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# 5.2 Power functions and polynomial functions (Page 4/19) Page 4 / 19 ## Identifying the degree and leading coefficient of a polynomial function Identify the degree, leading term, and leading coefficient of the following polynomial functions. $\begin{array}{ccc}\hfill f\left(x\right)& =& 3+2{x}^{2}-4{x}^{3}\hfill \\ \hfill g\left(t\right)& =& 5{t}^{2}-2{t}^{3}+7t\hfill \\ h\left(p\right)\hfill & =& 6p-{p}^{3}-2\hfill \end{array}$ For the function $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ the highest power of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is 3, so the degree is 3. The leading term is the term containing that degree, $\text{\hspace{0.17em}}-4{x}^{3}.\text{\hspace{0.17em}}$ The leading coefficient is the coefficient of that term, $\text{\hspace{0.17em}}-4.$ For the function $\text{\hspace{0.17em}}g\left(t\right),\text{\hspace{0.17em}}$ the highest power of $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}5,\text{\hspace{0.17em}}$ so the degree is $\text{\hspace{0.17em}}5.\text{\hspace{0.17em}}$ The leading term is the term containing that degree, $\text{\hspace{0.17em}}5{t}^{5}.\text{\hspace{0.17em}}$ The leading coefficient is the coefficient of that term, $\text{\hspace{0.17em}}5.$ For the function $\text{\hspace{0.17em}}h\left(p\right),\text{\hspace{0.17em}}$ the highest power of $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}$ so the degree is $\text{\hspace{0.17em}}3.\text{\hspace{0.17em}}$ The leading term is the term containing that degree, $\text{\hspace{0.17em}}-{p}^{3}.\text{\hspace{0.17em}}$ The leading coefficient is the coefficient of that term, $\text{\hspace{0.17em}}-1.$ Identify the degree, leading term, and leading coefficient of the polynomial $\text{\hspace{0.17em}}f\left(x\right)=4{x}^{2}-{x}^{6}+2x-6.$ The degree is 6. The leading term is $\text{\hspace{0.17em}}-{x}^{6}.\text{\hspace{0.17em}}$ The leading coefficient is $\text{\hspace{0.17em}}-1.$ ## Identifying end behavior of polynomial functions Knowing the degree of a polynomial function is useful in helping us predict its end behavior. To determine its end behavior, look at the leading term of the polynomial function. Because the power of the leading term is the highest, that term will grow significantly faster than the other terms as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ gets very large or very small, so its behavior will dominate the graph. For any polynomial, the end behavior of the polynomial will match the end behavior of the power function consisting of the leading term. See [link] . Polynomial Function Leading Term Graph of Polynomial Function $f\left(x\right)=5{x}^{4}+2{x}^{3}-x-4$ $5{x}^{4}$ $f\left(x\right)=-2{x}^{6}-{x}^{5}+3{x}^{4}+{x}^{3}$ $-2{x}^{6}$ $f\left(x\right)=3{x}^{5}-4{x}^{4}+2{x}^{2}+1$ $3{x}^{5}$ $f\left(x\right)=-6{x}^{3}+7{x}^{2}+3x+1$ $-6{x}^{3}$ ## Identifying end behavior and degree of a polynomial function Describe the end behavior and determine a possible degree of the polynomial function in [link] . As the input values $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ get very large, the output values $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ increase without bound. As the input values $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ get very small, the output values $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ decrease without bound. We can describe the end behavior symbolically by writing In words, we could say that as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values approach infinity, the function values approach infinity, and as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values approach negative infinity, the function values approach negative infinity. We can tell this graph has the shape of an odd degree power function that has not been reflected, so the degree of the polynomial creating this graph must be odd and the leading coefficient must be positive. Describe the end behavior, and determine a possible degree of the polynomial function in [link] . As It has the shape of an even degree power function with a negative coefficient. ## Identifying end behavior and degree of a polynomial function Given the function $\text{\hspace{0.17em}}f\left(x\right)=-3{x}^{2}\left(x-1\right)\left(x+4\right),\text{\hspace{0.17em}}$ express the function as a polynomial in general form, and determine the leading term, degree, and end behavior of the function. Obtain the general form by expanding the given expression for $\text{\hspace{0.17em}}f\left(x\right).$ $\begin{array}{ccc}\hfill f\left(x\right)& =& -3{x}^{2}\left(x-1\right)\left(x+4\right)\hfill \\ & =& -3{x}^{2}\left({x}^{2}+3x-4\right)\hfill \\ & =& -3{x}^{4}-9{x}^{3}+12{x}^{2}\hfill \end{array}$ The general form is $\text{\hspace{0.17em}}f\left(x\right)=-3{x}^{4}-9{x}^{3}+12{x}^{2}.\text{\hspace{0.17em}}$ The leading term is $\text{\hspace{0.17em}}-3{x}^{4};\text{\hspace{0.17em}}$ therefore, the degree of the polynomial is 4. The degree is even (4) and the leading coefficient is negative (–3), so the end behavior is Cos45/sec30+cosec30= Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2) exercise 1.2 solution b....isnt it lacking I dnt get dis work well what is one-to-one function what is the procedure in solving quadratic equetion at least 6? Almighty formula or by factorization...or by graphical analysis Damian I need to learn this trigonometry from A level.. can anyone help here? yes am hia Miiro tanh2x =2tanhx/1+tanh^2x cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3 proof AUSTINE sebd me some questions about anything ill solve for yall cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb favour how to solve x²=2x+8 factorization? x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN) Manifoldee x=2x+8 Manifoldee ×=2x-8 minus both sides by 2x Manifoldee so, x-2x=2x+8-2x Manifoldee then cancel out 2x and -2x, cuz 2x-2x is obviously zero Manifoldee so it would be like this: x-2x=8 Manifoldee then we all know that beside the variable is a number (1): (1)x-2x=8 Manifoldee so we will going to minus that 1-2=-1 Manifoldee so it would be -x=8 Manifoldee so next step is to cancel out negative number beside x so we get positive x Manifoldee so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1) Manifoldee so -1/-1=1 Manifoldee so x=-8 Manifoldee Manifoldee so we should prove it Manifoldee x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India mantu lol i just saw its x² Manifoldee x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8) Manifoldee I mean x²=2x+8 by factorization method Kristof I think x=-2 or x=4 Kristof x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia Mohamed i am in Cliff hii Amit how are you Dorbor well Biswajit can u tell me concepts Gaurav Find the possible value of 8.5 using moivre's theorem which of these functions is not uniformly cintinuous on (0, 1)? sinx helo Akash hlo Akash Hello Hudheifa which of these functions is not uniformly continuous on 0,1
Login Copyright (c) 2013 by KJAL Conditions of Use Privacy Policy Return to Blogmeister KJAL -- Blogmeister KJAL The students' blogs have been transferred to 8th grade. by KJAL teacher: Rye 8th Team Blog Entries List 25, 50, all Title: Math Letter / Module 1 (10/23/09) Description: Student letter to parents/guardians describing math in Module 1. Article posted October 26, 2009 at 01:52 PM GMT • comment • Reads 7709 Module 1 Dear Dad, I am going to be telling you about module 1 that we learned in math class I will tell you about 3 different parts of it. probability, pemdas, and frequency tables. First I will tell you about probability then pemdas then frequency tables. OK, lets get going. We learned about probability and how to find it out soo... lets say you had a bag of jolly ranchers and there were lets say 10 blue raspberry 7 cherry, 9 grape, 5 watermelon, and 3 apple we put them all in the paper bag. There are 34 jolly ranchers in all. You have 2 trails one 34 times the second 68 you pull 5 blue, 4 cherry, 4 grape, 3 watermelon, and 1 apple. So probability for this example would be for trial on for lets say cherry 7/34 ,blue 10/34,apple 3/34, grape 9/34, 5/34 for watermelon. So the probability is how many out of your total would you choose that flavor. So that's all about probability now lets move on to pemdas. Pemdas is actually parentheses exponents, multiplication, division, addition subtraction. This is the order you would do an equation like lets use this equation. 9x(6+5)-9 first do the 6+5 and that's 11 then 9x11= 99 then 99-9=90. pemdas is so you get the right order its the way to find the answer and that's pemdas! Frequency tables are used for surveys like if you were taking a survey on favorite colors and you have to have 3 columns one would be color the other would be tally's and and the last column would be frequency they separate all the information we collect. So after that we know we can use a frequency tables often. So now that I've told you about these strategies maybe you can use them in your daily life. Love, brco Article posted October 26, 2009 at 01:52 PM GMT • comment • Reads 7709 Article posted October 28, 2009 at 01:28 PM GMT • comment • Reads 1220 Dear Mom and Dad, My class has just finished learning about module 1 in the textbook. In Module 1 we reviewed many subjects including the order of operations, bar and line graphs, and probability. Module 1 allowed us to re-enforce our knowledge about many basic subjects. During Module 1 we conducted many experiments so that we would have a firm understanding of various subjects. The order of operations is a rule that allows us to organize complex math problems. When following the rules of the order of operations, this is the order in which you must solve the math problem, parenthesis – exponents – multiplication – division – addition – and subtraction. When you come across a math problem with either multiplication and division or addition and subtraction, it is important to solve the problem from left to right. We learned about the order of operations in class by evaluating various problems in the math textbook. In the beginning of module 1 we learned about when to use a bar or line graph. It is important to use a bar graph when graphing data that can be divided into distinct categories. You have to use a line graph when the data changes over time. We learned about the different types of graphs by collecting and graphing information during class. We were also given a take home quiz where we had to find a graph, and analyze it. Knowing when to use a bar or line graph is very important. Probability is a measurement used to predict the chance of something happening. There are two types of probability; theoretical, and experimental. Theoretical probability is the chance of something happening according to theory. Experimental probability is the chance of something happening according to your experiment. We learned about these subjects by finding the experimental and theoretical probability of many experiments, including flipping coins, tossing dice, and spinning a spinner. We used data tables to show our information. Thank you for taking the time to read about what I have been learning about in math class! Thanks Again, LRAS Article posted October 28, 2009 at 01:28 PM GMT • comment • Reads 1220 Article posted November 10, 2009 at 05:13 PM GMT • comment • Reads 1185 Dear Dad, For this module all we really did was review the same exact stuff that we learned last year, mostly just patterns and the order of operations, and a little algebra. However, we did learn the more advanced vocabulary that we did not use last year.Overall, we learned pretty much nothing. If a had to pick one favorite part of the module than it would probably be the order of operations, just because I got a little bit of a challenge in it. Besides that, nothing new was actually learned, we learned about PEMDAS, which means that we go in the order of Parentheses, Exponents, Multiplication and Division, and finally Addition and Subtraction. My least favorite part of the unit was definitely learning about patterns, because it was way too long of a chapter for something that we learned last year. All we really did anyway was learn what everything was called, which was the first thing we ever learned last year. We also learned about probability, which was a little bit interesting, but not a whole lot. There are basically two types of probability: theoretical probability is what will probably happen, and experimental probability is what did happen.On the subject of probability, we did a small project on finding the probability of a random event, like picking a gumball out of a bag, or flipping a coin 100 times. What my partner and I did was very simple, and stupid. We built a pair of industrial tweezers out of two 12 inch rulers, and taped them together. We then filled a bag with skittles and gumballs, and proceded to pick a candy 40 times. In the end, we figured that the more you do an experiment, the closer the experimental probability gets to the theoretical probability. Other people did experiments like ours, such as picking guitar picks and spinning a colored wheel. The one thing that the experiments taught us was that the more you conduct an experimentm the closer the experimental probability gets to the theoretical.Throughout the module we also did portfolio points, or practice activites that had to accumulate 100 points or more throughout the module. That was pretty much it for module 1, and next module will be much more interesting, because it is based completely on problem solving. Sincerely, LHDR Article posted November 10, 2009 at 05:13 PM GMT • comment • Reads 1185 Article posted November 10, 2009 at 07:04 PM GMT • comment • Reads 1176 Dear Mom, In this letter you will learn what I have learned in the 7th grade math module #1 from line and bar graphs to, order of operations and everything in between. I have learned about line and bar graph,tables,and frequency charts. We had to cut graphs out of the newspaper and tell what the graph was saying. I found this unit easy. Now we have moved onto learning about Venn Diagrams and problem solving. We used Venn Diagrams in other classes too, so it is interesting that we are also using them in math. LJSH Article posted November 10, 2009 at 07:04 PM GMT • comment • Reads 1176 Article posted October 28, 2009 at 01:29 PM GMT • comment • Reads 1230 Dear Mom, In Math we have been working on very importing things that were going to use in the future. like bar graph/line graph, Probability, frequency table. In the letter I'm going to show you about all of those things and talk about what we did. The first thing I'm going to talk about is bar graph/line graph. We used frequency tables to show information on are graph. And then we had to decide whether we should use a bar or line we learn that we use line graph hen you show long time and bar graph to show category's like favorite color or sport. There are many ways you could use a line/bar graph this is just some. The next thing I'm going to talk about is probability. Probability is use when you need to find a chance of some thing happening. For example if there was 1 red cube, 2blue cubes, and 1 orange cube the chance of getting a red is ¼, blue is ½, and orange is ¼ , so there a 50% chance that it will be blue and a 50% chance that it will be red or orange. Probability can be very useful to predict the future but it is not always 100% accurate its more like 70% accurate. The 3rd and finally thing I'm going to talk about is about frequency tables. Frequency tables can be used in every thing I talk about. You use a frequency table to show data like how many times you get heads or tales. You tally how many times you get heads and tails. After that you could put it on a graph to see it clearly. There are lots of other ways you could use frequency tables. Hope you enjoy my letter there was a lot I learn in this module this is just 3 things out of 10 things I learn like order of operations. Thanks For Reading H.R.M.A My Code Name Article posted October 28, 2009 at 01:29 PM GMT • comment • Reads 1230 Article posted October 28, 2009 at 06:07 PM GMT • comment • Reads 1133 Dear, Mom and Nana I have learned a lot of new things this year in math class. In module one I learned exponents and what the exponential form and standard form was and also what the difference between them was. Exponential form is when you have a number and you have a smaller number above, which clarifies that it's in exponential form. Standard form is when you have the number written out. Example: 2x2x2x2=16. My favorite thing I learned about was probability. Probability is a number from 0- 1 that tells us how likely an event is to happen. I learned what theoretical probability was and what experimental probability was. Theoretical probability is whats suppose to happen. Experimental probability is what the experiment turns out to be. Another one of my favorite things I learned was PEMDAS. In math class our teacher had us listen to a math song that was about pemdas. Most of our class mates didn't like the song but I liked it. I thought it was very interesting. Those are all the things I learned this year in module one. Thanks for reading. Article posted October 28, 2009 at 06:07 PM GMT • comment • Reads 1133 Article posted November 5, 2009 at 02:45 PM GMT • comment • Reads 1143 Dear Mom, I am going to tell you what I learned in7th grade math class. I feel like I understand it better then last years class. The bar graph assignment was something I enjoyed. We had to find a bar graph in the newspaper and then you had to answer questions about the graph. I picked out a graph that was easy to read and the questions were easy for me to answer. The other thing I like doing are frequency tables, because it is fun for me and I am good at that. A frequency table is when you count up the number of times something happens and then you make a chart. For example, you could pick different colors of marbles out of a bag. Next,you would record them on a table and then you can see how many times you pulled certain colors out of the bag. The other thing I like is term and term#.I like it because it is easy for me this year, and I am getting better at it .I really like exponents too, because there are fun and easy for me. An exponent is the little number that tells you how many times to multiply a number by itself. I dislike the algebra part because the numbers confuse me. I will do the problem sometimes and then I get it wrong. It does not happen all the time but I need help sometimes to finish these problems. I hope you like the letter I wrote to you. I hope you have an idea of what I am doing in math this year! Love, lske Article posted November 5, 2009 at 02:45 PM GMT • comment • Reads 1143 Article posted October 26, 2009 at 01:46 PM GMT • comment • Reads 1221 -To Whom it May Concern- I will be explaining to you what we have been learning in the past module in Math class.(module 1) The first major thing we learned about is graphs. We learned the workings of a proper line & bar graph inside-out. By the end, we could graph almost any data that was given to us. Bar graphs are better for comparing multiple amounts of data. Line graphs are best for seeing the growth/shrinkage of data over time, such as the stocks. Next, was probability. We studied this by using cards, dice, coins, and computer simulations to test what should happen, and what really happened. What could happen Is called theoretical probability. What did happen is called experimental probability. Say that you have a bag of fifty marbles, 25 of them are black, and 25 of them are red. You pick 25 of them. Theoretical probability says that out of every two draws, one should be red. P(r)=12.5/25=1/2. What really happened= thirteen out of the 25 draws turned out to be red. That's very close to the what the theoretical probability said it should be. Last was the order of operations. The order of operations are: Parentheses, exponents, multiplication, division, addition, and subtraction, or PEMDAS for short. To study PEMDAS we did problems with a mix of parentheses and exponents thrown in. To complete the problem correctly, you have to abide by the order of operations. We also created our own problems using PEMDAS. Those were the main things we've been doing here in module 1. Looking forward to module 2. Sincerely, LRSE Article posted October 26, 2009 at 01:46 PM GMT • comment • Reads 1221 Article posted October 28, 2009 at 02:54 PM GMT • comment • Reads 1094 Dear, mom and dad I am sending this letter to you about what I have been learning in math class last semester. These things have been really easy for me. The fun things for me have been easy but interesting like algebra and exponents. The things that have not been fun this semester are finding the difference between the term and term number and putting it into an algebraic term. So that is what I am talking about in my letter. In last semester it has been really easy I don't know why but my guess is that its going to be like this all year. I am not very challenged int this class but some of the things are interesting. These are the things we studied in the last semester analyzing graphs, making frequency tables, putting patterns and sequences into tables and graphs, finding areas of a square, finding the volume of a cube, finding experimental and theoretical probabilities. It seems like we did a lot but if your in the class its not very much work because we have math class every day for at least 45 minutes a day. The difference between theoretical and experimental probability is theoretical probability is what should happen like say there are four colors on a spinner wheel and you spin it twenty times you should get one color five times and experimental probability is what does happen so same scenario so you will probably get one color more times than another color. The one thing we probably spent the most time in in math class is experimenting with graphs because graphs are one of the biggest things in math for instance the world wide stock exchange which is a easy way to make money watching a graph. That's what I learned in math class this semester. -from, hhma Article posted October 28, 2009 at 02:54 PM GMT • comment • Reads 1094 Article posted November 5, 2009 at 02:45 PM GMT • comment (3) • Reads 1190 Dear Mom and Dad, In math class we just finished Module 1 in our textbook. In the module we learned all kinds of new things. The ones I like the most were probability, creating bar graphs using frequency tables and order of operations. We did many activities involving probability. Like picking cubes out of a brown paper bag, rolling dice and others. The one I liked the most was the one when we either worked alone or with a partner. I worked with a partner and we had to create a portability experiment. We called it bit size math. We picked skittles out of a bag 140 times. We made a frequency table of our results. The class also learned how to do create bar graphs using frequency tables. We did a take home quiz which we had to find a graph and answer questions about it. Like what is the graph about, what information was displayed on the vertical axis and what decision can the graph help a person make. We also created frequency tables with information that we surveyed from the class. Then we created bar graph based on the information we found from the class. We also learned more about order of operations. I was familiar with order of operations already but I was introduced to brackets. Ms. Harte let us listen to a song to help us remember the order of operations. One of ways to help you remember it is PEMDAS p for parenthesis, e for exponents, m for multiplication, d for division, a for addition and s for subtraction. The class did many small activities involving order of operations. Over all throughout this first module I think I learned a lot. Love, HJOL Article posted November 5, 2009 at 02:45 PM GMT • comment (3) • Reads 1190 Article posted October 26, 2009 at 01:46 PM GMT • comment • Reads 1175 Dear parents, In math class we just finished module 1 and I am going to tell you the stuff we learned in module 1. Well before I start, lets just say that you are going to be very proud about the things that I have learned. Well in the beginning of the year we were learning about frequency tables. They are very easy to make, just put the subjects in a spot on the table then count out how many there are and tally them then count how many there are for each one and right down the number. Sometimes they can even be fun to make. Later we learned about probability it isn't the greatest topic to learn about but it was okay. We learned about experimental and theoretical probability. We played probability simulator games on the computer which was a little fun. We also played a game called the difference game which is about probability. It wasn't the easiest topic to learn about and not that fun to learn about either but it was alright. In the end we learned about the order of operations also known as PEMDAS which is really fun to do and fun to learn about. It is really easy to do the problems but it is also really fun. We listened to this weird but catchy song about PEMDAS. I think the order of operations is probably the most fun thing that we learn about in math. Well that's just a few things we learned about in module 1. Its a great start for the year and I know that it is going to get harder but if it fun to learn then it wont be as hard as it would normally be. Have a good day and I hoped you enjoyed my letter and liked what I learned about in math class. Sincerely, HRTY Article posted October 26, 2009 at 01:46 PM GMT • comment • Reads 1175 Article posted October 28, 2009 at 01:25 PM GMT • comment (1) • Reads 1257 Dear Sister, In math class I learned how do a lot. We have done a lot with number sentences like order of operations and how to make equations replacing numbers with letters. The other thing we learned was probability. I learned how to do algebra by taking shape patterns and number sequences and finding a equation using N and T. An example would be T=6xN+1. I also learned how to add exponents to the equation. For example you could take the 6 and make it 6 to the 2nd power. We also learned probability. We learned how to pick cubes out of a bag. We learned how to figure out theoretical probability and how to make a frequency table. The problems we did gave us the amount of items and we had to calculate what the probability that each category. The last thing we learned was order of operations we learned how to use brackets and parenthesis and exponents. I think the order of operations should really be pemdas because you do brackets before multiplication and division. I hope you understand the explanations I have given you. Sincerely, GRKE Article posted October 28, 2009 at 01:25 PM GMT • comment (1) • Reads 1257 Article posted October 29, 2009 at 01:41 PM GMT • comment • Reads 1143 Parent-Teacher Letter Dear mom, we have been learning all about module 1 in our math textbooks. So in these next 3 paragraphs, I will be explaining all about frequency tables, sequences, and number tricks. So the first thing we learned is a frequency table. A frequency table is a table that has to do with how frequent (many times) a thing happens or has. For example, if I was taking a survey on how many have pets, I might do how many people have dogs, cats, fish, no pets, and other. While I was surveying, I would have done tallies to keep me on track. After I was done, I would have tallied up the tallies and in another column, I would put the total amount of tallies in that column. So to make a frequency table, you collect the data, tally it, and put the total amount in the table, and that's how you make a frequency table. The next thing we learned is how to do sequences. A sequence is a pattern that keeps going like 2,4,6,8,10,12,14,16,18,20, etc. To make a sequence, all you have to do is simply make a pattern and to keep it going. For example, if I made a short pattern, (10,20,30) I could keep it going to make a sequence. (10,20,30,40,50,60,70,80,90,100 etc.) So