yarns, cords, harness cords, special sewing threads, shoe threads, and so on are cabled. In case of direct count system, resultant count is calculate by using following formula N= N1+N2+N3…………………. In case of indirect count system, resultant count is calculate by using following formula. ## Exercise: Problem-1: If the weight of 200 meter cotton yarn is 4.724 grams, what will be its count in Ne, Tex and Denier? Problem-2: If a skein of viscose filament yarn weighs 1.70 g and its length is 70 m, what is the denier of that yarn? Problem-3: If 150 yards of jute yarn weighs 3 ounces, what is its count? Problem-4: If the weight of 1 lea cotton yarn is 30 grains, what will be the count? Problem-5: if the weight of 1 lea cotton yarn is 2.16 gm, what will be its count in Ne, tex, denier and Metric count? Problem-6: what length of yarn is contained in 1.2 kg of yarn of Ne 30? Problem-7: What is the resultant count of 40/3 jute yarn? Problem-8: If 3 ply yarn is prepared with 30, 25 and 20 cotton yarn, what will be the yarn count? Problem-9: What is the resultant count of 30/2 cotton yarn? Problem-10: If one single yarn count is 60 and two-ply count is 35, what is the other single yarn count? Sharing is caring! ### 2 thoughts on “Yarn Count Measurement in Shortcut Easy Method with 10 Exercise” This site uses Akismet to reduce spam. Learn how your comment data is processed.
# NCERT Solutions for Exercise 9.2 Class 9 Maths Chapter 9 - Areas Of Parallelograms And Triangles NCERT Solutions for Class 9 Maths chapter 9 exercise 9.2 has six questions based on the notion of a parallelogram with the same base and parallels. The questions consist of proving the area of two figures equal, finding any side, etc. NCERT book Exercise 9.2 Class 9 Maths has various standard questions that help to have in-depth knowledge of the topic. NCERT syllabus Class 9 Maths chapter 9 exercise 9.2. also has one problem-based question that gives a practical understanding of the topic, Latest : Trouble with homework? Post your queries of Maths and Science with step-by-step solutions instantly. Ask Mr AL NCERT solutions for Class 9 Maths exercise 9.2 focuses on the parallelograms on the same base, and between the same parallels, its solution gives a new perspective to solve the questions. The solution of this exercise provides the students with confidence in this topic. A parallelogram is a figure in which two opposite sides are parallel to each other whose diagonals are not equal. In Class 9 Maths chapter 9 the following exercises are also present. • Areas Of Parallelograms And Triangles Exercise 9.1 • Areas Of Parallelograms And Triangles Exercise 9.3 • Areas Of Parallelograms And Triangles Exercise 9.4 ## Areas Of Parallelograms And Triangles Class 9 Chapter 9 Exercise: 9.2 Q1 In Fig. , ABCD is a parallelogram, and . If , and , find AD. We have, AB = 16 cm, AE = 8 cm and CF = 10 cm Since ABCD is a parallelogram, therefore, AB = DC = 16 cm We know that, area of parallelogram (ABCD) = base . height = CD AE = (16 8 ) therefore area of parallelogram = AD CF = 128 Thus the required length of AD is 12.8 cm. Q2 If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that Join GE and HE, Since F, F, G, H are the mid-points of the parallelogram ABCD. Therefore, GE \|\| BC \|\|AD and HF \|\| DC \|\| AB. It is known that if a triangle and the parallelogram are on the same base and between the same parallel lines. then the area of the triangle is equal to the half of the area of the parallelogram. Now, EFG and \|\|gm BEGC are on the same base and between the same parallels EG and BC. Therefore, ar ( EFG) = ar (\|\|gm BEGC)...............(i) Similarly, ar ( EHG) = 1/2 . ar(\|\|gm AEGD)..................(ii) By adding eq (i) and eq (ii), we get ar ( EFG) + ar ( EHG) = 1/2 (ar (\|\|gm BEGC) + ar(\|\|gm AEGD)) ar (EFGH) = 1/2 ar(ABCD) Hence proved Q3 P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that . We have, ABCD is a parallelogram, therefore AB \|\| CD and BC \|\| AD. Now, APB and \|\|gm ABCD are on the same base AB and between two parallels AB and DC. Therefore, ar ( APB) = 1/2 . ar(\|\|gm ABCD)...........(i) Also, BQC and \|\|gm ABCD are on the same base BC and between two parallels BC and AD. Therefore, ar( BQC) = 1/2 . ar(\|\|gmABCD)...........(ii) From eq(i) and eq (ii), we get, ar ( APB) = ar( BQC) Hence proved. Q4 (i) In Fig. , P is a point in the interior of a parallelogram ABCD. Show that [ Hint : Through P, draw a line parallel to AB.] We have a \|\|gm ABCD and AB \|\| CD, AD \|\| BC. Through P, draw a line parallel to AB Now, APB and \|\|gm ABEFare on the same base AB and between the same parallels EF and AB. Therefore, ar ( APB) = 1/2 . ar(ABEF)...............(i) Similarly, ar ( PCD ) = 1/2 . ar (EFDC) ..............(ii) Now, by adding both equations, we get Hence proved. Q4 (ii) In Fig. , P is a point in the interior of a parallelogram ABCD. Show that [ Hint: Through P, draw a line parallel to AB.] We have a \|\|gm ABCD and AB \|\| CD, AD \|\| BC. Through P, draw a line parallel to AB Now, APD and \|\|gm ADGHare on the same base AD and between the same parallels GH and AD. Therefore, ar ( APD) = 1/2 . ar(\|\|gm ADGH).............(i) Similarily, ar ( PBC) = 1/2 . ar(\|\|gm BCGH)............(ii) By adding the equation i and eq (ii), we get Hence proved. Q5 In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that (i) (ii) (i) Parallelogram PQRS and ABRS are on the same base RS and between the same parallels RS and PB. Therefore, ............(i) Hence proved (ii) AXS and \|\|gm ABRS are on the same base AS and between same parallels AS and RB. Therefore, ar ( AXS) = 1/2 . ar(\|\|gm ABRS)............(ii) Now, from equation (i) and equation (ii), we get Hence proved. Q6 A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it? We have a field in the form of parallelogram PQRS and a point A is on the side RS. Join AP and AQ. The field is divided into three parts i.e, APS, QAR and PAQ. Since APQ and parallelogram, PQRS is on the same base PQ and between same parallels RS and PQ. Therefore, ............(i) We can write above equation as, ar (\|\|gm PQRS) - [ar ( APS) + ar( QAR)] = 1/2 .ar(PQRS) from equation (i), Hence, she can sow wheat in APQ and pulses in [ APS + QAR] or wheat in [ APS + QAR] and pulses in APQ. ## More About NCERT Solutions for Class 9 Maths Exercise 9.2 Class 9 Maths chapter 9 exercise 9.2 is about the parallelogram in the same base and same parallels. It has six questions, among which one question is application-based, all in all, the exercise is packed with standard questions aiming to give students in-depth knowledge. Exercise 9.2 Class 9 Maths -Parallelogram is a geometric figure in which the opposite sides are parallel; using this very concept, the questions on this exercise could be tackled efficiently. The NCERT solutions for Class 9 Maths exercise 9.2 mainly focuses on the various properties exhibited by the parallelogram. Exercise 9.2 Class 9 Maths primary focus is to enlighten and give good practice to the topic; that is why it has a nice set of standard questions. Also Read\| Areas Of Parallelograms And Triangles Class 9 Notes ## Benefits of NCERT Solutions for Class 9 Maths Exercise 9.2 • The NCERT solutions for Class 9 Maths exercise 9.2 assists students in solving and revising all of the problems in this exercise. • If you go over the NCERT solutions for Class 9 Mathematics chapter 9 exercise 9.2, you will be able to gain more marks, and if you practice it completely, it will help you score well in maths in examinations. • Parallelograms on the same base and between the same parallels are the basis of exercise 9.2 Class 9 Maths. Also, See • NCERT Solutions for Class 9 Maths Chapter 9 • NCERT Exemplar Solutions Class 9 Maths Chapter 9 ## NCERT Solutions of Class 10 Subject Wise • NCERT Solutions for Class 9 Maths • NCERT Solutions for Class 9 Science ## Subject Wise NCERT Exemplar Solutions • NCERT Exemplar Class 9 Maths • NCERT Exemplar Class 9 Science
Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs! Online Quiz (WorksheetABCD) Questions Per Quiz = 2 4 6 8 10 Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation3.8 Whole Numbers {0,1,2,3,..........} is called the set of Whole Numbers and is denoted by 'W'. Properties of Whole Numbers in Addition Closure Property: For any two whole numbers "a" and "b" their sum (a+b) is also a natural number. This is called the closure property of addition for whole numbers. Commutative Property: For any two whole numbers "a" and "b" a+b = b+a. This property is called commutative property of addition for whole numbers. Associative Property: For any three whole numbers a, b, c a+(b+c) = (a+b)+c. This property is called the associative property of addition for whole number. For every whole number a, a+0 = 0+a = a. 0 is called the additive identity in the set of all whole numbers. Properties of Whole Numbers in Multiplication Closure Property: For any two whole numbers a and b their product ab is also a whole number. This property is called the closure property of multiplication for whole number. Commutative Property: For any two whole numbers a, b ab =ba. This is called the commutative property of multiplication in whole numbers. Associative property: For any three whole numbers a, b, c a(bc) = (ab)c. This is called the associative property of multiplication in whole numbers. Multiplicative Identity: For any whole number a, a1 = 1*a = a. Here 1 is called the multiplicative identity in the set of all whole numbers. Directions: Answer the following questions. Also, write five examples of your own for each property. Q 1: For any two whole numbers a, b then a+b = b+a, name the property.Identity PropertyCommutative Property.None of theseAssociative Property Q 2: Set of whole numbers are denoted by ______.ZNWI Q 3: 3+(4+5) = (4+5)+3, is an example for _____of whole number.Associative Property of MultiplicationAssociative Property of AdditionCommutative Property of MultiplicationCommutative Property of Addition Q 4: What is the multiplicative identity in whole number.10None of these-1 Q 5: {0,1,2,3,.........} is called set of _______.Natural NumbersIntegersWhole NumbersPrime Numbers Q 6: 0 is called the _____ in the set of all whole numbers.None of theseAdditive IdentityMultiplicative Identity Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!
# Topic 6: Circular Motion and Gravitation ## 6.1 Circular motion Period — Time taken for one complete revolution Frequency — Number of complete revolutions per unit time, 1/T. Angular displacement — Angle, in radians, that a body has travelled around the centre of circular motion. Angular velocity — How fast an object revolves around a particular point. For the purposes of this syllabus, the centripetal force is the net force, which acts in the direction of the centre of the circular motion (perpendicular to the velocity). The centripetal force is provided by a sum of all the forces acting on the body undergoing circular motion. The centripetal acceleration is the acceleration of a body undergoing such a motion. Such a body has a constant speed, but the direction of its velocity is constantly changing. Hence, the centripetal acceleration is perpendicular to the velocity. By considering the velocity at time t, and the velocity at an infinitesimal time step later, one can show that the centripetal acceleration is a = ω2r = v2/r, where ω is the angular velocity, r is the distance between the object and the centre of rotation, and v is its velocity. ## 6.2 Gravitation Newton’s Law of Gravitation states that every body attracts every other body with a force proportional to the product of their masses and inversely proportional to the distance between them. F = G m1 m2 r-2. Here G is the constant of proportionality, the Gravitational Constant. The gravitational field strength is g = G m1 r-2. This can be expressed in force per unit mass (F m-1) or in terms of acceleration, also called the acceleration due to gravity. On earth, this is 9.81 ms-2. The gravitational potential energy of an object in a gravitational field is the work done by an external agent to bring that object from infinity to that point. U = – G m1 m2 r-1. The gravitational potential at a particular point is the work per unit mass that an external agent would have to do to bring a mass from infinity to that point. V =- G m1 r-1. Maybe give some examples. Kepler’s 3rd and geostationary orbits
# Difference between revisions of "Math 22 Product Rule and Quotient Rule" ## The Product Rule The derivative of the product of two differentiable functions is equal to the first function times the derivative of the second plus the second function times the derivative of the first. ${\displaystyle {\frac {d}{dx}}[f(x)\cdot g(x)]=f'(x)g(x)+f(x)g'(x)}$ Example: Find derivative of 1) ${\displaystyle f(x)=(x+1)(x^{2}+3)}$ Solution: ${\displaystyle f'(x)={\frac {d}{dx}}[(x+1)](x^{2}+3)+(x+1){\frac {d}{dx}}[(x^{2}+3)]}$ ${\displaystyle =(1)(x^{2}+3)+(x+1)(2x)=x^{2}+3+2x^{2}+2x=3x^{2}+2x+3}$ 2) ${\displaystyle f(x)=(4x+3x^{2})(6-3x)}$ Solution: ${\displaystyle f'(x)={\frac {d}{dx}}[(4x+3x^{2})](6-3x)+(4x+3x^{2}){\frac {d}{dx}}[(6-3x)]}$ ${\displaystyle =(4+6x)(6-3x)+(4x+3x^{2})(-3)=24+36x-12x-18x^{2}-12x-9x^{2}=-27x^{2}+12x+24}$ 3) ${\displaystyle f(x)=(e^{2}+x^{2})(4x+5)}$ Solution: ${\displaystyle f'(x)={\frac {d}{dx}}[(e^{2}+x^{2})](4x+5)+(e^{2}+x^{2}){\frac {d}{dx}}[(4x+5)]}$ ${\displaystyle =(2x)(4x+5)+(e^{2}+x^{2})(4)=8x^{2}+10x+4e^{2}+8x^{2}=-16x^{2}+10x+4e^{2}}$ ## The Quotient Rule The derivative of the quotient of two differentiable functions is equal to the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. ${\displaystyle {\frac {d}{dx}}[{\frac {f(x)}{g(x)}}]={\frac {f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}}}$ Example: Find derivative of 1) ${\displaystyle f(x)={\frac {x}{x-5}}}$ Solution: ${\displaystyle f'(x)={\frac {{\frac {d}{dx}}[x](x-5)-(x){\frac {d}{dx}}[(x-5)]}{(x-5)^{2}}}}$ ${\displaystyle ={\frac {(1)(x-5)-x(1)}{(x-5)^{2}}}={\frac {-5}{(x-5)^{2}}}}$ 2) ${\displaystyle f(x)={\frac {x^{2}}{x+3}}}$ Solution: ${\displaystyle f'(x)={\frac {{\frac {d}{dx}}[x^{2}](x+3)-(x^{2}){\frac {d}{dx}}[(x+3)]}{(x+3)^{2}}}}$ ${\displaystyle ={\frac {(2x)(x+3)-(x^{2})(1)}{(x+3)^{2}}}={\frac {2x^{2}+6x-x^{2}}{(x+3)^{2}}}={\frac {x^{2}+6x}{(x+3)^{2}}}}$ 3) ${\displaystyle f(x)={\frac {x^{2}+6x+5}{2x-1}}}$ Solution: ${\displaystyle f'(x)={\frac {{\frac {d}{dx}}[x^{2}+6x+5](2x-1)-(x^{2}+6x+5){\frac {d}{dx}}[(2x-1)]}{(2x-1)^{2}}}}$ ${\displaystyle ={\frac {(2x+6)(2x-1)-(x^{2}+6x+5)(2)}{(2x-1)^{2}}}={\frac {4x^{2}+12x-2x-6-2x^{2}-12x-10}{(2x-1)^{2}}}={\frac {2x^{2}-2x-16}{(2x-1)^{2}}}}$
# How do you find the center and radius for x^2 + y^2 +14x = 0? Oct 25, 2016 #### Explanation: The standard form for the equation of a circle is: ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$ Substitute ${\left(y - 0\right)}^{2}$ for ${y}^{2}$ ${x}^{2} + 14 x + {\left(y - 0\right)}^{2} = 0$ Using the pattern ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$, we observe that we must begin the process of completing the square by adding ${h}^{2}$ to both sides: ${x}^{2} + 14 x + {h}^{2} + {\left(y - 0\right)}^{2} = {h}^{2}$ For this circle, the ${h}^{2}$ term is, also the ${r}^{2}$ term, because the we will soon find that the center is offset to the left the same distance as the radius. We can find the value $h$ by setting the right side of the pattern equal to the first 3 terms in the equation: ${x}^{2} - 2 h x + {h}^{2} = {x}^{2} + 14 x + {h}^{2}$ The square terms cancel: $- 2 h x = 14 x$ #h = -7 and h^2 = 49 Substitute ${\left(x - - 7\right)}^{2}$ for the corresponding terms on the left and ${\left(x - - 7\right)}^{2} + {\left(y - 0\right)}^{2} = 49$ The radius should be represented as a positive number squared. ${\left(x - - 7\right)}^{2} + {\left(y - 0\right)}^{2} = {7}^{2}$ The center is found by observation $\left(- 7 , 2\right)$
# The 4 Major Math Concepts Your Kids Learn in PreK & Kindergarten These streamlined topics will give you the leg up to (finally!) understand what your youngsters are learning in math. By Jennifer Hogan Mar 29, 2016 Ages 3-6 Mar 29, 2016 There are so many different topics our children learn throughout the year, just keeping up with each night’s new piece of homework and its latest ideas can feel very overwhelming. As parents, we don’t get to see the major idea behind the “everyday” work and it can be frustrating to understand where each skill is going. In this first blog post of a continuing series, I will be highlighting the major math concepts that are taught at the different grade levels so we, as parents, can help to build and support these ideas at home. Here are the four major math concepts taught in pre-kindergarten and kindergarten, along with exercises you can practice with your children to help reinforce their learning. 1. Counting. Students are beginning their experience with numbers through counting, number names and written numerals. Students are learning to count objects and understand a one-to-one correspondence. They are also starting to compare different sets of objects and use appropriate language. • Touch different objects and count out loud. • Move objects from one group to another. • Count a set of objects and “see” or “write” that corresponding number. • Start to use comparing words: more than, less than, the same as. 2. Addition & Subtraction. This is the very early stage of adding and subtracting. The focus should be on developing an understanding of addition as “putting together and adding to,” and subtracting as “taking apart and taking from.” Students do not need to write equations at this young an age, but are encouraged to begin using them. • Tell stories about adding and subtracting. For example, for addition: Two bunnies sat on the grass. Three more bunnies hopped there. How many bunnies are on the grass now? For subtraction: Five apples were on the table. I ate two apples. How many apples are on the table now? Draw pictures about putting together and taking apart. • Count to 10 and break apart numbers (decompose – math’s fancy word for “breaking apart”) into different combinations. For example, 5 can be seen as: Image Credit: http://kindercraze.com 3. Measurement & Data. Young children are beginning to describe and compare their physical world. They are starting to classify, sort and group objects into categories. • Compare two different objects using appropriate language. For example: “John is taller than Sarah.” “This tree is shorter than that tree.” “My bag is heavier than your bag.” • Sort objects by color, size, material, etc. • Describe their physical world with directional words: in front of, behind, on top of, next to, below, etc. 4. Geometry. Students are starting to look at and compare two-dimensional (flat) and three-dimensional (solid) shapes. They are using appropriate language to recognize different shapes and talk about their attributes. • Find 2-D shapes in the world: squares, circles, triangles, rectangles, and hexagons. • Find 3-D shapes in the world: cubes, cones, cylinders, and spheres. • Count the different number of sides, vertices, angles, etc.Model different shapes using clay, sticks, pipe-cleaners, etc. See all blog posts in this series for information on what your child will learn in math class from preschool all the up through 8th grade. The Learning Toolkit Blog Age 6 Age 5 Age 4 Age 3 Measurement Math Counting and Numbers
# How do you evaluate the integral int dx/(1-cos3x)? Oct 12, 2017 $\int \frac{\mathrm{dx}}{1 - \cos 3 x} = - \frac{1}{3} \cot \left(\frac{3 x}{2}\right) + C$ #### Explanation: Substitute $t = 3 x$: $\int \frac{\mathrm{dx}}{1 - \cos 3 x} = \frac{1}{3} \int \frac{\mathrm{dt}}{1 - \cos t}$ Use now the trigonometric identity: $\cos t = \frac{1 - {\tan}^{2} \left(\frac{t}{2}\right)}{1 + {\tan}^{2} \left(\frac{t}{2}\right)}$ So that: $\frac{1}{1 - \cos t} = \frac{1}{1 - \frac{1 - {\tan}^{2} \left(\frac{t}{2}\right)}{1 + {\tan}^{2} \left(\frac{t}{2}\right)}}$ $\frac{1}{1 - \cos t} = \frac{1 + {\tan}^{2} \left(\frac{t}{2}\right)}{\left(1 + {\tan}^{2} \left(\frac{t}{2}\right)\right) - \left(1 - {\tan}^{2} \left(\frac{t}{2}\right)\right)}$ $\frac{1}{1 - \cos t} = \frac{1 + {\tan}^{2} \left(\frac{t}{2}\right)}{1 + {\tan}^{2} \left(\frac{t}{2}\right) - 1 + {\tan}^{2} \left(\frac{t}{2}\right)}$ $\frac{1}{1 - \cos t} = \frac{1 + {\tan}^{2} \left(\frac{t}{2}\right)}{2 {\tan}^{2} \left(\frac{t}{2}\right)}$ And as: $1 + {\tan}^{2} \alpha = 1 + {\sin}^{2} \frac{\alpha}{\cos} ^ 2 \alpha = \frac{{\cos}^{2} \alpha + {\sin}^{2} \alpha}{\cos} ^ 2 \alpha = \frac{1}{\cos} ^ 2 \alpha = {\sec}^{2} \alpha$ we get: $\int \frac{\mathrm{dt}}{1 - \cos t} = \int {\sec}^{2} \frac{\frac{t}{2}}{2 {\tan}^{2} \left(\frac{t}{2}\right)} \mathrm{dt}$ Substitute now: $u = \tan \left(\frac{t}{2}\right)$ $\mathrm{du} = \frac{1}{2} {\sec}^{2} \left(\frac{t}{2}\right) \mathrm{dt}$ and we have: $\int {\sec}^{2} \frac{\frac{t}{2}}{2 {\tan}^{2} \left(\frac{t}{2}\right)} \mathrm{dt} = \int \frac{\mathrm{du}}{u} ^ 2 = - \frac{1}{u} + C$ and undoing the substitution: $\int \frac{\mathrm{dt}}{1 - \cos t} = - \frac{1}{\tan} \left(\frac{t}{2}\right) + C = - \cot \left(\frac{t}{2}\right) + C$ $\int \frac{\mathrm{dx}}{1 - \cos 3 x} = - \frac{1}{3} \cot \left(\frac{3 x}{2}\right) + C$
# Percent Change Formula When there is a change in the values of some quantity, then we need to know the change value. Also sometimes it becomes more important to determine the percent change in the values. Students can compute it with the help of a simple formula. Sometimes to know the discounts change in price value or the change in prices of products or income, the percentage change formula proves much important.Â We express it in terms of percentage and it is the change in new value with respect to the old value.Â Further, the change in value is divided by the original value and then multiplied by 100. In this topic, we will discuss the percent change formula with examples. Let us learn it! ## Percent Change Formula ### What Is Percent Change? The percentage error value is also very important under the experimental calculations. This formula is the absolute value of the difference of both values i.e. measured value and the theoretical value. Further, this change is divided by the theoretical value. Next, multiply it by 100. Percentage change is a simple but important mathematical concept that represents the degree of change over time. It is used for many areas like finance, often to represent the price change of security. ### Understanding the Concept: Percentage change can be applied to any quantity or term that we measure. Let us say we are tracking the quoted price of some security. If the price increased, then use the value as (New Price – Old Price). If the price decreased, then use the value as (Old Price – New Price). This formula is used to track the prices of individual securities as well as the large market indexes. Also, it can be used for comparing the values of different currencies. ### Percentage Change Formula: $$Percent Change = \frac{(Old Value â€“New Value)} {Old Value}\times 100$$ Steps to Calculate: The simple step by step procedure is as below: 1. First, calculate the change (subtract old value from the new value) 2. Then divide that change by the old value (we may get a decimal number) 3. Finally, convert that to a percentage (by multiplying by 100 and adding a “%” sign) Here, it must be noted that the change value must be positive. Thus when the new value is greater than the old value, then it is a percentage increase, otherwise, it is a decrease. ### Solved examples forÂ Percent Change Formula Q.1: A selling price for a pair of socks change from â‚¹75 to â‚¹90. Calculate the percentage change in the selling price. Solution: Step 1: Calculate the change i.e. New value- old value = 90-75 =15 Step 2: Divide that change value by the old value, $$= \frac{ 15} {75} =\frac{1} {5}$$ Step 3: Convert that change to a percentage change value, $$= \frac {1} {5} \times 100$$ = 20 % Hence the percentage change in the selling price will be 20%. Q.2: What will be the percentage change in the weight of Ram, if he reduces his weight to 77 kg from 82 kg? Solution: Old weight = 82 Kg andÂ New Weight = 77 Kg Change in weight =Â Old value â€“ New value = 82 -77 = 5 Using the formula as below: $$Percent Change = \frac{(Old Value â€“New Value)} {Old Value}\times 100$$ =$$\frac{ 5} { 82 } \times 100$$ = 6.4 % Hence the percent change in his weight will be 6.4% 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 ## Browse ### 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
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 12.6: Extension: Tessellating Polygons Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives • Tessellating regular polygons ## What is a Tessellation? You have probably seen tessellations before. Examples of a tessellation are: a tile floor, a brick or block wall, a checker or chess board, and a fabric pattern. Tessellation: A tiling over a plane with one or more figures such that the figures fill the plane with no overlaps and no gaps. Notice the hexagon (cubes, first tessellation) and the quadrilaterals fit together perfectly. If we keep adding more, they will entirely cover the plane with no gaps or overlaps. We are only going to worry about tessellating regular polygons. To tessellate a shape it must be able to exactly surround a point, or the sum of the angles around each point in a tessellation must be 360\begin{align}360^\circ\end{align}. The only regular polygons with this feature are equilateral triangles, squares, and regular hexagons. Example 1: Draw a tessellation of equilateral triangles. Solution: In an equilateral triangle each angle is 60\begin{align}60^\circ\end{align}. Therefore, six triangles will perfectly fit around each point. Extending the pattern, we have: Example 2: Does a regular pentagon tessellate? Solution: First, recall that there are 540\begin{align}540^\circ\end{align} in a pentagon. Each angle in a regular pentagon is 540÷5=108\begin{align}540^\circ \div 5 = 108^\circ\end{align}. From this, we know that a regular pentagon will not tessellate by itself because 108\begin{align}108^\circ\end{align} times 2 or 3 does not equal 360\begin{align}360^\circ\end{align}. Tessellations can also be much more complicated. Check out http://www.mathsisfun.com/geometry/tessellation.html to see other tessellations and play with the Tessellation Artist, which has a link at the bottom of the page. ## Review Questions 1. You were told that equilateral triangles, squares, and regular hexagons are the only regular polygons that tessellate. Tessellate a square. Add color to your design. 2. What is an example of a tessellated square in real life? 3. How many regular hexagons will fit around one point? (First, recall how many degrees are in a hexagon, and then figure out how many degrees are in each angle of a regular polygon. Then, use this number to see how many of them fit around a point.) 4. Using the information from #2, tessellate a regular hexagon. Add color to your design. 5. You can also tessellate two regular polygons together. Try tessellating a regular hexagon and an equilateral triangle. First, determine how many of each fit around a point and then repeat the pattern. Add color to your design. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:
# 4.2 Modeling with linear functions (Page 6/9) Page 6 / 9 Access this online resource for additional instruction and practice with linear function models. ## Key concepts • We can use the same problem strategies that we would use for any type of function. • When modeling and solving a problem, identify the variables and look for key values, including the slope and y -intercept. See [link] . • Check for reasonableness of the answer. • Linear models may be built by identifying or calculating the slope and using the y -intercept. • The x -intercept may be found by setting $\text{\hspace{0.17em}}y=0,$ which is setting the expression $\text{\hspace{0.17em}}mx+b\text{\hspace{0.17em}}$ equal to 0. • The point of intersection of a system of linear equations is the point where the x - and y -values are the same. See [link] . • A graph of the system may be used to identify the points where one line falls below (or above) the other line. ## Verbal Explain how to find the input variable in a word problem that uses a linear function. Determine the independent variable. This is the variable upon which the output depends. Explain how to find the output variable in a word problem that uses a linear function. Explain how to interpret the initial value in a word problem that uses a linear function. To determine the initial value, find the output when the input is equal to zero. Explain how to determine the slope in a word problem that uses a linear function. ## Algebraic Find the area of a parallelogram bounded by the y -axis, the line $\text{\hspace{0.17em}}x=3,$ the line $\text{\hspace{0.17em}}f\left(x\right)=1+2x,$ and the line parallel to $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ passing through $\text{\hspace{0.17em}}\left(\text{2},\text{7}\right).$ 6 square units Find the area of a triangle bounded by the x -axis, the line $\text{\hspace{0.17em}}f\left(x\right)=12–\frac{1}{3}x,$ and the line perpendicular to $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ that passes through the origin. Find the area of a triangle bounded by the y -axis, the line $\text{\hspace{0.17em}}f\left(x\right)=9–\frac{6}{7}x,$ and the line perpendicular to $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ that passes through the origin. 20.01 square units Find the area of a parallelogram bounded by the x -axis, the line $\text{\hspace{0.17em}}g\left(x\right)=2,$ the line $\text{\hspace{0.17em}}f\left(x\right)=3x,$ and the line parallel to $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ passing through $\text{\hspace{0.17em}}\left(6,1\right).$ For the following exercises, consider this scenario: A town’s population has been decreasing at a constant rate. In 2010 the population was 5,900. By 2012 the population had dropped 4,700. Assume this trend continues. Predict the population in 2016. 2,300 Identify the year in which the population will reach 0. For the following exercises, consider this scenario: A town’s population has been increased at a constant rate. In 2010 the population was 46,020. By 2012 the population had increased to 52,070. Assume this trend continues. Predict the population in 2016. 64,170 Identify the year in which the population will reach 75,000. For the following exercises, consider this scenario: A town has an initial population of 75,000. It grows at a constant rate of 2,500 per year for 5 years. Find the linear function that models the town’s population $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ as a function of the year, $\text{\hspace{0.17em}}t,$ where $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is the number of years since the model began. $P\left(t\right)=75,000+2500t$ Find a reasonable domain and range for the function $\text{\hspace{0.17em}}P.$ If the function $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ is graphed, find and interpret the x - and y -intercepts. (–30, 0) Thirty years before the start of this model, the town had no citizens. (0, 75,000) Initially, the town had a population of 75,000. write down the polynomial function with root 1/3,2,-3 with solution if A and B are subspaces of V prove that (A+B)/B=A/(A-B) write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°) Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4 what is the answer to dividing negative index In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c. give me the waec 2019 questions the polar co-ordinate of the point (-1, -1) prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x tanh`(x-iy) =A+iB, find A and B B=Ai-itan(hx-hiy) Rukmini what is the addition of 101011 with 101010 If those numbers are binary, it's 1010101. If they are base 10, it's 202021. Jack extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve 1+cos²A/cos²A=2cosec²A-1 test for convergence the series 1+x/2+2!/9x3
# Lowest Common Multiple (L.C.M) Highest Common Factor (H.C.F) of Whole Numbers. Subject : Mathematics Term : First Term Week: Week Three Class : JSS 1 Previous lesson : The pupils have previous knowledge of Whole Numbers Continued: Problems solving in quantitative aptitude reasoning using large numbers Topic : Whole Numbers Continued: Problems solving in quantitative aptitude reasoning using large numbers Behavioural objectives : At the end of the lesson, the pupils should be able to • Write out the multiples of numbers • write out the lowest common multiple of numbers (LCM) • calculate the factors of numbers • Write out common factors of numbers • solve questions on Highest Common Factors ( HFC) Instructional Materials : • Wall charts • Pictures • Related Online Video • Flash Cards • Abacus • Numeric Table Chart Methods of Teaching : • Class Discussion • Group Discussion • Explanation • Role Modelling • Role Delegation Reference Materials : • Scheme of Work • Online Information • Textbooks • Workbooks • 9 Year Basic Education Curriculum • Workbooks Content : WEEK THREE TOPIC: LEAST COMMON MULTIPLE AND HIGHEST COMMON FACTOR OF WHOLE NUMBERS CONTENT • Multiples • Common Multiples • Least Common Multiples (LCM) • Highest Common Factors (HCF). Multiples Multiples mean finding the product of a positive integer with another positive integer. Simply put, when a whole number multiples another whole number, the result obtained is called the multiple of either of those numbers. The first four multiples of 12 are 12 x 1 = 12 12 x 2 = 24 12 x 3 = 36 12 x 4 =48 Thus, we write 12, 24, 36 and 48 as the first for multiples of 12. Note: Every whole number has an infinite number of multiples Every whole number has a finite number of factors. Example 1 Find the next five multiple of the following whole numbers. (a) 4 (b) 8 (c ) 11. Solution In these questions, the numbers are not to be included because it reads next. (a) 4, 4 x 1 = 4 (not included) 4 x 2 = 8 4 x 3 = 12 4 x 4 = 16 4 x 5 = 20 4 x 6 = 24 :. The next five multiples of 4 are 8, 12, 16, 20 and 24. (b) 8, 8 x 1 =8 multiple of 11 8 x 2 =16 11 x 1 =11 8 x 3 =24 11 x 2 = 22 8 x 4 =32 11 x 3 = 33 8 x 5 =40 11 x 4 = 44 8 x 6 = 48 11 x 5 = 55 11 x 6 = 66 the next five multiples of the next five multiples of 11 are 22, 33,44,55 and 66 8 are 16, 24, 32, 40 and 48. Example 2 Which of the following numbers 18, 20, 27 36 and 50 are 1. multiples of 2 2. multiples of 3 3. multiples of 4. Solution When a number can be divided exactly by another number it means the quotient is a multiple of the divisor. 18, 20, 27 36, 50 1. multiples of 2 are 18, 20, 27 36, 50 2. multiples of 3 are 18, 27 and 36 3. multiple of 4 are 20 and 36. EVALUATION 1. 1.Find the next 7 multiples of the following numbers. (a) 15 (b) 25 (c ) 13. 1. Which of the following whole numbers 37, 68, 51, 128, 85 and 187 are (a) multiples of 2 (b) multiples of 3 (c) multiples of 5 (d) multiples of 17. Common multiples When two or more numbers have a multiple in common, then the numbers is known as a common multiple. Example Find the first two common multiples of 4.6 and 8. Solution Their multiple are as shown below; 4 = 4, 8, 12, 16, 20 (24) 28,32, 36, 40, 44, (48) 52,56,60,64…… 6 = 6, 12, 18,(24), 30, 36, 42, (48), 54, 60, 66, 72. 8= 8, 16,(24), 32, 40, (48), 56, 64, 72, Considering the three whole numbers, their first two common multiples are 24 and 48. Examples Write down three common multiples of the following sets of numbers (a) 5 and 6 (b) 3, 10 and 15. Solution (a) First three common multiples of 5 and 6. 5 = 5, 10, 15, 20, 25, (30),35, 40 45, 50, 55 (60) 65, 70,75,80 85,(90) 95,100, 105,110,115,120. 6= 6, 12,18,24, (30),36, 42,48,54,(60),66,72,78,84, (90),96,102,106,114,120. :.The first three common multiples of 5 and 6 are, 30 60 and 90. (b) First three common multiples of 3, 10 and 15 3= 3,6,9,12,15,18,21, 24,27, (30),33,36,39,42,45,81,84,87,(90),93,96. 10= 10, 20, (30), 40, 50, (60) 70, 80 (90),100,110, etc 15 = 15, (30), 45,(60), 75, (90), 105, 120. :. The first three common multiples of 3, 10, and 15 are 30, 60 and 90. EVALUATION Find the first four common multiples of the following sets of numbers (a) 4 and 7 (b) 2,5 and 7 (c ) 3, 6, 9. 1. Essential Mathematic for JSS1 by AJS Oluwasanmipg 32 2. New General Mathematics for JSS1 by M.F Macraeetalpg 26-27. Least Common Multiples (LCM) You can find the least common multiples of two or more numbers by listing as many multiples as you need until you have one that is common to both or all the numbers. For instance to find the LCM of 24 and 15 Multiples of 24 = 24, 48, 72,96, 120……… Multiples of 15 = 15, 30, 45, 60, 75,90, 105, 120. Although the numbers will have many common multiples but, looking at what we are after, that is the least of the common multiples, the answer will be 120. :. LCM of 24 and 15 = 120. Rather than writing out a long list of multiples for each number, you can use the prime factors method to find the LCM. This is the method we are going to apply. Example 1 Find the LCM of the following whole numbers: (a) 24 and 15 (b) 8 and 45 ( c ) 16 and 18 (d) 90, 105 and 210. Solution (a) The LCM of 24 and 15 2 24 15 2 8 45 2 12 15 2 4 45 2 6 15 2 2 45 3 2 5 3 1 45 5 1 1 3 1 15 3 1 5 5 1 1 LCM = 2 x 2 x 2 x 3 x 5= 120 Lcm = 2 x 2 x 2 x 3 x 3 x 5 = 360 :.LCM of 24 and 15 = 120 ;. LCM of 8 and 45 = 360. (c) LCM of 16 and 18 (d) LCM of 90, 105 and 210 2 16 18 2 90 105 210 2 8 9 3 45 105 105 2 4 9 3 15 35 35 2 2 9 5 5 35 35 2 1 9 7 1 7 7 3 1 3 1 1 1 3 1 1cc LCM = 2 x 2 x 2 x x 2 x3 x3 = 144 LCM = 2 x 3 x 3 x 5 x 7 = 630 :. LCM of 16 and 18 = 144 :. LCM of 90, 105 and 210 is = 630. Given that the numbers are expressed as a product of prime factors, the lcmis the product of the prime factors of the numbers without double counting. Example 2 Find the LCM of the following .Leave your answers in prime factors. (a) 2 x 2 x 3, (b )2 x 2 x 5 2 x 2 x 2 x 5 3 x 5 x 7 2 x 2 x5 2 x 3 x 3 x 3 2x 2 x 3 x 3 x 5 3 x 5 x 5 x 7. Solution (a) 2 x 2 x 3 2 x 2 x 2 x 5 2 x 2 x 5 2 x 2 x 3 x 3 x 5 LCM = 2 x 2 x2 x 3 x 3 x 5. (b) 2 x 3 x 3 3x 5x 7 2 x 3 x 3 x 3 3 x 5 x 5 x 7 LCM = 2 x 3 x 3 x 3 x 5 x 5 x 7. EVALUATION 1.Find the LCM of the following (a) 4,6, and 9 (b) 6, 8, 10 and 12 (c ) 9, 10,12 and 15 (d) 108 and 360. 2. Find the Lcm of the following leaving your answers in index form. (a) 2 x 2 x 2 x 3 x 3 (b) 3 x 3 x 5 2 x 3 x 5 x5 2 x 3 x 7 2 x 2 x 3 x 3 x 5 2 x 5 x 5 x 7 (c) 23 x 32 x 5 3 x 53 x 72 24 x 3 x 72, 32 x 52 x 73 Highest Common Factor Highest common factor (HCF) of two or more numbers is the largest number that divides exactly into all the numbers. Example 1 Find the HCF of 21 and 84. Solution 3 21 2 84 7 7 2 42 1 3 21 7 7 1 21 = 3 x 7 84 = 2 x 2 x 3 x 7 HCF = 3 x 7 = 21 Example 2 Find the HCF of 195 and 330. Solution 3 195 5 65 1. 13 11 HCF of 195 and 330 195 = 3 x 5 x 13 330 = 2 x 3x5x11 HCF = 3 x 5 = 15 1. 330 2. 165 5 55 11 11 1 Example 3 Find the HCF of 288, 180 and 106. leave your answer in index form. Solution 2 288 2 180 2 108 2 144 2 90 2 54 2 72 3 45 3 27 2 36 3 15 3 9 2 18 5 5 3 3 3 9 1 1 3 3 1 288 = 2 x 2 x 2 x2x2 x 3 x 3 180 = 2×2 x 3 x 3 108 = 2×2 x3 x 3×3 HCF = 2x 2x 3 x 3 = 22 x 32 …….index form = 36 (ordinary form). Example 4 Find the HCF of the following . Leave the answers in prime factors and use index notation. (a) 23 x 32 x 7 22 x 3 x 52 22 x 33 x 5 (b) 23 x 52 x 7 22 x 32 x 5 33 x 53 x 72 Solution (a)23 x 32 x 7 = 2 x 22 x 3 x 3 x7 22 x 3 x 52 = 22 x 3 x 52 22 x 33 x 5 = 22x 3 x 32 x 5 HCF = 22 x 3 index form 4 x 3 = 12. (b) 23 x 52 x 7 = 23x 52 x 7 22 x 32 x 5 = 22 x 32x 5 33 x 53 x 72= 33 x 53 x 72 The factor that is common is in 5 :. HCF = 5 EVALUATION 1.Findthe HCF of the following; • 12 and 24 • 16 and 40 • 6 and 15 • 13 and 31 • 13 and 36 (a) 160, 96 and 224 (b) 189, 279and 108 (c) 126, 234 and 90. 2. Find the HCF of the following .Leave your answers in prime factors and use index notation. 22x 33 x 5 21 x 34 x 5 2 x 35 x 72 (b) 23 x 33 x 53 24 x 3 x 52 x 7 25 x 32 x 5 x 72 1. New General Mathematics for jss 1 by m.Fmacraeetalpg 25-26 2. Essential Mathematics for jss1 by AjSOluwasanmi, pg 31. WEEKEND ASSIGNMENT 1. The value of 23 x 32 is (a) 1291 (b) 658 (c) 729 (d)7 36 (e) 54 2. The LCM of 12 and 15 is (a) 90 (b) 60 (c) 30 (d) 120 (e) 180 3. The HCF of 63and 90 is (a) 7 (b) 3 (c) 12 (d) 6 (e) 9 4. The first three common multiples of 3 and 11 are (a) 3, 33, 66, (b) 11, 33, 66 ( c ) 33, 66, 99 (d) 33, 44, 55 (e) 33, 22, 11. 5. Which of the following whole numbers 22, 11, 54, 35, 40, 75, and 105 is /are multiples of 5? (a) 11, 22, 35 (b) 35, 40, 75, 105 ( c ) 54, 35, 40, 75, 105 (d) 35, 54, 40, 75 (e) 105,75,40,35,54. THEORY 1. Give the first five multiples of the following I. 5 II 7 III. 11 B Write down four common multiples of the following sets of numbers I. 3, 4 and 5 II. 3, 10 and 15. 2a. Find the LCM of I. 9, 24, 32, and 90 II. 23 x 32 x 5 x 7 3 x 53 x 72 24 x 3 x 72 32 x 52 x 73 b. Find the HCF of i. 126, 234 and 90 ii. 23 x 33 x 53 b. 63, 42, and 21 24 x 3 x 52 x 7 25 x 32 x 5 x 72 Presentation The topic is presented step by step Step 1: The class teacher revises the previous topics Step 2. He introduces the new topic Step 3: The class teacher allows the pupils to give their own examples and he corrects them when the needs arise Conclusion The class teacher wraps up or conclude the lesson by giving out short note to summarize the topic that he or she has just taught. The class teacher also goes round to make sure that the notes are well copied or well written by the pupils. He or she does the necessary corrections when and where the needs arise.