to make sequence, all you need to do is make a pattern and keep it going. And if you want to, you could make the sequence into shapes if you wanted to. The last thing we learned of was number tricks. Number tricks are tricks that you can do with any number or numbers and come out with the same number every time you do the trick with different numbers. For example, I'll give you a number trick and you can figure it out for yourself. (Think of a number, add 7, multiply by 5, subtract 5, divide by 5, subtract your original number, add 4, subtract 3, and your answer should be 7. So number tricks are nothing but little tricks so that you'll come out with the same thing every single time. So over-all, we learned all about frequency tables, sequences, and number tricks in module 1. All of the students are very good at these math lessons and we are moving on to module 2 and we hope that our progress will improve. Article posted October 29, 2009 at 01:41 PM GMT • comment • Reads 1143 Article posted October 26, 2009 at 01:52 PM GMT • comment • Reads 1102 MODULE ONE Dear parents, 10/22/09 I have been learning many important math facts throughout the last module in my math class. I will give you three examples of what I've been learning. In the first few weeks of school, I reviewed bar graphs and line graphs. I also analyzed a graph to understand how a graph works a bit more. Each graph has a horizontal axis and a vertical axis. It is better to use a line graph if you have a lot of information. This is because with a line, it can go up and down quite easily. With a bar graph, you can only have so much information displayed because it would get all jumbled up, and you would run out of room. So if you have more information, you should use a line graph, and if you have not too much info you should use a bar graph. I analyzed a graph by taking a graph from a newspaper, and answering questions about why a line graph or bar graph was used. So basically I studied each part of a line and bar graph during this module. I have learned more about probability throughout studying this module. I now know more about theoretical probability and experimental probability. The way that I have discovered this is by doing certain experiments and finding out the probability of each event. For example, I used a spinner to determine the theoretical and experimental probability of spinning each color on the spinner. Say there was the color red on the spinner, and it took up ¼ of the spinner. Then the theoretical probability of spinning red would be ¼. If I span the spinner 50 times, and only got red 35 times, the experimental probability would be 35/50 or 7/10. We also did an experiment with probability. What I did is I put 24 different colored skittles in a bag. I put in 6 yellow, 5 red, 4 green, 5 purple, and 5 oranges. I figured out the theoretical probability and the experimental probability of pulling out each color. The most recent thing that I learned about was the order of operations, also known as PEMDAS. In the word PEMDAS, the P stands for parentheses, which means that any equation in parentheses would be done first, then the E stands for exponents, M is multiplication, D is division, A is addition, and S is subtraction. The way that we learned the order of operations was by practicing each day different problems involving the order of operations. Even though the order of operations was not new to me, I still learned even more about it than I have ever known. In this first half of the first semester, I have learned so much already! This module has been great! Sincerely, CSMA Article posted October 26, 2009 at 01:52 PM GMT • comment • Reads 1102 Article posted October 28, 2009 at 02:52 PM GMT • comment (1) • Reads 1178 Dear Dad, Hi Dad! It's Rhbr! I'm at school right now writing this letter for a math project. The idea is to write about everything we've learned about this year so far in math class with Mrs. Harte. We kind of did a lot of reviewing at first, but it was pretty useful because I didn't remember a lot of it. It was also nice because that way we got to sort of start of the year a little easy at first. One of the first things we kicked off the year with was analyzing bar, point and line graphs. A lot of times we would look at why a bar graph is a bar graph and not a line graph and vice versa, based on the information the graphs displayed. After we did that for a while we worked a lot on frequency tables, which was something new. We did pattens, sequences, and algebraic equations to figure out the rule, which we displayed in the table. We also did a lot of exponents-- squaring area and cubing volume, as well as a little bit of binary numbers, which was confusing at first but then it got simpler. The practice for exponents was useful, too. But I think they're really annoying and tricky. Another interesting thing we worked on was number tricks. There was one where you punch in your phone number on a calculator and you perform a lot of tricks and because you multiply it by a fraction, it shrinks when you multiply and grows when you divide, but it took a while for us to realize that. Another part of the number tricks was that we would get a worksheet with different number tricks on it and we would solve them with 2 different base numbers, and then use a system of blocks (representing your base number) and circles (representing 1) and solve it. Later, we started using algebraic equations to solve them. The algebra made more sense to me. After that we moved onto probability, which was another review. We worked with experimental versus theoretical probability in most of what we did. We did some experiments with the color of blocks or names in a bag, and there was a website that we looked at with spinners and what colors appeared more depending on how many trials there were. We also did a partner project on theoretical versus experimental probability, which in my case was about the colors of M&M's vs. Skittles, which was fun. And right now we're finishing up a unit on order of operations, which was kind of interesting because I learned about brackets and fractions in it. SYS (That's texting talk for Sincerely, Your Son.) Article posted October 28, 2009 at 02:52 PM GMT • comment (1) • Reads 1178 Article posted October 28, 2009 at 01:26 PM GMT • comment • Reads 1138 Dear Mom and Dad, We have learned a lot in math class! From graphs to order of operations, here's what we've learned so far this year. We have learned about bar and line graphs. We used bar graphs for things such as surveys, data, and frequency charts. We used line graphs for things like data over a period of time. For example, stocks would be on a line graph, but favorite type of dog would be on a bar graph. To learn more about these, we had a assignment to find a graph and tell as much as we could about it. It was really fun! Another thing we learned about was the order of operations, or PEMDAS. PEMDAS stands for Parenthesis, Exponents, Multiplication, Division, Addition, and Subtraction. That's the order we do operations. We even sang a song about it! A sentence in parenthesis would come before a multiplication sentence, but an addition problem would come after division. Brackets ([]) extend parenthesis if there is a problem that goes in parenthesis, but has another parenthesis in them. Here's an example of brackets. [3+ ( 5+5)] +2 Another subject we learned about was probability. Probability is the chance of something happening. For example, if I said that there were ten candies in a bag and five of them were blue and five of them were red, the probability of picking red would be 5/10, or ½. We also learned about two other types of probability, experimental probability and theoretical probability. Theoretical probability is what should happen. If I asked what the chances were of picking a red candy out of a box that had 3 red, 5 blue, and 2 yellow, the theoretical probability is 3/10. If I asked you to pick a random candy in a bag without looking ten times, you might get a different answer, which is experimental probability. Experimental probability is what actually happens. If you did it ten times, you may have gotten a blue seven times, a red twice and a yellow once. Math class has been fun! I don't think I've disliked anything. I hope you found this letter useful. I can't wait to see what we do next! Sincerely, LSGA Article posted October 28, 2009 at 01:26 PM GMT • comment • Reads 1138 Article posted October 28, 2009 at 01:27 PM GMT • comment • Reads 1140 Dear Dad, In this letter, I am going to explain what we have been doing in math class so far. Well, we have just finished module one. Most of it was just a review from last year, so it was pretty simple. The main things we went over were analyzing bar and line graphs, patterns and sequences, probability, and the Order of Operations. We also did a little bit of algebra. My favorite part of the module was doing experiments on probability. There are two types of probability: theoretical and experimental. Theoretical probability is what should happen in an experiment, and experimental probability is what does happen. So, say I was going to do an experiment with picking marbles from a bag, and I had 4 red marbles, 4 blue marbles, and 4 green marbles. Then the theoretical probability of picking red would be four twelfths or one third. But, the experimental probability might be something different. Okay so, maybe I did the experiment 24 times, and I got 9 blue marbles, 7 red marbles, and 8 green marbles. Then the experimental probability of getting red would be seven twenty-fourths. We did a project like this where we got to choose our topic. My project was egg-picking. I drew 4 different faces on 12 eggs. There were 3 happy eggs, 3 sad eggs, 3 surprised eggs, and 3 angry eggs. My least favorite part of the module was probably analyzing bar and line graphs because it was boring and easy. I mean, all we did was look at a graph and answer questions about it. There's not much more to say about analyzing graphs so I'm going to talk about the Order of Operations next. The Order of Operations is a way to figure out problems with more than one operation. So if you had a problem that was 4+(1-2)/3, you couldn't just do it from right to left to right. There's a certain order and that order is the Order of Operations. So, what is that order you might ask? Well, it's: Parentheses Exponents Multiplication and Division Additon and Subtraction OR PEMDAS! A good way to remember it is Please Excuse My Dear Aunt Sally. Or as a fellow class mate once said: Please Execute My Darn Aunt Sally Well, that's all I have to say about this Module! Bye For Now, ~HSJO~ Article posted October 28, 2009 at 01:27 PM GMT • comment • Reads 1140 Article posted October 28, 2009 at 02:46 PM GMT • comment • Reads 1220 Dear Sister, I'm writhing to you today because its for a assignment for math class and I wanted to let you know what I'm doing in class. What I've been learning in math this module is how to make frequency tables and the order of operations. Frequency tables give you information for bar graphs or line graphs. A table could be about one thing or five different things like colors and coins. With colors it could have lots of different colors or categories in the table. Where in coins every coin is the same. You can also have a sequence in a table by having terms and term number to make an equation like T=N+2. I am also learning about the order of operations. The order of operations is spelled out to help students remember it. Its called PEMDAS! Teachers also sometimes get songs about the subject we are learning. PEMDAS stands for Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction which are called the operations. You might ask yourself why the put it in as one word? They do that because of the order they have to go in. Order of Operations! The last thing we have been learning about is bar graphs and line graphs. How are these two graphs different? These graphs are different because a bar graph can have more than one categories when a line graph has one category and is usually about something over a certain period of time. Sincerely, DSLA Article posted October 28, 2009 at 02:46 PM GMT • comment • Reads 1220 Article posted December 3, 2009 at 07:16 PM GMT • comment • Reads 1126 Dear, Brother I am writing this to you because my math teacher thinks we should write a letter about what we have been learning in module one of our math book. So we have been learning about, bar graphs, line graphs, analyzing a graph, frequency tables, number tricks, variables, algebra, patterns and sequences, tables and graphs, equations, terms and term numbers in frequency tables, exponents, exponential form, standard form, area, perimeter, volume, probability (theoretical and experimental),- we put names in a bag, cubes in a bag, a computer simulation, our own experiment that we got to make up, and we learned more about the order of operations, and we listened to a song about it. My favorite part about the module was number tricks, probability, and order of operations. I liked number tricks because they are fun to do and I liked how the number tricks always work and that you could start with a number and then it would end up being your age in the end. Try doing this, start with your age, multiply by 4,divide by 2, subtract your age and what is your answer? I liked probability because it was really easy for me, and it was fun to do. The order of operations has always been something that I liked to do, I like it when I am challenged because I like to see how good i am at it and to see how fast I can go. But the order of operations is just fun for me, and a lot of the problems are really easy for me. I didn't like the graphs that we had to do. The first thing we did with graphs was we had a take home quiz and we had to find a graph so that we could fill out the sheet that my teacher gave me. I didn't like it because it was a little bit boring and I already knew just about everything. I also didn't like exponents. I didn't like exponents because I don't like doing the big numbers like that, and I don't like how you have to have it in exponential or standard form. But other than that math and school are going well, and I have an A in math right now. Love, dsno Article posted December 3, 2009 at 07:16 PM GMT • comment • Reads 1126 Article posted October 26, 2009 at 02:34 PM GMT • comment • Reads 1108 Dear mom and dad, I am writing this letter to tell you what we have done in math class. We have studied a lot of subjects in module 1 and I want to tell you about them. We have studied bar graphs, line graphs, number tricks, patterns and sequences, exponents, probability, and order of operations. I will explain three subjects that we learned and hopefully you will understand them. When we were studying exponents we found the area of a cube, the volume of a cube and square roots. We also studied exponential form and standard form. Exponential form is when a multiplication problem is just simplified. If the problem is 5x5x5x5 you could simplify it by doing 5 to the 4th power which is a 5 with a little four right after it. Standard form for the problem above would be 5x5x5x5 and exponential form would be 5 to the 4th power. The little number is the amount of time the first number appears in the problem. When we were studying Probability in class we put a name in a bag and we picked them out one by one and we had to find the probability of what the chance a girl will be pick a chance that a boy will be pick and other chances. We also put three different colored cubes in a bag and we had to decide how many of each color is in the bag based on probability. We did computer simulations about the probability of picking rolling or flipping a certain outcome. We played games that had to do with probability and we created our own game. There are two types of probability and they are theoretical probability and experimental probability. Theoretical probability is what should happen and experimental probability is what does happen. We also learned the order of operations. If is the order you should go in when you have a problem that is complicated. I remember it by using a mnemonic device that is PEMDAS. It stands for Parentheses, exponents, multiplication and division, then addition and subtraction. That is how I remember the order of operations. When we were studying the order of operations Mrs. Harte (our math teacher) showed us a really funny song. That is what we did in math class. These are some things that we have learned in math class. We have learned a lot of new things and have done a lot of fun projects. I hope you like my letter. Sincerely, Your daughter Article posted October 26, 2009 at 02:34 PM GMT • comment • Reads 1108 Article posted October 28, 2009 at 02:50 PM GMT • comment • Reads 1168 Dear Mom, Mom we have been learning a lot in math. We have covered so much that I want to tell you all about what we have been learning in 7th grade math. The first thing we learned was Bar graphs and line graphs. Also how to analyze graphs. Learning if we should make it a line graph or a bar graph. Another thing is exponents, I like doing exponents, they are not bad. Exponents are shortening a really long problem, say two to the fifth power. We have been learning about probability also. There is a couple of things we did on probability, we put names in a bag of our class and pulled each one out and found its probability. There was theoretical and experimental probability. We have done lots of equations, like term and term numbers. Another thing that we have been learning in math is sequences, and finding what will be the next sequence. When we are learning something new we usually get to play a game to understand what is going on and help us with key concepts. Also we are learning about squares, cubes, and finding the area. With cubes it is finding the volume. We have learned so much in math class and we are going to keep going, we are all done with module 1. I can not wait to tell you about what will be happening in module 2. Love, BSVI Article posted October 28, 2009 at 02:50 PM GMT • comment • Reads 1168 Article posted October 28, 2009 at 01:29 PM GMT • comment • Reads 1102 Dear Parents, In my math class in module one I learned a lot of things, mainly divided up into 6 areas. Graphs, number sequences, number tricks, exponents, probability, and order of operations. In the graphs unit we learned about bar graphs, line graphs, and how to tell the difference between them. It was a unit that bled into the second area, number patterns. Number patterns are just patterns in numbers. For example, 0, 3, 6, 9, 12, 15, 18, 21. What is the pattern? You simply add 3, although the patterns we did in class were much more difficult so don't expect our math class to be in in cakeland or whatever. The third area we covered was the number tricks area. Take a number, add 3, multiply it by 2, add four, divide by 2, and subtract your original number. Is your answer 5? If its not, then you did it wrong! The fourth area was exponents, which is a number that tells you how many times to multiply a number by itself, but you probably already know how they work, so I won't dwell on that. The fourth area we covered, which was pretty important, was probability. Suppose you roll a die. How likely is it that you roll a one? That is called probability, and is extended to all random events. Our class did experiments with that, and I worked with ahem my partner on our “eggsperiment”. The final area, last but not least, was order of operations, which is the order in which you do longer equations. 9+8x2 would equal 25, not 34, because of the order of operations, and PEMDAS, an acronym for parentheses, exponents, multiplication, division, addition, and subtraction, which is the order in which yo do the problem. I hope you had a good time learning about what I am doing in this unit, but I can just to you about it at home so I won't write any more. Sincerely, BHMA Article posted October 28, 2009 at 01:29 PM GMT • comment • Reads 1102 Article posted October 28, 2009 at 01:23 PM GMT • comment (1) • Reads 1175 Dear Mom, In math class we have learned a lot from module one in the math books, and now I am going to teach you some things I've learned. I have picked three topics that I liked more than some, but they are all pretty good. The first topic that I liked was probability. Say that there were ten marbles in a bag, and there were four different colored marbles. There were six white marbles, two red marbles, one black marble, and one blue marble. A question could be... What is the probability of picking a red marble? The answer would be...2/10 of a chance of picking a red marble. Ms. Harte did something with us like that, but she used names on pieces of paper and different colored cubes. Another topic that we learned, and or practiced was the order of operations. I liked that because it was fun to multiply, divide, add, and subtract. We used PEMDAS to help us remember it a little better. P is for parenthesis...E is for exponents...M is for multiplication...D is for division...A is for addition...and S is for subtraction. An example would be......4 x (3 + 8) – 5 = 39. Parenthesis comes first, exponents comes second, multiplication and division...you do it from left to right, and addition and subtraction comes from left to right. We also worked on algebra, I liked that because it was fun to learn and do. An example for the algebra would be...T=4N + 6. The T is the term, the 4n is how term numbers, so the term number four times, and then plus six. These are some topics from module one that I liked. And hopefully I can teach you. Sincerely, WJAU Article posted October 28, 2009 at 01:23 PM GMT • comment (1) • Reads 1175 Article posted November 5, 2009 at 03:35 PM GMT • comment • Reads 1273 Dear Mom, This module we have learned many new and interesting things like probability, number tricks, frequency tables, and more. One of the first things we learned about math were line and bar graphs. We did a lot of things with graphs. I also learned when to use a bar graph or line graph. You use a bar graph for categories and a line graph to show trends over time . Then we got a take home quiz called Analyze a Graph. We had to cut out a line or bar graph out of the newspaper or a magazine, glue it onto a piece of paper, and answer questions about it. Next, we started studying frequency tables. We did some questions in the book about them like what they are, how to make them, and how they help us. We also started the unit on probability. We studied theoretical probability, experimental probability, did a probability project, and too many more things to count! We did something called names in a bag where Mrs. Harte but all the names of the people in the class in a bag and whoever was picked first and whoever was picked last got 10 math points. The winners were PJCH, who was picked first and WRCL, who was picked last. Even though I didn't win, it was a lot of fun. We also did something called cubes in a bag where Mrs. Harte put cubes in a bag and we recorded what she picked each time on a frequency table and at the end, we had to guess how many cubes there were of each color based on the information on the frequency table. Also in probability, we did an experiment. I did mine with WRCL. It was about picking Jolly Ranchers out of a bag and finding both the theoretical and experimental probability of each flavor. I was a lot of fun and we got a yummy treat after! I really enjoyed studying Module 1. I learned a lot and I had fun! From, SHBR Article posted November 5, 2009 at 03:35 PM GMT • comment • Reads 1273 Article posted October 26, 2009 at 01:53 PM GMT • comment (1) • Reads 1190 Module One Dear parents, 10/22/09 In math we are learning to do a lot of things. Some things are totally new and other things I know most about. Some of my favorite things that we learned are line graphs, bar graphs, and order of operations. Some of my least favorite things are probability, and volume of a cube. My favorite thing about a line graph was taking it from a frequency table and then making a graph out of it. We graphed a lot of things that were in out math books, but some things were not in our books. We learned about bar graphs at the same time that we were learning about line graphs too. The bar graphs were almost the same as the line graphs only instead of using a dot you use bars Another one of my favorite things was order of operations. The order of operations is P.E.M.D.A.S. It stands for Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. For example (2+4) x 300- 250+ 80, you would first do 2+4 because it is in parentheses. Second since there is no exponents you would do the multiplication or division which would be the x300. Lastly you would do addition or subtraction so -250 or +80. That is the Order of Operations. I liked almost everything that we did in math class but some things I did not like. One of them was the number tricks. I thought that they were kind of hard at first, but then I began to understand them more. Once I started to understand them then I liked them a little more. I also think that I didn't like then because it seemed like we stayed on that subject for a while (but we really didn't). So far we have learned a lot of things in module one in math class. Some things were totally new and other things we just reviewed. Most of the things were fun like PEMDAS, and the line and bar graphs. Thank you for reading what I have learned in Module one in math class! Sincerely, LHAL Article posted October 26, 2009 at 01:53 PM GMT • comment (1) • Reads 1190 Article posted October 23, 2009 at 02:12 PM GMT • comment • Reads 1204 Dear Dad This year so far math has been mostly a review for me. From line graphs to order of operations I've really only learned two things. Math has been really easy and only two problems really challenged me. The first things we learned about was bar graphs, line graphs, and frequency tables. Bar graphs and line graphs I knew about and that was very easy. Frequency tables were slightly harder because it was a concept we worked less on in 5th and 6th grade. We worked on # tricks after that and while it might have been a bit harder for the rest of the kids, I learned about it with my parents and Mrs. Harte. After that we did patterns which was in the same boat as the #tricks. Next we had exponents, probability, and order of operations. I'm going to skip probability and give it it's own paragraph. Exponents seemed ridiculously easy (though not as easy as the graphs). We had spent a lot of time on exponents in 6th grade. Order of operations was harder (a little bit) because we learned about brackets and fraction symbols. We spent a lot of time on probability and learned a bit about simulations. We also did things with a bag of cubes and a bag of names. We learned about theoretical and experimental probability. We also did a project on probability. My project was a lot harder than everybody elses. I took a 4,6,10,and 20 sided dice., then I put them all in a bag and randomly chose and rolled one dice. After that I came up with the probability of rolling every # 1-20 So, so far through this year it has basically been one big review. Sincerely, LRTU Article posted October 23, 2009 at 02:12 PM GMT • comment • Reads 1204 Article posted October 26, 2009 at 01:54 PM GMT • comment • Reads 1163 Dear Sister, I have sent you this letter because I want to tell you what I have/am learning in Math class. We have learned what type of information to make a graph out of. Like, if you were doing the change in weather in cities, you would make it a line graph. And if you were doing favorite types of ice cream it would be a bar graph. And we've learned to analyze the data from graphs too. We've worked on sequences and patterns and learned how to make them into tables and how to graph them. How to turn them into algebra equations. Like T=n+3. So we'd take block patterns and make a sequence for them. We've worked on probability like, putting cubes in a bag and tallying how many times a certain colored cube comes up. And we would make tables out of that information. Another probability game that we played was when ms. Harte had everyone in my classes names in a bag and shook it and gave math points to whom ever was picked first and last out of the bag. I wasn't chosen. That's some of what we've been learning in math class this term. I'm pretty sure you learned it last year or it refreshed your memory. Your sister, Drco Article posted October 26, 2009 at 01:54 PM GMT • comment • Reads 1163 Article posted October 28, 2009 at 02:49 PM GMT • comment • Reads 1090 Dear Mom, I have learned so much in math this year for module one. I already knew a lot of them but some of them I have gotten a lot better at. Ms. Harte helps us with a lot of things that we get stuck, so, so far I haven't had any problems in difficult areas. Like last year I really didn't understand how to do the order of operations. This year we got to even got to hear a song thaw was awesome. Order of operations got a lot easier and my test scores for it were really good. The brackets for me were new but I got the hang of them because of all the activities we did. Frequency table last year were also very hard for me. So we had much practice using graphs and certain activities to understand probability in so many different ways. My friend and I worked on a project together about experimental probability and theoretical probability. I have gotten a lot of good test scores because of only really doing reviews. I look forward to you reading this. Sincerely, Your Son Article posted October 28, 2009 at 02:49 PM GMT • comment • Reads 1090 Article posted October 28, 2009 at 02:50 PM GMT • comment • Reads 1150 Dear Dad, In Math class my grade has just finished Module 1. It was a good unit with lots of learning and games. We went through some really fun things too. Some of my favorites were our probability unit, computing with exponents, and the order of operations unit. Our probability unit was fun. We had games that we had played to help us solve probability. One game that we played was putting all of our names in a bag. Then, we pulled out the names one-by-one. The first and the last names got 10 Math points. I was second to last and so close. Every time we took a name out of the bag we would say the fraction out loud. Then when it was just another person and me, everyone stated the probability of who it would be. We also worked on experimental and theoretical probability. We did an experiment on it. I chose to do a project on dice. I would role two dice and take the smaller number away from the bigger one. I would use the sum for my tally. I showed what I thought would happen (theoretical) and what did happen (experimental.) When we did computing with exponents I really learned a lot. It really cleared up my understanding of exponents. We did a lot of exponent work sheets. For example 5 to the second power is 5 times 5, which is 25. This unit really helped me practice and nail down the technique. We did drills so that we would know how to write in Standard and Exponential form. Standard form is when you take the exponent and and write in regular form (Example = 4 to the 2nd power is 16.) Exponential form is when you take a number sequence and write it as an exponent (Example = 3 x 3 x 3 x 3 is 3 to the fourth power.) This unit was good and very helpful. The Order of Operations unit was my favorite unit. We learned the order of operation last year, so we reviewed it this year, and learned more advanced things. We learned how to do the order of operations in a fraction. We also earned how to do the order of operations wit exponents. In class we would do fun races in with larger problems. The factions were tricky at first, but once you got them down they became fun. The order of operations is Parenthesis, Exponents, Multiplication, Division, Addition, and Subtraction. -Otherwise know is PEMDAS. We had a song about the order of operations and PEMDAS, it was really funny. Overall this Module was good and brought on challenges and fun activities. I hope you enjoyed this letter covering Probability, Exponents, and the Order of Operations. Sincerely, BSLI Article posted October 28, 2009 at 02:50 PM GMT • comment • Reads 1150 Article posted November 5, 2009 at 03:30 PM GMT • comment • Reads 1185 Dear Mom and Dad, In math we learned about graphs. I made a bar graph on my computer. I counted how many colors of cubes I had. I had red, yellow, orange, blue and black cubes. I opened up Excel on my computer and clicked on graph, and when I put my numbers and colors in, it made a graph. I am learning subtraction, too. Subtraction is kind of hard. I am practicing with blocks, money and counting pictures of things on worksheets. I am having a hard time remembering how to figure out what the answer is in subtraction. I am also practicing with money. I know how much pennies, nickels, dimes, and quarters are worth. I know that there are five pennies in a nickel, ten in a dime, and twenty-five in a quarter. I am using the coins to practice addition and subtraction. It’s a little hard sometimes, but I’m doing my best! Ms. S has been helping me practice my shapes. I know triangles, squares, rectangles, hexagons, and even more! We play a game and I have to tell her the name of all the shapes. I try to get the answer right. I make some mistakes, but I work through it. The most important thing I have learned in math class is that it’s OK to make mistakes. I know I can fix them. When you make mistakes and fix them it’s called problem solving. I know that everyone makes mistakes, and that’s part of life. It’s all a part of learning. Love, JSHA Article posted November 5, 2009 at 03:30 PM GMT • comment • Reads 1185 Article posted October 28, 2009 at 01:27 PM GMT • comment • Reads 1227 Dear Mom, In Math we have been learning about a lot of things but I'm going to tell you about three of them. They are probability, Order of Operations and number tricks. The number tricks are so fun and I actually made my own. In the problems you pick a number and go down the line and it tells you to either add subtract multiply or divide by a certain number. Sometimes it comes out to be your same number that you picked before or maybe they pick a certain number to finish on. We also did Order of operations. An Easier way to say it is PEMDAS which stands for Parentheses, Exponents, Multiplication, Division,Addition, and Subtraction. That means that you do Parentheses first then Exponents then Multiplication or Division. Next you do either Addition or Subtraction. Its like that because you go from left to right and pemdas should really be spelled like this P because you go from left to right with m,d and the left to right E with a,s. MD AS We also did probability. One thing that I want to talk about is the computer simulated games for probability. On the computer if you went on a certain website it would calculate it for you. You could use coins,spinners,dice,and cards. I was very fun and useful. Those were some of the things that we learned in module 1 of math. Love, NJMA Article posted October 28, 2009 at 01:27 PM GMT • comment • Reads 1227 Article posted October 28, 2009 at 01:28 PM GMT • comment • Reads 1159 Dear mum, Hello. We have been learning a lot in math! We are learning theoretical and experimental probability. What that is, is theoretical probability means to know what a fraction is out of an amount of large numbers for example, if I have 30 guitar picks, and 10 of them were red, 9 of them were blue, and 11 of them were green, then my probability would be 20 over 60 for red, 18 over 60 for blue, and 22 over 60 for green because we would double the numbers. For an experimental probability, that would be you actually doing the experiment and then writing down what you got in a fraction form, for example, if I were to pick out of the same bag 60 times, and I got 20 reds, 30 greens and 10 blues, then it would be, 20 over 60 for red, 30 over 60 for green and 10 over 60 for blues. We would then compare our experiment probability with our theoretical probability and say the differences that we see. In this case, the differences are that we got 10 over 60 for blue and our theoretical probability was 8 away. For the green, we got 30 over 60, but for our theoretical probability it was 8 more than our theoretical probability. For the red we actually got the same amount for both our theoretical and experiment probability. We have also been learning exponents. Exponents are simply only exponential form and standard form. Standard form is when you write out a whole problem. For example, 3x3x3 would be in standard form when you find the answer. Exponencial form is when you write it out in a shorter way. For example, 3 to the 3rd power. Then, you would solve it. It would be 18 for the answer. We also worked on our graphs. We used bar graphs and line graphs. We worked on figuring out which graph to work on with different information. We took a take-home-test on graph's. We found out all of the information on them, and past it in about a week later. We looked at the map to find the vertical an horizontal sides name. It was a very good module. Sincerely, Whab Article posted October 28, 2009 at 01:28 PM GMT • comment • Reads 1159 Article posted December 3, 2009 at 07:05 PM GMT • comment • Reads 1334 Dear Mom and Dad, In the first module we did a lot of stuff similar to what I've been doing. It was kind of a review of the end of last year with some new fun things. We did a lot with with making tables into different types of graphs. The type of table we used to organize our data is called a frequency table. We used bar graphs and line graphs to turn those tables into something else. Bar graphs are good when you have a number representing on the vertical axis and usually a description or noun on the horizontal axis . Line graphs are a good way to put the stocks of a company into a graph, so you can see the increase and decrease. In the beginning of the year we learned a lot about number tricks. They are a fun way to use math. We used them with shapes like circles and squares. We used them with variables and a little bit of algebra. One type was when you would take your home phone number and you would end up with it again. Number tricks was a fun way to do math. We did a lot with probability. We learned about experimental probability and theoretical probability. Experimental probability is the amount of something actually happened. Theoretical probability is the amount of something that should happen. We used probability with, “Names In The Bag.” We all put our names in the bag and the first and the last people to get picked won. We counted the probability that a person would get picked. The probability was getting bigger after every pick. In the first module we did a lot of things that make math fun. We did a lot more stuff than these things. I have to let you know the, “Math Song” was awesome. Everybody hated it, but me and Mark. The song was on the Order of Operations. I hope we do more fun things in the next module. Love, WSRE Article posted December 3, 2009 at 07:05 PM GMT • comment • Reads 1334 Article posted October 28, 2009 at 05:51 PM GMT • comment • Reads 1070 Dear Dad, This is a letter about module 1 in math the paragraphs are gonna be kinda bad though because there has to be 3 of them. But I still have to do it so here goes... In the early part of the module we learned about things like line graphs,analyzing a graph, and frequency tables . We also learned about patterns/sequences. Some examples of this are. that take-home quiz about graphs where you had to analyze a line graph. For frequency tables we had those txtbk. questions where you had to draw a frequency table . For sequences/patterns we had a few homeworks. Later on in the module we learned about , variables,writing equations like T=N2-1, area, volume,and probability. Some examples of activities to help us understand these subjects are. For variables we used them for patterns and writing equations. For writing equations we used them to see if a pattern was always repeated and didn't stop or if it would break. Area wasn't really new to me but we did have a couple slightly challenging ones. Volume was definitely more challenging and we had some very hard problems. For probability we used a virtual spinner,deck of cards,coins and dice; we also did a probability project in which we had to give theoretical and experimental probability (I did mine w/ cat food!). Lastly near the end of the module we learned about P.E.M.D.A.S. and problem solving. Here are some activities we used to enhance our learning experience. For P.E.M.D.A.S. we heard the order of operations song :(,we had to do a few worksheets and we also had a test on the order of operations. For problem solving all we really did were worksheets and text book questions. Sincerely, psja. Article posted October 28, 2009 at 05:51 PM GMT • comment • Reads 1070 Article posted October 22, 2010 at 05:30 PM GMT • comment • Reads 980 Dear Mom and Dad, This year in math class I have learned some new things. The first thing I learned in module one was how to interpret bar and line graphs, as well as learning how to make frequency tables. A frequency table is just a table where you record your data/ frequency that you find from your research or experiment. I have made data tables before so frequency tables were a review. The way you interpret graphs is pretty simple, all you have to do is look at a graph and extract important information that could help you answer any questions you have about a certain topic. A few things you need to know, you need to know what information is displayed on each axis, a line graph shows a set of data that changes over time, a bar graph shows a set of data recorded on one specific time survey. In section 2 we covered several topics. We learned how to represent number and shape sequences with graphs and algebraic equations. Here's an example, Term number 1 2 3 4 5 Term 1 4 9 16 25 Algebraic equation T=2N This equation tells you to multiply the term number by itself (term # x term #) to get your term. Using o an equation you can predict other terms in the sequence, this is another thing we learned. We also learned about exponents and how to use them. An exponent is used instead of multiplying a number by itself over and over again, here's an example, instead of writing out 5x5x5x5x5x5x5x5 you can turn this into an exponent which would be 5 to the eighth power. After learning how to use exponents we were able to find the volume of a cube by multiplying the length of one edge times 3 to get the volume. Volume is the area or space inside of a 3 dimensional object. If you call the length of one edge E the formula in exponential form would be E to the third power. Next is probability which is what chance you have at getting something. Here's an example , you have a normal 6 sided die, what is the probability (chance) of rolling an even number . The probability would be 3/6 because there are three evens out of a total of six possible outcomes. I did a project on probability with my friend ARJO, our project was called Pizza Party. For Pizza Party we cut out different types of pizza out of colored paper, then we placed the paper pizza in a bag, next we picked the different pieces and recorded how many times we picked each type of pizza, this was our experimental probability. I enjoyed this project and it helped me understand how probability works. Just recently we went over expressions and representations. The thing we worked on most was the order of operations, the order of operations is a set of steps on how to solve multiple step equations. The way to remember these operations is “PEMDAS” Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. This is the way to evaluate a complicated equation. If the problem looks like this, 40-10x5, you would do 10x5 first and then 40-50 which is -10. You read a problem from left to right, like a book. I also had a more in depth review on graphs and how to use them, I already explained this in the first paragraph. Most of this module was a review and was fairly easy for me. My favorite part was the order of operations and the algebra, I enjoyed these sections because they were the only part that was a bit of a challenge. I enjoy working out long, complicated problems and these allowed me to be challenged a little. I hope you enjoyed my little letter on what I’ve learned in module 1! Sincerely, AJJA Article posted October 22, 2010 at 05:30 PM GMT • comment • Reads 980 Article posted October 22, 2010 at 04:30 PM GMT • comment • Reads 970 Dear Parents, In Math we just finished module 1. Some of the things we learned were bar and line graphs, tallying data, sequences, exponents, squares, cubes, and Algebra. My favorite thing we learned is the exponents because I understood them the most and they are fun to do. Writing out the problem with something different than just the old boring numbers. My least favorite thing we learned was the Algebra because it was hard to understand it and it was just difficult but in the end I started to understand it more so it was all good. We did a probability project that helped me get what it was. I did it with a partner and some people did the project alone. My partner was AJJA. Our project was pretty cool it was called Pizza Party my favorite part was the name, it sounds just so fun. The quizzes we took were really short because they were on small parts in the module. The big test is in the end of the module like the one we just took. In class our teacher played this song on the order of operations to help us memorize what to do in the equation first. I remember by P.E.M.D.A.S. P stands for parentheses so you would do what is in the parentheses first. E stands for Exponents then you would do the exponents next, then M stands for multiplication that’s next after the parentheses and exponents. Next is D, it stands for division that is 4th in P.E.M.D.A.S. Next is A is for addition that is second to last and last but not least S is subtraction that is what P.E.M.D.A.S. stands for. Another thing to remember it is “Please Excuse My Dear Aunt Sally.” So that is some of the most important things we learned in our first module in math class. Thanks for the help when I needed it too. Your Son, ARJO Article posted October 22, 2010 at 04:30 PM GMT • comment • Reads 970 Article posted November 2, 2010 at 03:46 PM GMT • comment • Reads 978 Dear Mother, How are you? Oh good, just checking. I would like to talk all about what I have been working on in math class. To be more specific, Module 1. We learned about data displays, sequences and exponents, probability, problem solving, and expressions and representations. I hope you enjoy my overlook of math class. So to begin, I will talk about when we started working on tallying data, which is all about intervals, vertical axis, horizontal axis, frequency, bar and line graphs. We would collect data and make frequency tables, Frequency tables are when you make tallies into a table. It makes looking at data much easier. It has helped me a lot. A interval is a step between grid lines on a scale. A bar graph is a visual display of data that falls into distinct categories. A horizontal axis is a horizontal line one a graph that is labeled with the categories or with a scale and a frequency table is a table that shows how often each data item occurs. That was all we learned in the first part of Module 1. Next we learned more about exponents and sequences. This was my favorite section of Module 1 because I enjoy exponents and trying to figure out sequences. So in this section, we talked about sequences called terms. A term of a sequence is a number indicating the position of a term in a sequence. As we learn more about this, we carry on to a bit of algebra. Like finding patterns and such. When we find a pattern, we make it into a algebraic problem like T (for term) N (for number sequence) and then add the other numbers to make it an equation. T= 6N is an example of showing how to set it up. This means T= 6xN. After that we went over probability . I didn't really enjoy this section as much as I enjoyed the others. I don't enjoy anything to do with fractions. So anyway, probability is a number from 0 to 1 that tells how likely an event occurs. We did a project on probability. My group used gummy bears (thanks for buying them! ). We put the gummy bears in a lunch box, and tallied the number of times we got a certain color. We made a frequency table to show the data we collected. We added the theoretical probability and experimental probability. Experimental Probability is a probability determined by repeating an experimental a number of times observing the results. Theoretical Probability is a probability that can be determined without actually doing an experiment. If the outcomes of an experiment are equally likely. The next thing we went over was one of my favorites. Order of Operations. Order of Operations is the correct order in which to perform the mathematical operations in an expression. How we learned it was PEMDAS. Parenthesis, Exponents, Multiplication, Division, Adding, and Subtracting. You go from left to right when doing the equation. So here is an example of an equation. 7(6+7)2-5=1178. So how you figure this out is remember PEMDAS. 6+7 is in parenthesis so you do that first. There is an exponent over that so when you get 6+7 (which is 13) square that number and you should get 169. Because multiplication is next in the order of operations, you would do multiplication. 169X7 come out to be 1183. Next comes subtracting. You subtract 5 from 1183 and get 1178. That's basically all there is to PEMDAS and order of operations. I hope you kind of understand all about Module 1 and the first 2 months of math class! :) - AHKEA Article posted November 2, 2010 at 03:46 PM GMT • comment • Reads 978 Article posted October 22, 2010 at 06:53 PM GMT • comment • Reads 998 October 22, 2010 Dear Dad, I have just finished module 1 in math. I learned lots of things withing this module. I am about to tell you the topics we learned and some activities we did with them. We started off learning about graphs and when you should use a bar graph or a line graph. The time to use a bar graph is when you have information where there's categories involved. The time to use a line graph is when you have recorded information over time. The next thing that we all learned to do was frequency tables. When we first came to school we got a list of topics that we will learn in this module and frequency tables was one of them. When I saw that I was confused and wondered what that topic was about. Then we started to learn about them and it was a lot of fun. With the frequency tables we learned how to make them and record information on them. Another activity that we did was number tricks. Meaning that you take a number, add 2, subtract 4, Multiply by 7, and so on. We did 2 work sheets on that so that it would be easier to transition into algebra. I really liked algebra because it's like a puzzle to me. Mom says that when I get into high school the problems will seem even more fun and like a puzzle. I hope she's right! One of the other things that we did was we were to module a number sequence then have a word sentence, table, and graph to go with it. We did a couple of things with those and the transition into our next topic. The next topic was to find the nth number of a number sequence. Most of these transitions you will notice, relates to somewhat extent to each other. I think that these transitions helped me get the concept even before we started the topic. The next thing we learned about was to compute with exponents. Computing with exponents was one that I liked. Yet at the same time after a while I think it kind of gets annoying. Plus if you told me to compute a problem with exponents I wouldn't quite understand it until you explained it more. The next two topics are to do with probability. One is to use fractions from 0-1 to estimate probability. The first work sheet was called Introduction to Probability. 0 meant not possible and 1 meant certain. Then they asked you a bunch of different questions like will you have cereal for breakfast this week. And other different things like that. Then from there we transitioned into experimental and theoretical probability. We used spinners, dice, decks of cards, everything we could get our hands on for probability. The big project fro probability was a poster. You had to bring in your own supplies and you had to have a table with the theoretical and experimental probability with a frequency table. I used letter dice and was very happy with my end result. The last topic so far that we learned in math was order of operations. Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction or as you my know it as Please Excuse My Dear Aunt Sally or even PEMDAS. I had already learned all that last year so it was an review but the good thing about it was that I perfected it. I really liked math this past module and I can't wait to find out what I'm going to learn in the next! Yours Truly, BREM Article posted October 22, 2010 at 06:53 PM GMT • comment • Reads 998 Article posted November 2, 2010 at 03:57 PM GMT • comment • Reads 874 Dear Mom, In math class, we have just finished module 1, and I have taken away a lot from it. We studied many different subjects including interpreting bar and line graphs, constructing frequency tables, creating bar graphs using frequency tables, using variables to solve number tricks, predicting the nth term of a number sequence, computing exponents, probability, and the order of operations. I thought most of this was interesting to learn, but some of it caused me trouble. In this letter I explain how we learned what we did and explain the subject to you. Probability is on subject I loved learning about because I thought it was easy. It is basically learning about the numbers of chance. Probability explains what the outcomes of an experiments probably will be. For example, my good friend and I did and experiment on the probability of gummy bears. In this case, we put about 100 gummy bears into a plastic bag and picked one gummy out 100 times, and with that we were able to find the experimental probability of what color the gummy bear will be. With 25 red, 25 yellow, 25 orange, ad 25 green, the probability of picking out a red would be 25/100 because there are 25 red and 100 chances of picking a red. As we learned this subject, things got harder to deal with but I payed attention in class so everything worked out in the end! :) Another unit in module one we explored would be a unit full of numbers and patterns and number tricks, which mystified almost all of the people in my class. We first leaned how to decipher a pattern by studying the numbers and shapes in the term and finding what number will some next. Usually, I found the pattern really quick, but some of them were so hard I had to give up and work on another.. For example, if each term was double the term number plus 2, the basic equation that solves every term would be 2N + 2. (N = term). After learning number and shape patterns, we moved on to number tricks, and this part of module 1 was very interesting! The way these work is simple but confusing, too. When you begin, you think of any number you want, unless it say otherwise. Then you are put through many mathematical events witch ends with the number you started with, or one number every time. These were so much fun! But I tried making one, and it is so hard! Line and bar graphs also made their way back from second grade! :) But this time learned how to decide to use a bar or a line graph. This is how we know. If you would like to show how data progresses or changes over time, the right choice would be a line graph. If you would like to just show the data that doesn't have many changes that are important, and bar graph would suit that set of data. We also learned how to analyze a graph by being tested with a take home quiz. We answered a series of questions. And when using graphs, we had to make a frequency table, also. Frequency tables were a way to collect data using tallies, and writing the final frequency on one of the columns. Most of this was a review, but I also learned something in module 1.This was Exponents in the order of operations. (if your confused this means how to use exponents in an equation when it is necessary to use the order of operations) When dealing with the order of operations, the equations are usually long, and have many numbers, we use the order of operations to make them easier to solve. But Exponents are hard for me, so using them in an even bigger equation was tough. I first figured out the exponent, by multiplying the number as many times as it says, then use PEMDAS, which is the order we are supposed to do our work in.