New Maths Content Take a look below at some of the new resources we’ve picked out for you to try – you can see the rest in the New Content area when you log in. Please note that the content listed below is arranged according to the National Curriculum in England. Maths Activities Key Stage 1 Are You Shore?: #32816 – This Activity focuses on the following positional language: left, right top, middle, bottom, on top of, above, below, under, between. Plaice Value: #32782 – Identify and represent numbers from 0 to 100, using place value blocks. Lower Key Stage 2 Carnival Graphs: #34914 – Interpret and present data in bar charts and time graphs to solve problems. Change of Scene: #34667 – Converting between different units of measure including: time, length, volume and mass. Fractions to Go!: #34638 – Use diagrams to recognise and show equivalent fractions with small denominators. Hole Number Fishing: #34629 – Round decimals with one decimal place to the nearest whole number. In Safe Hands: #34499 – Tell and write the time from an analogue, 12- hour and 24- hour clock. On the Rocks: #34545 – Fill in the missing numbers in the number patterns. Answer the questions about the number patterns. Pieces of Cake: #34542 – Students are taught to recognise the decimal equivalence of the fractions ¼, 1/2, and ¾. Pizza Box Fractions: #34532 – Students answer a selection of different fraction problems. Questions include ordering unit fractions and fractions with the same denominator, recognising equivalent fractions, finding a fraction of a discrete set of objects and adding and subtracting fractions. Shapes in the Sand: #34548 – Identify lines of symmetry in 2-D shapes in different orientations. Thirds of a Feather: #34929 – Help Sten to find non-unit fractions of different quantities. This Scampi Happening: #32783 – Identify and represent numbers from 0 to 1,000, using place value blocks. Time Bandit: #34926 – Convert time between analogue and digital 12 and 24-hour clocks. Time to Chill: #34820 – Count back through zero to include negative numbers. Upper Key Stage 2 Bargain Hunter: #34617 – Recall and use equivalences between simple fractions, decimals and percentages, including in different contexts. Folding Space: #34923 – Identify the different nets of 3-D shapes and their features. Match Day: #34938 – Identify and describe the position of a reflected shape. Out on a Limb: #34676 – Help Sten build a treehouse by calculating lengths and perimeters. Calculations involve the conversion of units between centimetres and metres, and include decimals with up to 3 decimal places. Party Problems: #34886 – Calculate with measurements. Positional Play: #34935 – Identify and describe the position of a shape following a translation. Ratio Rations: #34620 – Solve shop themed problems, involving the relative size of 2 quantities, where missing values can be found using multiplication and division facts. Reference given to ratio and the use of a:b notation. Secret Formula: #34889 – Use given formulae to find the perimeter and area of triangles, squares, rectangles, parallelograms and regular polygons and the volume of cubes and cuboids. Shed Shapes: #34932 – Calculate the perimeter of composite shapes. Sun, Sand and Shopping: #34911 – Use estimation to check if Sten and his friends have enough money to buy the items they want in the beach shop. Then use estimation strategies in different real life scenarios with the summer holiday theme. Learn Screens Lower Key Stage 2 Novel Addition – Chapter 1: #34893 – An introduction into formal written methods of columnar addition. This Learn Screen does not look at how to carry numbers. Novel Addition – Chapter 2: #34894 – An introduction into formal written methods of columnar addition. This Learn Screen looks at how to carry numbers. Novel Subtraction – Chapter 1: #34895 – An introduction into formal written methods of columnar subtraction. This Learn Screen does not look at exchanging. Novel Subtraction – Chapter 2: #34896 – An introduction into formal written methods of columnar subtraction. This Learn Screen looks at exchanging. Upper Key Stage 2 Bargains Galore: #34892 – Use knowledge of fractions and decimals to help work out a percentage of a whole number.
# Law of Sines ## Theorem Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$. Then: $\dfrac a {\sin A} = \dfrac b {\sin B} = \dfrac c {\sin C} = 2 R$ where $R$ is the circumradius of $\triangle ABC$. ## Proof 1 Construct the altitude from $B$. From the definition of sine: $\sin A = \dfrac h c$ and $\sin C = \dfrac h a$ Thus: $h = c \sin A$ and: $h = a \sin C$ This gives: $c \sin A = a \sin C$ So: $\dfrac a {\sin A} = \dfrac c {\sin C}$ Similarly, constructing the altitude from $A$ gives: $\dfrac b {\sin B} = \dfrac c {\sin C}$ $\blacksquare$ ## Proof 2 Construct the circumcircle of $\triangle ABC$, let $O$ be the circumcenter and $R$ be the circumradius. Construct $\triangle AOB$ and let $E$ be the foot of the altitude of $\triangle AOB$ from $O$. By the Inscribed Angle Theorem: $\angle ACB = \dfrac {\angle AOB} 2$ From the definition of the circumcenter: $AO = BO$ From the definition of altitude and the fact that all right angles are congruent: $\angle AEO = \angle BEO$ Therefore from Pythagoras's Theorem: $AE = BE$ and then from Triangle Side-Side-Side Congruence: $\angle AOE = \angle BOE$ Thus: $\angle AOE = \dfrac {\angle AOB} 2$ and so: $\angle ACB = \angle AOE$ Then by the definition of sine: $\sin C = \map \sin {\angle AOE} = \dfrac {c / 2} R$ and so: $\dfrac c {\sin C} = 2 R$ The same argument holds for all three angles in the triangle, and so: $\dfrac c {\sin C} = \dfrac b {\sin B} = \dfrac a {\sin A} = 2 R$ $\blacksquare$ ## Proof 3 ### Acute Case Let $\triangle ABC$ be acute. Construct the circumcircle of $\triangle ABC$. Let its radius be $R$. Construct its diameter $BX$ through $B$. By Thales' Theorem, $\angle BAX$ is a right angle. $\angle AXB = \angle ACB$ Then: $\ds \sin \angle AXB$ $=$ $\ds \dfrac {AB} {BX}$ Definition of Sine of Angle $\ds \leadsto \ \$ $\ds \sin \angle ACB$ $=$ $\ds \dfrac c {2 R}$ $\ds \leadsto \ \$ $\ds 2 R$ $=$ $\ds \dfrac c {\sin C}$ The same construction can be applied to each of the remaining vertices of $\triangle ABC$. Hence the result. $\Box$ Let $\triangle ABC$ be obtuse. As for the acute case, construct the circumcircle of $\triangle ABC$. Let its radius be $R$. Construct its diameter $BX$ through $B$. By Thales' Theorem, $\angle BCX$ is a right angle. We note that $\Box ABXC$ is a cyclic quadrilateral. $\angle BXC = 180 \degrees - A$ Hence using a similar argument to the acute case: $\ds a$ $=$ $\ds 2 R \sin \angle BXC$ $\ds$ $=$ $\ds 2 R \map \sin {180 \degrees - A}$ $\ds$ $=$ $\ds 2 R \sin A$ and the result follows. $\blacksquare$ ## Also presented as Some sources do not include the relation with the circumradius, but instead merely present: $\dfrac a {\sin A} = \dfrac b {\sin B} = \dfrac c {\sin C}$ ## Also known as The Law of Sines is also known as the sine law, sine rule or rule of sines. ## Historical Note The Law of Sines was documented by Nasir al-Din al-Tusi in his work On the Sector Figure, part of his five-volume Kitāb al-Shakl al-Qattā (Book on the Complete Quadrilateral).
# Polygon Properties What is a Polygon? A closed, 2-dimensional figure with three or more straight line segments connected. It is a closed plane figure where line segments are joined together. Though the sides do not cross each other but they meet at every vertex. Circles, quadrilaterals and pentagons are all examples of polygons but circle is not a polygon as it is not joined by any straight line. ### Types of polygons Different kinds of polygons differ on the basis length of sides and size of angles. Some of the different types of polygons are: Regular polygons: All angles are equal and are of same length. Regular polygons are equiangular as well as equilateral. Equiangular: In case of equiangular polygon, all angles are equal. Equilateral: In case of equilateral polygon, all sides are equal. Convex: When a straight line is drawn through a convex polygon, it crosses over to two sides at the most. However, the interior angles formed are always less than 180 degrees. Concave: A concave polygon shows somewhat different characteristics as if you draw a straight line through the concave polygon that crosses more than two sides then at least one of the interior angles must be more than 180 degrees. ### Different formulas regarding polygons There are formulas to determine the area of a regular polygon. If you consider N as the number of sides and S as the length from center to a corner then: Area of a regular polygon = (1/2) N sin(360Â°/N) S2 Another formula to calculate the sum of the interior angles of a polygon = (N â€“ 2) x 180Â°. There are also some formulas to calculate the number of diagonals in a polygon. The formula to calculate the number of diagonals is 1/2 N(N-3) The number of triangles to be created in a polygon is also derived by the formula (N â€“ 2) ### Different terms related to polygons Diagonal: The diagonal is the line that connects two vertices. However, diagonals are not sides of a polygon. Exterior Angle: When two adjacent sides form an angle outside the polygon then it is called an exterior angle. Interior Angle: When two adjacent sides form an angle inside the polygon, then it is called an interior angle. Side: The lines that make the polygon are called sides. Vertex: Vertex is the common point where two sides of a polygon meet. The plural of vertex is vertices. ### Special types of polygons There are also some special types of polygons which includes the special quadrilaterals and the special triangles. The list of special quadrilaterals includes polygons like rhombus, parallelogram, square, rectangle and the trapezoid. The list of special triangles includes right, equilateral, scalene, isosceles, obtuse and acute triangles. Each polygon bears a special name according to the number of sides. The following table will let you know of the names of different polygons based on their sides: Name Sides N-gon n Triangle 3 Pentagon 5 Hexagon 6 Heptagon 7 Octagon 8 Decagon 10 Dodecagon 12 Some of the other names which have been proposed are: Name Sides Nonagon, Enneagon 9 Undecagon, Hendecagon 11 Tridecagon, Triskaidecagon 13 Icosagon 20 Triacontagon 30 Tetracontagon 40 Pentacontagon 50 Hexacontagon 60 Heptacontagon 70 Octacontagon 80 Enneacontagon 90 Hectogon, Hecatontagon 100 Chiliagon 1,000 Myriagon 10,000 Further, if you want to name a polygon with number of sides like 86 or 73, then try to follow these rules. You can combine the prefixes and suffixes mentioned below to name polygons with different number of sides. Thus, an 86 sided Polygon would be called Octacontakaihexagon and a 73 sided polygon would be called Heptacontakaitrigon. However, most people prefer to use the naming format “n-gon” so a 86 sided polygon can be simply called 86-gon. slot bonus new member 100
Courses Courses for Kids Free study material Offline Centres More Store # What is the approximate percentage increase in the total employees of the companies from 2013 to 2016?A.50%B.10.5%C.9.5%D.9.2% Last updated date: 13th Jun 2024 Total views: 411.9k Views today: 5.11k Verified 411.9k+ views Hint: From the graph , first calculate the total number of employees in both the companies act 2013 and the total number of employees in both the companies at 2016. Then calculate the percentage increase using the formula $\dfrac{{increase{\text{ }}in{\text{ }}number}}{{{\text{original value}}}}100$ Step 1 : From the graph , The number of employees at company P in the year 2013 is 25000 The number of employees at company Q in the year 2013 is 45000 Therefore the total number of workers in both the companies act 2013 = $25000 + 45000 = 70000$ Step 2: Now from the graph, The number of employees at company P in the year 2016 is 50000 The number of employees at company Q in the year 2016 is 55000 Therefore the total number of workers in both the companies at 2016 = $50000 + 55000 = 105000$ Step 3 : Now we are asked to find the percentage increase of total employees of both the companies. Increase in percentage =$\dfrac{{increase{\text{ }}in{\text{ }}number}}{{{\text{original value}}}}100$ Here the increase in number is the difference between the number of employees in both the companies at 2016 and the total number of employees of both the companies act 2013. And the original value is the number of employees in 2013. Therefore , Percentage increase $= \dfrac{{105000 - 70000}}{{70000}}100 \\ = \dfrac{{35000}}{{70000}}100 \\ = \dfrac{1}{2}*100 \\ = 50\% \\$ Therefore the increase in percentage of the total number of employees in both the companies is 50% The correct option is A. Note: Steps used to calculate the percentage increase: 1. Calculate the difference (increase) between the two numbers you are comparing. 2.Increase = New Number - Original Number. 3. Divide the increase by the original number and multiply the answer by 100. 4.% increase = Increase ÷ Original Number × 100.
Coordinate Proofs Ηλεκτρονική - Συσκευές 10 Οκτ 2013 (πριν από 4 χρόνια και 7 μήνες) 97 εμφανίσεις Name: Math 2 Date: Coordinate Proofs Objective: Prove geometric theorems using coordinate methods. Introduction to Coordinate P roofs Over the past few lessons, we have studied how to perform all the basic kinds of coordinate calculations: finding slopes, distances, midpoints , and points forming other ratios. While applying these methods to particular diagrams, we often saw outcomes such as slopes turning out to be equal, distances turning out to be equal or in a particular ratio, and midpoints of segments turning out to be th e same point. These occurrences were often an indication that there is a geometric theorem (a provable fact about geometry) that applied to that type of diagram. We are now turning our attention to identifying these theorems and writing proofs of them, usi ng the machinery of coordinate calculations. (Working with coordinates is just one of the possible approaches to geometric proof. We will pursue other proof techniques in the next unit of the course: Unit 4, Deductive Geometry.) The idea of a coordinate p roof is to verify a geometric theorem using the relevant coordinate calculations. For example: Facts about equal lengths and other length relationships can be proved by calculating lengths using the distance formula and comparing th em. Facts about parallel lines or perpendicular lines can be proved by calculating slopes using and comparing them (equal slopes indicate parallel; slopes having a product of 1 indicate perpendicular). Facts concerning midpoints can be proved by using the midpoint coordinates in appropriate distance or slope calculations. Facts concerning points forming other ratios can be proved using the formulas we devised at our last class. For example, the point that’s of the way from ( x 1 , y 1 ) to ( x 2 , y 2 ) can be calculated as ( x 1 + x 2 , y 1 + y 2 ). For other fractions, replace and with any k and (1 k ). Usually we will be trying to prove a theorem about all shapes of a particular kind (examples: all triangles, or about all right triangles). An important strategy for writing coordinate proofs is to work with a s hape whose coordinates are variable (so that the proof applies to all shapes of that kind) but that is placed at a location in the coordinate plane that makes the coordinate calculations relatively easy. For example, when proving theorems about triangles, it is often easiest to put one vertex at the origin and one side along an axis. You’ll see such a setup used in the first problem. For other types of shapes as well, choosing a convenient location makes proofs easier. For today’s assignment, each problem w ill specify where the shape is located. Eventually, you will be expected to make these decisions on your own. Name: Math 2 Date: Problems Directions: Complete on separate paper. You may wish to use graph paper but it is not required. 1. Consider XYZ with vertices at (0, 0), ( a , 0), and ( b , c ) respectively. Let L , M , and N be the midpoints of the sides, as shown. a. Calculate the coordinates of L , M , and N . b. A segment between midpoints of two sides of a triangle, such as ML , is called a midsegme nt . Calculate the lengths of midsegment ML and side XY . How do they compare? c. Calculate the slopes of midsegment ML and side XY . How do they compare? d. midsegments that you have proved in parts b and c . e. Now state a all the sides of LMN in relation to all the sides of XYZ . 2. For proofs about isosceles triangles (triangles with two equal sides), it’s easiest to put the triangle’s line of symmetry along an axis. a. Draw a graph of a triangle with vertices at ( a , 0), ( a , 0), and (0 , b ). b. Calculate distances to verify that this triangle is an isosceles triangle. c. Find the coordinates of the midpoints of the two equal sides. d. In triangles, a segment from a vertex to the midpoint of the opposite side is called a median . Prove tha t in an isosceles triangle, two of the medians have equal lengths. Name: Math 2 Date: 3. in the diagram. a. a parallelogram, what must coordinate d equal (in terms of the other variables a , b , and/or c )? Hint: In parallelograms, opposite sides always have equal lengths. Make lengths WX and ZY be equal. For the rest of this problem, assume that d is as you stated in part a a parallelogram. b. Calculate the coordinates of the midpoint o f diagonal WY . c. Calculate the coordinates of the midpoint of diagonal ZX . d. State the theorem you have just proved about the diagonals of a parallelogram. 4. in the diagram. Note that although the coordinate labels ar e the same as before, we are no longer d that was in problem 24a . a. Explain why this diagram is an appropriate setup for trapezoids. b. Let M be the midpoint of ZW and N be the midpoint of YX . Calculate the s lope of MN and the distance MN . c. How does distance MN relate to distances ZY and WX ? d. State the theorem(s) you have just proved about trapezoids. Name: Math 2 Date: 5. Here is an appropriate coordinate general. It does not assume t hat the properties. Draw the midpoints of the four sides and connect them to form the
Ordinal Numbers for intermediate learners Imagine that you are on a list and you want to know where you are standing. To refer to that number, we use 'ordinal numbers.' Why Do We Use Ordinal Numbers We use ordinal numbers to show the position of something on a list or in a certain order. How to Write Ordinal Numbers: 1-10 We usually add 'th' at the end of numbers in order to turn them into ordinal numbers. But one, two, and three are exceptions. Take a look at the list below: Written Numeral first 1st second 2nd third 3rd fourth 4th fifth 5th sixth 6th seventh 7th eighth 8th ninth 9th tenth 10th Tip! Pay attention to the spelling of ordinal numbers. The spelling of some numbers changes when we add 'th'. For example: Five → fifth Eight → eighth Nine → ninth How to Write Ordinal Numbers: 11-20 Just like ordinal numbers from four to ten, ordinal numbers from 11 to 20 get 'th' at the end. Take a look at this list: Written Numeral eleventh 11th twelfth 12th thirteenth 13th fourteenth 14th fifteenth 15th sixteenth 16th seventeenth 17th eighteenth 18th nineteenth 19th twentieth 20th Tip! You can see that the spelling of some numbers changes. For example: twelve → twelfth Twenty → twentieth How to Write Ordinal Numbers: Compound When we want to write compound ordinal numbers, we use this form: the first part (cardinal number) + hyphen + second part (ordinal number) Written Numeral twenty-first 21st twenty-second 22nd twenty-third 23rd twenty-fourth 24th twenty-fifth 25th twenty-sixth 26th twenty-seventh 27th twenty-eighth 28th twenty-ninth 29th How to Write Ordinal Numbers: The Tens As you saw before, when forming ordinal numbers, the letter 'y' at the end of the number changes into 'ie'. Take a look at the list below: Written Numeral tenth 10th twentieth 20th thirtieth 30th fortieth 40th fiftieth 50th sixtieth 60th seventieth 70th eightieth 80th ninetieth 90th Comments Loading recaptcha You might also like Possessives In order to add to your bookmarks you must sign in to your account Possessives are used to indicate possessions. In this lesson, we will discuss the different types of possessives, including nouns, determiners, and pronouns. Numbers In order to add to your bookmarks you must sign in to your account If math describes how the world works, numbers are the alphabet that makes math possible. In this lesson, we will discover all about them. Cardinal Numbers In order to add to your bookmarks you must sign in to your account When we want to refer to quantity, we use numbers. These numbers are called 'cardinal numbers.' Roman Numerals In order to add to your bookmarks you must sign in to your account The ancient numbers that are still being used, we all have definitely seen them, let's learn them properly now. When to Spell Out Numbers In order to add to your bookmarks you must sign in to your account There are many types of ways to say a number but do you know when and where to use each of them? In this lesson, we will discover more about numbers. Fractions, Decimals and Percentages In order to add to your bookmarks you must sign in to your account A fraction is a particular form of writing numbers. In this lesson, we will learn how to use them. Download LanGeek app
# Interactive Classroom - Mr. Muchka's Classroom Main Idea and New Vocabulary Key Concept: Rational Numbers Example 1: Write a Fraction as a Decimal Example 2: Write a Fraction as a Decimal Example 3: Real-World Example Example 4: Write Decimals as Fractions Example 5: Write Decimals as Fractions Express rational numbers as decimals and decimals as fractions. rational number terminating decimal repeating decimal Write a Fraction as a Decimal Write as a decimal. means 3 16. 0.18 75 Divide 3 by 16. Write a Fraction as a Decimal Answer: 0.1875 Write as a decimal. A. 0.55 B. 0.5625 C. 0.916 _ D. 1.7 Write a Fraction as a Decimal Write as a decimal. can be written as . Divide 35 by 11 and add a negative sign. 3 33 20 11 90 88 2 Write a Fraction as a Decimal Answer: Write A. B. C. 5.5 D. 5.7 as a decimal. FRUIT When Juliana went strawberry picking, 28 of the 54 strawberries she picked weighed less than 2 ounces. To the nearest thousandth, find the fraction of the strawberries that weighed less than 2 ounces. To find the fraction of the strawberries that weighed less than 2 ounces, divide the number of strawberries that weighed less than 2 ounces, 28, by the total number of strawberries, 54. 28 54 ENTER 0.5185185185 Look at the digit to the right of the thousandths place. Since the digit is 5, round up. Answer: The fraction of the strawberries that weighed less than 2 ounces was 0.519. SCHOOL In Mrs. Townleys eighth grade science class, 4 out of 22 students did not turn in their homework. To the nearest thousandth find the fraction of students who did not turn in their homework. A. 0.094 B. 0.148 C. 0.182 D. 0.252 Write a Fraction as a Decimal Write 0.32 as a fraction. 0.32 0.32 is 32 hundredths. Simplify. Answer: Write 0.65 as a fraction. A. B. C. D. Write Decimals as Fractions ALGEBRA Write simplest form. as a mixed number in Assign a variable to the value . Let N = 2.777... . Then perform operations on N to determine its fractional value. Write Decimals as Fractions N 10(N) = 2.777... = 10(2.777...) Multiply each side by 10 because 1 digit repeats. 10N = 27.777... Multiplying by 10 moves the decimal point 1 place to the right. __ 25 9 Answer: N = 2.777... Subtract N = 0.777... to eliminate the repeating part. N = Divide each side by 9. 9N = 25 Simplify. __ Write 0.45 as a fraction in simplest form. A. B. C. D. ## Recently Viewed Presentations • You now have 30 seconds left 10 9 8 7 6 5 4 3 2 1 STOP 2. A census-taker knocks on a door and asks the woman inside how many children she has and how old they are. "I... • insight into strategy, policy and practice within, between and across institutions We offer a rich source of research, data and evidence and resources on impactful solutions across HE in the UK and Ireland, and also in HE sectors in other... • and is the part of grammar shared by speakers and listeners in communication. In 1957, American linguist Chomsky proposed the transformational-generative grammar (TG), thus providing a model of the description of human languages. The GOAL of TG is . to... • Times MS PGothic Arial Times New Roman Blank Presentation 1_Blank Presentation PowerPoint Presentation A food chain is a model that shows a sequence of feeding relationships. PowerPoint Presentation PowerPoint Presentation PowerPoint Presentation A food web shows a complex network of... • Course Overview Course Overview Objective Topics Course Materials Assignments Guidelines Objectives A better understanding of issues that could make a project succeed or fail To learn about managing the software development process To learn about the responsibilities of project managers... • BALLISTICSand FIREARMS " A well regulated militia, being necessary to the security of a free state, the right of the people to keep and bear arms, shall not be infringed." ... In the field of forensic science, forensic ballistics is... • OtolaryngolClin North Am. 2014 Oct;47(5):763-78. Viral or bacterial infections may result in sialadenitis from direct inoculation of the d uctal epithelium or through ascending oral cavity infections. • An exception was the election of Congressman Vito Marcantonio in East Harlem district in New York. Marcantonio was first elected in 1934 as a Republican and was defeated for reelection. In 1940, he was elected on the ALP line, and...
# How do you find the reciprocal of 3/8? Oct 24, 2016 $\frac{8}{3}$ #### Explanation: The color(blue)"product" ("multiplication") of a number and it's reciprocal equals 1. Hence the reciprocal of a number $= \frac{1}{\text{number}}$ Thus the reciprocal of $\frac{3}{8} = \frac{1}{\frac{3}{8}} = \frac{8}{3}$ Note : $\frac{3}{8} \times \frac{8}{3} = 1$
# If a triangle with side lengths in the ratio 3:4:5 is inscribed in a circle of radius 3, how do you find the area? ##### 1 Answer Jun 16, 2015 Find area of right triangle #### Explanation: The hypotenuse = 2R = 6 Since the side ratio is 3: 4: 5, the new ratio of the actual sides is: 3.6: 4.8: 6. The 2 legs of the right triangle are: 3.6 and 4.8 The area is: $s = \frac{3.6 \left(4.8\right)}{2} = 8.64$
# Math Expressions Grade 5 Unit 3 Lesson 11 Answer Key Solve Division Problems ## Math Expressions Common Core Grade 5 Unit 3 Lesson 11 Answer Key Solve Division Problems Math Expressions Grade 5 Unit 3 Lesson 11 Homework Unit 3 Lesson 11 Solve Division Problems Math Expressions Question 1. Consider the division problem $$\frac{1}{2}$$ ÷ 3. Describe a situation this division could represent. Draw a diagram to represent the division. Then find the solution. 1/2 / 3 = 0.16. Explanation: In the above-given question, given that, the factors are 1/2 and 3. divide the factors. 1/2 = 0.5. 0.5 / 3 = 0.16. Write an equation. Then solve. Show your Work. Lesson 11 Answer Key Math Expressions Question 2. A rectangle has an area of 12 square feet and a length of 5 feet. What is its width? The width of the rectangle = 2.4 feet. Explanation: In the above-given question, given that, A rectangle has an area of 12 square feet and a length of 5 feet. area of the rectangle = l x w. where l = length. w = width. 12 = 5 x x. x = 12/5. x = 2.4. so the width of the rectangle = 2.4 feet. Unit 3 Lesson 11 Answer Key Math Expressions Question 3. A tortoise must walk $$\frac{1}{12}$$ mile to visit a friend. He plans to break the journey into four equal parts with breaks in between. How long will each part of his journey be? The length of each part of his journey = 3.92 miles. Explanation: In the above-given question, given that, A tortoise must walk $$\frac{1}{12}$$ mile to visit a friend. 1/12 – 4. 1/12 = 0.08. 0.08 – 4 = 3.92. so the length of each part of his journey = 3.92 miles. Division Questions For Grade 5 Math Expressions Question 4. Harry worked 7 hours last week. This is $$\frac{1}{3}$$ as many hours as Aidan worked. How many hours did Aidan work? The number of hours did Aidan work = 6.67 hours. Explanation: In the above-given question, given that, Harry worked 7 hours last week. This is $$\frac{1}{3}$$ as many hours as Aidan worked. 7 – 1/3. 1/3 = 0.33. 7 – 0.33 = 6.67. so the number of hours did Aidan work = 6.67 hours. Unit 3 Lesson 11 Math Expressions Question 5. Lin is a camp counselor. She is making small bags of trail mix for campers to take on a hike. She has 2 pounds of raisins and is putting $$\frac{1}{8}$$ pound in each bag. How many bags can she fill before she runs out of raisins? The number of bags can she fill before she runs out of raisins = 0.25. Explanation: In the above-given question, given that, Lin is a camp counselor. She is making small bags of trail mix for campers to take on a hike. She has 2 pounds of raisins and is putting $$\frac{1}{8}$$ pound in each bag. 2 x 1/8. 1/4 = 0.25. so the number of bags can she fill before she runs out of raisins = 0.25. Division Problems For Grade 5 Math Expressions Question 6. Mr. Ramirez bought $$\frac{1}{4}$$ pounds of cashews. He divided the cashews equally among his three children. How much did each child get? The quantity did each child get = 0.25. Explanation: In the above-given question, given that, Mr. Ramirez bought $$\frac{1}{4}$$ pounds of cashews. 1/4 = 0.25. 0.12 + 0.12 = 0.25. so the quantity did each child get = 0.25. Math Expressions Grade 5 Unit 3 Lesson 11 Remembering 1(1/8) + 4(2/3) = 5.7. Explanation: In the above-given question, given that, the fractions are 1(1/8) and 4(2/3). 1(1/8) = 9/8. 4(2/3) = 14/3. 9/8 = 1.1. 14/3 = 4.6. 1.1 + 4.6 = 5.7. Unit 3 Lesson 11 Practice Problems Answer Key Question 2. 6(1/4) – 4(5/6) = 1.4. Explanation: In the above-given question, given that, subtract the fractions. the fractions are 6(1/4) and 4(5/6). 6(1/4) = 25/4. 4(5/6) = 29/6. 25/4 = 6.2. 29/6 = 4.8. 6.2 – 4.8 = 1.4. Division Examples For Grade 5 Math Expressions Question 3. 9(1/3) + 7(8/9) = 17.1. Explanation: In the above-given question, given that, the fractions are 9(1/3) and 7(8/9). 9(1/3) = 28/3. 7(8/9) = 71/9. 28/3 = 9.3. 71/9 = 7.8. 9.3 + 7.8 = 17.1. 5(2/7) + 5(11/14) = 5.7. Explanation: In the above-given question, given that, the fractions are 5(2/7) and 5(11/14). 5(2/7) = 37/7. 5(11/14) = 81/14. 37/7 = 5.2. 81/14 = 5.7. 5.2 + 5.7 = 10.9. Question 5. 4 – 2(2/5) = 1.6. Explanation: In the above-given question, given that, subtract the fractions. the fractions are 4 and 2(2/5). 2(2/5) = 12/5. 12/5 = 2.4. 4 – 2.4 = 1.6. Question 6. 6(5/8) + 3(1/2) = 5.7. Explanation: In the above-given question, given that, the fractions are 6(5/8) and 3(1/2). 6(5/8) = 53/8. 3(1/2) = 7/2. 53/8 = 6.6. 7/2 = 3.5. 6.6 + 3.5 = 10.1. Predict whether the product will be greater than, less than, or equal to the second factor. Then compute the product. Question 7. $$\frac{5}{5}$$ . 9 = x Predict : x 9 Compute: x = _______ 9 = 9. x = 9. Explanation: In the above-given question, given that, the fractions are 5/5 and 9. multiply the fractions. 5/5 = 1. 1 . 9 = x. x = 9. Question 8. $$\frac{7}{8}$$ . 9 = x Predict : x 9 Compute: x = _______ 7.2 < 9. x = 7.2. Explanation: In the above-given question, given that, the fractions are 7/8 and 9. multiply the fractions. 7/8 = 0.8. 0.8 . 9 = x. x = 7.2. Question 9. 1$$\frac{3}{5}$$ . 9 = x Predict : x 9 Compute: x = _______ 5.4 < 9. x = 5.4. Explanation: In the above-given question, given that, the fractions are 3/5 and 9. multiply the fractions. 3/5 = 0.6. 0.6 . 9 = x. x = 5.4. Question 10. 1$$\frac{1}{2}$$ . $$\frac{4}{5}$$ = x Predict : x $$\frac{4}{5}$$ Compute: x = _______ 1.2 > 0.8. x = 1.2. Explanation: In the above-given question, given that, the fractions are 1(1/2) and 4/5. multiply the fractions. 3/2 = 1.5. 4/5 = 0.8. 1.5 x 0.8 = x. x = 1.2. Question 11. $$\frac{6}{6}$$ . $$\frac{4}{5}$$ = x Predict : x $$\frac{4}{5}$$ Compute: x = _______ 0.8 = 0.8. x = 0.8. Explanation: In the above-given question, given that, the fractions are 6/6 and 4/5. multiply the fractions. 6/6 = 1. 4/5 = 0.8. 1 . 0.8 = x x = 0.8. Question 12. $$\frac{2}{5}$$ . $$\frac{4}{5}$$ = x Predict : x $$\frac{4}{5}$$ Compute: x = _______ 32 > 4/5. x = 32. Explanation: In the above-given question, given that, the fractions are 2/5 and 4/5. multiply the fractions. 2/5 = 0.4. 4/5 = 0.8. 0.4 . 0.8 = x. x = 32. Divide. Question 13. 6 ÷ $$\frac{1}{4}$$ = ___ 6 ÷ 1/4 = 24. Explanation: In the above-given question, given that, divide the fractions. the fractions are 6 and 1/4. 1/4 = 0.25. 6 ÷ 0.25 = 24. Question 14. 2 ÷ 3 = ___ 6 ÷ 3 = 2. Explanation: In the above-given question, given that, divide the fractions. the fractions are 6 and 3. 6 ÷ 3 = 2. Question 15. 10 ÷ 3 = ___ 10 ÷ 3 = 3.3. Explanation: In the above-given question, given that, divide the fractions. the fractions are 10 and 3. 10 ÷ 3 = 3.3. Question 16. 200 ÷ $$\frac{1}{4}$$ = ___ 200 ÷ 1/4 = 800. Explanation: In the above-given question, given that, divide the fractions. the fractions are 200 and 1/4. 1/4 = 0.25. 200 ÷ 0.25 = 800. Question 17. $$\frac{1}{4}$$ ÷ 8 = ___ 1/4 ÷ 8 = 0.03. Explanation: In the above-given question, given that, divide the fractions. the fractions are 8 and 1/4. 1/4 = 0.25. 8 ÷ 0.25 = 0.03. Question 18. $$\frac{1}{7}$$ ÷ 6 = _____ 6 ÷ 1/7 = 42.8. Explanation: In the above-given question, given that, divide the fractions. the fractions are 6 and 1/7. 1/7 = 0.14. 6 ÷ 0.14 = 42.8. Question 19. Stretch Your Thinking Harrison is playing a board game that has a path of loo spaces. After his first turn, he is $$\frac{1}{5}$$ of the way along the spaces. On his second turn, he moves $$\frac{1}{4}$$ fewer spaces than he moved on his first turn. On his third turn, he moves 1$$\frac{1}{4}$$ times as many spaces than he moved on his first turn. What space is he on after three turns? The space after three runs = 1.7. Explanation: In the above-given question, given that, Harrison is playing a board game that has a path of loo spaces. After his first turn, he is $$\frac{1}{5}$$ of the way along the spaces. On his second turn, he moves $$\frac{1}{4}$$ fewer spaces than he moved on his first turn. On his third turn, he moves 1$$\frac{1}{4}$$ times as many spaces than he moved on his first turn. 1/5 + 1/4 + 5/4. 1/4 = 0.25. 1/5 = 0.2. 5/4 = 1.25. 1.25 + 0.2 + 0.25 = 1.7. so the space after three runs = 1.7.
# HW: Pg. 341-342 #13-61 eoo. ## Presentation on theme: "HW: Pg. 341-342 #13-61 eoo."— Presentation transcript: HW: Pg #13-61 eoo Quiz 1 Pg. 344 #13-26 Vocabulary Polynomial Long Division: When you divide a polynomial ______ by a divisor _____, you get a quotient polynomial ______ and a remainder polynomial _____. This can be written as: Remainder Theorem: If a polynomial _____ is divided by ______, then the remainder is __________. Synthetic Division: Only use the ___________ of the polynomial and the _____________ must be in the form _________. Factor Theorem: A polynomial _____ has a factor ________ if and only if ___________. Use polynomial long division EXAMPLE 1 SOLUTION Write polynomial division in the same format you use when dividing numbers. Include a “0” as the coefficient of x2 in the dividend. At each stage, divide the term with the highest power in what is left of the dividend by the first term of the divisor. This gives the next term of the quotient. ) Use polynomial long division EXAMPLE 1 3x2 + 4x – 3 x2 – 3x + 5 quotient x2 – 3x + 5 3x4 – 5x3 + 0x2 + 4x – 6 ) Multiply divisor by 3x4/x2 = 3x2 3x4 – 9x3 + 15x2 Subtract. Bring down next term. 4x3 – 15x x 4x3 – 12x2 + 20x Multiply divisor by 4x3/x2 = 4x Subtract. Bring down next term. –3x2 – 16x – 6 –3x2 + 9x – 15 Multiply divisor by – 3x2/x2 = – 3 –25x + 9 remainder Use polynomial long division EXAMPLE 1 3x4 – 5x3 + 4x – 6 x2 – 3x + 5 = 3x2 + 4x – 3 + –25x + 9 ANSWER CHECK You can check the result of a division problem by multiplying the quotient by the divisor and adding the remainder. The result should be the dividend. (3x2 + 4x – 3)(x2 – 3x + 5) + (–25x + 9) = 3x2(x2 – 3x + 5) + 4x(x2 – 3x + 5) – 3(x2 – 3x + 5) – 25x + 9 = 3x4 – 9x3 + 15x2 + 4x3 – 12x2 + 20x – 3x2 + 9x – 15 – 25x + 9 = 3x4 – 5x3 + 4x – 6 ) Use polynomial long division with a linear divisor EXAMPLE 2 x2 + 7x quotient + 7 x – 2 x3 + 5x2 – 7x ) x3 – 2x2 Multiply divisor by x3/x = x2. 7x2 – 7x Subtract. 7x2 – 14x Multiply divisor by 7x2/x = 7x. 7x + 2 Subtract. 7x – 14 Multiply divisor by 7x/x = 7. remainder 16 ANSWER x3 + 5x2 – 7x +2 x – 2 = x2 + 7x + 7 + 16 for Examples 1 and 2 GUIDED PRACTICE Divide using polynomial long division. (2x2 – 3x + 8) + –18x + 7 x2 + 2x – 1 ANSWER (x2 – 3x + 10) + –30 x + 2 ANSWER Use synthetic division EXAMPLE 3 SOLUTION – – – –21 – –16 2x3 + x2 – 8x + 5 x + 3 = 2x2 – 5x + 7 – 16 ANSWER Factor a polynomial EXAMPLE 4 SOLUTION Because x + 2 is a factor of f (x), you know that f (–2) = 0. Use synthetic division to find the other factors. – –4 – –16 3 – – Use the result to write f (x) as a product of two Factor a polynomial EXAMPLE 4 Use the result to write f (x) as a product of two factors and then factor completely. f (x) = 3x3 – 4x2 – 28x – 16 Write original polynomial. = (x + 2)(3x2 – 10x – 8) Write as a product of two factors. = (x + 2)(3x + 2)(x – 4) Factor trinomial. for Examples 3 and 4 GUIDED PRACTICE Divide using synthetic division. Factor the polynomial completely given that x – 4 is a factor. x2 + x – 4 + 11 x + 3 ANSWER ANSWER (x – 4)(x –3)(x + 1) 4x2 + 5x + 2 + 9 x – 1 ANSWER ANSWER (x – 4)(x –2)(x +5) Standardized Test Practice EXAMPLE 5 SOLUTION Because f (3) = 0, x – 3 is a factor of f (x). Use synthetic division. –2 – –60 Use the result to write f (x) as a product of two factors. Then factor completely. f (x) = x3 – 2x2 – 23x + 60 = (x – 3)(x2 + x – 20) = (x – 3)(x + 5)(x – 4) The zeros are 3, –5, and 4. The correct answer is A. ANSWER for Example 5 GUIDED PRACTICE Find the other zeros of f given that f (–2) = 0. ANSWER 3 and –3 ANSWER 1 and –7 HOmework: Pg. 356 #15-35 eoo
Descriptive Statistics # 7 Display Data ### Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs One simple graph, the stem-and-leaf graph or stemplot, comes from the field of exploratory data analysis. It is a good choice when the data sets are small. To create the plot, divide each observation of data into a stem and a leaf. The leaf consists of a final significant digit. For example, 23 has stem two and leaf three. The number 432 has stem 43 and leaf two. Likewise, the number 5,432 has stem 543 and leaf two. The decimal 9.3 has stem nine and leaf three. Write the stems in a vertical line from smallest to largest. Draw a vertical line to the right of the stems. Then write the leaves in increasing order next to their corresponding stem. For Susan Dean’s spring pre-calculus class, scores for the first exam were as follows (smallest to largest): 33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100 Stem-and-Leaf Graph Stem Leaf 3 3 4 2 9 9 5 3 5 5 6 1 3 7 8 8 9 9 7 2 3 4 8 8 0 3 8 8 8 9 0 2 4 4 4 4 6 10 0 The stemplot shows that most scores fell in the 60s, 70s, 80s, and 90s. Eight out of the 31 scores or approximately 26% were in the 90s or 100, a fairly high number of As. Try It For the Park City basketball team, scores for the last 30 games were as follows (smallest to largest): 32; 32; 33; 34; 38; 40; 42; 42; 43; 44; 46; 47; 47; 48; 48; 48; 49; 50; 50; 51; 52; 52; 52; 53; 54; 56; 57; 57; 60; 61 Construct a stem plot for the data. Stem Leaf 3 2 2 3 4 8 4 0 2 2 3 4 6 7 7 8 8 8 9 5 0 0 1 2 2 2 3 4 6 7 7 6 0 1 The stemplot is a quick way to graph data and gives an exact picture of the data. You want to look for an overall pattern and any outliers. An outlier is an observation of data that does not fit the rest of the data. It is sometimes called an extreme value. When you graph an outlier, it will appear not to fit the pattern of the graph. Some outliers are due to mistakes (for example, writing down 50 instead of 500) while others may indicate that something unusual is happening. It takes some background information to explain outliers, so we will cover them in more detail later. The data are the distances (in kilometers) from a home to local supermarkets. Create a stemplot using the data: 1.1; 1.5; 2.3; 2.5; 2.7; 3.2; 3.3; 3.3; 3.5; 3.8; 4.0; 4.2; 4.5; 4.5; 4.7; 4.8; 5.5; 5.6; 6.5; 6.7; 12.3 Do the data seem to have any concentration of values? NOTE The leaves are to the right of the decimal. The value 12.3 may be an outlier. Values appear to concentrate at three and four kilometers. Stem Leaf 1 1 5 2 3 5 7 3 2 3 3 5 8 4 0 2 5 5 7 8 5 5 6 6 5 7 7 8 9 10 11 12 3 Try It The following data show the distances (in miles) from the homes of off-campus statistics students to the college. Create a stem plot using the data and identify any outliers: 0.5; 0.7; 1.1; 1.2; 1.2; 1.3; 1.3; 1.5; 1.5; 1.7; 1.7; 1.8; 1.9; 2.0; 2.2; 2.5; 2.6; 2.8; 2.8; 2.8; 3.5; 3.8; 4.4; 4.8; 4.9; 5.2; 5.5; 5.7; 5.8; 8.0 Stem Leaf 0 5 7 1 1 2 2 3 3 5 5 7 7 8 9 2 0 2 5 6 8 8 8 3 5 8 4 4 8 9 5 2 5 7 8 6 7 8 0 The value 8.0 may be an outlier. Values appear to concentrate at one and two miles. A side-by-side stem-and-leaf plot allows a comparison of the two data sets in two columns. In a side-by-side stem-and-leaf plot, two sets of leaves share the same stem. The leaves are to the left and the right of the stems. (Figure) and (Figure) show the ages of presidents at their inauguration and at their death. Construct a side-by-side stem-and-leaf plot using this data. Ages at Inauguration Ages at Death 9 9 8 7 7 7 6 3 2 4 6 9 8 7 7 7 7 6 6 6 5 5 5 5 4 4 4 4 4 2 2 1 1 1 1 1 0 5 3 6 6 7 7 8 9 8 5 4 4 2 1 1 1 0 6 0 0 3 3 4 4 5 6 7 7 7 8 7 0 0 1 1 1 4 7 8 8 9 8 0 1 3 5 8 9 0 0 3 3 Presidential Ages at Inauguration President Age President Age President Age Washington 57 Lincoln 52 Hoover 54 J. Adams 61 A. Johnson 56 F. Roosevelt 51 Jefferson 57 Grant 46 Truman 60 Madison 57 Hayes 54 Eisenhower 62 Monroe 58 Garfield 49 Kennedy 43 J. Q. Adams 57 Arthur 51 L. Johnson 55 Jackson 61 Cleveland 47 Nixon 56 Van Buren 54 B. Harrison 55 Ford 61 W. H. Harrison 68 Cleveland 55 Carter 52 Tyler 51 McKinley 54 Reagan 69 Polk 49 T. Roosevelt 42 G.H.W. Bush 64 Taylor 64 Taft 51 Clinton 47 Fillmore 50 Wilson 56 G. W. Bush 54 Pierce 48 Harding 55 Obama 47 Buchanan 65 Coolidge 51 Presidential Age at Death President Age President Age President Age Washington 67 Lincoln 56 Hoover 90 J. Adams 90 A. Johnson 66 F. Roosevelt 63 Jefferson 83 Grant 63 Truman 88 Madison 85 Hayes 70 Eisenhower 78 Monroe 73 Garfield 49 Kennedy 46 J. Q. Adams 80 Arthur 56 L. Johnson 64 Jackson 78 Cleveland 71 Nixon 81 Van Buren 79 B. Harrison 67 Ford 93 W. H. Harrison 68 Cleveland 71 Reagan 93 Tyler 71 McKinley 58 Polk 53 T. Roosevelt 60 Taylor 65 Taft 72 Fillmore 74 Wilson 67 Pierce 64 Harding 57 Buchanan 77 Coolidge 60 Another type of graph that is useful for specific data values is a line graph. In the particular line graph shown in (Figure), the x-axis (horizontal axis) consists of data values and the y-axis (vertical axis) consists of frequency points. The frequency points are connected using line segments. In a survey, 40 mothers were asked how many times per week a teenager must be reminded to do his or her chores. The results are shown in (Figure) and in (Figure). Number of times teenager is reminded Frequency 0 2 1 5 2 8 3 14 4 7 5 4 Try It In a survey, 40 people were asked how many times per year they had their car in the shop for repairs. The results are shown in (Figure). Construct a line graph. Number of times in shop Frequency 0 7 1 10 2 14 3 9 Bar graphs consist of bars that are separated from each other. The bars can be rectangles or they can be rectangular boxes (used in three-dimensional plots), and they can be vertical or horizontal. The bar graph shown in (Figure) has age groups represented on the x-axis and proportions on the y-axis. By the end of 2011, Facebook had over 146 million users in the United States. (Figure) shows three age groups, the number of users in each age group, and the proportion (%) of users in each age group. Construct a bar graph using this data. 13–25 65,082,280 45% 26–44 53,300,200 36% 45–64 27,885,100 19% Try It The population in Park City is made up of children, working-age adults, and retirees. (Figure) shows the three age groups, the number of people in the town from each age group, and the proportion (%) of people in each age group. Construct a bar graph showing the proportions. Age groups Number of people Proportion of population Children 67,059 19% Retirees 131,662 38% The columns in (Figure) contain: the race or ethnicity of students in U.S. Public Schools for the class of 2011, percentages for the Advanced Placement examine population for that class, and percentages for the overall student population. Create a bar graph with the student race or ethnicity (qualitative data) on the x-axis, and the Advanced Placement examinee population percentages on the y-axis. Race/ethnicity AP examinee population Overall student population 1 = Asian, Asian American or Pacific Islander 10.3% 5.7% 2 = Black or African American 9.0% 14.7% 3 = Hispanic or Latino 17.0% 17.6% 4 = American Indian or Alaska Native 0.6% 1.1% 5 = White 57.1% 59.2% 6 = Not reported/other 6.0% 1.7% Try It Park city is broken down into six voting districts. The table shows the percent of the total registered voter population that lives in each district as well as the percent total of the entire population that lives in each district. Construct a bar graph that shows the registered voter population by district. District Registered voter population Overall city population 1 15.5% 19.4% 2 12.2% 15.6% 3 9.8% 9.0% 4 17.4% 18.5% 5 22.8% 20.7% 6 22.3% 16.8% Below is a two-way table showing the types of pets owned by men and women: Dogs Cats Fish Total Men 4 2 2 8 Women 4 6 2 12 Total 8 8 4 20 Given these data, calculate the conditional distributions for the subpopulation of men who own each pet type. Men who own dogs = 4/8 = 0.5 Men who own cats = 2/8 = 0.25 Men who own fish = 2/8 = 0.25 Note: The sum of all of the conditional distributions must equal one. In this case, 0.5 + 0.25 + 0.25 = 1; therefore, the solution “checks”. ### Histograms, Frequency Polygons, and Time Series Graphs For most of the work you do in this book, you will use a histogram to display the data. One advantage of a histogram is that it can readily display large data sets. A rule of thumb is to use a histogram when the data set consists of 100 values or more. A histogram consists of contiguous (adjoining) boxes. It has both a horizontal axis and a vertical axis. The horizontal axis is labeled with what the data represents (for instance, distance from your home to school). The vertical axis is labeled either frequency or relative frequency (or percent frequency or probability). The graph will have the same shape with either label. The histogram (like the stemplot) can give you the shape of the data, the center, and the spread of the data. The relative frequency is equal to the frequency for an observed value of the data divided by the total number of data values in the sample.