( parentheses, exponents, multiplication, division, addition, and subtraction) And then I figured it out! Thanks for listening, CJIV Article posted November 2, 2010 at 03:57 PM GMT • comment • Reads 874 Article posted October 22, 2010 at 06:49 PM GMT • comment • Reads 975 Dear Mom and Dad, Our class just finished working on Module 1. In Module 1 I learned lots of helpful things. There were a lot of worksheets and problems to solve. Here are a couple of the things that we did that helped me learn all of these important things that we did in Module 1. I learned how to make a frequency table and what the differences are between a bar graph and a line graph. I also learned how to analyze graphs. When learning about analyzing graphs we had a take home quiz where we had to pick a graph from a newspaper and analyze it. I also learned how to make a bar graph using the data I collected from a frequency table. I learned how to use variables to show pattern rules, and how to use symbols and algebra to explain how a number trick works. I also learned how to get a rule for a sequence and figure out the nth term. My favorite thing in this Module was doing patterns and finding rules for the patterns so that you could find the nth term. I also liked the things we did on probability. I did an experiment on probability with MHHO where we picked Starbursts out of a bag. I also worked on probability with spinners and flipping one coin, two coins, or three coins. I enjoyed these activities a lot. I also liked that section because it refreshed my memory on theoretical and experimental probability. I also refreshed my memory on Order of Operations and doing exponents. During the Order of Operations section we learned PEMDAS, parenthesis, exponents, multiplication, division, addition, and subtraction, which is the order of operations. We also listened to a song on the order of operations. We also had a test which I hope I aced, and a bunch of worksheets that I enjoyed doing because they were mostly very easy and fun. I had a great time learning these things in this module! Sincerely, CHBE Article posted October 22, 2010 at 06:49 PM GMT • comment • Reads 975 Article posted November 2, 2010 at 07:10 PM GMT • comment • Reads 940 Dear Mom and Dad, We just finished off Module one in our Math textbooks. There were a lot of things to learn. We learned about the order of operations, probability, frequency tables, and much, much more! The first thing we learned about was Graphs. We did a lot of work with interpreting graphs. We did several pages in the book where we had to answer questions based on a graph. We also learned when to use a line graph and when to use a bar graph. You use a line graph to show changes over time and you use a bar graph when you are putting things into categories. We even took a take home quiz where we had to answer several questions using a graph we found in the news paper. Next we learned about number tricks and patterns. The number tricks were really fun and it was amazing that they worked. I even made my own. We used algebra and shapes to figure out how they worked each time. The patterns were kind of tricky. Some of them were really easy to figure out the but others I couldn't find it on my own. To find the nth term we used algebraic equations like T= 5N. It was really fun. The next thing we did was frequency tables. We used several sets of data and created a table with it. After frequency tables came exponents. I think exponents are really fun but it sometimes gets annoying how you have keep multiplying large numbers. We used exponents to find the volume of a cube and we used them in the order of operations. The next thing we did was Probability. We did an experiment about probability where we got a partner and picked something out of a bag. My partner and I did jolly ranchers and we called pour experiment jolly pickings. We also did a lottery where our Math teacher put our names in a bag and picked them out. The first and last person picked won 10 math points and the second and the second to last person picked won 5 math points. Everyone else won nothing. The last thing we did was order of operations. Order of operations is by far my favorite thing we did. It is very easy for me. I like how there are a lot of steps. We did a lot of order of operations problems and we were all so good at it that we had the test on it in the same week that we learned about it. This is what we did in Module one. Sincerely, CJKA Article posted November 2, 2010 at 07:10 PM GMT • comment • Reads 940 Article posted October 22, 2010 at 02:05 PM GMT • comment • Reads 986 Dear Mom, I have learned SO MUCH from Module 1! Module 1 is basically the first part of the textbook. It doesn't take up all of the trimester, only about half. This is what I have learned: Graphs: This is the first thing that I learned in Module 1. Bar graphs and line graphs. We have worked with both of them, making them, seeing examples, even looking for them in articles. Tables: A huge part of Module 1 was learning to make frequency tables. We mostly made these throughout the whole module. You get data from a probability experiment and put them on these. Number tricks: I didn't enjoy this part very much. During this section of the module, we learned ways to start with one number and always end with the same result using blocks or algebra. Number sequences: This is basically solving patterns. For instance, 2, 4, 6, and so on, we would find out the next number, or term, 8. We would also need to find the rule, in this case, term number plus itself equals the term. Exponents: In module 1, we learned a little bit about exponents. Not too big ones, but things like two to the fourth power, or sixteen. We learned how to write it in standard form, or a normal number, as well as the exponential form. Probability: Goodness, this was a big part of Module 1! This was basically learning the probability of certain events and putting it into fractions and frequency tables, like picking a certain color marble out of a bunch. We even created our own instance in a project and recorded the data. We also learned how to find the theoretical probability as well as the experimental probability. Order of Operations: My favorite part. I just love solving the mathematical signs as well as the parentheses. It's so fun! Sincerely, FHRO Article posted October 22, 2010 at 02:05 PM GMT • comment • Reads 986 Article posted October 22, 2010 at 04:26 PM GMT • comment • Reads 924 Dear Mom and Dad, Let me tell you what I learned in Module 1 this year in math class. There were so many worksheets, homework, and tests. During these months I've learned a lot about probability, patterns, tables, exponents, and Order of Operations. Hopefully you'll find this interesting and helpful. First I learned bar and line graphs. There are so many graphs I didn't know about! We did a one or two worksheets on that and then we moved to frequency tables. During the rest of the Module we used frequency tables so it was a big “must know” kind of thing. Frequency tables are when you have an object that you want to put in an organizer, preferably a table. You tally how many objects and then in another column you write your frequency. A frequency is how many of the object you have. We did a couple of worksheets on that as well. Later I learned pattern problems. You would have a worksheet and while solving a problem you would also be unlocking a pattern. Or there would be a number trick. You would have a worksheet and a calculator. While your following the problem on your calculator you will come up with the same number or there would be a trick on it. We also did a couple worksheets on that as well. We took a quiz on that whole chapter. Then I learned exponents. Exponents are so confusing, or at least were. There would be a number say five and the problem would say five time five time five times three. So instead of writing that all out you would say five to the third power time three. We took a quiz on that and a couple of worksheets then we used it for the rest of the Module as well. Also I learned probability, my favorite thing to learn. We did a big project with candy and math. I had a partner and it was a lot of fun. We did a lot of worksheets on it and took one or two quizzes on it. That chapter was a lot of fun and learning. After we did the project we did a couple of more worksheets. Finally I learned Order of Operations. It was a little difficult at first, but I got the hang of it. We took a quiz on that and we had worksheets galore. It was a quick chapter but it felt like a lifetime. I was confused half the time and couldn't understand it. I eventually got it but I still have a little difficulty with it. There were so much that I have learned in the past months and I've had a lot of fun while doing so. Math is a fun subject and I enjoy going to it everyday. Love, HJME Article posted October 22, 2010 at 04:26 PM GMT • comment • Reads 924 Article posted October 22, 2010 at 04:28 PM GMT • comment • Reads 880 Dear Mum, This year in math class we are learning some new and interesting things that we didn't know last year. We learned about patterns, and how we can figure out how to solve them. After we mastered the trick, we moved on to figuring out how we can solve them with algebra. We used T to represent the term, and we used N to represent the term number. For example, if T was 5 and N was 2, the algebraic expression could be T = 2N + 1. After algebra we moved on to exponents. Exponents are the little numbers above the bigger numbers used to represent a long multiplication sequences. For example, four to the sixth power would be the same thing as four times four times four times four times four times four. Exponents can also be used when solving a pattern. When we were done with exponents we worked on this thing called probability. Probability is the chances that something will happen, on a scale of one to zero. One is certain that it will happen, zero is that it's impossible. All the in between is just that, the in between. For example, if I had a bag half filled with blue marbles, and half filled with green marbles that have purple polka-dots, the probability of picking a yellow marble are zero. The probability of picking a green with purple polka-dots or a blue marble is one. The probability of picking just a blue marble is ½. Next we learned about the order of operations. The order of operations is the way to solve problems that have more than one operation. It goes like this; parenthesis, exponents, multiplication, division, addition, subtraction. An acronym you can use to remember this is PEMDAS. I prefer to say Please Excuse My Dear Aunt Sally. Those are pretty much the most important things that we have learned this year in math class, but I'm sure we'll learn much more. Sincerely, HJMA Article posted October 22, 2010 at 04:28 PM GMT • comment • Reads 880 Article posted October 22, 2010 at 02:15 PM GMT • comment • Reads 967 Dear Mom, In math class we are learning about many useful things. In the beginning of the module, we learned mostly about graphs. We learned about how to graph data, construct graphs, and how and when to use a frequency table. We also learned what the difference is between the the two graphs and which one to use at the appropriate time. You use a bar graph when you are graphing data from a survey or pole and you use a line graph when you are graphing data over a specific time period. We also learned how to make a model sequence with word sentences, tables, and graphs. This unit was mostly about turning sentences into tables and graphs. It was pretty easy. After that we learned about using variables to solve number tricks. Using variables can make problems so much easier. We also learned how to predict the nth term in a number sequence (example: T=n+4-2 so the 15nth term is 17).Learning variables helped a lot with that. Next we learned about computing with exponents. We had done this last year so it was the easiest for me and we spent the least time on it. Then we learned about probability. We learned about experimental and theoretical probability and about estimating probability using 0 through 1 (example: 1/3). We did a project on this last week. I worked on it with my friend. My friend and I put 50 star burst’s in a bag. We put a certain amount of each color and recorded the theoretical probability. Then we picked out of the bag a certain amount of times and then we recorded the experimental probability. The last thing we learned was the order of operations. In order to do this right all that you have to know is P.E.M.D.A.S. This stands for Parentheses Exponents Multiplication and Division Addition and Subtraction. We have been having a lot of fun in math class this year. Sincerely, Your daughter Article posted October 22, 2010 at 02:15 PM GMT • comment • Reads 967 Article posted October 22, 2010 at 06:10 PM GMT • comment (1) • Reads 951 Dear Mom and Dad, We are just finishing up Module 1 in our math book in our math class. I think we learned lots of interesting things and I hope I teach you something that maybe you didn't even know. I know you will be pleased with what we have been learning. Now, here is a brief overview of what we did during this module. First we learned what the difference is between bar and line graphs. We found out that bar graphs are used when you have to group information into categories. Line graphs are used when you have information that extends over a long period of time. Next we learned about probability. Our teacher told us the two different types of probability. One type is theoretical probability. Theoretical probability is what is supposed to happen in an experiment. The other type of probability is experimental probability. Experimental probability is what the outcome of the experiment is. In class we would compare these to probabilities and see how they differ. Then we learned how to predict the nth term in a term number sequence. There would be a term and a term number. The term number is the number of the term in the sequence. The term is the number in the number pattern or sequence. We used an algebra equation to figure out what the next term would be. We also did some stuff on algebra equations. We would use n, then find what n would be in that certain equation. The last thing we learned is about order of operations. Order of operations is pretty self-explanatory. It is the order that the you do the operations go in .We were taught that you have to follow PEMDAS. PEMDAS stands for parentheses, exponents, multiplication or division, and addition or subtraction. You always do equations from left to right. So you will always do the problems left to right using PEMDAS. This is briefly all we have done for this first Module. I thought is was very fun and I definitely know I learned a lot. I can't wait to learn more in the upcoming modules. Hope you like this, and hopefully I taught you guys something. Bye for now! Love, KJCH Article posted October 22, 2010 at 06:10 PM GMT • comment (1) • Reads 951 Article posted October 22, 2010 at 06:41 PM GMT • comment • Reads 951 Dear Mom and Dad, We just finished Module 1 in our mathematics textbook. There were a lot of worksheets and problems to solve. First, we learned how to make frequency tables. Just tally the amount of times something happened in one column, then write the number of tallies in another. To make a frequency graph, just make a sideways bar graph. We also learned how to decide whether to use a bar or line graph. Use a bar graph when you want to put data into categories, and a line graph when showing data over time. Next, we moved onto probability. CHBE and I worked on a project where we picked Starbursts candy out of a bag, then made a table to show experimental and theoretical probability. Experimental probability is what happened in the experiment. The theoretical probability is what was supposed to happen in the experiment. We also did experiments with spinners and dice to show probability. After probability, we studied number patterns. Our teacher showed us all kinds of cool number tricks that seemed unreal. We used algebra to show how the tricks worked. We also used algebra to show the relation from term to term number in a table, so we could find the nth figure. Then, we studied order of operations. Our math teacher played us a funny rap about PEMDAS. P for parentheses, E for exponents, M for multiplication, D for division, A for addition and S for subtraction, you can call it PEMDAS! It was awesome. We did some worksheets. To go with that, we learned how to calculate exponents. At the end of the Module, we had a big test. Love, MHHO Article posted October 22, 2010 at 06:41 PM GMT • comment • Reads 951 Article posted October 22, 2010 at 02:15 PM GMT • comment • Reads 891 October 21, 2010 Dear Mom & Dad, Our class just finished Module one. Here are some of the sections we learned about. We have done a lot in Module one. We started the school year learning about interpreting bar and line graphs. During that section we had a take home quiz where I had to find a graph in the newspaper. While learning about bar graphs we also learned how to make frequency tables that relate to the graph. In module one we also used variables to solve number tricks, in this section there was some tricky problems. We also learned about predicting the nth term of a number sequence. I found that algebra is a lot easier than it looks. Our class also learned how to change problems into standard and exponential form. Sometimes exponents turned out to be really big numbers. We learned to find the experimental and theoretical probability of an event. This section was one of my favorites because we worked on a project where we had a lot of fun while learning. My partner and I did the probability of different colored M&M's out of a bag. Lastly we learned about the order of operations. During this section we learned a cool song to remember the order of operations. As you can see our class learned a lot in Module one. Now I'm excited to start Module two. Your Daughter, NRAL Article posted October 22, 2010 at 02:15 PM GMT • comment • Reads 891 Article posted November 28, 2010 at 10:30 PM GMT • comment • Reads 889 Dear Mom and Dad, We just finished Module 1 in our mathematics textbook. There were a lot of worksheets and problems to solve. First, we learned how to make frequency tables. Just tally the amount of times something happened in one column, and then write the number of tallies in another. To make a frequency graph, just make a sideways bar graph. We did several pages in the book where we had to answer questions based on a graph. We also learned when to use a line graph and when to use a bar graph. You use a line graph to show changes over time and you use a bar graph when you are putting things into categories. We even took a take home quiz where we had to answer several questions using a graph we found in the news paper. Next, we moved onto probability. I worked on a project where I picked Gummy Bears out of a bag, then made a table to show experimental and theoretical probability. Experimental probability is what happened in the experiment. The theoretical probability is what was supposed to happen in the experiment. We also did experiments with spinners and dice to show probability. After probability, we studied number patterns. Our teacher showed us all kinds of cool number tricks that seemed unreal. We used algebra to show how the tricks worked. We also used algebra to show the relation from term to term number in a table, so we could find the nth figure. Once I got the hang of it, the problems were very simple to solve. Then, we studied order of operations. Our math teacher played us a funny rap about PEMDAS. P for parentheses, E for exponents, M for multiplication, D for division, A for addition and S for subtraction, you can call it PEMDAS! It was awesome. We did some worksheets. To go with that, we learned how to calculate exponents. At the end of the Module, we had a big test. Sincerely, RJAV Article posted November 28, 2010 at 10:30 PM GMT • comment • Reads 889 Article posted November 2, 2010 at 06:56 PM GMT • comment • Reads 944 Dear Mom and Dad, Module one was a lot of fun! We learned about probability, algebra, the order of operations, exponents, and patterns. I liked the patterns unit because I like figuring out complicated math formulas. It requires a knowledge of algebra so I was exited! I love patterns and figuring out life sized problems. I also liked the algebra unit because I like using complex equations, and when I do algebra I feel like I’m one of those people on TV with the long equations written on chalkboards. It actually is fun to do algebra because it challenges my logic ability. I somehow always seem to do algebra the complicated way. The exponent unit was very easy! All it is, is multiplying it by itself the number of times it says in the small number next to the bigger number. Did you know that one to any power is one? Same with zero, zero to any power is one as well. I didn't like the probability unit because I felt it too repetitive. Well I do like probability I just felt that the unit just went on about spinners and dice and coin flips. We were supposed to do the Stock Market Simulation this module but it was postponed until early November. Here's an example of something we did this module, try to figure this out! 1 2 3 4 5 8 16 24 32 40. SRKE Article posted November 2, 2010 at 06:56 PM GMT • comment • Reads 944 Article posted October 22, 2010 at 07:09 PM GMT • comment • Reads 871 Dear Mom, I just got done learning all about Module 1 in my math class! I have learned so many cool and important things in this Module. One of the many things I’ve learned about is bar graphs and line graphs, and when to use them. Did you know you use a bar graph when you need to display different categories? Also, you use a line graph to display information over an amount of time. Learning about graph was one of my favorite things about this module because I caught on quickly and it was easy for me to learn. Another thing I learn in this Module is, how to make a frequency chart. I loved using the frequency charts because it helped me a lot with my probability. It was always fun using this chart. For example, say I had a wheel and the colors on this wheel were blue,red, and green. Now say I spin this wheel 20 times and come out with the results, 5 for red. 10 for blue, and 5 for green. I would make tallys on my chart and then after I was done. I would have a section on my chart labeled frequency, in that section, I would write down the number of tallys and the number of times I did it. Such as this: 5/20 Guess what else I learned! I learned about one of my favorite things, PROBABILITY! I worked a lot on this part of Module 1, and it was my favorite part. Some of the things I learned in this part was how to determine the experimental and theoretical probability. Which isn't as hard as it seems. You see, for theoretical probability all you have to do is list the number of the item you put in the bag and how many things are in the bag. For example, I put in 10 cards in a bag, I put in 3 red cards so I would write it like this : 3/10. For experimental probability it's very similar. I'll explain it through example. Say I have the same bag with the same 10 cards. Inside that bag there are 3 red cards and 7 black cards. I make a frequency chart and pull cards out of the bag 20 times. I pull out a black card 15 times, so the experimental probability would be 15/20. Also, I learned about Algebra. One of the things we learned in this part is how to identify the rule of a sequence of numbers or patterns. For example, If the pattern is 1,4,9 the rule would be T=N2. I got this by looking at the term number and then the pattern. The last thing we learn about was the order of operations, other wise known as PEMDAS or Please Excuse My Dear Aunt Sally. The order of operations consists of parentheses, exponents, multiplication, division, addition and subtraction. You go in order except for multiplication, division, addition, and subtraction. For those you go from left to right. Order of Operations was probably the easiest for me to learn because I love getting to the bottom of math problems. Sincerely, LSHO Article posted October 22, 2010 at 07:09 PM GMT • comment • Reads 871 Article posted October 22, 2010 at 02:15 PM GMT • comment • Reads 909 Dear Dad, I learned a lot while learning Module 1. And I can't believe I learned all of this in this small amount of time. And now I'm going to tell you what I learned. I learned about bar and line graphs. Like how you do and line graph when your graphing something over a period of time, and you do a bar graph when you are graphing data from a survey. I learned how to do theoretical and experimental probability. I learned even more about probability then I did last year. I also learned how to make a frequency table with tallies. We even did a project with frequency tables and probability. I did the project with NRAL. We put 100 M&Ms in a paper bag and pulled out an M&M 200 times. For each color we picked out we put a tally on the frequency table under that color. I also learned number tricks. They were so fun they were my favorite part of Module 1. We would start out with any number we wanted and follow the steps and if you u did a different number it would come out the same every time. I learned how to to use letters in a equation like this T=3N+5. That went a long with finding out the next pattern term. That was fun too. I also learned exponents, but I didn't like it. It was my least favorite part of Module 1. Because I don't like multiplying things so many times. The last thing I learned was the order of operations. It goes parentheses, exponents, multiplication, division, addition, and subtraction. It's also known as “PEMDAS.” I did well in that. I can't believe I learned so much in that small amount of time!:) Your Daughter, LJTA Article posted October 22, 2010 at 02:15 PM GMT • comment • Reads 909 Article posted October 22, 2010 at 05:21 PM GMT • comment • Reads 896 Dearest Father, This year in math, I've learned about graphing, probability and the order of operations in Module 1.We've learned how to interpret bar and line graphs, this helps when I'm recording data. I learned to use a bar graph when the data is separated into different categories and a line graph when you have one category of data that changes over time. We also took information from frequency tables, which we also learned how to make. These tables are made with data that was already taken. The frequency table consist of three columns, one for the subject category, one for the for tallies and one for the frequency. After this we learned how to how to make frequency tables and bar and line graphs we moved on to using variables to solve number tricks. We used algebra to solve these number tricks. This section in Module 1 was more complicated but it became easy after many worksheets. In Module 1 we also learned how to use sequences, word sentences, tables and graphs. We did many worksheets and math point pages on this subject. Modeling these graphs, sequences, tables and word sentences was not a challenge for me. Next, we learned to find the nth term of a number sequence, this was some serious algebra. This was very interesting to learn because I haven’t studied number sequences before. Next the powers flowered and we started using exponents in the next section of Module 1. Exponents are a short way of writing a multiplication problem. We wrote both standard