(Remember, frequency is defined as the number of times an answer occurs.) If: • f = frequency • n = total number of data values (or the sum of the individual frequencies), and • RF = relative frequency, then: For example, if three students in Mr. Ahab’s English class of 40 students received from 90% to 100%, then, <!–<newline count=”1″/>–>f = 3, n = 40, and RF = = = 0.075. 7.5% of the students received 90–100%. 90–100% are quantitative measures. To construct a histogram, first decide how many bars or intervals, also called classes, represent the data. Many histograms consist of five to 15 bars or classes for clarity. The number of bars needs to be chosen. Choose a starting point for the first interval to be less than the smallest data value. A convenient starting point is a lower value carried out to one more decimal place than the value with the most decimal places. For example, if the value with the most decimal places is 6.1 and this is the smallest value, a convenient starting point is 6.05 (6.1 – 0.05 = 6.05). We say that 6.05 has more precision. If the value with the most decimal places is 2.23 and the lowest value is 1.5, a convenient starting point is 1.495 (1.5 – 0.005 = 1.495). If the value with the most decimal places is 3.234 and the lowest value is 1.0, a convenient starting point is 0.9995 (1.0 – 0.0005 = 0.9995). If all the data happen to be integers and the smallest value is two, then a convenient starting point is 1.5 (2 – 0.5 = 1.5). Also, when the starting point and other boundaries are carried to one additional decimal place, no data value will fall on a boundary. The next two examples go into detail about how to construct a histogram using continuous data and how to create a histogram using discrete data. The following data are the heights (in inches to the nearest half inch) of 100 male semiprofessional soccer players. The heights are continuous data, since height is measured. 60; 60.5; 61; 61; 61.5 63.5; 63.5; 63.5 64; 64; 64; 64; 64; 64; 64; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5 66; 66; 66; 66; 66; 66; 66; 66; 66; 66; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5 68; 68; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69.5; 69.5; 69.5; 69.5; 69.5 70; 70; 70; 70; 70; 70; 70.5; 70.5; 70.5; 71; 71; 71 72; 72; 72; 72.5; 72.5; 73; 73.5 74 The smallest data value is 60. Since the data with the most decimal places has one decimal (for instance, 61.5), we want our starting point to have two decimal places. Since the numbers 0.5, 0.05, 0.005, etc. are convenient numbers, use 0.05 and subtract it from 60, the smallest value, for the convenient starting point. 60 – 0.05 = 59.95 which is more precise than, say, 61.5 by one decimal place. The starting point is, then, 59.95. The largest value is 74, so 74 + 0.05 = 74.05 is the ending value. Next, calculate the width of each bar or class interval. To calculate this width, subtract the starting point from the ending value and divide by the number of bars (you must choose the number of bars you desire). Suppose you choose eight bars. NOTE We will round up to two and make each bar or class interval two units wide. Rounding up to two is one way to prevent a value from falling on a boundary. Rounding to the next number is often necessary even if it goes against the standard rules of rounding. For this example, using 1.76 as the width would also work. A guideline that is followed by some for the width of a bar or class interval is to take the square root of the number of data values and then round to the nearest whole number, if necessary. For example, if there are 150 values of data, take the square root of 150 and round to 12 bars or intervals. The boundaries are: • 59.95 • 59.95 + 2 = 61.95 • 61.95 + 2 = 63.95 • 63.95 + 2 = 65.95 • 65.95 + 2 = 67.95 • 67.95 + 2 = 69.95 • 69.95 + 2 = 71.95 • 71.95 + 2 = 73.95 • 73.95 + 2 = 75.95 The heights 60 through 61.5 inches are in the interval 59.95–61.95. The heights that are 63.5 are in the interval 61.95–63.95. The heights that are 64 through 64.5 are in the interval 63.95–65.95. The heights 66 through 67.5 are in the interval 65.95–67.95. The heights 68 through 69.5 are in the interval 67.95–69.95. The heights 70 through 71 are in the interval 69.95–71.95. The heights 72 through 73.5 are in the interval 71.95–73.95. The height 74 is in the interval 73.95–75.95. The following histogram displays the heights on the x-axis and relative frequency on the y-axis. Try It The following data are the shoe sizes of 50 male students. The sizes are continuous data since shoe size is measured. Construct a histogram and calculate the width of each bar or class interval. Suppose you choose six bars. 9; 9; 9.5; 9.5; 10; 10; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5 12; 12; 12; 12; 12; 12; 12; 12.5; 12.5; 12.5; 12.5; 14 Smallest value: 9 Largest value: 14 Convenient starting value: 9 – 0.05 = 8.95 Convenient ending value: 14 + 0.05 = 14.05 The calculations suggests using 0.85 as the width of each bar or class interval. You can also use an interval with a width equal to one. Create a histogram for the following data: the number of books bought by 50 part-time college students at ABC College. The number of books is discrete data, since books are counted. 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1 2; 2; 2; 2; 2; 2; 2; 2; 2; 2 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3 4; 4; 4; 4; 4; 4 5; 5; 5; 5; 5 6; 6 Because the data are integers, subtract 0.5 from 1, the smallest data value and add 0.5 to 6, the largest data value. Then the starting point is 0.5 and the ending value is 6.5. Next, calculate the width of each bar or class interval. If the data are discrete and there are not too many different values, a width that places the data values in the middle of the bar or class interval is the most convenient. Since the data consist of the numbers 1, 2, 3, 4, 5, 6, and the starting point is 0.5, a width of one places the 1 in the middle of the interval from 0.5 to 1.5, the 2 in the middle of the interval from 1.5 to 2.5, the 3 in the middle of the interval from 2.5 to 3.5, the 4 in the middle of the interval from _______ to _______, the 5 in the middle of the interval from _______ to _______, and the _______ in the middle of the interval from _______ to _______ . • 3.5 to 4.5 • 4.5 to 5.5 • 6 • 5.5 to 6.5 Calculate the number of bars as follows: where 1 is the width of a bar. Therefore, bars = 6. The following histogram displays the number of books on the x-axis and the frequency on the y-axis. Using this data set, construct a histogram. Number of hours my classmates spent playing video games on weekends 9.95 10 2.25 16.75 0 19.5 22.5 7.5 15 12.75 5.5 11 10 20.75 17.5 23 21.9 24 23.75 18 20 15 22.9 18.8 20.5 Some values in this data set fall on boundaries for the class intervals. A value is counted in a class interval if it falls on the left boundary, but not if it falls on the right boundary. Different researchers may set up histograms for the same data in different ways. There is more than one correct way to set up a histogram. #### Frequency Polygons Frequency polygons are analogous to line graphs, and just as line graphs make continuous data visually easy to interpret, so too do frequency polygons. To construct a frequency polygon, first examine the data and decide on the number of intervals, or class intervals, to use on the x-axis and y-axis. After choosing the appropriate ranges, begin plotting the data points. After all the points are plotted, draw line segments to connect them. A frequency polygon was constructed from the frequency table below. Frequency distribution for calculus final test scores Lower bound Upper bound Frequency Cumulative frequency 49.5 59.5 5 5 59.5 69.5 10 15 69.5 79.5 30 45 79.5 89.5 40 85 89.5 99.5 15 100 The first label on the x-axis is 44.5. This represents an interval extending from 39.5 to 49.5. Since the lowest test score is 54.5, this interval is used only to allow the graph to touch the x-axis. The point labeled 54.5 represents the next interval, or the first “real” interval from the table, and contains five scores. This reasoning is followed for each of the remaining intervals with the point 104.5 representing the interval from 99.5 to 109.5. Again, this interval contains no data and is only used so that the graph will touch the x-axis. Looking at the graph, we say that this distribution is skewed because one side of the graph does not mirror the other side. Try It Construct a frequency polygon of U.S. Presidents’ ages at inauguration shown in (Figure). Age at inauguration Frequency 41.5–46.5 4 46.5–51.5 11 51.5–56.5 14 56.5–61.5 9 61.5–66.5 4 66.5–71.5 2 The first label on the x-axis is 39. This represents an interval extending from 36.5 to 41.5. Since there are no ages less than 41.5, this interval is used only to allow the graph to touch the x-axis. The point labeled 44 represents the next interval, or the first “real” interval from the table, and contains four scores. This reasoning is followed for each of the remaining intervals with the point 74 representing the interval from 71.5 to 76.5. Again, this interval contains no data and is only used so that the graph will touch the x-axis. Looking at the graph, we say that this distribution is skewed because one side of the graph does not mirror the other side. Frequency polygons are useful for comparing distributions. This is achieved by overlaying the frequency polygons drawn for different data sets. We will construct an overlay frequency polygon comparing the scores from (Figure) with the students’ final numeric grade. Frequency distribution for calculus final test scores Lower bound Upper bound Frequency Cumulative frequency 49.5 59.5 5 5 59.5 69.5 10 15 69.5 79.5 30 45 79.5 89.5 40 85 89.5 99.5 15 100 Frequency distribution for calculus final grades Lower bound Upper bound Frequency Cumulative frequency 49.5 59.5 10 10 59.5 69.5 10 20 69.5 79.5 30 50 79.5 89.5 45 95 89.5 99.5 5 100 #### Constructing a Time Series Graph Suppose that we want to study the temperature range of a region for an entire month. Every day at noon we note the temperature and write this down in a log. A variety of statistical studies could be done with these data. We could find the mean or the median temperature for the month. We could construct a histogram displaying the number of days that temperatures reach a certain range of values. However, all of these methods ignore a portion of the data that we have collected. One feature of the data that we may want to consider is that of time. Since each date is paired with the temperature reading for the day, we don‘t have to think of the data as being random. We can instead use the times given to impose a chronological order on the data. A graph that recognizes this ordering and displays the changing temperature as the month progresses is called a time series graph. To construct a time series graph, we must look at both pieces of our paired data set. We start with a standard Cartesian coordinate system. The horizontal axis is used to plot the date or time increments, and the vertical axis is used to plot the values of the variable that we are measuring. By doing this, we make each point on the graph correspond to a date and a measured quantity. The points on the graph are typically connected by straight lines in the order in which they occur. The following data shows the Annual Consumer Price Index, each month, for ten years. Construct a time series graph for the Annual Consumer Price Index data only. Year Jan Feb Mar Apr May Jun Jul 2003 181.7 183.1 184.2 183.8 183.5 183.7 183.9 2004 185.2 186.2 187.4 188.0 189.1 189.7 189.4 2005 190.7 191.8 193.3 194.6 194.4 194.5 195.4 2006 198.3 198.7 199.8 201.5 202.5 202.9 203.5 2007 202.416 203.499 205.352 206.686 207.949 208.352 208.299 2008 211.080 211.693 213.528 214.823 216.632 218.815 219.964 2009 211.143 212.193 212.709 213.240 213.856 215.693 215.351 2010 216.687 216.741 217.631 218.009 218.178 217.965 218.011 2011 220.223 221.309 223.467 224.906 225.964 225.722 225.922 2012 226.665 227.663 229.392 230.085 229.815 229.478 229.104 Year Aug Sep Oct Nov Dec Annual 2003 184.6 185.2 185.0 184.5 184.3 184.0 2004 189.5 189.9 190.9 191.0 190.3 188.9 2005 196.4 198.8 199.2 197.6 196.8 195.3 2006 203.9 202.9 201.8 201.5 201.8 201.6 2007 207.917 208.490 208.936 210.177 210.036 207.342 2008 219.086 218.783 216.573 212.425 210.228 215.303 2009 215.834 215.969 216.177 216.330 215.949 214.537 2010 218.312 218.439 218.711 218.803 219.179 218.056 2011 226.545 226.889 226.421 226.230 225.672 224.939 2012 230.379 231.407 231.317 230.221 229.601 229.594 Try It The following table is a portion of a data set from www.worldbank.org. Use the table to construct a time series graph for CO2 emissions for the United States. CO2 emissions Year Ukraine United Kingdom United States 2003 352,259 540,640 5,681,664 2004 343,121 540,409 5,790,761 2005 339,029 541,990 5,826,394 2006 327,797 542,045 5,737,615 2007 328,357 528,631 5,828,697 2008 323,657 522,247 5,656,839 2009 272,176 474,579 5,299,563 ### Uses of a Time Series Graph Time series graphs are important tools in various applications of statistics. When recording values of the same variable over an extended period of time, sometimes it is difficult to discern any trend or pattern. However, once the same data points are displayed graphically, some features jump out. Time series graphs make trends easy to spot. ### How NOT to Lie with Statistics It is important to remember that the very reason we develop a variety of methods to present data is to develop insights into the subject of what the observations represent. We want to get a “sense” of the data. Are the observations all very much alike or are they spread across a wide range of values, are they bunched at one end of the spectrum or are they distributed evenly and so on. We are trying to get a visual picture of the numerical data. Shortly we will develop formal mathematical measures of the data, but our visual graphical presentation can say much. It can, unfortunately, also say much that is distracting, confusing and simply wrong in terms of the impression the visual leaves. Many years ago Darrell Huff wrote the book How to Lie with Statistics. It has been through 25 plus printings and sold more than one and one-half million copies. His perspective was a harsh one and used many actual examples that were designed to mislead. He wanted to make people aware of such deception, but perhaps more importantly to educate so that others do not make the same errors inadvertently. Again, the goal is to enlighten with visuals that tell the story of the data. Pie charts have a number of common problems when used to convey the message of the data. Too many pieces of the pie overwhelm the reader. More than perhaps five or six categories ought to give an idea of the relative importance of each piece. This is after all the goal of a pie chart, what subset matters most relative to the others. If there are more components than this then perhaps an alternative approach would be better or perhaps some can be consolidated into an “other” category. Pie charts cannot show changes over time, although we see this attempted all too often. In federal, state, and city finance documents pie charts are often presented to show the components of revenue available to the governing body for appropriation: income tax, sales tax motor vehicle taxes and so on. In and of itself this is interesting information and can be nicely done with a pie chart. The error occurs when two years are set side-by-side. Because the total revenues change year to year, but the size of the pie is fixed, no real information is provided and the relative size of each piece of the pie cannot be meaningfully compared. Histograms can be very helpful in understanding the data. Properly presented, they can be a quick visual way to present probabilities of different categories by the simple visual of comparing relative areas in each category. Here the error, purposeful or not, is to vary the width of the categories. This of course makes comparison to the other categories impossible. It does embellish the importance of the category with the expanded width because it has a greater area, inappropriately, and thus visually “says” that that category has a higher probability of occurrence. Time series graphs perhaps are the most abused. A plot of some variable across time should never be presented on axes that change part way across the page either in the vertical or horizontal dimension. Perhaps the time frame is changed from years to months. Perhaps this is to save space or because monthly data was not available for early years. In either case this confounds the presentation and destroys any value of the graph. If this is not done to purposefully confuse the reader, then it certainly is either lazy or sloppy work. Changing the units of measurement of the axis can smooth out a drop or accentuate one. If you want to show large changes, then measure the variable in small units, penny rather than thousands of dollars. And of course to continue the fraud, be sure that the axis does not begin at zero, zero. If it begins at zero, zero, then it becomes apparent that the axis has been manipulated. Perhaps you have a client that is concerned with the volatility of the portfolio you manage. An easy way to present the data is to use long time periods on the time series graph. Use months or better, quarters rather than daily or weekly data. If that doesn’t get the volatility down then spread the time axis relative to the rate of return or portfolio valuation axis. If you want to show “quick” dramatic growth, then shrink the time axis. Any positive growth will show visually “high” growth rates. Do note that if the growth is negative then this trick will show the portfolio is collapsing at a dramatic rate. Again, the goal of descriptive statistics is to convey meaningful visuals that tell the story of the data. Purposeful manipulation is fraud and unethical at the worst, but even at its best, making these type of errors will lead to confusion on the part of the analysis. ### References Burbary, Ken. Facebook Demographics Revisited – 2001 Statistics, 2011. Available online at http://www.kenburbary.com/2011/03/facebook-demographics-revisited-2011-statistics-2/ (accessed August 21, 2013). “9th Annual AP Report to the Nation.” CollegeBoard, 2013. Available online at http://apreport.collegeboard.org/goals-and-findings/promoting-equity (accessed September 13, 2013). “Overweight and Obesity: Adult Obesity Facts.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/obesity/data/adult.html (accessed September 13, 2013). Data on annual homicides in Detroit, 1961–73, from Gunst & Mason’s book ‘Regression Analysis and its Application’, Marcel Dekker “Timeline: Guide to the U.S. Presidents: Information on every president’s birthplace, political party, term of office, and more.” Scholastic, 2013. Available online at http://www.scholastic.com/teachers/article/timeline-guide-us-presidents (accessed April 3, 2013). “Presidents.” Fact Monster. Pearson Education, 2007. Available online at http://www.factmonster.com/ipka/A0194030.html (accessed April 3, 2013). “Food Security Statistics.” Food and Agriculture Organization of the United Nations. Available online at http://www.fao.org/economic/ess/ess-fs/en/ (accessed April 3, 2013). “Consumer Price Index.” United States Department of Labor: Bureau of Labor Statistics. Available online at http://data.bls.gov/pdq/SurveyOutputServlet (accessed April 3, 2013). “CO2 emissions (kt).” The World Bank, 2013. Available online at http://databank.worldbank.org/data/home.aspx (accessed April 3, 2013). “Births Time Series Data.” General Register Office For Scotland, 2013. Available online at http://www.gro-scotland.gov.uk/statistics/theme/vital-events/births/time-series.html (accessed April 3, 2013). “Demographics: Children under the age of 5 years underweight.” Indexmundi. Available online at http://www.indexmundi.com/g/r.aspx?t=50&v=2224&aml=en (accessed April 3, 2013). Gunst, Richard, Robert Mason. Regression Analysis and Its Application: A Data-Oriented Approach. CRC Press: 1980. “Overweight and Obesity: Adult Obesity Facts.