and exponential forms in this Module. When we finished exponents we started learning about probability. We did many experiments with flipping coins and picking little different colored squares out of a bag. I also did a project with my friend on probability. We found the experimental and theoretical probability. We picked jelly beans out of a bag, then we wrote about our results in a paragraph and table. I enjoyed this project because it was really active and I got to eat jelly beans! Our experiment was called Jelly Belly. The last thing we learned in Module 1 is the order of operations. In Module 1, we learned an easy way to remember the order of operations: PEMDAS (parentheses, exponents, multiplication, division, addition, subtraction). The order goes left to right in the PEMDAS order. I learned a lot of useful and handy information in Module 1. Sincerley, MHCA Article posted October 22, 2010 at 05:21 PM GMT • comment • Reads 896 Article posted October 22, 2010 at 07:14 PM GMT • comment • Reads 957 Dear daddy, October 22nd 2010 So far I have learned a lot in math class. We have done so many different things. I will tell you three thing we did the one I enjoyed the most the least and the middle. The thing I enjoyed the most would probably be the patterns. We learned how to find the pattern and predict some future term in the pattern. A term is the pattern. The term number is the number of the pattern. I found it very fun and I got it right away. The thing that was okay is the bar graphs and line graphs. We did a lot with graphs last year, but I still learned a lot of things. I learned how to get information out of them. I learned how to read them too. I also learned how to find out if you should use a line graph or a bar graph. The thing I least enjoyed was the frequency tables. I found them quite easy. I learned what they were, how to get the information, how to read them, and how to make them. They are a lot like graphs. You have different columns. Lets say you wanted to get the frequency of the number of pets on your street. You would have a column for the type of pets. A column for the tally’s, for the number of pets, then you would have a column for frequency, you count the number of tally’s. I learned a lot of things this module. I hope I learn a lot more. Even if there is some review I know I will learn a least something new. From, SSMA Article posted October 22, 2010 at 07:14 PM GMT • comment • Reads 957 Article posted November 2, 2010 at 04:30 PM GMT • comment • Reads 920 Article posted November 2, 2010 at 04:30 PM GMT • comment • Reads 920 Article posted November 2, 2010 at 07:08 PM GMT • comment • Reads 908 10-21-10 Dear Mom, Recently in math class we have been learning all new and interesting elements of math. At the beginning of the module we started off reviewing last year's work. Then we moved into patterns and learned how to find the rule to a pattern and express it in algebra. Say if there were three numbers in a row and they were all random numbers I could figure them out and write the rule in algebraic form. We also learned how to make our own patterns. After learning about all the tricks and gadgets that come with patterns we moved into algebra alone. Algebra was very interesting because instead of using all numbers and math symbols we used letters. Algebra is all expressed in letters and some numbers because it is a way find out a number if you don't know what it is. The algebra letter represents the number that you can't figure out. Frequency tables and graphs were next on the list. We learned that a frequency table was the outcome, tally, and the frequency all together. You would use a frequency table when there were different say colors that kids liked and different amounts like one color. A frequency table is a way to organize all that information. After it is all put into the frequency table you can either make a bar graph or line graph. It really depends on the type of data. If it was a survey you would use a bar graph to show the information, and if it was showing the change in data over time then you would use a line graph to measure the change in data. After we learned about tables and graphs we moved in to probability. We did a huge probability experiment when we chose some object to pick out of a bag and then we created a table and wrote a summary about our experiment. I learned a lot about experimental and theoretical probability and how to compare the two together. I worked with PJPH and we did our experiment on Razzles. We also used spinners to determine what was going to happen. We would find the theoretical probability of what would happen and then once we had spun the spinner we found the experimental probability. Once we were almost done we still had to do the order of operations. This is just like PEMDAS which I learned about last year. It is when we took a number sentence and used PEMDAS order to find the correct answer. PEMDAS is a rhyme to remember that parenthesis come first and then exponents, multiplication and division are equal and so are addition and subtraction. You have to follow that order to get the true answer to the number sentence. We are now done with module one so I wanted to inform you of what has been happening here in math class. I have been learning so much new and old nut all exciting information as you can tell. I will write soon! I have to continue learning all this information! Talk later! xxo, FSCA!!!! Article posted November 2, 2010 at 07:08 PM GMT • comment • Reads 908 Article posted October 22, 2010 at 06:20 PM GMT • comment • Reads 935 Dear Nana, How are you doing? Tell Pop I said hi. In school I have been learning a lot of new stuff in math this module. I am going to tell you a little bit about that. We have been learning a lot of stuff so I am only going to share about the things I enjoyed. We learned how to make bar and line graphs and when to use them. You use a bar graph when you are taking a survey and you use a line graph when you are keeping track of time. I found I was good at this so it was fun. I also enjoyed learning how to do order of operations. I have found I enjoy this then most things we do even if I am not the best at it. During this project I learned what PEMDAS [Parentheses, exponents, multiplication, division, addition and subtraction.] To remember that we say please excuse my dear Aunt Sally we even heard a song about it by the way your a pain in my PEM DAS [just kidding but...] I have also learned that exponents are not as hard as I originally thought they were. I did not understand them at all I remember that I was always worried when they were on a test but now I think they are easy. Moving on, we have also been working hard on probability. I not the best at it but I understand it and know how to do them which is the most important. But I really liked the projects we did with it. One of my strongest subjects I have to say was making frequency tables I do not know why but they just come easy to me Maybe it is because I like to sort things but I don't know. But it makes me happy I don't have to worry about them on a test. I have had a lot of fun learning about new stuff all term I cant wait to learn more Love LRKE Article posted October 22, 2010 at 06:20 PM GMT • comment • Reads 935 Article posted November 2, 2010 at 07:02 PM GMT • comment • Reads 832 Dear Mother, My class just finished module 1 in our textbooks. Since it was the beginning of the year it was basically easy stuff and reviews but we still learned. Now that we are done with module 1 we can start to learn more advanced math. In module 1 there were six sections. Section 1 was about data displays, section 2 was about sequences and exponents, and in section 3 we learned about probability. Section 4 was about problem solving and section 5 and 6 were about problem solving and order of operations. In section 1 we learned about tallying data, and bar and line graphs. When you tally data you are supposed to gather the data you want and tally how many times it comes up. A bar graph is a graph that is used to display data that falls into distinct categories. A line graph is graph that shows how data changes over time. That is what we learned in section 1 in module 1. In section 2 we learned about modeling sequences and exponents, squares, and cubes. A sequence is an ordered list of numbers or objects called terms. We learned about term numbers and symbols also. Exponents show a pattern of repeated multiplication. Exponents make it simpler to write out as well. Instead of writing out 3.3.3.3 you could just write 34 or three to the fourth power. In section 3 we were taught about probability and theoretical probability. Probability is experiments that predict the outcome of an event. Probability is also a number from 0 to 1 that tells you how likely something will happen. Section 4 was about problem solving. The textbook gave us four steps to problem solving. Step 1: understand the problem. Step 2: Make a Plan. Step 3: Carry out the plan. Step 4: look back, or check out work. Those were all very helpful to me, especially when I got stuck on a difficult math equation. This was definitely an important chapter to me. Section 5 in the textbook taught the class about evaluating solutions and making connections. Section 6 was about the order of operations. The order of operations helps when you do long mathematical equations. It lets you break down long equations into sections to make it smaller and easier to figure out since it is better laid out. Since we started school that is what we have been learning for the first trimester. From, Kid Article posted November 2, 2010 at 07:02 PM GMT • comment • Reads 832 Article posted October 22, 2010 at 04:20 PM GMT • comment • Reads 904 Dear Mom, Yellow In math class, we have been in a section called module one and have learned many things in these first few months of school. Module one is now over and we will soon be starting module two. In module one, I have learned to make and solve algebra equations, make bar and line graphs and when to use them, probability, and the order of operations! Thanks to Mrs. Harte's excellent teaching skills, I now know that the probability of you giving me a raise in my allowance is 1/10. The thing about probability is it's the same as fractions! If there is a spinner with four equal areas, the probability that the spinner will land on area one would be ¼, the same as it's fraction. So it has ¼ chance of being spun, and it has a fraction value of ¼! It's so simple! That's not all though! I now know the definition of the word PEMDAS: Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction. This will enable me the godly power of never getting the order of operations mixed up. Look at this problem: 1+4x5-(10-8)=____. Can you guess the answer. It's 19! You would first do whatever is in the parentheses, then you would do exponents, but since there are no exponents, you would skip to multiplication and division. After that, you can finally do addition and subtraction from left to right, and... you're done! 19! Plus, those bar and line graphs will certainly help me with my job when I'm older! Bar graphs are used for one set of data that stays the same. Line graphs are for a set of data that changes over time. So if I collected data on a plant growing over a week, what graph do I use? LINE GRAPH! You should thank Mrs. Harte for her awesome teaching! From, MHJU. Article posted October 22, 2010 at 04:20 PM GMT • comment • Reads 904 Article posted November 3, 2010 at 02:56 PM GMT • comment • Reads 880 Dear Mom, In math class, we have been in a section called module one and have learned many things in these first few months of school. Module one is now over and we will soon be starting module two. In module one, I have learned to make and solve algebra equations, make bar and line graphs and when to use them, probability, and the order of operations! Thanks to Mrs. Harte's excellent teaching skills, I now know that the probability of you giving me a raise in my allowance is 1/10. The thing about probability is it's the same as fractions! If there is a spinner with four equal areas, the probability that the spinner will land on area one would be ¼, the same as it's fraction. So it has ¼ chance of being spun, and it has a fraction value of ¼! It's so simple! That's not all though! I now know the definition of the word PEMDAS: Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction. This will enable me the godly power of never getting the order of operations mixed up. Look at this problem: 1+4x5-(10-8)=____. Can you guess the answer. It's 19! You would first do whatever is in the parentheses, then you would do exponents, but since there are no exponents, you would skip to multiplication and division. After that, you can finally do addition and subtraction from left to right, and... you're done! 19! Plus, those bar and line graphs will certainly help me with my job when I'm older! Bar graphs are used for one set of data that stays the same. Line graphs are for a set of data that changes over time. So if I collected data on a plant growing over a week, what graph do I use? LINE GRAPH! You should thank Mrs. Harte for her awesome teaching! From, MHJU. Article posted November 3, 2010 at 02:56 PM GMT • comment • Reads 880 Article posted November 2, 2010 at 07:07 PM GMT • comment • Reads 840 Dear Dad, I just finished Module 1 of my math text book. We learned all about probability,exponents, order of operations, tables and graphs. We did a probability project on a type of items we choose then after you make a small poster all about the probability of the items you chose. Probability is a great project and skill. We did lots of experiments on it. We played games on the computer, we did the probability of cards and dice. They were great ways to learn probability. Exponents aren't really something you'll use a lot in life but its an easy way to show an equation when the numbers are the same and you're multiplying them all by each other. How exponents work is like this. So say you have the number 7, it's in this equation 7x7x7x7x7. With exponents all you have to write would be 75. Exponents are really just another way to show one number being multiplied by it's self several times. Order of Operations is how you evaluate an equation in the right order so you get the correct answer. There are a few different ways you can remember the order of operations the easiest one for me to remember is PEMDAS. PEMDAS stands for Parentheses Exponents Multiplication Division Addition Subtraction. Another way that people sometimes remember it is Please Excuse My Dear Aunt Sally. Well anyway Order of Operations makes any types of equations easier to understand and that is why math equations are easy and fun to do. Tables and Graphs help throughout your whole life. You use them in almost all subjects. Especially when you calculating population from this year to this year. Anyway, we had to make a table for our probability posters and we made graphs and tables for homework, schoolwork, quizzes, tests, and assignments. Tables and graphs are also a great skill to have for future reference and jobs. All the math skills we learned in Module 1 are all skills that we should know throughout are lives. I learned a lot about skills that I can use in the future and I can’t wait to tell you more about math. Article posted November 2, 2010 at 07:07 PM GMT • comment • Reads 840 Article posted November 3, 2010 at 03:02 PM GMT • comment • Reads 854 Dear Dad, I just finished Module 1 of my math text book. We learned all about probability,exponents, order of operations, tables and graphs. We did a probability project on a type of items we choose then after you make a small poster all about the probability of the items you chose. Probability is a great project and skill. We did lots of experiments on it. We played games on the computer, we did the probability of cards and dice. They were great ways to learn probability. Exponents aren't really something you'll use a lot in life but its an easy way to show an equation when the numbers are the same and you're multiplying them all by each other. How exponents work is like this. So say you have the number 7, it's in this equation 7x7x7x7x7. With exponents all you have to write would be 75. Exponents are really just another way to show one number being multiplied by it's self several times. Order of Operations is how you evaluate an equation in the right order so you get the correct answer. There are a few different ways you can remember the order of operations the easiest one for me to remember is PEMDAS. PEMDAS stands for Parentheses Exponents Multiplication Division Addition Subtraction. Another way that people sometimes remember it is Please Excuse My Dear Aunt Sally. Well anyway Order of Operations makes any types of equations easier to understand and that is why math equations are easy and fun to do. Tables and Graphs help throughout your whole life. You use them in almost all subjects. Especially when you calculating population from this year to this year. Anyway, we had to make a table for our probability posters and we made graphs and tables for homework, schoolwork, quizzes, tests, and assignments. Tables and graphs are also a great skill to have for future reference and jobs. All the math skills we learned in Module 1 are all skills that we should know throughout are lives. I learned a lot about skills that I can use in the future and I can’t wait to tell you more about math. Article posted November 3, 2010 at 03:02 PM GMT • comment • Reads 854 Article posted October 22, 2010 at 02:13 PM GMT • comment • Reads 907 Dear Mom, October 21,2010 In math class there are different modules, so here is what I learned it module 1. I learned how to make bar and line graphs. And how bar graphs show the data that fall into different categories. And line graphs show the data over time. We did a experiment with that where we did how many and types of pets in the classroom. I also learned about probability we took dice and rolled them and we found out the experimental and theoretical probability was. We also did a project on that where we either spun a spinner of picked different items out of a baggy. We also learned how to do order of operations. I learned a trick too on order of operations its called pemdas. If you don’t know the rule for order of operations then you do parenthesis then exponents then multiplication and division then addition and subtraction you do all those thing in order from left to right. We learned how to do frequency tables with different things like favorite pets. You first put the topics then tally then frequency. We also learned how to format the frequency to a bar graph or line graph. We learned patterns in shapes and numbers and how to turn them into equation we did many experiments on patterns in shapes and numbers like trying to find out how many will be in the 100th figure and the 30th when there were only 4 examples. Love, Your Daughter Article posted October 22, 2010 at 02:13 PM GMT • comment • Reads 907 Article posted November 4, 2011 at 03:23 PM GMT • comment • Reads 357 Hi Momma and Daddy, In Math class we have been working on Module 1. If you're wondering what a module is I'll explain. A module is the way our textbook is divided in our book there are modules in all. This Friday we just finished up Module 1. You might be wondering what did we learn in Module 1, well I'll explain.... In the first section of module 1 we learned about analyzing graphs. By analyzing a graph we can tell what intervals are represented on the vertical axis and the horizontal axis. Also by using our data we can decide whether to make a line graph or a bar graph. We also learned about tallying for graphs as well. In the second section of module 1 we learned about sequences and beginning algebra. We learned how to predict patterns by finding relationships between the term number and term. We also learned how to draw pictures to help and represent the different sequences. We also reviewed how to find the volume of a cube by using height times width times length. In section 3 of module 1 about probability. In this unit we did a project. I choose to work with Lily as you know we chose gumballs out of a bag and then recorded the experimental probability. We learned how to reduce experimental probability as well as theoretical probability. Before we went to section 4 we went to section 6. In section 6 we evaluated problems with order of operations. We were reminded of PEMDAS. Parentheses, Exponents, Multiplication or division, addition or subtraction. Then we learned about writing algebra problems. Right now we are learning about problem solving. What we are trying to learn is to solve and problem and show all your work. Sometimes it's better to show your work then necessarily getting the right answer. It may be difficult but it is really fun once you get the hang of it. I'm glad you enjoyed reading about my Math class. Thank you for reading!!! Love, CJBR Article posted November 4, 2011 at 03:23 PM GMT • comment • Reads 357 Article posted November 4, 2011 at 03:09 PM GMT • comment • Reads 426 Dear Dad, Module 1 was a lot longer than I thought, and there are more modules to come. So here's what I learned. Graphs: We did a quick re-visit on graphs, line and bar. We could read them and turn them into Frequency Tables: Frequency tables are data tables that show how often something occurs. Like age, we use tally marks to show it. And that's the beginning of Algebra: 9=Z x 3, algebra replaces numbers in a term with letters, so a student can figure out what the missing number is. Exponents: 32, no you do NOT do 3 x 2. it means to multiply the 3 twice or 3 x 3 and that equals 9 not 6 ,you find this in Order of Operations: 3 - 4 x 32 PEMDAS, Parenthesis, Exponents ,Multiplication, Division, Addition, and Subtraction. The is the order of what operation you do first. There are no parentheses in this equation so we go to the exponent. 3 x 3= 9, then the multiplication 9 x 4= 36, after that we do the subtraction 36-3. And it all equals 33. These things are kind of the basics to get us warmed up for a long year ahead ,the next module is problem solving so keep an eye out. Article posted November 4, 2011 at 03:09 PM GMT • comment • Reads 426 Article posted November 3, 2011 at 04:07 PM GMT • comment • Reads 437 Dear Mom & Dad This in math class we learned about interpret and line graphs, construct frequency tables, create bar graphs using a frequency tables, using variables to solve number tricks, model a number sequence with a word sentence table and graph. In probability I worked with a partner her name is hsgr. We did an awsome job . It was on gum. We pulled it out of a bag and we had to find the experimental probability and the Theoretical probability. We had to make a poster. We also did order of operation. PEMDAS Please Excuse My Dead Aunt Sally. Or Parentheses Exponents Multiplication Division Addition and Subtraction. Article posted November 3, 2011 at 04:07 PM GMT • comment • Reads 437 Article posted November 3, 2011 at 03:57 PM GMT • comment • Reads 450 Dear Mom and Dad, My math class just finished up Module1 and I'm going to tell you about it. First we worked on probability. We learned how to predict what the results of our experiments would be based on the theoretical probability. We did probability experiments with a partner. My partner was Jaco. Another thing we learned about was exponents. We learned how to translate exponential form to standard standard form and standard to exponential. Lastly, my class learned about the order of operations. We learned how to evaluate problems with parentheses, exponents, multiplication and division, and addition and subtraction. In that order is how you solve the problem. A way my class was taught to remember it was from PEMDAS. Sincerely, Carmelo Article posted November 3, 2011 at 03:57 PM GMT • comment • Reads 450 Article posted November 3, 2011 at 04:21 PM GMT • comment • Reads 426 In Module 1 we learned about graphs,sequences, exponents,probability,problem solving and order of operations. For sequences, we figured out patterns and did math magic ticks. We would study the first four terms in a pattern and then would try to figure out what the 10th,20th and 100th terms in the sequence were. We would write an algebraic formula to figure out any term in the sequence. The math magic tricks were really fun but challenging. You would pick a number perform a couple of operations, then the result would be your original number or a different number. We would do the trick several times and if the results are a different number than the original number, it should be the same number every time you solve it. Then we would explain steps using squares and circles and algebra. For probability we did experiments and online simulations. We did a bunch of different experiments using dice, cards and then came up with our own experiment. For my experiment I used toothpicks and a compass. I tossed 2 toothpicks with one end cut off over a compass 160 times and recorded which way they pointed closest to. The name of my experiment was Which Way? The theoretical probability was 40 for North, South, East and West. The Experimental probability was 44 for North, 40 for South, 43 for east and 33 for West. We also used a probability simulation online where the computer rolled a dice or drew a card as many times as we wanted it to and it made a graph and fractions. Article posted November 3, 2011 at 04:21 PM GMT • comment • Reads 426 Article posted November 3, 2011 at 04:04 PM GMT • comment • Reads 465 Dear Mom and Dad I am writing this paragraph to let you know about what I have been doing in math class this past month. We just finished module 1 and had a big test on it I think I did pretty good hopefully. In this module we learned about a lot of new things such as........ interpreting bar graphs to the order of operations. My favorite thing we did was that we did a section on finding the theoretical and experimental probability of an event. We had to get a partner and decide on a thing to base our project off of we choose M & M's we then had to make a chart and summary on this topic. Another really fun project that we did was a unit on number tricks it showed me another way of doing math. We did many worksheets and games to help us learn how to do them right. During the unit on number tricks we had to learn how to change the numbers into variables which I think will come in much handy next year. The last thing I want to mention to you guys was that we did a project on attendance sheets where we took all of them from the year before when I was in 6th grade and it showed that the most people were out on Monday’s I was really surprised I thought that most people would be out on Tuesdays. from, your daughter Article posted November 3, 2011 at 04:04 PM GMT • comment • Reads 465 Article posted November 3, 2011 at 05:56 PM GMT • comment • Reads 407 Dear Mother, I am going to tell three things I have done in math class. One thing I did was a probability experiment where we had to pick a certain thing that contains a bunch of little things; like a deck of cards, box of candy etc. You could also pick a partner and my partner was BJBA and we choose to do our experiment with spree's. We had 5 of 4 different colors red, yellow, orange, purple. We picked one out of a bag 100 times. That was what we did for our experiment. Another experiment I did was in class we had to do a coin flipping experiment and there were 3 tables. The first table was to see what a penny would land on heads or tails and we did that 50 times. The second table was with 2 pennies and to see if they were both heads, both tails, heads and tails, or tails and heads. We also did that 50 times. The third table was with 3 pennies and to see if they all landed on heads, all tails, tails heads tails, tails tails heads, heads heads tails, heads tails heads, heads tails tails & tails heads heads. And we did that another 50 times. Th last thing I am going to tell you is when we did order of operations and we had a bunch of pages on it some had exponents and parenthesis and some had just a lot of numbers on them but that was it. Article posted November 3, 2011 at 05:56 PM GMT • comment • Reads 407 Article posted November 3, 2011 at 04:12 PM GMT • comment • Reads 427 Dear Mom and Dad, We are doing a lot of fun and exciting projects and activities in math class. One of the fun and exciting things we are doing is the order of operations. We have done a lot of worksheets and we read about it a lot in the textbooks. One thing I find interesting about it is that brackets and parentheses are pretty much the same exact thing, the only two differences are that parentheses go first and parentheses are inside of brackets. Those are some things we are doing to help us learn about the order of operations in math class. Another interesting thing we are doing in math class is frequency tables. It is practically a chart. We did a math points project on it but I didn't do it because I thought that it would take way to long. I've learned that you can make frequency charts on any thing. It can range from how many pets 5 people have to how long 500 people live. That is some things I've learned about frequency tables. That is all for this letter thank you for reading. From, The most amazing child ever. Article posted November 3, 2011 at 04:12 PM GMT • comment • Reads 427 Article posted October 28, 2011 at 04:16 PM GMT • comment (2) • Reads 462 Dear Mom and Dad, We just finished Module 1 in math class. We learned about probability, analyzing a graph, algebra equations for patterns, and order of operations. For the probability unit we did a number of things. We played games that we recorded the results of and we also did a lot of games involving dice. This all led up to a mini project that I did with my friend Carmelo. We put 20 baseball cards in a hat and did 100 random draws. This was my favorite probability activity. For the order of operations unit we did a lot of worksheets. We learned how to use brackets and parentheses in our equations. This was definitely my favorite unit. I think that I'm pretty good with multi-step problems. This trimester we had to get 100 math points. There was one class period Ms. Hart let me do math points. I did a ton of them and handed them in. The total came out to 176. We also did a lot of algebra equation by looking at patterns. We did a lot of worksheets and activities for this. Sincerely, Jaco Article posted October 28, 2011 at 04:16 PM GMT • comment (2) • Reads 462 Article posted November 3, 2011 at 03:01 PM GMT • comment • Reads 421 Dear Mom, We have just completed module one in our math books. We did a lot of graphing and tables, algebra, probability and order of operations. I really want you to know more about what we learned so I will tell you. Graphing was the first section in the module. We learned how to analyze a graph and how to choose the right graph for each set of data. We also learned what a frequency table was and how to use it. The second section was algebra and patterns. We learned how to predict the nth term in a pattern by coming up with an equation, (T=n). Also we learned that algebra isn't as big, bad, and scary as we thought. We spent a few weeks on probability and did some really fun experiments. Our math teacher went around and had each of us pick a cube out of a bag and tally what color. We did individual projects too, for mine I put a dice in a mug and picked the dice out. What ever number my thumb was on was the number I tallied. Probability was my second favorite module. To wrap up the unit we revisited the order of operations. The order is, parentheses, exponents, multiplication and division, and addition and subtraction, or P.E.M.D.A.S. To help us remember the order we say, Please Excuse My Dear Aunt Sally. If you don't know what exponents are the are a number with a little number in the corner which stands for how many times you multiply the number by its self, for example, 22. Love, Your Daughter Article posted November 3, 2011 at 03:01 PM GMT • comment • Reads 421 Article posted November 4, 2011 at 02:55 PM GMT • comment (1) • Reads 438 Dear Parents, Hello parents of mine. I am learning about a lot in math class this year in module 1. I would like to tell you about some of the things we are learning about in math. This year in module 1 we learned about how to do algebraic problems like : (2-1+ 4)x6= or 11x(1+4)-3=. We also learned about how to do exponents 2 to the power of 2 or 100 to the power of 4. The other things we learned about were like how to find different things on patterns. I also was learning about how to find the nth form of a pattern. We take a test on all we learned in the module when we finish the module we also take lots of pre tests and quizzes we do them usually after a unit ends. We will have math points due after the test usually. I do a lot of the pattern and algebra ones because they are more math points. We are learning to solve challenging and unique problems that require a chart or diagram. Sincerely, HJPA Article posted November 4, 2011 at 02:55 PM GMT • comment (1) • Reads 438 Article posted November 3, 2011 at 03:55 PM GMT • comment • Reads 331 Dear Mom and Dad, One thing I learned in module 1 is the order of operations. I learned that parentheses come first. After parentheses come exponents. Then multiplication and/or division, and then addition and/or subtraction. The fraction line means divide. I also learned about exponents. Any number to the second power (squared 2) means multiply by itself. If anything is to the third power (cubed 3), you multiply it by itself and then the product by the base number. With ten, the exponent is the number of zeros. 10 3 = 1000 10 4 = 10000 10 5 = 100000 I learned about probability. If there are 10 coins in a bag, and three of them are pennies, there is a 30% theoretical probability of picking a penny. I also learned about graphing. If both axis on a graph have numbers in them, then you should use a point graph. If both axis have words, then you should use a bar graph. Sincerely RJCH Article posted November 3, 2011 at 03:55 PM GMT • comment • Reads 331 Article posted November 3, 2011 at 03:55 PM GMT • comment • Reads 337 Dear Mom and Dad, One thing I learned in module 1 is the order of operations. I learned that parentheses come first. After parentheses come exponents. Then multiplication and/or division, and then addition and/or subtraction. The fraction line means divide. I also learned about exponents. Any number to the second power (squared 2) means multiply by itself. If anything is to the third power (cubed 3), you multiply it by itself and then the product by the base number. With ten, the exponent is the number of zeros. 10 3 = 1000 10 4 = 10000 10 5 = 100000 I learned about probability. If there are 10 coins in a bag, and three of them are pennies, there is a 30% theoretical probability of picking a penny. I also learned about graphing. If both axis on a graph have numbers in them, then you should use a point graph. If both axis have words, then you should use a bar graph. Sincerely RJCH Article posted November 3, 2011 at 03:55 PM GMT • comment • Reads 337 Article posted November 3, 2011 at 05:57 PM GMT • comment (1) • Reads 453 Dear Mom and Dad, This module in math (the first one) has been good. We have learned a ton of stuff and were tested on each subject we learned about. The first thing we learned about was how to make graphs and frequency tables. We learned this by recording lots of different data and placing it on graphs and frequency tables. That was a good review to make sure we have the knowledge we need. Another thing we learned was how to find the Nth term in a sequence. We practiced by frequently doing these problems and learning some cool rules and tricks along the way from our awesome teacher Mrs. Heart. We learned a lot more things but I'm only want and need to tell you one more. The last thing we learned about was probability. This unit was fun because we did lot's of physical experiments in the class making it more interesting. The big experiment we did was each of us pick a group of objects and experiment to see what the probability of picking them were, after we did that we recorded the information on a frequency table and put it on a poster, we also wrote a summary of what we did. This was cool because I got to see everybody's work. Article posted November 3, 2011 at 05:57 PM GMT • comment (1) • Reads 453 Article posted November 4, 2011 at 07:09 PM GMT • comment • Reads 438 Dear mom and dad, we have been working on module one. In that we have been working on exponents like 4 to the powerof 6 =4096, 4 6 4x4x4x4x4x4. Exponents can be hard like 9 5, but I know them pretty well. We have also been working on probability of stuff like this, what is the probability of spinning a D. the experimental is 1/2 We have been working on frequency tables to like, the numbers on the left is how many times the letter got picked. And that is what we have been working on in math so far. Sincerely, your son Article posted November 4, 2011 at 07:09 PM GMT • comment • Reads 438 Article posted November 3, 2011 at 03:00 PM GMT • comment • Reads 419 Dear Parents, I am writing you a letter about module 1 from math class. One thing I learned in this module is what experimental and theoretical probability are. Theoretical probability is what you think is going to happen before it happens. It is when you predict something using logical ways. Experimental probability is what actually happened. After you do the experiment the data that you have is the experimental probability. Another thing that I learned is the order of the operations. The order is pemdas or in other words parentheses, exponents, multiplication, division, addition, then subtraction. We learned that if it is addition and subtraction is all that is left then you just do what is first. One last thing I learned is exponents. They look like this 2 4. The little number is an exponent. Instead of being 2x4 it is actually 2x2x2x2. Most people thing it is 2x4 but they are wrong. They made exponents to shorten the equation. That is what I learned in module 1. Article posted November 3, 2011 at 03:00 PM GMT • comment • Reads 419 Article posted November 3, 2011 at 04:09 PM GMT • comment (2) • Reads 428 To Whom It May Concern, This year in math class we have been learning about a lot of things. We learned about graphing and frequency tables, probability, and order of operations. For probability I did a project with my partner, JRVI where we experimented with the probability of pulling four different kinds of candy from a bag. After, we ate ALL of the candy. I had a lot of fun with the order of operations unit because I find solving those kind of problems and work on the order of numbers is really interesting and fun to do. We did a lot of worksheets and math points that were about order of operations. What we did with graphing and frequency tables was in the beginning of the year which was a really super unit. At the beginning of the year we made a frequency table of how many times students in sixth, seventh and eight grade were absent last year. I had I lot of fun in math class during module one!!! Love, The Golden Child(SSAM) Article posted November 3, 2011 at 04:09 PM GMT • comment (2) • Reads 428 Article posted November 3, 2011 at 03:47 PM GMT • comment • Reads 474 Dear Mom and Dad, In Math, class we just finished Module 1. A Module is consisted of many sections of different math that we will learn. Let me tell you about some of the things that I have learned within this Module. One of the things we learned about was graphing. We learned how to extract information from graphs, and also how to make graphs, both bar and line graphs. We also learned when to use a specific graph for different things. Along with graphing, we also learned more about probability. We did a project for that section, and used tally’s to find the frequency, and we found the theoretical and experimental probability. My project was about the color of Tic-Tacs that we pulled out of a bag. Last of all, we learned about the order of operations and exponents. We had lots of practice upon this to get good at it, and we also were taught how to deal with exponents in the order of operations. I hope that you learned a little about some of the different math that we have been learning, Sincerely, Your Daughter. Article posted November 3, 2011 at 03:47 PM GMT • comment • Reads 474 Article posted November 3, 2011 at 05:57 PM GMT • comment • Reads 458 Math in Module 1 Dear Parents, This year in Math module 1 I have learned many fun and new things. Some of the things I have learned are frequency tables, patterns, and probability. A frequency table is basically a table with three vertical columns. The first column is what ever the tables about, the second column is for the tally, and the third column is for the frequency or total. Patterns are new to me this year. I have learned how to find the algebraic equation to patterns. Sometimes it takes a while to find out the pattern but once you get it is easy. Many times it how many cubes are in each figure. In Probability we have added on to what we already learned in 6th grade. I have learned the difference between theoretical probability and experimental probability. I have done a project with my friend on probability that gives the summary, experimental and theoretical probability. I have had lots of fun and learned a lot in Module 1 and cant wait till module 2. Sincerely, HHSC Article posted November 3, 2011 at 05:57 PM GMT • comment • Reads 458 Article posted November 4, 2011 at 06:43 PM GMT • comment • Reads 455 Hello mom, I am writing you a letter while I’m in math class to tell you about what we've done so far this year in mathematics class. The classes are colored and my class is red. Every week we have a new math log and each day we put down what our homework is for one week. One thing we did was we had a ripped page from a phone book and we took tally of the last digit of the first 100 phone numbers and put them on a frequency table. That was in the beginning of the year; one of the more recent things we've done was just a few weeks ago. Me and a friend took a bag of (honey dew) doughnut munchkin things and randomly chose one doughnut out of the bag at a time we then recorded the frequency of a chocolate, glazed and what we call the “eridy” doughnut on a frequency table; which we then put on a plain white piece of paper. We decorated the piece of paper we did this all on, with pretty colors and pictures of donuts. We also did number tricks which we solved using variables. You must be wondering what a number trick is... here are a few examples: choose a number, add 3, multiply by 2, add 4, divide by 2, subtract original number. Here's another one: Choose a number, add the next larger number, add 7, divide by 2, subtract 4... what are your answers? Those are some things we have learned in module one this year... we are soon going on to a new module which is module 2. I hope you enjoyed learning a bit about what we learned about in math class this year. Love, your daughter. Article posted November 4, 2011 at 06:43 PM GMT • comment • Reads 455 Article posted November 3, 2011 at 04:01 PM GMT • comment (2) • Reads 516 Dear parents, This module we learned about graphing, patterns, square numbers, order of operations and exponents. One of the activities we did in the first section was we learned how to make a frequency table of all the absences in the jr. high of 2010, and make a graph. We got into partners and we counted up the absences for each grade and predicted what day would have the most absentees. The most absences believe it or not were on Tuesday. I had a little trouble with the visual part of the patterns but it was okay. With the square numbers we did a few worksheets and learned how to apply it to geometry. That section was a little boring but I learned a lot because of it. In the third section, we took a long time to learn about PEMDAS, which is the order of operations. First P or parenthesis, then E or exponents then after that M for multiplication, D for division, A for addition and last but not least S for subtraction. Finally, we finished the module with exponents. It was a little confusing but finally we made it through. Thank you for your love and support! I love you! Love, KSMO Article posted November 3, 2011 at 04:01 PM GMT • comment (2) • Reads 516 Article posted November 4, 2011 at 07:15 PM GMT • comment • Reads 460 Dear Mom and Dad, I have finished Module 1 in Math Class, we have covered many interesting things. For example, we have learned about exponents, theoretical and experimental probability, order of operations, frequency tables, a little bit of algebra, and much more. My favorite topic so far has been order of operations. It's really fun to try to solve math problems with PEMDAS. For frequency tables we did many fun activities. One of them was we had to look at the attendance sheets for every grade last year. We then had to record every single absence into a frequency table. A frequency table is a chart that shows tallys of whatever category you choose. For order of operations we did a lot of math problems on the Eno board and paper. One of the activities we did was we had to copy down math problems from the board. We then figured out the problems using PEMDAS and then got chosen to do them on the Eno board. It was really helpful to see the whole problem explained in front of you. For finding the experimental and theoretical probability of an event we did a lot of creative work. My favorite activity for learning probability was creating a poster that shows the results of an experiment you and your partner came up with. For example my partner and I picked a random coin out of a bag, recorded our results into a frequency table, and found the theoretical and experimental probability. We then transferred this information on to a creative poster that is now hanging up in Mrs. Harte's classroom. Article posted November 4, 2011 at 07:15 PM GMT • comment • Reads 460 Article posted November 3, 2011 at 03:47 PM GMT • comment • Reads 375 Dear Mom and Dad, I am going to tell you what we did in module 1 since we just finished it. In module 1 of our textbooks Math Thematics we covered many different topics. We learned about how to interpret a bar graph and line graph by recording the data of last years attendance sheets. We covered the nth term by using patterns. We would have to predict things like the hundredth term. Instead of doing the pattern one hundred times, you use the nth term and replace the N with 100 and there you go. We learned how to compute with exponents by finding out the standard form and the exponential form to figure out the answer. To find the theoretical and experimental probability of an event. We learned this by doing a project with a friend. The project was to figure out a way to find out the probability of a subject of our chose. My partner and I chose picking gum out of a bag. We also had to find out the experimental probability and compere the two. From, HSGR Article posted November 3, 2011 at 03:47 PM GMT • comment • Reads 375 Article posted November 4, 2011 at 07:22 PM GMT • comment • Reads 447 Dear Mom, In math we have finally finished Module 1. By the way, a module is a whole section that has a bunch of different lessons in it. Our first lesson was... Bar and line graphs. I liked this section, because it was fairly easy, and kind of fun. Our home work that we had on this asked us questions about what was selling the most, to what kids like to eat the most at lunch. Next lesson was... Patterns and predictions. This was a little more challenging for me. You would have a table (math table) with numbers in it. You then have to figure out what the pattern is, and complete the table. Our next lesson was... Exponents, squares, and cubes. Let's start with exponents. Exponents are for example 3², or 3 to the 2nd power. So for this exponent the answer would be 9. I got that by multiplying 3*3. Those are 3 of our lesson from module 1. Hope you enjoyed!! Love your amazing daughter, Article posted November 4, 2011 at 07:22 PM GMT • comment • Reads 447 Login Copyright (c) 2013 by KJAL Conditions of Use Privacy Policy Return to Blogmeister
# Permutation And Combination Permutation and combination are the ways to represent a group of objects by selecting them in a set and forming subsets. It defines the various ways to arrange a certain group of data. When we select the data or objects from a certain group, it is said to be permutations, whereas the order in which they are represented is called combination. Both concepts are very important in Mathematics. Permutation and combination are explained here elaborately, along with the difference between them. We will discuss both the topics here with their formulas, real-life examples and solved questions. Students can also work on Permutation And Combination Worksheet to enhance their knowledge in this area along with getting tricks to solve more questions. ## What is Permutation? In mathematics, permutation relates to the act of arranging all the members of a set into some sequence or order. In other words, if the set is already ordered, then the rearranging of its elements is called the process of permuting. Permutations occur, in more or less prominent ways, in almost every area of mathematics. They often arise when different orderings on certain finite sets are considered. ## What is a Combination? The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. In smaller cases, it is possible to count the number of combinations. Combination refers to the combination of n things taken k at a time without repetition. To refer to combinations in which repetition is allowed, the terms k-selection or k-combination with repetition are often used. Permutation and Combination Class 11 is one of the important topics which helps in scoring well in Board Exams. ## Permutation and Combination Formulas There are many formulas involved in permutation and combination concepts. The two key formulas are: ### Permutation Formula A permutation is the choice of r things from a set of n things without replacement and where the order matters. nPr = (n!) / (n-r)! ### Combination Formula A combination is the choice of r things from a set of n things without replacement and where order does not matter. Learn how to calculate the factorial of numbers here. ## Difference Between Permutation and Combination Permutation Combination Arranging people, digits, numbers, alphabets, letters, and colours Selection of menu, food, clothes, subjects, team. Picking a team captain, pitcher and shortstop from a group. Picking three team members from a group. Picking two favourite colours, in order, from a colour brochure. Picking two colours from a colour brochure. Picking first, second and third place winners. Picking three winners. ## Uses of Permutation and Combination A permutation is used for the list of data (where the order of the data matters) and the combination is used for a group of data (where the order of data doesn’t matter). ## Solved Examples on Permutation and Combinations Example 1: Find the number of permutations and combinations if n = 12 and r = 2. Solution: Given, n = 12 r = 2 Using the formula given above: Permutation: nPr = (n!) / (n-r)! =(12!) / (12-2)! = 12! / 10! = (12 x 11 x 10! )/ 10! = 132 Combination: $$_{n}C_{r} = \frac{n1}{r!(n-r)!}$$ $$\frac{12!}{2!(12-2)!} = \frac{12!}{2!(10)!} = \frac{12\times 11\times 10!}{2!(10)!} = 66$$ Example 2: In a dictionary, if all permutations of the letters of the word AGAIN are arranged in an order. What is the 49th word? Solution: Start with the letter A The arranging the other 4 letters: G, A, I, N = 4! = 24 First 24 words Start with the letter G arrange A, A, I and N in different ways: 4!/2! = 12 Next 12 words Start with the letter I arrange A, A, G and N in different ways: 4!/2! = 12 Next 12 words This accounts up to the 48th word. The 49th word is “NAAGI”. Example 3: In how many ways a committee consisting of 5 men and 3 women, can be chosen from 9 men and 12 women? Solution: Choose 5 men out of 9 men = 9C5 ways = 126 ways Choose 3 women out of 12 women = 12C3 ways = 220 ways The committee can be chosen in 27720 ways. ## Permutation and Combination – Practise Questions Question 1: In how many ways can the letters be arranged so that all the vowels come together? Word is “IMPOSSIBLE.” Question 2: In how many ways of 4 girls and 7 boys, can be chosen out of 10 girls and 12 boys to make the team? Question 3: How many words can be formed by 3 vowels and 6 consonants taken from 5 vowels and 10 consonants? From the above discussion, students would have gained certain important aspects related to this topic. To gain further understanding of the topic, it would be advisable that students should work on sample questions with solved examples. To learn more about different maths concepts, register with BYJU’S today. ## Frequently Asked Questions – FAQs ### What do you mean by permutations and combinations? A permutation is an act of arranging the objects or numbers in order. Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter. ### Give examples of permutations and combinations. The example of permutations is the number of 2 letter words which can be formed by using the letters in a word say, GREAT; 5P_2 = 5!/(5-2)! The example of combinations is in how many combinations we can write the words using the vowels of word GREAT; 5C_2 =5!/[2! (5-2)!] ### What is the formula for permutations and combinations? The formula for permutations is: nPr = n!/(n-r)! The formula for combinations is: nCr = n!/[r! (n-r)!] ### What are the real-life examples of permutations and combinations? Arranging people, digits, numbers, alphabets, letters, and colours are examples of permutations. Selection of menu, food, clothes, subjects, the team are examples of combinations. ### Write the relation between permutations and combinations. The formula for permutations and combinations are related as: nCr = nPr/r!
AP State Syllabus AP Board 7th Class Maths Solutions Chapter 9 Construction of Triangles Ex 2 Textbook Questions and Answers. ## AP State Syllabus 7th Class Maths Solutions 9th Lesson Construction of Triangles Exercise 2 Question 1. Draw ΔCAR in which CA = 8 cm, ∠A = 60° and AR = 8 cm. Measure CR, ∠R and ∠C. What kind of triangle is this? Solution: CA = 8 cm, ∠A = 60°, AR = 8 cm Step. -1: Draw a rough sketch of a triangle and label it with the given measurements. Step -2: Draw a line segment CA of length 8 cm. Step -3: Draw a ray $$\overrightarrow{\mathrm{AX}}$$ making an angle 60° with CA. Step -4: Draw an arc of radius 8 cm fromA which cuts $$\overrightarrow{\mathrm{AX}}$$ at C. Step -5: Join C, R to get the required Δ CAR. CR = 8 cm, ∠C = 60° and ∠R = 60°. ∴ This is an equilateral triangle. Question 2. Construct ΔABC in which AB = 5cm, ∠B = 45° and BC = 6cm. Solution: AB = 5cm, ∠B = 45° and BC = 6cm. Step -1: Draw a rough sketch of a triangle and label it with the given measurements. Step -2: Draw a line segment AB of length 5cm. Step -3: Draw a ray $$\overrightarrow{\mathrm{BY}}$$ making an angle 45° with AB. Step -4: Draw an arc of radius 6 cm from B, which cuts $$\overrightarrow{\mathrm{BY}}$$ at C. Step -5: Join A, B to get the required ΔABC. Question 3. Construct ΔPQR such that ∠R = 100°, QR = RP = 5.4 cm. Solution: ∠R= 100°,QR= RP = 5.4cm. Step -1: Draw a rough sketch of a triangle and label it with the given measUrements. Step -2: Draw a line segment QR of length 5.4 cm. Step -3: Draw a ray $$\overrightarrow{\mathrm{RX}}$$ making an angle 100° with QR. Step -4: Draw an arc of radius 5.4 cm from R, which cuts $$\overrightarrow{\mathrm{RX}}$$ at P. Step -5: Join P, Q to get the required ΔPQR Question 4. Construct ΔTEN such that TE = 3 cm, ∠E = 90° and NE = 4 cm. Solution: TE = 3cm, ∠E = 90°, NE = 4cm. Step -1: Draw a rough sketch of the triangle and label it with the given measurements. Step -2: Draw a line segment TE of length 3 cm. Step -3: Draw a ray $$\overrightarrow{\mathrm{EX}}$$ making an angle 90° with TE. Step -4: Draw an arc of radius 4 cm from E, which cuts $$\overrightarrow{\mathrm{EX}}$$ at N. Step -5: Join N, T to get the required ΔTEN.