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/obesity/data/adult.html (accessed September 13, 2013). ### Chapter Review A stem-and-leaf plot is a way to plot data and look at the distribution. In a stem-and-leaf plot, all data values within a class are visible. The advantage in a stem-and-leaf plot is that all values are listed, unlike a histogram, which gives classes of data values. A line graph is often used to represent a set of data values in which a quantity varies with time. These graphs are useful for finding trends. That is, finding a general pattern in data sets including temperature, sales, employment, company profit or cost over a period of time. A bar graph is a chart that uses either horizontal or vertical bars to show comparisons among categories. One axis of the chart shows the specific categories being compared, and the other axis represents a discrete value. Some bar graphs present bars clustered in groups of more than one (grouped bar graphs), and others show the bars divided into subparts to show cumulative effect (stacked bar graphs). Bar graphs are especially useful when categorical data is being used. A histogram is a graphic version of a frequency distribution. The graph consists of bars of equal width drawn adjacent to each other. The horizontal scale represents classes of quantitative data values and the vertical scale represents frequencies. The heights of the bars correspond to frequency values. Histograms are typically used for large, continuous, quantitative data sets. A frequency polygon can also be used when graphing large data sets with data points that repeat. The data usually goes on y-axis with the frequency being graphed on the x-axis. Time series graphs can be helpful when looking at large amounts of data for one variable over a period of time. ### For the next three exercises, use the data to construct a line graph.In a survey, 40 people were asked how many times they visited a store before making a major purchase. The results are shown in (Figure).Number of times in storeFrequency142103164654 In a survey, several people were asked how many years it has been since they purchased a mattress. The results are shown in (Figure). Years since last purchase Frequency 0 2 1 8 2 13 3 22 4 16 5 9 Several children were asked how many TV shows they watch each day. The results of the survey are shown in (Figure). Number of TV shows Frequency 0 12 1 18 2 36 3 7 4 2 The students in Ms. Ramirez’s math class have birthdays in each of the four seasons. (Figure) shows the four seasons, the number of students who have birthdays in each season, and the percentage (%) of students in each group. Construct a bar graph showing the number of students. Seasons Number of students Proportion of population Spring 8 24% Summer 9 26% Autumn 11 32% Winter 6 18% Using the data from Mrs. Ramirez’s math class supplied in (Figure), construct a bar graph showing the percentages. David County has six high schools. Each school sent students to participate in a county-wide science competition. (Figure) shows the percentage breakdown of competitors from each school, and the percentage of the entire student population of the county that goes to each school. Construct a bar graph that shows the population percentage of competitors from each school. High school Science competition population Overall student population Alabaster 28.9% 8.6% Concordia 7.6% 23.2% Genoa 12.1% 15.0% Mocksville 18.5% 14.3% Tynneson 24.2% 10.1% West End 8.7% 28.8% Use the data from the David County science competition supplied in (Figure). Construct a bar graph that shows the county-wide population percentage of students at each school. Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Complete the table. Data value (# cars) Frequency Relative frequency Cumulative relative frequency What does the frequency column in (Figure) sum to? Why? 65 What does the relative frequency column in (Figure) sum to? Why? What is the difference between relative frequency and frequency for each data value in (Figure)? The relative frequency shows the proportion of data points that have each value. The frequency tells the number of data points that have each value. What is the difference between cumulative relative frequency and relative frequency for each data value? To construct the histogram for the data in (Figure), determine appropriate minimum and maximum x and y values and the scaling. Sketch the histogram. Label the horizontal and vertical axes with words. Include numerical scaling. Answers will vary. One possible histogram is shown: Construct a frequency polygon for the following: 1. Pulse rates for women Frequency 60–69 12 70–79 14 80–89 11 90–99 1 100–109 1 110–119 0 120–129 1 2. Actual speed in a 30 MPH zone Frequency 42–45 25 46–49 14 50–53 7 54–57 3 58–61 1 3. Tar (mg) in nonfiltered cigarettes Frequency 10–13 1 14–17 0 18–21 15 22–25 7 26–29 2 Construct a frequency polygon from the frequency distribution for the 50 highest ranked countries for depth of hunger. Depth of hunger Frequency 230–259 21 260–289 13 290–319 5 320–349 7 350–379 1 380–409 1 410–439 1 Find the midpoint for each class. These will be graphed on the x-axis. The frequency values will be graphed on the y-axis values. Use the two frequency tables to compare the life expectancy of men and women from 20 randomly selected countries. Include an overlayed frequency polygon and discuss the shapes of the distributions, the center, the spread, and any outliers. What can we conclude about the life expectancy of women compared to men? Life expectancy at birth – women Frequency 49–55 3 56–62 3 63–69 1 70–76 3 77–83 8 84–90 2 Life expectancy at birth – men Frequency 49–55 3 56–62 3 63–69 1 70–76 1 77–83 7 84–90 5 Construct a times series graph for (a) the number of male births, (b) the number of female births, and (c) the total number of births. Sex/Year 1855 1856 1857 1858 1859 1860 1861 Female 45,545 49,582 50,257 50,324 51,915 51,220 52,403 Male 47,804 52,239 53,158 53,694 54,628 54,409 54,606 Total 93,349 101,821 103,415 104,018 106,543 105,629 107,009 Sex/Year 1862 1863 1864 1865 1866 1867 1868 1869 Female 51,812 53,115 54,959 54,850 55,307 55,527 56,292 55,033 Male 55,257 56,226 57,374 58,220 58,360 58,517 59,222 58,321 Total 107,069 109,341 112,333 113,070 113,667 114,044 115,514 113,354 Sex/Year 1870 1871 1872 1873 1874 1875 Female 56,431 56,099 57,472 58,233 60,109 60,146 Male 58,959 60,029 61,293 61,467 63,602 63,432 Total 115,390 116,128 118,765 119,700 123,711 123,578 The following data sets list full time police per 100,000 citizens along with homicides per 100,000 citizens for the city of Detroit, Michigan during the period from 1961 to 1973. Year 1961 1962 1963 1964 1965 1966 1967 Police 260.35 269.8 272.04 272.96 272.51 261.34 268.89 Homicides 8.6 8.9 8.52 8.89 13.07 14.57 21.36 Year 1968 1969 1970 1971 1972 1973 Police 295.99 319.87 341.43 356.59 376.69 390.19 Homicides 28.03 31.49 37.39 46.26 47.24 52.33 1. Construct a double time series graph using a common x-axis for both sets of data. 2. Which variable increased the fastest? Explain. 3. Did Detroit’s increase in police officers have an impact on the murder rate? Explain. ### Homework (Figure) contains the 2010 obesity rates in U.S. states and Washington, DC. State Percent (%) State Percent (%) State Percent (%) Alabama 32.2 Kentucky 31.3 North Dakota 27.2 Alaska 24.5 Louisiana 31.0 Ohio 29.2 Arizona 24.3 Maine 26.8 Oklahoma 30.4 Arkansas 30.1 Maryland 27.1 Oregon 26.8 California 24.0 Massachusetts 23.0 Pennsylvania 28.6 Colorado 21.0 Michigan 30.9 Rhode Island 25.5 Connecticut 22.5 Minnesota 24.8 South Carolina 31.5 Delaware 28.0 Mississippi 34.0 South Dakota 27.3 Washington, DC 22.2 Missouri 30.5 Tennessee 30.8 Florida 26.6 Montana 23.0 Texas 31.0 Georgia 29.6 Nebraska 26.9 Utah 22.5 Hawaii 22.7 Nevada 22.4 Vermont 23.2 Idaho 26.5 New Hampshire 25.0 Virginia 26.0 Illinois 28.2 New Jersey 23.8 Washington 25.5 Indiana 29.6 New Mexico 25.1 West Virginia 32.5 Iowa 28.4 New York 23.9 Wisconsin 26.3 Kansas 29.4 North Carolina 27.8 Wyoming 25.1 1. Use a random number generator to randomly pick eight states. Construct a bar graph of the obesity rates of those eight states. 2. Construct a bar graph for all the states beginning with the letter “A.” 3. Construct a bar graph for all the states beginning with the letter “M.” 1. Example solution for using the random number generator for the TI-84+ to generate a simple random sample of 8 states. Instructions are as follows. • Number the entries in the table 1–51 (Includes Washington, DC; Numbered vertically) • Press MATH • Arrow over to PRB • Press 5:randInt( • Enter 51,1,8) Eight numbers are generated (use the right arrow key to scroll through the numbers). The numbers correspond to the numbered states (for this example: {47 21 9 23 51 13 25 4}. If any numbers are repeated, generate a different number by using 5:randInt(51,1)). Here, the states (and Washington DC) are {Arkansas, Washington DC, Idaho, Maryland, Michigan, Mississippi, Virginia, Wyoming}. Corresponding percents are {30.1, 22.2, 26.5, 27.1, 30.9, 34.0, 26.0, 25.1}. Suppose that three book publishers were interested in the number of fiction paperbacks adult consumers purchase per month. Each publisher conducted a survey. In the survey, adult consumers were asked the number of fiction paperbacks they had purchased the previous month. The results are as follows: Publisher A # of books Freq. Rel. freq. 0 10 1 12 2 16 3 12 4 8 5 6 6 2 8 2 Publisher B # of books Freq. Rel. freq. 0 18 1 24 2 24 3 22 4 15 5 10 7 5 9 1 Publisher C # of books Freq. Rel. freq. 0–1 20 2–3 35 4–5 12 6–7 2 8–9 1 1. Find the relative frequencies for each survey. Write them in the charts. 2. Use the frequency column to construct a histogram for each publisher’s survey. For Publishers A and B, make bar widths of one. For Publisher C, make bar widths of two. 3. In complete sentences, give two reasons why the graphs for Publishers A and B are not identical. 4. Would you have expected the graph for Publisher C to look like the other two graphs? Why or why not? 5. Make new histograms for Publisher A and Publisher B. This time, make bar widths of two. 6. Now, compare the graph for Publisher C to the new graphs for Publishers A and B. Are the graphs more similar or more different? Explain your answer. Often, cruise ships conduct all on-board transactions, with the exception of gambling, on a cashless basis. At the end of the cruise, guests pay one bill that covers all onboard transactions. Suppose that 60 single travelers and 70 couples were surveyed as to their on-board bills for a seven-day cruise from Los Angeles to the Mexican Riviera. Following is a summary of the bills for each group. Singles Amount(?) Frequency Rel. frequency 51–100 5 101–150 10 151–200 15 201–250 15 251–300 10 301–350 5 Couples Amount(?) Frequency Rel. frequency 100–150 5 201–250 5 251–300 5 301–350 5 351–400 10 401–450 10 451–500 10 501–550 10 551–600 5 601–650 5 1. Fill in the relative frequency for each group. 2. Construct a histogram for the singles group. Scale the x-axis by ?50 widths. Use relative frequency on the y-axis. 3. Construct a histogram for the couples group. Scale the x-axis by ?50 widths. Use relative frequency on the y-axis. 4. Compare the two graphs: 1. List two similarities between the graphs. 2. List two differences between the graphs. 3. Overall, are the graphs more similar or different? 5. Construct a new graph for the couples by hand. Since each couple is paying for two individuals, instead of scaling the x-axis by ?50, scale it by ?100. Use relative frequency on the y-axis. 6. Compare the graph for the singles with the new graph for the couples: 1. List two similarities between the graphs. 2. Overall, are the graphs more similar or different? 7. How did scaling the couples graph differently change the way you compared it to the singles graph? 8. Based on the graphs, do you think that individuals spend the same amount, more or less, as singles as they do person by person as a couple? Explain why in one or two complete sentences. Singles Amount(?) Frequency Relative frequency 51–100 5 0.08 101–150 10 0.17 151–200 15 0.25 201–250 15 0.25 251–300 10 0.17 301–350 5 0.08 Couples Amount(?) Frequency Relative frequency 100–150 5 0.07 201–250 5 0.07 251–300 5 0.07 301–350 5 0.07 351–400 10 0.14 401–450 10 0.14 451–500 10 0.14 501–550 10 0.14 551–600 5 0.07 601–650 5 0.07 1. See (Figure) and (Figure). 2. In the following histogram data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where both boundary values are included). 3. In the following histogram, the data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where values on both boundaries are included). 4. Compare the two graphs: • Both graphs have a single peak. • Both graphs use class intervals with width equal to ?50. • The couples graph has a class interval with no values. • It takes almost twice as many class intervals to display the data for couples. 3. Answers may vary. Possible answers include: The graphs are more similar than different because the overall patterns for the graphs are the same. 5. Check student’s solution. 6. Compare the graph for the Singles with the new graph for the Couples: • Both graphs have a single peak. • Both graphs display 6 class intervals. • Both graphs show the same general pattern. 1. Answers may vary. Possible answers include: Although the width of the class intervals for couples is double that of the class intervals for singles, the graphs are more similar than they are different. 7. Answers may vary. Possible answers include: You are able to compare the graphs interval by interval. It is easier to compare the overall patterns with the new scale on the Couples graph. Because a couple represents two individuals, the new scale leads to a more accurate comparison. 8. Answers may vary. Possible answers include: Based on the histograms, it seems that spending does not vary much from singles to individuals who are part of a couple. The overall patterns are the same. The range of spending for couples is approximately double the range for individuals. Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows. # of movies Frequency Relative frequency Cumulative relative frequency 0 5 1 9 2 6 3 4 4 1 1. Construct a histogram of the data. 2. Complete the columns of the chart. Use the following information to answer the next two exercises: Suppose one hundred eleven people who shopped in a special t-shirt store were asked the number of t-shirts they own costing more than ?19 each. The percentage of people who own at most three t-shirts costing more than ?19 each is approximately: 1. 21 2. 59 3. 41 4. Cannot be determined c If the data were collected by asking the first 111 people who entered the store, then the type of sampling is: 1. cluster 2. simple random 3. stratified 4. convenience Following are the 2010 obesity rates by U.S. states and Washington, DC. State Percent (%) State Percent (%) State Percent (%) Alabama 32.2 Kentucky 31.3 North Dakota 27.2 Alaska 24.5 Louisiana 31.0 Ohio 29.2 Arizona 24.3 Maine 26.8 Oklahoma 30.4 Arkansas 30.1 Maryland 27.1 Oregon 26.8 California 24.0 Massachusetts 23.0 Pennsylvania 28.6 Colorado 21.0 Michigan 30.9 Rhode Island 25.5 Connecticut 22.5 Minnesota 24.8 South Carolina 31.5 Delaware 28.0 Mississippi 34.0 South Dakota 27.3 Washington, DC 22.2 Missouri 30.5 Tennessee 30.8 Florida 26.6 Montana 23.0 Texas 31.0 Georgia 29.6 Nebraska 26.9 Utah 22.5 Hawaii 22.7 Nevada 22.4 Vermont 23.2 Idaho 26.5 New Hampshire 25.0 Virginia 26.0 Illinois 28.2 New Jersey 23.8 Washington 25.5 Indiana 29.6 New Mexico 25.1 West Virginia 32.5 Iowa 28.4 New York 23.9 Wisconsin 26.3 Kansas 29.4 North Carolina 27.8 Wyoming 25.1 Construct a bar graph of obesity rates of your state and the four states closest to your state. Hint: Label the x-axis with the states. ### Key Terms Frequency the number of times a value of the data occurs Histogram a graphical representation in xy form of the distribution of data in a data set; x represents the data and y represents the frequency, or relative frequency. The graph consists of contiguous rectangles. Relative Frequency the ratio of the number of times a value of the data occurs in the set of all outcomes to the number of all outcomes
# Divide By 7 : Basic Division Test! Trivia Quiz Approved & Edited by ProProfs Editorial Team The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes. \| By Llshepler L Llshepler Community Contributor Quizzes Created: 33 \| Total Attempts: 14,860 Questions: 13 \| Attempts: 270 Settings This is a 1 minute timed divided by 7 test. You must get 100% accuracy in order to move on to the next division test. • 1. ### 0 ÷ 7 = Explanation When dividing any number by 0, the result is always 0. This is because division can be thought of as the process of sharing or distributing a certain quantity equally among a given number of groups. In this case, we are trying to distribute 0 into 7 groups. Since there is nothing to distribute, each group will have 0. Therefore, the answer to 0 divided by 7 is 0. Rate this question: • 2. • 3. ### 14 ÷ 7 = Explanation The given equation 14 ÷ 7 = 2 is a division problem. When 14 is divided by 7, it results in 2. This means that if we distribute 14 equally into 7 groups, each group will have 2. Therefore, the answer to the equation is 2. Rate this question: • 4. ### 21 ÷ 7 = Explanation The given equation is a division problem, where 21 is divided by 7. When we divide 21 by 7, we get 3 as the quotient. This means that if we distribute 21 equally among 7 groups, each group will have 3. Therefore, the answer to the equation is 3. Rate this question: • 5. ### 28 ÷ 7 = Explanation The equation 28 ÷ 7 means dividing 28 by 7. When we divide 28 by 7, we get the quotient of 4. This means that if we distribute 28 equally into 7 groups, each group will have 4. Therefore, the answer to the equation is 4. Rate this question: • 6. ### 35 ÷ 7 = Explanation The division symbol "÷" represents the operation of dividing. In this case, we are dividing 35 by 7. When we divide 35 by 7, we get the quotient of 5. Therefore, the answer is 5. Rate this question: • 7. ### 42 ÷ 7 = Explanation The given equation is a division problem. When we divide 42 by 7, we are essentially asking how many times 7 can go into 42. Since 7 can go into 42 exactly 6 times (7 x 6 = 42), the answer is 6. Rate this question: • 8. ### 49 ÷ 7 = Explanation The given equation 49 ÷ 7 is asking for the quotient when 49 is divided by 7. Since 7 goes into 49 evenly, the answer is 7. Rate this question: • 9. ### 56 ÷ 7 = Explanation The given question is asking to divide 56 by 7. When we divide 56 by 7, we get the quotient as 8. Therefore, the correct answer is 8. Rate this question: • 10. ### 63 ÷ 7 = Explanation The given question is asking for the result of dividing 63 by 7. When we divide 63 by 7, we get the quotient of 9. Therefore, the correct answer is 9. Rate this question: • 11. ### 70 ÷ 7 = Explanation The division symbol "÷" indicates that we need to divide the number 70 by the number 7. When we divide 70 by 7, we get the quotient of 10. Therefore, the answer to the given question is 10. Rate this question: • 12. ### 77 ÷ 7 = Explanation The given equation 77 ÷ 7 can be solved by dividing 77 by 7, which equals 11. Rate this question: • 13. ### 84 ÷ 7 = Explanation The given equation is asking for the result of dividing 84 by 7. When we divide 84 by 7, we get 12. Rate this question: Quiz Review Timeline + Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness. • Current Version • Mar 20, 2023 Quiz Edited by ProProfs Editorial Team • Mar 19, 2009 Quiz Created by Llshepler Related Topics × Wait! Here's an interesting quiz for you.
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ## Algebra (all content) ### Course: Algebra (all content)>Unit 10 Lesson 10: Introduction to factorization # Intro to factors & divisibility Sal explains what it means for a polynomial to be a factor of another polynomial, and what it means for a polynomial to be divisible by another polynomial. ## Want to join the conversation? • But how did he get 10x in the first place? In the first problem, he only multiplied the similar terms (3 times -2, x times x etc.) But in the second one he multiplied everything • When multiplying binomials, think of it as doing the distributive property. Multiply each term by each term. So x * x = x^2, while 3 * 7 = 21. But, x * 7 =7x, while 3 * x = 3x. So, x^2 +7x + 3x + 21. Simplifying that, you add the 3x to the 7x to equal 10x. The final answer is, x^2 + 10x + 21 • Hi, Just to get some clarity, What is the difference between binomial, polynomial, trinomial, etc? • A monomial is a polynomial with 1 term. A binomial is a polynomial with 2 terms. A trinomial is a polynomial with 3 terms. • can decimals be factors or is it just integers? as an example, you can definitely say that 3 is a factor of 6, but can you say that 2.5 is a factor of 5? • Great question! When we talk about factors of whole numbers, we are looking for whole numbers. 2.5 is not a whole number, so it is not a factor of 5. Negative integers, like -1 and -5, are not factors either, because they are not whole numbers. The only factors of 5 are 1 and 5, making 5 prime. • Why did he put 10x at ?? (1 vote) • When you multiply (x+3)(x+7), you would get x squared plus 7x plus 3x plus 21. You have 3x and 7x, which add together to get 10x. • What exactly is the purpose of factorizing something? • i dont understand • we know that (a+b)(a-b)=a^2-b^2 This means that, a+b is a factor of a^2-b^2, ALSO a^2-b^2 is divisible by a+b but dividing a^2-b^2 by a+b does not give a-b plz help me to understand this concept in term of this formula THANK YOU. (1 vote) • Look, if a=1&b=2, a^2-b^2=-3, and -3/a+b=-3/3=-1, a-b=-1, so (a^2-b^2)/(a+b)=a-b. (Replace unknowns with numbers) • If x^2 + 10x + 21 is a trinomial, then is x^2 + 3x + 7x + 21 a quadrinomial, or still a trinomial, since it can be simplified to the former? • In its expanded form, it is a quadrinomial because it has 4 terms at that point. However, it is still a quadratic expression. • what is the difference between polynomials, and binomials? • A polynomial is simply a fancy word to describe a math statement with several terms. A mononomial is the opposite - it is a statement with only one term. A binomial is a type of polynomial that contains exactly two terms. All of these are pretty simple to remember because they use the same Greek/Latin roots that many other math terms use. (Poly means many, mono means one, and bi means two). Here are some examples of all of these concepts. Polynomial: - 4x^2 + 5x + 6 - x + a + c - 433 + 19b + 17c + 25a Mononomial: - 5x - a - 19 - 7 Binomial: - x + 7 - 4x + 9 - 9a + b Hope that helps! (1 vote) • Is not (3xy)(-2x^2y^3) the same as [(xy) + (xy) + (xy)][(-x^2y^3) + (-x^2y^3)] ?