# ISEE Middle Level Math : How to multiply fractions ## Example Questions ### Example Question #51 : How To Multiply Fractions Multiply the following: Explanation: To multiply fractions, we will multiply the numerators together, then we will multiply the denominators together. Note that we do NOT need to find a common denominator. ### Example Question #52 : How To Multiply Fractions Multiply the following: Explanation: To multiply fractions, we will multiply the numerators together, then we will multiply the denominators together. Note that we do NOT have to find a common denominator. So, we get ### Example Question #53 : How To Multiply Fractions Multiply the following: Explanation: To multiply fractions, we will multiply the numerators together, then we will multiply the denominators together. Note that we do NOT need to find a common denominator. So, we will multiply: ### Example Question #54 : How To Multiply Fractions Multiply the following: Explanation: To multiply fractions, we will multiply the numerators together, then we will multiply the denominators together. Note that we do NOT need to find a common denominator. So, Before we multiply, we can simplify to make things easier. The 8 and the 4 can both be divided by 4. So, Now, we can multiply straight across. We get ### Example Question #55 : How To Multiply Fractions Multiply the following: Explanation: To multiply fractions, we will multiply the numerators together, then we will multiply the denominators together. Note that we do NOT need to find a common denominator. So, We must write 7 as a fraction. We know that whole numbers can be written as fractions over 1. So, Now, we can multiply straight across. We get ### Example Question #56 : How To Multiply Fractions Multiply the following: Explanation: To multiply fractions, we will multiply the numerators together, then we will multiply the denominators together. Note that we do NOT need to find a common denominator. So, in the problem We will first write 18 as a fraction. We know that whole numbers can be written as a fraction over 1. So, we get Now, before we multiply, we can simplify to make things easier. The 3 and the 18 can both be divided by 3. We get Now, we can multiply straight across. We get ### Example Question #57 : How To Multiply Fractions Multiply the following: Explanation: To multiply fractions, we will multiply the numerators together and the denominators together. We get ### Example Question #58 : How To Multiply Fractions Multiply the following: Explanation: To multiply, we will multiply the numerators together and the denominators together. We get ### Example Question #59 : How To Multiply Fractions Multiply the following: Explanation: To multiply, we will multiply the numerators together, then we will multiply the denominators together. So, we get ### Example Question #60 : How To Multiply Fractions Multiply the following:
# The Sides of a Triangle Are 50 Cm, 78 Cm and 112 Cm. the Smallest Altitude is - Mathematics MCQ The sides of a triangle are 50 cm, 78 cm and 112 cm. The smallest altitude is • 20 cm • 30 cm • 40 cm • 50 cm #### Solution The area of a triangle having sides aband s as semi-perimeter is given by, A = sqrt(s(s-a)(s-b)(s-c)), where s = (a+b+c)/2 Therefore the area of a triangle, say A having sides 50 cm, 78 cm and 112 cm is given by a = 50 cm ; b = 78 cm ; c = 112 cm s = (50+78+112)/2 s = 240/2 s = 120 cm A = sqrt(120 (120-50)(120-78)(120-112)) A = sqrt((120(70)(42)(8)) A = sqrt(2822400) A = 1680 cm2 The area of a triangle, having p as the altitude will be, $\text{ Area} = \frac{1}{2} \times \text{ base } \times \text{ height }$ Where, A = 1680 cm2 We have to find the smallest altitude, so will substitute the value of the base AC with the length of each side one by one and find the smallest altitude distance i.e. p Case 1 AC = 50 cm 1680 = 1/2 (50 xx p) 1680 xx 2 = 50 xx p p = (1680 xx 2)/50 p = 67.2 cm Case 2 AC = 78 cm 1680 = 1/2 (78 xx p) 1680 xx 2 = 78 xx p p = (1680xx2)/78 p = 43 cm Case 3 Ac = 112 cm 1680 =1/2 (112 xx p) 1680 xx 2 = 112 xx p p = (1680 xx 2 )/112 p = 30 cm Is there an error in this question or solution? Chapter 17: Heron’s Formula - Exercise 17.4 [Page 24] #### APPEARS IN RD Sharma Mathematics for Class 9 Chapter 17 Heron’s Formula Exercise 17.4 \| Q 5 \| Page 24 Share
# Texas Go Math Grade 3 Lesson 3.2 Answer Key Compare Fractions with the Same Numerator Refer to our Texas Go Math Grade 3 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 3 Lesson 3.2 Answer Key Compare Fractions with the Same Numerator. ## Texas Go Math Grade 3 Lesson 3.2 Answer Key Compare Fractions with the Same Numerator Essential Question How can you compare fractions with the same numerator? To compare fractions with the same numerator, all you have to do is compare the denominators. The fraction with the bigger denominator is smaller. Unlock the Problem Josh is at Enzoâ€™s Pizza Palace. He can sit at a table with 5 of his friends or at a different table with 7 of his friends. The same-size pizza is shared equally among the people at each table. At which table should Josh sit to get more pizza? â€¢ Including Josh, how many friends will be sharing a pizza at each table? It is given that Josh is at Enzoâ€™s Pizza Palace. He can sit at a table with 5 of his friends or at a different table with 7 of his friends Hence, from the above, We can conclude that The number of friends that will be sharing a pizza at each table including Josh is: 6 friends (or) 8 friends â€¢ What will you compare? We will compare the total number of pieces that are shared by all the friends since all the friends will get only 1 piece Model the problem. There will be 6 friends sharing Pizza A or 8 friends sharing Pizza B. So, Josh will get either $$\frac{1}{6}$$ or $$\frac{1}{8}$$ of a pizza. • Shade $$\frac{1}{6}$$ of Pizza A. • Shade $$\frac{1}{8}$$ of Pizza B . • Which piece of pizza is larger? $$\frac{1}{6}$$ >Â $$\frac{1}{8}$$ So, Josh should sit at the table with 6 friends to get more pizza. Math talk Mathematical Processes Suppose Josh wants two pieces of one of the pizzas above. Is $$\frac{2}{6}$$ or $$\frac{2}{8}$$ of the pizza a greater amount? Explain how you know. It is given that Josh wants two pieces of one of the pizzas. The given two pieces of pizza are: $$\frac{2}{6}$$ and $$\frac{2}{8}$$ Now, From the above fractions, We can observe that The numerators are the same The fraction with the biggest denominator is smaller So, $$\frac{2}{6}$$ > $$\frac{2}{8}$$ Hence, from the above, We can conclude that The $$\frac{2}{6}$$ of the pizza has a greater amount Question 1. Which pizza has more pieces? ____ The more pieces a whole is divided into, ___ the pieces are. The given two pieces of pizza are: $$\frac{2}{6}$$ and $$\frac{2}{8}$$ Now, From the above fractions, We can observe that The numerators are the same The fraction with the biggest denominator is smaller So, $$\frac{2}{6}$$ > $$\frac{2}{8}$$ Hence, from the above, We can conclude that The pizza that has 8 pieces has more pieces The more pieces a whole is divided into, the greater the pieces are. Compare Fractions Same Numerator Lesson 3.2 Answer Key Question 2. Which pizza has fewer pieces? ___ The fewer pieces a whole is divided into, the __ the pieces are. The given two pieces of pizza are: $$\frac{2}{6}$$ and $$\frac{2}{8}$$ Now, From the above fractions, We can observe that The numerators are the same The fraction with the biggest denominator is smaller So, $$\frac{2}{6}$$ > $$\frac{2}{8}$$ Hence, from the above, We can conclude that The pizza that has 6 pieces has fewer pieces The fewer pieces a whole is divided into, the lesser the pieces are. Use fraction strips. On Saturday, the campers paddled $$\frac{2}{8}$$ of their planned route down the river. On Sunday, they paddled $$\frac{2}{3}$$ of their route down the river. On which day did the campers paddle farther? Compare $$\frac{2}{8}$$ and $$\frac{2}{3}$$. • Place a âœ“next to the fraction strips that show more parts in the whole. • Shade $$\frac{2}{8}$$. Then shade $$\frac{2}{3}$$. • So, $$\frac{2}{8}$$ <Â $$\frac{2}{3}$$ So, The campers paddled farther on Sunday Share and Show Question 1. Shade the models to show $$\frac{1}{6}$$ and $$\frac{1}{4}$$. Then compare the fractions. $$\frac{1}{6}$$ $$\frac{1}{4}$$ The given fractions are: $$\frac{1}{6}$$ and $$\frac{1}{4}$$ So, The representation of the given fractions are: Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that Compare. Write <, >, or . Question 2. $$\frac{1}{8}$$ $$\frac{1}{3}$$ The given fractions are: $$\frac{1}{8}$$ and $$\frac{1}{3}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that Question 3. $$\frac{3}{4}$$ $$\frac{3}{8}$$ The given fractions are: $$\frac{3}{4}$$ and $$\frac{3}{8}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that Question 4. $$\frac{2}{6}$$ $$\frac{2}{3}$$ The given fractions are: $$\frac{2}{6}$$ and $$\frac{2}{3}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that Go Math Grade 3 Lesson 3.2 Compare Fraction with Same Numerator Question 5. $$\frac{3}{6}$$ $$\frac{3}{6}$$ The given fractions are: $$\frac{3}{6}$$ and $$\frac{3}{6}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that Problem Solving Math Talk Mathematical Processes Explain why $$\frac{1}{2}$$ is greater than $$\frac{1}{4}$$? The given fractions are: $$\frac{1}{2}$$ and $$\frac{1}{4}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that $$\frac{1}{2}$$ is greater than $$\frac{1}{4}$$ Question 6. H.O.T. Write Math Zach has a piece of pie that is $$\frac{1}{4}$$ of a pie. Max has a piece of pie that is of a pie. Maxâ€™s piece is smaller than Zachâ€™s piece. Explain how this could happen. Draw a picture to show your answer. It is given that Zach has a piece of pie that is $$\frac{1}{4}$$ of a pie. Max has a piece of pie that is of a pie. Maxâ€™s piece is smaller than Zachâ€™s piece. Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that The representation of a piece of the pie of Max and Zach is: Unlock the Problem Question 7. Quinton and Hunter are biking on trails in Katy Trail State Park. They hiked $$\frac{5}{6}$$ mile in the morning and $$\frac{5}{8}$$ mile in the afternoon. Did they bike a greater distance in the morning or in the afternoon? a. What do you need to know? ____ It is given that Quinton and Hunter are biking on trails in Katy Trail State Park. They hiked $$\frac{5}{6}$$ mile in the morning and $$\frac{5}{8}$$ mile in the afternoon Hence, from the above, We can conclude that You need to know which fraction of the mile is greater b. The numerator is 5 in both fractions, so compare $$\frac{1}{6}$$ and $$\frac{1}{8}$$. Explain. The given fractions are: $$\frac{5}{6}$$ and $$\frac{5}{8}$$ Now, We know that, The fraction that has the same numerator with the bigger denominator is smaller. Hence, from the above, We can conclude that $$\frac{1}{6}$$ > $$\frac{1}{8}$$ c. What plan or strategy will you use? The strategy you will use to find is: The fraction that has the same numerator with the bigger denominator is smaller. d. Complete the sentences. In the morning, the boys biked $$\frac{5}{6}$$Â mile. In the afternoon, the boys hiked $$\frac{5}{8}$$ mile. The boys hiked a greater distance in the afternoon. So, $$\frac{5}{6}$$ >Â $$\frac{5}{8}$$ Question 8. Multi-Step Sense or Nonsense? James ate $$\frac{2}{4}$$ of his pancake. David ate $$\frac{2}{3}$$ of his pancake. Both pancakes are the same size. Who ate more of his pancake? James said he knows he ate more because four is greater than three. Does his answer make sense? Shade the models. Then use math language to explain your answer. It is given that James ate $$\frac{2}{4}$$ of his pancake. David ate $$\frac{2}{3}$$ of his pancake. Both pancakes are the same size It is also given that James said he knows he ate more because four is greater than three Now, The given fractions are: $$\frac{2}{4}$$ and $$\frac{2}{3}$$ Now, We know that, The fraction that has the same numerator with the bigger denominator is smaller. So, The representation of the models of James and David is: Hence, from the above, We can conclude that $$\frac{2}{4}$$ < $$\frac{2}{3}$$ The answer of James does not make sense Use models to compare. Fill in the bubble for the correct answer choice. Question 9. Ben, Lara, and Liz had pizza for lunch. Ben had a slice of the Mushroom pizza. Lara and Liz each ate one slice of the Olive pizza. Which fraction shows how much pizza Ben ate? (A) $$\frac{1}{6}$$ (B) $$\frac{2}{6}$$ (C) $$\frac{1}{8}$$ (D) $$\frac{2}{8}$$ It is given that Ben, Lara, and Liz had pizza for lunch. Ben had a slice of the Mushroom pizza. Lara and Liz each ate one slice of the Olive pizza Now, The given figures are: Now, From the above, We can observe that The Mushroom pizza has 6 pieces and 1 piece is eaten by Ben Hence, from the above, We can conclude that The fraction that shows the amount of pizza Ben ate is: Question 10. Which symbol makes this statement true? (A) < (B) > (C) = (D) none The given fractions are: $$\frac{2}{6}$$ and $$\frac{2}{8}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that The symbol that makes the given statement true is: Question 11. Representations Multi-Step Three more friends joined Ben, Lara, and Liz for lunch. Mike and Tom each had one slice of the Olive pizza. Grace had a slice of the Mushroom pizza. After all of the friends had eaten lunch, which statement is true about the slices of Olive and Mushroom pizza that was left? (A) $$\frac{2}{6}$$ < $$\frac{2}{8}$$ (B) $$\frac{3}{6}$$ < $$\frac{3}{8}$$ (C) $$\frac{4}{6}$$ > $$\frac{4}{8}$$ (D) $$\frac{5}{6}$$ = $$\frac{5}{8}$$ It is given that Three more friends joined Ben, Lara, and Liz for lunch. Mike and Tom each had one slice of the Olive pizza. Grace had a slice of the Mushroom pizza and all of the friends had eaten lunch Now, The given figures are: Now, From the given information, The fraction of the olive pizza is: $$\frac{4}{8}$$ The fraction of the mushroom pizza is: $$\frac{4}{6}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that The statement that is true about the slices of olive and mushroom pizza that was left is: Texas Test Prep Question 12. Before taking a hike, Kate and Dylan each ate part of the same-size granola bars. Kate ate $$\frac{1}{3}$$ of her bar. Dylan ate $$\frac{1}{2}$$ of his bar. Which of the following correctly compares the amounts of granola bars that were eaten? (A) $$\frac{1}{3}$$ > $$\frac{1}{2}$$ (B) $$\frac{1}{2}$$ < $$\frac{1}{3}$$ (C) $$\frac{1}{2}$$ > $$\frac{1}{3}$$ (D) $$\frac{1}{3}$$ = $$\frac{1}{2}$$ It is given that Before taking a hike, Kate and Dylan each ate part of the same-size granola bars. Kate ate $$\frac{1}{3}$$ of her bar. Dylan ate $$\frac{1}{2}$$ of his bar Now, The given fractions are: $$\frac{1}{3}$$ and $$\frac{1}{2}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that The statement that correctly compares the amounts of granola bars that were eaten is: ### Texas Go Math Grade 3 Lesson 3.2 Homework and Practice Answer Key Question 1. Shade the models to show $$\frac{1}{8}$$ and $$\frac{1}{6}$$. Then compare the fractions. $$\frac{1}{8}$$ $$\frac{1}{6}$$ The given fractions are: $$\frac{1}{8}$$ and $$\frac{1}{6}$$ Now, The representation of the given fractions in the given models is: Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that Question 2. Shade the models to show $$\frac{3}{4}$$ and $$\frac{3}{6}$$. Then compare the fractions. $$\frac{3}{4}$$ $$\frac{3}{6}$$ The given fractions are: $$\frac{3}{4}$$ and $$\frac{3}{6}$$ Now, The representation of the given fractions in the given models is: Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that Compare. Write <, >, or =. Question 3. $$\frac{1}{3}$$ $$\frac{1}{4}$$ The given fractions are: $$\frac{1}{3}$$ and $$\frac{1}{4}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that $$\frac{1}{6}$$ $$\frac{1}{2}$$ The given fractions are: $$\frac{1}{6}$$ and $$\frac{1}{2}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that Question 5. $$\frac{7}{8}$$ $$\frac{7}{8}$$ The given fractions are: $$\frac{7}{8}$$ and $$\frac{7}{8}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that Question 6. $$\frac{2}{3}$$ $$\frac{2}{6}$$ The given fractions are: $$\frac{2}{3}$$ and $$\frac{2}{6}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that Problem Solving Question 7. Gina and Russ both order the same size sandwich. Gina eats $$\frac{2}{4}$$ of her sandwich. Russ eats $$\frac{2}{6}$$ of his sandwich. Who eats more of the sandwich? It is given that Gina and Russ both order the same size sandwich. Gina eats $$\frac{2}{4}$$ of her sandwich. Russ eats $$\frac{2}{6}$$ of his sandwich Now, The given fractions are: $$\frac{2}{4}$$ and $$\frac{2}{6}$$ Now, We know that, The fraction with the bigger denominator is smaller. So, $$\frac{2}{4}$$ > $$\frac{2}{6}$$ Hence, from the above, We can conclude that Gina eats more of the sandwich Question 8. Karina makes crafts to sell at the fair. She makes $$\frac{1}{2}$$ of the crafts on Saturday and $$\frac{1}{4}$$ of the crafts on Sunday. On which day did she make fewer crafts? It is given that Karina makes crafts to sell at the fair. She makes $$\frac{1}{2}$$ of the crafts on Saturday and $$\frac{1}{4}$$ of the crafts on Sunday Now, The given fractions are: $$\frac{1}{2}$$ and $$\frac{1}{4}$$ Now, We know that, The fraction with the bigger denominator is smaller. So, $$\frac{1}{2}$$ > $$\frac{1}{4}$$ Hence, from the above, We can conclude that Karina made fewer crafts on Sunday Texas Test Prep Lesson Check Question 9. Which symbol makes this statement true? (A) < (B) > (C) = (D) none The given fractions are: $$\frac{4}{6}$$ and $$\frac{4}{8}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that The symbol that makes the given statement true is: Question 10. Which symbol makes the statement true? (A) < (B) > (C) = (D) none The given fractions are: $$\frac{2}{4}$$ and $$\frac{2}{3}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that The symbol that makes the given statement true is: Question 11. Hal finished $$\frac{2}{8}$$ of his math problems. Aaron finished $$\frac{2}{4}$$ of his math problems. Which statement is correct? (A) $$\frac{2}{8}$$ < $$\frac{2}{4}$$ (B) $$\frac{2}{8}$$ = $$\frac{2}{4}$$ (C) $$\frac{2}{4}$$ < $$\frac{2}{8}$$ (D) $$\frac{2}{8}$$ > $$\frac{2}{4}$$ It is given that Hal finished $$\frac{2}{8}$$ of his math problems. Aaron finished $$\frac{2}{4}$$ of his math problems Now, The given fractions are: $$\frac{2}{8}$$ and $$\frac{2}{4}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that The correct statement is: Sharon ate $$\frac{4}{4}$$ of one orange and $$\frac{4}{8}$$ of a second orange. Which statement is correct? (A) $$\frac{4}{8}$$ > $$\frac{4}{4}$$ (B) $$\frac{4}{8}$$ < $$\frac{4}{4}$$ (C) $$\frac{4}{4}$$ = $$\frac{4}{8}$$ (D) $$\frac{4}{4}$$ < $$\frac{4}{8}$$ It is given that Sharon ate $$\frac{4}{4}$$ of one orange and $$\frac{4}{8}$$ of a second orange. Now, The given fractions are: $$\frac{4}{4}$$ and $$\frac{4}{8}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that The correct statement is: Question 13. Multi-Step Jordan ate 3 slices of cheese pizza. Then Len ate two slices of veggie pizza and 1 slice of cheese pizza. Which statement is true about the amount of leftover pizza? (A) $$\frac{4}{6}$$ < $$\frac{4}{8}$$ (B) $$\frac{4}{8}$$ > $$\frac{4}{6}$$ (C) $$\frac{4}{6}$$ > $$\frac{4}{8}$$ (D) $$\frac{4}{6}$$ = $$\frac{4}{8}$$ It is given that Jordan ate 3 slices of cheese pizza. Then Len ate two slices of veggie pizza and 1 slice of cheese pizza Now, The given figures are: Now, From the above, We can observe that The total number of slices of veggie pizza is: 6 slices The total number of slices of cheese pizza is: 6 slices Now, From the given information, The total number of slices of pizzas eaten by Jordan is: $$\frac{4}{6}$$ The total number of slices of pizza eaten by Len is: $$\frac{4}{8}$$ Now, We know that, The fraction with the bigger denominator is smaller. Hence, from the above, We can conclude that Scroll to Top Scroll to Top
# Exponential Growth – Examples and Practice Problems Exponential functions can be used to model population growth scenarios or other situations that follow patterns with growth at fixed rates. There are formulas that can be used to find solutions to most problems related to exponential growth. Here, we will look at a summary of exponential growth and the formulas that can be used to solve these types of problems. In addition, we will look at several examples with answers of exponential growth in order to learn how to apply these formulas. ##### ALGEBRA Relevant for Exploring examples of exponential growth. See examples ##### ALGEBRA Relevant for Exploring examples of exponential growth. See examples ## Summary of exponential growth Exponential growth is a pattern of data that shows larger increases over time, creating the curve of an exponential function. For example, if a bacteria population starts with 2 in the first month, then with 4 in the second month, 16 in the third month, 256 in the fourth month, and so on, it means that the population grows exponentially with a power of 2 every month. The following formula is used to model exponential growth. If a quantity grows by a fixed percentage at regular intervals, the pattern can be described by this function: We recall that the original exponential function has the form . In the original growth formula, we have replaced b with . So, in this formula we have: • initial value. This is the starting amount before growth. • growth rate. This is represented as a decimal. • time interval. This is the time that has passed. Most naturally occurring events continually grow. For example, bacteria continue to grow over a 24-hour period. Bacteria don’t wait until the end of 24 hours to reproduce all at the same time. To model the continuous growth that occurs naturally such as populations, bacteria, etc., we use the exponential ee can be thought of as a universal constant that represents growth possibilities using a continuous process. Furthermore, using e we can also represent growth measured periodically over time. Therefore, if a quantity is continually growing with a fixed percentage, we can use the following formula to model this pattern: In this formula we have: • final value. This is the amount after growth. • initial value. This is the amount before growth. • exponential. e is approximately equal to 2718… • continuous growth rate. It is also called the constant of proportionality. • elapsed time. ## Exponential growth – Examples with answers The following examples use the formulas detailed above and some variations to find the solution. It is recommended that you try to solve the exercises yourself before looking at the answer. ### EXAMPLE 1 A population of bacteria grows according to the function , where t is measured in minutes. How many bacteria will there be after 4 hours (240 minutes)? This is continuous growth, so we have the formula . We can recognize the following data: Therefore, we have: Therefore, there will be 12 151 bacteria after 4 hours. ### EXAMPLE 2 A population of bacteria grows according to the function , where t is measured in minutes. When will the population reach 50 000? Here, we have the same formula as the previous exercise, but now we have to find the time knowing the final quantity. We can recognize the following data: Therefore, we have: Therefore, the population of bacteria will become 50 000 after 310.73 minutes. ### EXAMPLE 3 We can model the population of a community with the formula . Here, A represents population and t represents time in years. What is the population after 10 years? We already have a given formula: . We have to calculate the population using time . Therefore, we substitute to get: Therefore, the population in the community after 10 years will be 10 513. ### EXAMPLE 4 The population of a certain community was 10 000 in 1980. In 2000, it was found to have grown to 20 000. Form an exponential function to model the population of community P that changes through time t. When we have continuous population growth, we can model the population with the general formula , where represents the initial population, λ is the exponential growth constant and t is time. Using the given information, we have to find the constant λ to complete the formula. Therefore, we have: Thus, we can model the population growth of the community with the formula . ### EXAMPLE 5 The population growth of a small city is modeled with the function . When did the population reach 37 500 if in 1980 the population was 12 500? We can substitute the values in the formula with the given information: Now, we have to solve for time: Thus, the city’s population reached 37 500 in 1989. ### EXAMPLE 6 One type of bacteria doubles every 5 minutes. Assuming we start with one bacterium, how many bacteria will we have at the end of 96 minutes? We know that bacteria grow continuously, so we have to use the formula: The bacteria doubles every 5 minutes, so after 5 minutes, we will have 2. We use this to find the value of k: Now, we form the equation using this value of k and solve using the time of 96 minutes: ## Exponential growth – Practice problems Practice using the exponential growth formulas with the following exercises. Solve the problems and select an answer. Check your answer to verify that you selected the correct one.