# Difference between revisions of "2018 AMC 8 Problems/Problem 19" ## Problem 19 In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid? $[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("+",(0,0)); draw(shift(1,0)box); label("-",(1,0)); draw(shift(2,0)box); label("+",(2,0)); draw(shift(3,0)box); label("-",(3,0)); draw(shift(0.5,0.4)box); label("-",(0.5,0.4)); draw(shift(1.5,0.4)box); label("-",(1.5,0.4)); draw(shift(2.5,0.4)box); label("-",(2.5,0.4)); draw(shift(1,0.8)box); label("+",(1,0.8)); draw(shift(2,0.8)box); label("+",(2,0.8)); draw(shift(1.5,1.2)*box); label("+",(1.5,1.2)); [/asy]$ $\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$ ## Solution 1 You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns: +--+, -++-, ----, ++++, -+-+, +-+-, ++--, --++. There are 8 patterns and so the answer is $\boxed{\textbf{(C) } 8}$ -NinjaBoi2000 ## Solution 2 The sign of the next row on the pyramid depends on previous row. There are two options for the previous row, - or +. There are three rows to the pyramid that depend on what the top row is. Therefore, the ways you can make a + on the top is $2^3=8$, so the answer is $\boxed{\textbf{(C) } 8}$ ## Solution 3 There is also a pretty simple approach to this problem. Since in the bottom row you can either have 4 of the same signs, 3 of the same signs and one of another, and 2 of the same signs and one of the other, this can be thought of as the 4th Row of the Pascal’s Triangle, which is $1 4 6 4 1$. Since 3 of one sign and 1 of the other doesn’t work, all you need to add is $1 + 6 + 1 = 8$, so the answer is $\boxed{\textbf{(C) } 8}$
Perimeter and Area Circumference and Area of a Circle Terms Associated with Circles The circumference of a circle is its "perimeter," or the distance around its edge. If we broke the circle and bent it into one flat line, the length of this line would be its circumference: Circumference of a Circle The diameter of a circle is a line segment from one point on the edge of the circle to another point on the edge, passing through the center of the circle. It is the longest line segment that cuts across the circle from one point to another. There are many different diameters, but they all have the same length: Diameters of a Circle The radius of a circle is a line segment from the center of the circle to a point on the edge of the circle. It is half of a diameter, and thus its length is half the length of the diameter. Again, there are many radii, but they all have the same length. In the following diagram, a , b , and c are all radii: The area of a circle is the total number of square units that fill the circle. The area of the following circle is about 13 units. Note that we count fractional units inside the circle as well as whole units. Area of a Circle Formula for the Circumference of a Circle Mathematicians have discovered a special number, called pi (represented by Π ), which is the ratio of the circumference of any circle to the length of its diameter. Π is roughly equal to 3.14--most scientific calculators have a " Π " button that will produce more digits. Π is a non- terminating, non- repeating decimal; thus, Π is an irrational number. Since Π is the ratio of the circumference to the diameter, Π = c/d ; c = Π×d ; and d = c/Π ; where c and d are the circumference and the diameter, respectively. The most important equations to remember are the last two. Thus, to find the circumference of a circle, multiply the diameter by Π . If you know only the radius (a more likely scenario), multiply the radius by 2 to find the diameter: c = 2×Π×r . To find the diameter of a circle, divide the circumference by Π . Use 3.14 for Π . Try it! Find a pan, trash can, or other large circular object. Measure around the edge, and then measure the diameter. The circumference divided by the diameter should be roughly equal to Π . Formula for the Area of a Circle Interestingly enough, Π is also the ratio between the area of a circle and the square of its radius. Thus, Π = A/r 2 ; A = Π×r 2 ; and r = . The most important equation to remember is the middle equation, A = Π×r 2 . Thus, to find the area of a circle, square the radius and multiply by Π . If the radius is unknown but the diameter is known, divide the diameter by 2 to find the radius. Examples: What is the circumference of a circle with diameter 5? c = d×Π = 5×3.14 = 15.7 What is the circumference of a circle with radius 3? d = 3×2 = 6;c = d×Π = 6×3.14 = 18.8 What is the area of a circle with radius 3? A = Π×r 2 = 3.14×32 = 28.3 What is the area of a circle with diameter 5? r = 5/2 = 2.5;A = Π×r 2 = 3.14×2.52 = 19.6 What is the diameter of a circle with circumference 11? d = c/Π = 11/3.14 = 3.50 What is the radius of a circle with circumference 11? r = d /2 = 3.50/2 = 1.75 What is the area of a circle with circumference 11? A = Π×r 2 = 3.14×1.752 = 9.62 Take a Study Break Star Trek gets SEXY Chris Pine and Zoe Saldana heat up the red carpet! Are you afraid of relationships? Auntie SparkNotes can help! Sexy starlet style See every single look from the Met Gala! Geeky Actors: Then and Now Travel back in time! Villains We Want These Actresses to Play From super cute to super bad! 10 Movies Better Than Their Books What do you think? How To Look Like J-Law... When you don't look like J-Law. 12 Scientific Inaccuracies in Into Darkness What did Star Trek get wrong? The Book Read What You Love, Anywhere You Like Get Our FREE NOOK Reading Apps
Solving Linear Diophantine Equations with Linear Congruences # Solving Linear Diophantine Equations with Linear Congruences Recall that a linear diophantine equation is an equation of the form $ax + by = c$ for some $a, b, c \in \mathbb{Z}$ where $(x, y)$ is a solution if $x, y \in \mathbb{Z}$ satisfies this equation. So, let $a, b, c \in \mathbb{Z}$ and consider $ax + by = c$. Then this equation can be rewritten as $ax = c - by$. Thus $ax \mid (c - by)$, i.e.: (1) \begin{align} by \equiv c \pmod a \end{align} Or similarly, we can rewrite $ax + by = c$ as $by = c - ax$. Thus $by \mid (c - ax)$, i.e.: (2) \begin{align} \quad ax \equiv c \pmod b \end{align} Upon solving these linear congruences we will simultaneously solve the corresponding linear diophantine equations. Let's look at an example. Suppose that we want to solve $9x + 10y = 11$. Then, consider the following linear congruence: (3) \begin{align} \quad 9x & \equiv 11 \pmod {10} \\ \quad 9x & \equiv 1 \pmod {10} \end{align} By inspection we see that $9(9) = 81 \equiv 1 \pmod 10$ and so: (4) \begin{align} \quad 9(9x) \equiv 9(1) \pmod {10} \\ \quad x \equiv 9 \pmod {10} \end{align} Thus $x = 9 + 10t$ where $t \in \mathbb{Z}$. Substituting this into our linear diophantine equation gives us: (5) \begin{align} \quad 9(9 + 10t) + 10y &= 11 \\ \quad 81 + 90t + 10y &= 11 \\ \quad y &= -7 - 9t \end{align} So the solutions to $9x + 10y = 11$ are given by $x = 9 + 10t$ and $y = -7 - 9t$ for $t \in \mathbb{Z}$.
# Lesson 7 One Hundred Percent Let’s solve more problems about percent increase and percent decrease. ### 7.1: Notice and Wonder: Double Number Line What do you notice? What do you wonder? ### 7.2: Double Number Lines For each problem, complete the double number line diagram to show the percentages that correspond to the original amount and to the new amount. 1. The gas tank in dad’s car holds 12 gallons. The gas tank in mom’s truck holds 50% more than that. How much gas does the truck’s tank hold? 2. At a movie theater, the size of popcorn bags decreased 20%. If the old bags held 15 cups of popcorn, how much do the new bags hold? 3. A school had 1,200 students last year and only 1,080 students this year. What was the percentage decrease in the number of students? 4. One week gas was \$1.25 per gallon. The next week gas was \$1.50 per gallon. By what percentage did the price increase? 5. After a 25% discount, the price of a T-shirt was \12. What was the price before the discount? 6. Compared to last year, the population of Boom Town has increased 25%. The population is now 6,600. What was the population last year? ### 7.3: Representing More Juice Two students are working on the same problem: A juice box has 20% more juice in its new packaging. The original packaging held 12 fluid ounces. How much juice does the new packaging hold? • Here is how Priya set up her double number line. • Here is how Clare set up her double number line. Do you agree with either of them? Explain or show your reasoning. Clare's diagram could represent a percent decrease. Describe a situation that could be represented with Clare's diagram. ### 7.4: Protecting the Green Sea Turtle Green sea turtles live most of their lives in the ocean, but come ashore to lay their eggs. Some beaches where turtles often come ashore have been made into protected sanctuaries so the eggs will not be disturbed. 1. One sanctuary had 180 green sea turtles come ashore to lay eggs last year. This year, the number of turtles increased by 10%. How many turtles came ashore to lay eggs in the sanctuary this year? 2. At another sanctuary, the number of nesting turtles decreased by 10%. This year there were 234 nesting turtles. How many nesting turtles were at this sanctuary last year? ### Summary We can use a double number line diagram to show information about percent increase and percent decrease: The initial amount of cereal is 500 grams, which is lined up with 100% in the diagram. We can find a 20% increase to 500 by adding 20% of 500: \begin{align}500+(0.2)\boldcdot 500 &= (1.20)\boldcdot 500\\&=600\end{align} In the diagram, we can see that 600 corresponds to 120%. If the initial amount of 500 grams is decreased by 40%, we can find how much cereal there is by subtracting 40% of the 500 grams: \begin{align}500−(0.4)\boldcdot 500 &= (0.6)\boldcdot 500\\&=300\end{align} So a 40% decrease is the same as 60% of the initial amount. In the diagram, we can see that 300 is lined up with 60%. To solve percentage problems, we need to be clear about what corresponds to 100%. For example, suppose there are 20 students in a class, and we know this is an increase of 25% from last year. In this case, the number of students in the class last year corresponds to 100%. So the initial amount (100%) is unknown and the final amount (125%) is 20 students. Looking at the double number line, if 20 students is a 25% increase from the previous year, then there were 16 students in the class last year. ### Glossary Entries • percentage decrease A percentage decrease tells how much a quantity went down, expressed as a percentage of the starting amount. For example, a store had 64 hats in stock on Friday. They had 48 hats left on Saturday. The amount went down by 16. This was a 25% decrease, because 16 is 25% of 64. • percentage increase A percentage increase tell how much a quantity went up, expressed as a percentage of the starting amount. For example, Elena had50 in the bank on Monday. She had $56 on Tuesday. The amount went up by$6. This was a 12% increase, because 6 is 12% of 50.
11 Components Big Ideas Formative Assessment Lesson End of Unit Assessments Browse Components Big Ideas Formative Assessment Lesson End of Unit Assessments Big Idea: The structure of polynomial graphs and equations gives insights into their roots. 1 week Evidence of Understanding • analyze connections between the zeros of the polynomial function and its graph • approximate the zeros of a polynomial function from its graph and verify them algebraically • discover and describe patterns for even or odd multiplicity of zeros • create a polynomial function (in factored form) from the zeros on its graph • verify a zero by algebraically showing that the product of the factors are equivalent to the polynomial’s function rule (in cases where the lead coefficient is one) • for zeros j, k, and p, show that (x - j)(x - k)(x - p) = x3 + bx2 + cx + d • add, subtract, and multiply complex numbers • generalize characteristics about the zeros of any polynomial function and use them to solve problems • make and evaluate predictions for possible roots of a polynomial from its equation • describe patterns and properties relating the quantities in a quadratic, cubic, or quartic equation and its roots • Example: students can discover, show, and explain that if 1 is a root for any quadratic, ax2 + bx + c, then the sum of a, b, and c must be zero • possible extension: discover and use the Rational Zeros Theorem to identify all the possible rational zeros of a function • explain why the Remainder Theorem can prove whether any value is a zero for the function • use long division to verify the roots of a polynomial with the Remainder Theorem • possible extension: understand how synthetic division is related to long division and can use it to also test and determine roots for any polynomial equation • compare the properties of two functions/equations according to their zeros/roots (or vice versa) • use characteristics of a polynomial equation to determine and justify methods for finding roots • solve for and state all the roots and corresponding factors using a variety of methods including analysis of a graph or table in addition to long division, factoring, etc. Develop conceptual understanding: multiplicity, factor, long division, Remainder Theorem, (+)Synthetic Division, (+)Fundamental Theorem of Algebra Supporting terms to communicate: roots, zeros, coefficient, perfect squares, difference of squares, imaginary, complex Core Resource A core resource supports multiple days of instruction. • Rational Roots This multiple-day resource supports students in making connections between the equation of a polynomial function and its roots. Resource: Rational Roots This multiple-day resource supports students in making connections between the equation of a polynomial function and its roots. All Resources From: • Unit 4
# Percents, Fractions, and Decimals Finding a Percent of a Number Lessons 6.6 & 6.7. ## Presentation on theme: "Percents, Fractions, and Decimals Finding a Percent of a Number Lessons 6.6 & 6.7."— Presentation transcript: Percents, Fractions, and Decimals Finding a Percent of a Number Lessons 6.6 & 6.7   Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.  Use ratio and rate reasoning to solve real-world and mathematical problems CCS: 6.RP.3.   Students will be able to:  Represent percents in different but related ways, such as a ratio, a fraction and a decimal  Use proportions to solve percentage problems  Use decimals to calculate daily transactions (finding sales tax or figuring tips on restaurant bills) Objectives   A percent is a ratio that compares a number to 100. It means “per 100.”  49 out of 100 is 49%. Vocabulary   Imagine a decimal point in the place of the percent sign, and move the decimal two spaces to the left (the same as dividing by 100). Writing Percents as Decimals 26%.26 40%.40.4 7%.07   Place the percent in a fraction with a denominator of 100.  Simplify the fraction. Writing Percents as Fractions 26% 26 100 13 50 75% 75 100 3 4   Move the decimal point two spaces to the right, and add a % symbol (this is the same as multiplying by 100). Writing Decimals as Percents.34 34%.19 19%.12512.5%.6 60% 1 100%   Divide the numerator by the denominator to get a decimal.  Change the decimal to a percent by moving the decimal point to the right (multiply by 100). Writing Fractions as Percents 6 25 0.2424%   Writing Percents as Fractions and Decimals Writing Percents as Fractions and Decimals Video Time Should we use decimals OR proportions (fractions) OR both???  Using a Proportion  Set up a proportion that uses the percent over 100.  Cross multiply to write an equation.  Solve the equation. To set up your proportion, think, “IS over OF equals PERCENT over 100.”  Example – What is 20% of 30? = 20 100 part whole 30 x = 100x 30(20) = 100x 600 100 = x6  Using a Decimal  Change the percent to a decimal.  Multiply that decimal by the number you are finding the percent of.  Example – What is 18% of 70? 18% = 0.18 0.18 x 70 = 12.6   Finding Percents of a Number Finding Percents of a Number  Using Proportions to find Percents of a Number Using Proportions to find Percents of a Number Video Time!  Homework Time! Reteaching 6.6 and 6.7 Handout Classwork Time! Math at the MallMath at the Mall -practice percentages and finding the best deal while shopping at a virtual mall Download ppt "Percents, Fractions, and Decimals Finding a Percent of a Number Lessons 6.6 & 6.7." Similar presentations
Area of triangle Chapter 4 Class 12 Determinants Concept wise In a 2D plane, For three points Β Β A (x 1 , y 1 ), Β B (x 2 , y 2 ), Β C (x 3 , y 3 ) ## Find area of triangle with vertices (3, 8), (-4, 2), (5, 1) Check solution - Example 17 ## Important Points There are some points to note:- 1. If Area of triangle = 0 , then the three points are collinear 2. If the value of determinant comes negative, we will take the positive value as area Example Therefore, Area = 45 square units 3. If area is given, We take both positive and negative value of determinant Example If Area = 3 square units Β Check the questions below to learn more Β Get live Maths 1-on-1 Classs - Class 6 to 12 ### Transcript Area of β = 1/2 \|β 8(π₯_1&π¦_1&[email protected]π₯_2&π¦_2&[email protected]π₯_3&π¦_3&1)\| For points in 3D plane A (x1, y1, z1) B (x2, y2 , z2) C (x3, y3 , z3) Area of βABC = 1/2 \|β 8(π₯_1&π¦_1&π§[email protected]π₯_2&π¦_2&π§[email protected]π₯_3&π¦_3&π§_3 )\| Example If Area = 1/2 \|β 8(9&2&[email protected]&β1&[email protected]&0&β2)\| = 1/2 Γ β90 = β45 Therefore, Area = 45 square units 1/2 \|β 8(1&3&[email protected]&0&[email protected]π&0&1)\| = Β± 3
# Search by Topic #### Resources tagged with Divisibility similar to Elevenses: Filter by: Content type: Stage: Challenge level: ### There are 54 results Broad Topics > Numbers and the Number System > Divisibility ### Elevenses ##### Stage: 3 Challenge Level: How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results? ### Take Three from Five ##### Stage: 4 Challenge Level: Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him? ### What Numbers Can We Make? ##### Stage: 3 Challenge Level: Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make? ### What Numbers Can We Make Now? ##### Stage: 3 Challenge Level: Imagine we have four bags containing numbers from a sequence. What numbers can we make now? ### Going Round in Circles ##### Stage: 3 Challenge Level: Mathematicians are always looking for efficient methods for solving problems. How efficient can you be? ### Eminit ##### Stage: 3 Challenge Level: The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M? ### Digital Roots ##### Stage: 2 and 3 In this article for teachers, Bernard Bagnall describes how to find digital roots and suggests that they can be worth exploring when confronted by a sequence of numbers. ### American Billions ##### Stage: 3 Challenge Level: Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3... ### Divisively So ##### Stage: 3 Challenge Level: How many numbers less than 1000 are NOT divisible by either: a) 2 or 5; or b) 2, 5 or 7? ### Remainders ##### Stage: 3 Challenge Level: I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number? ##### Stage: 3 Challenge Level: Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true. ### Gaxinta ##### Stage: 3 Challenge Level: A number N is divisible by 10, 90, 98 and 882 but it is NOT divisible by 50 or 270 or 686 or 1764. It is also known that N is a factor of 9261000. What is N? ### Flow Chart ##### Stage: 3 Challenge Level: The flow chart requires two numbers, M and N. Select several values for M and try to establish what the flow chart does. ### Digat ##### Stage: 3 Challenge Level: What is the value of the digit A in the sum below: [3(230 + A)]^2 = 49280A ### Dozens ##### Stage: 2 and 3 Challenge Level: Do you know a quick way to check if a number is a multiple of two? How about three, four or six? ### Powerful Factorial ##### Stage: 3 Challenge Level: 6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!? ### Counting Factors ##### Stage: 3 Challenge Level: Is there an efficient way to work out how many factors a large number has? ### Oh! Hidden Inside? ##### Stage: 3 Challenge Level: Find the number which has 8 divisors, such that the product of the divisors is 331776. ### Ewa's Eggs ##### Stage: 3 Challenge Level: I put eggs into a basket in groups of 7 and noticed that I could easily have divided them into piles of 2, 3, 4, 5 or 6 and always have one left over. How many eggs were in the basket? ### Legs Eleven ##### Stage: 3 Challenge Level: Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have? ### Differences ##### Stage: 3 Challenge Level: Can you guarantee that, for any three numbers you choose, the product of their differences will always be an even number? ### AB Search ##### Stage: 3 Challenge Level: The five digit number A679B, in base ten, is divisible by 72. What are the values of A and B? ### Square Routes ##### Stage: 3 Challenge Level: How many four digit square numbers are composed of even numerals? What four digit square numbers can be reversed and become the square of another number? ### Ben's Game ##### Stage: 3 Challenge Level: Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters. ### Mod 3 ##### Stage: 4 Challenge Level: Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3. ### Three Times Seven ##### Stage: 3 Challenge Level: A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why? ##### Stage: 3 Challenge Level: List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it? ### Expenses ##### Stage: 4 Challenge Level: What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time? ### Repeaters ##### Stage: 3 Challenge Level: Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13. ### Remainder ##### Stage: 3 Challenge Level: What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2? ### The Remainders Game ##### Stage: 2 and 3 Challenge Level: A game that tests your understanding of remainders. ### Factoring Factorials ##### Stage: 3 Challenge Level: Find the highest power of 11 that will divide into 1000! exactly. ### Fac-finding ##### Stage: 4 Challenge Level: Lyndon chose this as one of his favourite problems. It is accessible but needs some careful analysis of what is included and what is not. A systematic approach is really helpful. ### Knapsack ##### Stage: 4 Challenge Level: You have worked out a secret code with a friend. Every letter in the alphabet can be represented by a binary value. ### Big Powers ##### Stage: 3 and 4 Challenge Level: Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas. ### What an Odd Fact(or) ##### Stage: 3 Challenge Level: Can you show that 1^99 + 2^99 + 3^99 + 4^99 + 5^99 is divisible by 5? ### LCM Sudoku ##### Stage: 4 Challenge Level: Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it. ### 396 ##### Stage: 4 Challenge Level: The four digits 5, 6, 7 and 8 are put at random in the spaces of the number : 3 _ 1 _ 4 _ 0 _ 9 2 Calculate the probability that the answer will be a multiple of 396. ### Squaresearch ##### Stage: 4 Challenge Level: Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares? ### Just Repeat ##### Stage: 3 Challenge Level: Think of any three-digit number. Repeat the digits. The 6-digit number that you end up with is divisible by 91. Is this a coincidence? ### Obviously? ##### Stage: 4 and 5 Challenge Level: Find the values of n for which 1^n + 8^n - 3^n - 6^n is divisible by 6. ### Multiplication Magic ##### Stage: 4 Challenge Level: Given any 3 digit number you can use the given digits and name another number which is divisible by 37 (e.g. given 628 you say 628371 is divisible by 37 because you know that 6+3 = 2+7 = 8+1 = 9). . . . ### Transposition Fix ##### Stage: 4 Challenge Level: Suppose an operator types a US Bank check code into a machine and transposes two adjacent digits will the machine pick up every error of this type? Does the same apply to ISBN numbers; will a machine. . . . ### Check Codes ##### Stage: 4 Challenge Level: Details are given of how check codes are constructed (using modulus arithmetic for passports, bank accounts, credit cards, ISBN book numbers, and so on. A list of codes is given and you have to check. . . . ### Skeleton ##### Stage: 3 Challenge Level: Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum. ### Sixational ##### Stage: 4 and 5 Challenge Level: The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . . ### The Chinese Remainder Theorem ##### Stage: 4 and 5 In this article we shall consider how to solve problems such as "Find all integers that leave a remainder of 1 when divided by 2, 3, and 5." ### Odd Stones ##### Stage: 4 Challenge Level: On a "move" a stone is removed from two of the circles and placed in the third circle. Here are five of the ways that 27 stones could be distributed. ### N000ughty Thoughts ##### Stage: 4 Challenge Level: How many noughts are at the end of these giant numbers?
# Advance Mathematics Section 3.5 Objectives: ## Presentation on theme: "Advance Mathematics Section 3.5 Objectives:"— Presentation transcript: Define Even and Odd functions algebraically and graphically Sketch graphs of functions using shifting, and reflection f(-x) = -2 (-x)5 + 4( -x )3 +7(-x) f(-x) = ( -x )5 + ( -x )2 Even and Odd Functions Terminology Definition Example Type of Symmetry f is an even function f(-x) = f(x) y = f(x) = x2 w.r.t y-axis odd function f(-x) = - f(x) y = f(x ) = x3 Origin Example 9 Determine whether a function is even, odd or neither. a) f ( x ) = 3x4 + 5x2 – b) f( x ) = -2x5 +4x3 +7x c) f( x ) = x3 +x2 Substitute x by –x f(-x) = -2 (-x)5 + 4( -x )3 +7(-x) = 2x5 – 4x3 – 7x = - (-2x5 +4x3 +7 ) = - f(x) f(-x) = - f(x) f is odd Solution: Substitute x by –x f(-x) = ( -x )5 + ( -x )2 = - x5 + x2 f(-x) is not equal to f( x) nor –f(x). Therefore, f is neither. Substitute x by –x f( -x) = 3( -x )4 + 5 ( -x )2 - 4 = 3x4 + 5x2 – 4 = f(x) f( -x) = f(x) f is even. Example 10 a) c) b) d) Continue… Check whether the following graphs represent an even or odd functions or neither. a) c) The graph represents an even function The graph represents neither b) d) The graph represents neither The graph represents an odd function Example 11 Continue… Complete the graph of the following if b) Symmetric w.r.t origin a) Symmetric w.r.t y-axis c) Function is even d) Function is odd e) Symmetric w.r.t x-axis Graph of (e ) does not represent a function Vertical Shifting Example 12 y = f( x ) + c Up c units y = f ( x ) - c Down c units Example 12 Below is the graph of a function y = f ( x ). Sketch the graphs of y = f ( x ) + 1 b) y = f ( x ) - 2 y = f(x) + 1 y = f (x) y = f (x)-2 Horizontal Shifting Continued… y = f( x + c ) Left c units Right c units Example 13. Given the graph of a function y = f ( x ). Sketch the graphs of y = f ( x + 3 ) b) y = f ( x – 4 ) y = f ( x ) y = f ( x ) + 3 Horizontal Shift 3 units to the left y = f ( x ) y = f ( x ) - 4 Horizontal Shift 4 units to the right Left h units and Up k units Left h units and Down k units Example 14 Continued… Can you tell the effects on the graph of y = f ( x ) y = f( x + h ) + k y = f( x + h ) - k y = f( x - h ) + k y = f( x - h ) - k Left h units and Up k units Left h units and Down k units Right h units and Up k units Right h units and Down k units Example 15 Below is the graph of a function y = f ( x ). Sketch the graph of y = f ( x + 2 ) - 1 y = f( x + 2 ) - 1 y = f( x ) Example 16 Continued… Below is the graph of a function Sketch the graph of Solution: The graph of the absolute value is shifted 2 units to the right and 3 units down y = f( x ) y =f(x-2)-3 Vertical Compress by a factor 1/c Vertical Stretching y = cf( x) ( c> 1 ) Vertical Stretch by a factor c y = (1/c)f ( x) ( c > 1 ) Vertical Compress by a factor 1/c Note1 :When c > 1. Then 0 < 1/c < 1 Note 2 : c effects the value of y only. Example 17 Below is the graph of a function y = x2 . Sketch the graphs of y = 5 x2 2. y = (1/5)x2 x y = x2 y=5x2 y=1/5x2 2 10 .4 1 5 .2 -1 Example 18 If the point P is on the graph of a function f. Find the corresponding point on the graph of the given function. P ( 0, 5 ) y = f( x + 2 ) – 1 P ( 3, -1 ) y = 2f(x) +4 3) P( -2,4) y = (1/2) f( x-3) + 3 Solution: P ( 0,5). y = f( x + 2 ) – 1 shifts x two units to the left and shifts y one unit down. The new x =0 – 2 = -2, and the new y = 5 – 1 = 4. The corresponding point is ( -2, 4 ). 2) P(3,-1). y = 2f(x) +4 has no effect on x. But it doubles the value of y and shifts it 4 units vertically up. Therefore the new x = 3(same as before ), and the new value of y = 2 (-1 ) + 4 = 2. Therefore, the corresponding point is ( 3,2 ). P(-2, 4 ). y = (1/2) f( x-3) + 3 shifts x 3 units to the right and splits the value of y in half and then shifts it 3 units up. That is, the new value of y = (1/2)(4) + 3 = 5. Therefore, the corresponding point is ( 1, 5 ). Reflecting a graph through the x-axis y = -f( x) Reflection through the x-axis (x-axis acts as a plane mirror) Note: For any point P(x,y) on the graph of y = f(x), The graph of y = - f(x) does not effect the value of x, but changes the value of y into - y Example 19 Below is the graph of a function y = x2 . Sketch the graph of y = - x2 x y = x2 y = -x2 2 4 -4 1 -1 -2 Sketching a piece-wise function Definition: Piece-wise function is a function that can be described in more than one expression. Example 20 Sketch the graph of the function f if Solution: Graph y = 2x + 5 and take only the portion to the left of the line x = -1. The point (-1, 3 ) is included. Graph y = x2 and take only the portion where –1 < x < 1. Note: the points ( -1,1) and ( 1, 1 ) are not included Graph y = 2 and take only the portion to the right of x = 1. Note: y = 2 represents a horizontal line. The point (1, 2 ) is included. Sketching the graph of an equation containing an absolute Note: To sketch an absolute value function We have to remember that And hence, the graph is always above the x-axis. The part of the graph that is below the x-axis will be reflected above the x-axis. Example 21 Sketch the graph of y = g ( x ) = Solution: Strategy: 1. Graph y = f(x) = x2. 2. Graph y = f( x ) - 9 = x2 – 9 by shifting the graph of f 9 units down 3. Graph g(x) by keeping the portion of the graph y = f( x ) - 9 = x2 – 9 which is above the x-axis the same, and reflecting the portion where y < 0 with respect to the x-axis. 4. Delete the unwanted portion Example 22 Solution: A picture can replace 1000 words Below is the graph of y = f(x). Graph Solution: A picture can replace 1000 words Let the animation talk about itself Do all homework exercises in the syllabus
Grade Level: 4th Skill: Graphs and Charts Topic: Y-Coordinates Goal: Understand that the length of a vertical line segment equals the difference of the y-coordinates. Skill Description: Use subtraction to find the vertical length of the line on a coordinate grid. ## Building Blocks/Prerequisites ### Sample Problems (1) Graph the ordered pairs and connect the points. Find the length of the line segment. (6,12) and (6, 19) (7 units) (2) What is the length of the line segment with points (8, 41) and (8, 27) (14 units) (3) Use the coordinate grid and find the length of the line segment. [Graph the coordinates below on a coordinate grid] (4, 8) and (4, 19) (11 units) (4) Which ordered pairs has the vertical distance of 17 units: (7, 9) (7, 26) (9, 7) (26, 7) (9, 27) (9, 17) [b- (7,9) (7,26) ] (5) If the first point is located at (8, 7) where is the second point on the coordinate grid to have a vertical length of 4 units (down)? (8, 3) ### Learning Tips (1) Math Vocabulary: Ordered pair- (x, y) a pair of numbers used to locate a point on a coordinate grid. x-axis- the horizontal line on a coordinate grid (line that goes from left to right) y-axis- the vertical line on a coordinate grid (line that goes from top to bottom) x-coordinate- the first number of an ordered pair, which tells how far to move horizontally along the x-axis. y-coordinate- the second number in an ordered pair, which tells how far to move vertically along the y-axis Vertical line- a line that goes up and down from top to bottom For example, i.e.) \| \| \| \| \| \| Line segment- a line that has two endpoints i.e.) •---------------------• Coordinate grid- a map-like tool that is used to locate specific points. (2) Use a number line to explain length of the y-axis (Make it vertical on your example): ‹--׀--------•--------׀--------׀--------׀--------•--------׀--------׀--------׀--------׀---› 1 2 3 4 5 6 7 8 9 10 → → → → The length is 4 units (count the arrows from one number to the next). There are 4 units between 2 and 6. To find the vertical distance between two points on a coordinate grid, count units or subtract the y-coordinates (6 – 2 =4 units). (3) Color-code the y-coordinate. Most children get confused on finding the length of the horizontal and vertical distance because they subtract the wrong coordinates. In the beginning color-code the y-coordinates, so they understand that those are the two numbers that are to be subtracted (Hint: the x-coordinate should be the same number) Use a red marker to outline the x-coordinate. For example, What is the horizontal length of (3, 2) and (3, 4)? If they see the x-coordinates color coded it will help them grasp the concept of subtracting the x-coordinates to find the length. (4) Relate coordinate grids to Maps: Use a real map and teach your child how to use the coordinate grid on a map to find cities in the US. On most maps, the coordinates are letters and numbers. To find a location they will be given a coordinate of C-2 and the city will be labeled at that point. Once they understand how to use coordinates to locate locations, have them find the vertical distance between two cities. (5) Explain the parts of a coordinate grid: Draw a coordinate grid on large chart paper and identify the parts of the grid y-axis x-axis how to plot points (x-coordinate go across on the x-axis; y-coordinate go straight up from the x-coordinate to number on the y-axis) horizontal lines vertical lines Practice making vertical lines. Parent will give the child two ordered pairs have the child graph and connect the points. Discuss why it is a vertical line and not horizontal. (6) Practice simple subtractions facts (timed drills, flash cards) ### Extra Help Problems (1) Graph the ordered pairs and connect the points. Find the length of the line segment. (4, 5) and (4, 19) (14 units) (2) Graph the ordered pairs and connect the points. Find the length of the line segment. (5, 10) and (5, 17) (7 units) (3) Graph the ordered pairs and connect the points. Find the length of the line segment. (7, 4) and (7, 8) (4 units) (4) Graph the ordered pairs and connect the points. Find the length of the line segment. (9, 0) and (9, 12) (12 units) (5) Graph the ordered pairs and connect the points. Find the length of the line segment. (0, 5) and (0, 20) (15 units) (6) Graph the ordered pairs and connect the points. Find the length of the line segment. (1, 8) and (1, 13) (5 units) (7) What is the length of the line segment with points (10, 107) and (10, 88)? (19 units) (8) What is the length of the line segment with points (12, 41) and (12, 64)? (23 units) (9) What is the length of the line segment with points (15, 8) and (15, 21)? (13 units) (10) What is the length of the line segment with points (7, 34) and (7, 43)? (9 units) (11) What is the length of the line segment with points (6, 18) and (6, 37)? (19 units) (12) What is the length of the line segment with points (12, 30) and (12, 15)? (15 units) (13) Use the coordinate grid and find the length of the line segment. (Do not show ordered pairs, only the graph) [Graph (8,6) and (8,4)] (2 units) (14) Use the coordinate grid and find the length of the line segment. . (Do not show ordered pairs, only the graph) [Graph (11,1) and (11, 3)] (4 units) (15) Use the coordinate grid and find the length of the line segment. . (Do not show ordered pairs, only the graph) [Graph (3,2) and (3, 15)] (13 units) (16) Use the coordinate grid and find the length of the line segment. . (Do not show ordered pairs, only the graph) [Graph (10, 1) and (10, 6)] (5 units) (17) Use the coordinate grid and find the length of the line segment. . (Do not show ordered pairs, only the graph) [Graph (7, 14) and (7, 8)] (6 units) (18) Which ordered pairs has the vertical distance of 27 units (4, 13) and (4, 40) (13, 4) and (13, 40) (4, 8) and (4, 5) [ (a) (4, 13) and (4, 40)] (19) Which ordered pairs has the vertical distance of 4 units (11,4) and (4, 11) (11,8) and (4, 8) (8, 4) and (8, 11) [ (c) (8, 4) and (8, 11)] (20) Which ordered pairs has the vertical distance of 82 units (108, 12) and (16, 12) (12, 16) (12, 108) (16, 12) and (16, 108) [ (b)(12, 16) (12, 108)] (21) Which ordered pairs has the vertical distance of 32 units (20, 21) and (20, 53) (23, 20) and (53, 20) (23, 18) and (18, 46) [(a) (20, 21) and (20, 53)] (22) Which ordered pairs has the vertical distance of 15 units (10, 20) and (10, 25) (10, 15) and (10, 30) (15, 10) and (30, 10) [(b) (10, 15) and (10, 30)] (23) If the first point is located at (12, 1) where is the second point on the coordinate grid to have a vertical length of 8 units (up)? (12, 9) (24) If the first point is located at (2, 9) where is the second point on the coordinate grid to have a vertical length of 5 units (down)? (2, 4) (25) If the first point is located at (5, 8) where is the second point on the coordinate grid to have a vertical length of 11 units (up)? (5, 19)
{[ promptMessage ]} Bookmark it {[ promptMessage ]} Ch6 Exercises # Ch6 Exercises - IMPORTANT PROBABILITY RULES ILLUSTRATED... This preview shows pages 1–3. Sign up to view the full content. 1 IMPORTANT PROBABILITY RULES ILLUSTRATE D Solutions to the wrap-up problems - Probability 1) A die is rolled once. Each number is equally likely to face up. Find the probabilities of the following events. a) The number facing up is 6 b) The number facing up is not 6 c) The number facing up is odd d) The number facing up is less than 4 e) The number facing up is odd or is less than 4 f) The number facing up is odd or is even SOLUTION Define the events symbolically A = {6} B = {1, 3, 5} C = {1, 2, 3} D = {2, 4, 6} Calculate probability of each event above by summing up their constituent simple events a) P (A) = 1/6 b) We are looking for the complement of A. The complement rule states: 6 / 5 6 / 1 1 ) A ( P 1 ) A ( P = = = c) P (B) = 1/6 + 1/6 + 1/6 = 3/6= ½ d) P (C) = 1/6 + 1/6 + 1/6) = 3/6= ½ e) P (B U C) = P (B) + P(C) – P (B C) = ½ + ½ - 2/6 = 2/3 Note that {1, 3} are common to B and C. f) P (B U D) = P (B) + P (D) note that B and D are mutually exclusive = ½ + ½ = 1 This preview has intentionally blurred sections. Sign up to view the full version. View Full Document 2 2) 60% check email in the first hour of work. 70% check email in the last hour of work 4 0% check email in the first and last hour of work Find the probability that a worked selected randomly will check email in the first hour or last hour of work or both. SOLUTION Try it yourself! First define the relevant events and then use the additive rule. Are the relevant events mutually exclusive? 3) Hospital records show that 12% of all patients are admitted for surgical treatment, 16% are admitted for obstetrics, and 2% receive both obstetrics and surgical treatment. If a new patient is admitted to the hospital, what is the probability that the patient will be admitted either for This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]} ### Page1 / 6 Ch6 Exercises - IMPORTANT PROBABILITY RULES ILLUSTRATED... This preview shows document pages 1 - 3. Sign up to view the full document. View Full Document Ask a homework question - tutors are online
# McGraw Hill Math Grade 8 Lesson 15.2 Answer Key Customary Units of Liquid Volume Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 15.2 Customary Units of Liquid Volume to secure good marks & knowledge in the exams. ## McGraw-Hill Math Grade 8 Answer Key Lesson 15.2 Customary Units of Liquid Volume Exercises CONVERT Question 1. 72 c = ___________ qt 18 qt, Explanation: As 1 cup = 0.25 quart, So 72 c = 18 qt. Question 2. 12 qt = ____________ c 48 c, Explanation: As 1 qt = 4 cups, So 12 qt = 48 c. Question 3. 3 qt 1 pt = ____________ c 5.99605 c, Explanation: As 1 qt = 1.66535 pt, So 3 qt = 4.99605 pt, Therefore 3 qt 1 pt = 4.99605 pt + 1 pt = 5.99605 c. Question 4. 192 gal = _____________ qt 922.3296 qt, Explanation: As 1 gal = 4.8038 qt, So 192 gal = 192 X 4.8038 = 922.3296 qt. Question 5. 328 qt = ____________ gal 82 gal, Explanation: As 1 qt = 0.25 gal, So 328 qt = 82 gal. Question 6. 2 gal 1 qt = _______________ c 10.6076 c, Explanation: As 1 gal = 4.8038 qt, So 2 gal = 9.6076, Therefore 2 gal 1 qt = 9.6076 + 1 = 10.6076 c. Question 7. 6 qt 1 c = ____________ c 25 c, Explanation: As 1 qt = 4 c, So 6 qt = 24 c, Therefore 6 qt 1 c = 24 c + 1 c = 25 c. Question 8. 52 c = _____________ qt 13 qt, Explanation: As 1 c = 0.25 qt, So 52 c = 13 qt. Question 9. $$\frac{3}{4}$$ gal = _____________ c 12 c, Explanation: As 1 gal = 16 c, So $$\frac{3}{4}$$ gal = 12 c. Question 10. 66 pt = _______________ gal 9.907854 gal, Explanation: As 1 pt = 0.150119 gal, So 66 pt = 66 X 0.150119 = 9.907854 gal. Question 11. 4 qt 1 c = _____________ c 17 c, Explanation: As 1 qt = 4 c, So 4 qt = 16 c, Therefore 4 qt 1 c = 16 c + 1 c = 17 c. Question 12. 7$$\frac{1}{2}$$ qt = _____________ gal = ___________ C 1.875 gal, 30 c, Explanation: As 1 qt = 0.25 gal, So 7$$\frac{1}{2}$$ qt = $$\frac{15}{2}$$ qt = 1.875 gal. As 1 qt = 4 c, So 7$$\frac{1}{2}$$ qt = $$\frac{15}{2}$$ qt = 7.5 X 4 = 30 c. Question 13. 35 qt 3 c = ____________ gal 9 gal, Explanation: As 1 qt = 4 c, So 35 qt = 140 c, Therefore 35 qt 3 c = 140 c + 4 c = 144 c, As we know 35 qt 3 c is 144 c, As 1 c = 0.0625 gal, So 144 c = 9 gal. Question 14. 256 c = _______________ gal 16 gal, Explanation: As 1 c = 0.0625 gal, So 256 c = 16 gal. Question 15. 132 c = ___________ pt = ___________ qt 4461.6 pt, 33 qt, Explanation: As 1 c = 33.8 pt, So 132 c = 4461.6 pt, As 1 c = 0.25 qt, So 132 c = 33 qt. Question 16. 34 gal 1 qt = ____________ pt 273.6656 pt, Explanation: As 1 gal = 4.8038 qt, So 34 gal = 163.3292 qt, Therefore 34 gal 1 qt = 163.3292 qt + 1 qt = 164.3292 qt As we know 1 qt = 1.66535 pt, So 164.3292 qt = 273.6656. Question 17. Allisonâ€™s aquarium holds 25 gallons of water. One quart of water evaporated from the aquarium. How much water is left in the aquarium? 119.095 quarts is left in the aquarium, Explanation: Given Allisonâ€™s aquarium holds 25 gallons of water. One quart of water evaporated from the aquarium. So water is left in the aquarium is as 1 gallon = 4.8038 quart, 25 gallons = 25 X 4.8038 quart = 120.095 quarts, 120.095 quarts – 1 quarts = 119.095 quarts. Question 18. Bobby drinks 1 quart and 1 cup of water for every 3 miles he runs. If he runs 9 miles, how much water should he drink?

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