# How to Identify Triangles ••• Thinkstock/Comstock/Getty Images Print A triangle is a three-sided polygon. Knowing the rules and relationships between the various triangles helps to understand geometry. More importantly, for the high school student and the college-bound senior, this knowledge will help you save time on the all-important SAT tests. Measure the three sides of the triangle with a ruler. If all three sides are the same length, then it is an equilateral triangle, and the three angles contained by those sides are the same. So an equilateral triangle is also an equiangular triangle. An important point to remember is that, in this case, all three angles measure 60 degrees. Regardless of the length of the sides, each angle of the equiangular triangle will be 60 degrees. Cross-check by measuring the angles with the protractor. If each angle measures 60 degrees, then the triangle is equiangular and--by definition--equilateral. Label the triangle "isosceles" if only two sides are equal. Remember that the angles contained by the two equal sides (the base angles) will be equal to each other. So, if you know one base angle in an isosceles triangle, you can find the other two angles. For example, if one angle is 55 degrees, then the other base angle will be 55 degrees. The third angle will be 70 degrees, derived from 180 - (55+55). Conversely, if two angles are equal, then two sides will also be equal. Know that the equilateral triangle is a special case of the isosceles triangle since it has not two but all three sides and all three angles equal. A right triangle is also a special case of the isosceles triangle. The angles of the right isosceles triangle measure 90 degrees, 45 degrees and 45 degrees. If you know one angle, you can determine the other two. Learn that a right triangle has one 90-degree angle. The side opposite the 90-degree angle is the hypotenuse, and the other two sides are the legs of the triangle. The Pythagorean theorem relates to the right triangle and states that the square on the hypotenuse is equal to the sum of the squares on the other two sides. A special case of the right triangle is the 30-60-90 triangle. Look at the three angles of the triangle. If each angle is less than 60 degrees, then label the triangle an "acute" triangle. If even one angle measures more than 90 degrees, then the triangle is an obtuse triangle. The other two angles of the obtuse triangle will be less than 90 degrees. Learn these basic properties of triangles. They will help you save time when working on geometry problems. The sum of the angles of a triangle equals 180 degrees. So, if you know two angles, you can deduce the third. In special cases, knowing just one angle will give you the other two. If you know one interior angle, then you can find the exterior angle of the triangle by subtracting the interior angle from 180 degrees. For example, if the interior angle measures 80 degrees, the corresponding exterior angle will be 180 - 80 = 100 degrees. The largest side has the largest angle opposite it. It follows that the shortest side has the smallest angle opposite it. • Ruler • Protractor
# Marginal probability To explain what marginal probability is, we need a contingency table. A contingency table is a table in which we show frequency for 2 variables. One variable is used to categorize rows and the other is used to categorize columns. Suppose a company specializes in training students to pass the GED test. The company had 200 students last year. The following contingency table show their pass rates broken down by sex. In the table above, the variable that categorizes rows is the sex. This variable can be males or females. The variable that categorizes columns is the result. This variable can be pass or fail. Did you notice the followings? • The number of males + females = 102 + 98 = 200 • The number of students who passed + the number of students who failed = 114 + 86 = 200 • The number of females who failed the GED test is equal to 30 Now that you understand what a contingency table is, what is marginal probability ? In the table above, there are 4 events. • A student is a male • A student is a female • A student has passed • A student has failed The probability of each of these 4 events is called marginal probability or simple probability The 4 marginal probabilities can be calculated as follows P(A student is a male) = Number of males / Total number of students P(A student is a male) = 102 / 200 = 0.51 P(A student is a female) = Number of females / Total number of students P(A student is a female) = 98 / 200 = 0.49 P(A student has passed) = Number of students who passed / Total number of students P(A student has passed) = 114 / 200 = 0.57 P(A student has failed) = Number of students who failed / Total number of students P(A student has failed) = 86 / 200 = 0.43 The marginal probabilities are shown along the right side and along the bottom of the table below. ## Recent Articles 1. ### How To Find The Factors Of 20: A Simple Way Sep 17, 23 09:46 AM There are many ways to find the factors of 20. A simple way is to... 2. ### The SAT Math Test: How To Be Prepared To Face It And Survive Jun 09, 23 12:04 PM The SAT Math section is known for being difficult. But it doesn’t have to be. Learn how to be prepared and complete the section with confidence here.
Miscellaneous Category : 6th Class FACE VALUE The face value of a number does not change regardless of the place it occupies. Therefore, the face value of a number is the number itself. For example: 781 face value of 7 = 7. Face value of 8 = 8, Face value of 1 = 1. SUCCESSOR Successor of every number comes just after the number. It is obtained by adding 1 to the given number. For example: Successor of 28 = 28 + 1 = 29 Successor of 79 = 79 + 1 = 80. PREDECESSOR Predecessor of every number comes just before the number. It is obtained by subtracting 1 from the given number. For example: Predecessor of 28 = 28 - 1 = 27 Predecessor of 30 = 30 - 1 = 29 1. If two positive or two negative integers are added, we add their values without considering their signs and put common sign before the sum. +36 + 27 + 36 + 27 + 63 1. To add a positive and a negative integer, we calculate the difference in their numerical values regardless of their signs and put the sign of greater numerical value integer to the value of difference. +36 - 27 +36 -27 + 9 1. Closure property of addition: The sum of two integers is always an integer. Examples: (i) 4 + 3 = 7, which is an integer (ii) 4 + (-3) = 1, which is an integer (iii) - 4 + 3 = - 1, which is an integer (iv) - 4 + (-3) = -7, which is an integer 1. Commutative law of addition: If x and y are any two integers, then, x + y = y+ x Examples: (i) – 7 + 8 = 1 and 8 + (-7) = 1 $\therefore$ -7 + 8 = 8 + (-7) (ii) (-5) + (-8) = -13 and (-8) + (-5) = -13 $\therefore$ (-5) + (-8) = (-8) + (-5) 1. Associative law of addition: If x, y and z are any three integers then (x + y) + z = x + (y + z) Example: {(-5) + (- 6)} + 7 = -11 + 7 = -4 (-5) + {(- 6) + 7} = -5 + 1 = - 4 $\therefore$ {(-5) + (- 6)} + 7 = - 5 + {(-6) + 7} SUBTRACTION OF INTEGERS For any integers x and y, we define. (i) x - y = x + (additive inverse of y)$=x+(-y)$ (ii) x - (- y) = x + {additive inverse of (- y)} = x + y PROPERTIES OF SUBTRACTION OF INTEGERS 1. Closure property for subtraction: If x and y are any integer, then x - y is always an integer. Examples: (i) 3 - 5 = - 2, which is an integer (ii) (-3) - 6 = - 9, which is an integer 1. Subtraction of an integer is not commutative: Examples: (i) Consider the integers 2 and 4, we have (2 - 4) = 2 + (- 4) = -2 and (4 - 2) = 4 + (-2) = 2 $\therefore \,\,(2-4)\ne (4-2)$. (ii) Consider the integers (-5) and 3 we have (-5) -3 = (-5) + (- 3) = -8 and 3 - (-5) = 8 $\therefore \,\,(-5)-3\ne -3-(-5)$ 1. Subtraction of integers is not associative: Consider the integers 4, (-5) and (-6), we have. {4 - (-5)} - (-6) = (4 + 5) - (- 6) = 9 - (- 6) = 9 + 6 = 15 4 - {(-5) - (-6)} = 4 - {(-5) + 6} = 4 - 1 = 3 {4 - (-5)} - (-6) ^ 4 - {(-5) - (- 6)} MULTIPLICATION OF INTEGERS Rule 1: To find the product of two integers with unlike sign, first get their product regardless to their signs, then give minus sign to the product. Examples: $(-40)\times 9$ = - 360 Rule 2: To find the product of two integers with like signs, first find their product regardless to their signs then put plus sign to product. Examples: $3\times 5$ = + 15 PROPERTIES OF MULTIPLICATION OF INTEGERS 1. Closure properties for multiplication: The product of two integers is always an integer. Examples: (i) $3\times 2=6$, which is an integer (ii) $(-3)\,\times 2=-6,$ which is an integer 1. Commutative law of multiplication: For any two integers x and y, we have. $(x\times y)=(y\,\times x)$ Examples: (i) $2\times (-6)=-12$ and $(-6)\times 2=-12$ $\therefore \,\,2\times (-6)=(-6)\times 2$ (ii) $(-3)\times (-7)\,=21$ and $(-7)\times (-3)\,=21$ $\therefore \,(-3)\,\times (-7)\,=(-7)\,\times (-3)$ 1. Associative law of multiplication: For any integers x, y and z, we have Examples: Consider the integers 3, (- 4) and (-5), we have $\{3\times (-4)\}\times (-5)\,=-12\times (-5)=60$ and $3\times \{(-4)\times (-5)\}=3\times 20=60$ $\therefore$ $\{3\times (-4)\}\,\times (-5)\,=3\times \,\{(-4)\times (-5)\}$ DIVISION OF INTEGERS Rule 1: To find the division of two integers with unlike sign, first get their quotient regardless to their signs, then give minus sign to the product. Example: $-74\div 2$ $=\frac{-74}{2}$ = - 37 Rule 2: To find the division of two integers with like signs, first find their product regardless to their signs then put plus sign to quotient. Example: $48\pm 6$ $=\frac{48}{6}$ = 8 PROPERTIES OF DIVISION OF INTEGERS: 1. If x and y are integers then $(x\div y)$ is not necessarily an integer Example: (i) 12 and 5 are both integer but (12 - 5) is not an integer. (ii) (-12) and 5 are both integer. But [(-12) - 5)] is not an integer. 1. If x is an integer and x 0, then x - x = 1 Examples: (i) $10\div 10=1$ (ii) $(-5)\div (-5)=1$ 1. If x is an integer, then $(x\div 1)=x$ Examples: (i) $5\div 1=5$ (ii) $(-5)\div 1=(-5)$ 1. If x is an integer and $x\ne 0,$ then $(0\div x)=0$ but $(x\div 0)$ is not meaningful. Examples: (i) $0\div 9=0$ (ii) $0\div (-9)=0$ BODMAS RULE When a single expression contains many mathematical operations then BODMAS rules are used for multiplication of the expression the word BODMAS has been arranged according to the priority of the operations. BODMAS STANDS FOR B stands for Bracket O stands for of D stands for Division M stands for Multiplication S stands for Subtraction BRACKET Sometimes in complex expressions, we require a set of operations to be performed prior to the others. Here we make use of brackets. Brackets which are commonly used are: Bracket Name () Parentheses or common bracket {} Brace or curly bracket [ ] Square on box bracket ELIMINATION OF BRACKETS The solution of the expression is obtained by eliminating the brackets. For example: Steps to simplify the expression $[\{2+4)\times 5\}-1]:$ 1st step: Eliminate the common bracket by adding 2 and $4=[\{6\times 5\}-1]$ 2nd step: Eliminate the curly bracket by multiplying 6 and 5 = [30 - 1] 3rd step: Eliminate the box bracket by subtracting 30 and 1 = 29 The solution of the given expression is 29. DIVISION OF ALGEBRIC EXPRESSION To divide one polynomial to other (i) keep the polynomials which is to be divided in division form (ii) then divide first term of dividend by first term of divisor and write quotient (iii) then write the product of quotient x divisor, below the divident and subtract it from dividend (iv) repeat this process until the degree of remainder becomes less than divisor. For example: $2{{x}^{2}}+3x+1$ by $(x+1)$ Quotient of the division $=2x+1$ MULTIPLICATION OF FRACTIONS To multiply like or unlike fractions, we should first multiply the numerators and denominators and then, write the answer into lowest term of the resulting fraction. For example: To multiply $\frac{8}{3}$ and $\frac{24}{5}$ Multiplication of the fractions $=\frac{\text{Product}\,\text{of}\,\text{numerators}}{\text{Product}\,\text{of}\,\text{denominators}}$ $=\frac{8\times 24}{3\times 5}\,=\frac{192}{15}$ The HCF of 192 and 15 is 3, therefore, $\frac{192\div 3}{15\div 3}\,=\frac{64}{5}$ DIVISION OF FRACTIONS To divide the fractions, we should first multiply the dividend by reciprocal of divisor and then, write the answer into lowest term of the resulting fraction. For example: To divide $\frac{75}{6}$ by $\frac{18}{5}$ Division of fractions $=\frac{75}{6}\div \frac{18}{5}\,=\frac{75}{6}\,\times \frac{5}{18}\,=\frac{375}{108}$ The HCF of 375 and 108 is 3 therefore, $=\frac{375\div 3}{108\div 3}\,=\frac{125}{36}$ MULTIPLICATION OF DEC IMALS Step (i) Multiply the given decimals by omitting decimal point. Step (ii) Put the decimal point in the result by counting total number of decimal places for both the decimals. For example: The multiply 16.24 and 41.16 1624 x 4116 ________ 6684384 Now, the total decimal places in the given decimals = 4 So, resulting product = 668.4384 DIVISION OF DECIMALS Step (i) Decimal point of the divisor is moved to right until it becomes a whole number. Step (ii) Decimal point of the dividend is moved to right as same as the number of places have been moved in divisor. Step (iii) Multiply both divisor and dividend by a number to get whole number of necessary. Step (iv) Put decimal point in the quotient by counting total number of decimal places for both dividend and divisor For example: Division of 467.2 by 2.4 $=\frac{467.2}{2.4}=\frac{4672}{24}\times \frac{10}{10}\,=194.6$ PARALLELOGRAM Perimeter of parallelogram ABCD = 2 (Sum of two adjacent sides) = 2(AB + BC) Area of parallelogram ABCD = Base x Height = AB x CE For example: (i) the perimeter (ii) the area Solution: Perimeter = 2 (AB + BC) = 2 (5 + 3) = 16 cm Area $=AB\times CE$ $=5\times 4=20\,c{{m}^{2}}$ Other Topics LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL TODAY ONLY! You need to login to perform this action. You will be redirected in 3 sec
Question # Find number of value of $x\in \left[ -2\pi ,2\pi \right]$, If ${{\log }_{0.5}}\sin x=1-{{\log }_{0.5}}\cos x$. Hint: Use property of logarithm that is ${{\log }_{a}}a=1$and ${{\log }_{c}}a-{{\log }_{c}}b={{\log }_{c}}\left( \dfrac{a}{b} \right)$to simplify the given relation. Use a graphical approach to find values of â€˜xâ€™ for simplicity. Here, it is given that ${{\log }_{0.5}}\sin x=1-{{\log }_{0.5}}\cos x$, and then we need to determine all the values of x lying in$\left[ -2\pi ,2\pi \right]$. We have ${{\log }_{0.5}}\sin x=1-{{\log }_{0.5}}\cos x\ldots \ldots (1)$ As we know the property of the logarithm function that ${{\log }_{a}}a=1$ where $a>0$and $a\ne 1$. Or vice-versa is also true. It means we can replace â€˜1â€™ from equation (1) by ${{\log }_{0.5}}0.5$for the simplification of the problem. Hence, equation (1) can be written as ${{\log }_{0.5}}\sin x={{\log }_{0.5}}0.5-{{\log }_{0.5}}\cos x$ We can use property of logarithm ${{\log }_{c}}a-{{\log }_{c}}b={{\log }_{c}}\left( \dfrac{a}{b} \right)$, with the above equation and get ${{\log }_{0.5}}\sin x={{\log }_{0.5}}\left( \dfrac{0.5}{\cos x} \right)\ldots \ldots (2)$ As we know that â€˜aâ€™ should be equal to â€˜bâ€™ if ${{\log }_{c}}a={{\log }_{c}}b$. Hence, using the above property with equation (2), we get $\dfrac{\sin x}{1}=\dfrac{0.5}{\cos x}$ On cross-multiplying, we get $\sin x\cos x=\dfrac{1}{2}$or $2\sin x\cos x=1\ldots \ldots (3)$ As we know the trigonometric identity of $\sin 2x$as $\sin 2x=2\sin x\cos x$or vice-versa. Hence, equation (3) can be given as $\sin 2x=1\ldots \ldots (4)$ Now, we have to find â€˜xâ€™ in the interval$\left[ -2\pi ,2\pi \right]$. So we have $-2\pi \le x\le 2\pi$ Multiplying by â€˜2â€™ on each side we get $-4\pi \le 2x\le 4\pi$ Now, drawing graph of $\sin x$ from $-4\pi$ to $4\pi$, we get As we can observe that $y=\sin x$ has values of 1 at $\dfrac{\pi }{2},\dfrac{5\pi }{2},\dfrac{-3\pi }{2},\dfrac{-7\pi }{2}$. Now, we have the equation $\sin 2x=1$. Hence, $2x=\dfrac{-7\pi }{2},\dfrac{-3\pi }{2},\dfrac{\pi }{2}.\dfrac{5\pi }{2}$ Or $x=\dfrac{-7\pi }{4},\dfrac{-3\pi }{4},\dfrac{\pi }{4}.\dfrac{5\pi }{4}$ Note: One can get confusion between $y=\sin x$ and equation $\sin 2x=1$. Graph of $y=\sin x$ is representing the general relation between angles and values which is not related to equation$\sin 2x=1$. One can suppose â€˜2xâ€™ as â€˜tâ€™ as well for the simplicity, so we will get equation $\sin t=1$. Now, t will lie in $\left[ -4\pi ,4\pi \right]$ as $t=2x$; hence find all values of â€˜tâ€™ then find â€˜xâ€™ by using relation $x=\dfrac{t}{2}$.
# 1. (a) Compute the correct and necessary vector components... 1. (a) Compute the correct and necessary vector components between Point B and Point A needed to compute the azimuth BA ? You have the choice between a vector with two components ( ?N,?E ) or three components ( ?N,?E,?H ). (b) Explain why your choice in (a) is correct--either two components or three components; or does it even matter? Explain. 2. Compute the correct and necessary vector components between Point P and Point A needed to (a) compute the horizontal distance PA, and (b) the correct slope distance PA? 3. The approximate angular relationship between vectors PA and PB is indicated to be about 8 degrees. Use the dot product to confirm that the two vectors do not form an angle of 90 degrees at P. Take the answer to your computations and use it to explain why the two vectors do not form an angle of 90 Degrees. Document Preview: SUR 330 Homework 1 Vectors Sketch and Data Station P (Instrument Station) N = 12, 881.63'; E = 9799.86'; H = 5174.30' 1. (a) Compute the correct and necessary vector components between Point B and Point A needed to compute the azimuth BA? You have the choice between a vector with two components (?N,?E ) or three components (?N,?E,?H ). (b) Explain why your choice in (a) is correct--either two components or three components; or does it even matter? Explain. 2. Compute the correct and necessary vector components between Point P and Point A needed to (a) compute the horizontal distance PA, and (b) the correct slope distance PA? 3. The approximate angular relationship between vectors PAand PB is indicated to be about 8 degrees. Use the dot product to confirm that the two vectors do not form an angle of 90 degrees at P. Take the answer to your computations and use it to explain why the two vectors do not form an angle of 90 Degrees.4. Basic definition given in most linear algebra textbooks: Three points determine a plane. You are given three points above. Use them to determine the normal vector to the resulting plane. Matrices 5. Consider a ordinary spring that might be on the screen door of your house. Hookes law states that a linear relationship exists between the force applied to that spring (x) and the amount it stetches (y). Available experimental data for a particular spring (neglecting units) is : y x Since the relationship between the force and the resulting spring length is linear--the model is that of a straight line. The model for this least squares problem is y = mx + b. In order fill the matrices A and b in the matrix equation Ax =b, we rewrite the equation: mx +b =y To simplify this problem we assume that the force applied to the spring is a constant (exactly known). Y is the only measured quantity. We need to solve for the two variables: m = slope and b = y intercept. (a) Write... Attachments:
# EteMS KHOO SIEW YUEN SMK.HI-TECH 2005. ## Presentation on theme: "EteMS KHOO SIEW YUEN SMK.HI-TECH 2005."— Presentation transcript: EteMS KHOO SIEW YUEN SMK.HI-TECH 2005 Topic : Circuler Meassure Area of Sector Mathematic Skills 1. Determine - area of sector radius - the radius - the angle subtended center of a circle 2. Find the area of segment of circle Prior Knowledge The students already learnt how to convert measurement in radians to degrees and vice versa. how to determine the length of arc, radius and angle subtended at the centre of a circle. - how to find the perimeter of the segment of circle. Learning Outcomes: At the end of this lesson, students will be able to: find the area of sector radius, the radius and the angle subtended center of a circle. - answer the quiz 100% correctly. - answer at least 3 out of 5 questions given in the worksheet correctly. CONNECT TO LESSON PLAN Area of Sector Area Quiz Radius Angle Worksheets What is Area Sector of circle??? Like that??? And this is a circle with sector AOB. The area shaded blue color names as Area AOB. This is a circle with center O. O Diagram below shows a circle with 4cm radius and angle 270 o, find the shaded area. Given r=4cm, angle = 270 o r=4 cm O 270o 2. Find the angle of the sector AOB in radiant. = 270o x = 270o x Given r = r cm Perimeter = 90cm Arc AB + OA + AB = 90cm 50 + r + r = 90cm 2r = 40 r = 40 3. Diagram above shows a sector AOB with center O made from wire length 90cm, the length of arc OAB is 50cm, find : - the angle in radiant The area of sector AOB 4. A circle sector with area 25. 2cm2 and the angle at the centre is 1 4. A circle sector with area 25.2cm2 and the angle at the centre is 1.4rad. Find the radius of the circle. 5. Find the area of minor sector AOB. = 270o x Similar presentations
# If Sr Denotes the Sum of the First R Terms of an A.P. Then , S3n: (S2n − Sn) is - Mathematics MCQ If Sr denotes the sum of the first r terms of an A.P. Then , S3n: (S2n − Sn) is #### Options • n • 3n • 3 • none of these #### Solution Here, we are given an A.P. whose sum of r terms is Sr. We need to find (S_(3n))/(S_(2n) - S_n). Here we use the following formula for the sum of n terms of an A.P. S_n = n/2 [ 2a + (n -1 ) d] Where; a = first term for the given A.P. d = common difference of the given A.P. = number of terms So, first we find S3n, S_(3n) = (3n)/2 [ 2a + (3n - 1)d] =(3n)/2 [2a + 3nd - d ] .................(1) Similarly, S_(2n) = (2n)/2 [ 2a + (2n - 1 ) d ] = (2n)/2 [2a + 2nd -d] .................(2) Also, S_n = n/2 [ 2a + (n-1) d] =n/2 [2a + nd - d ] So, using (1), (2) and (3), we get, (S_(3n))/(S_(2n) - S_n) = ((3n)/2 [2a + 3nd - d])/((2n)/2 [ 2a + 2nd - d ] - n/2 [ 2a + nd - d ]) Taking n/2 common, we get, (S_(3n))/(S_(2n) - S_n) =(3[2a + 3nd - d])/(2[2a + 2nd - d ]- [2a + nd - d]) =(3[2a + 3nd - d])/(4a + 4nd - 2d - 2a - nd + d) =(3[2a + 3nd - d])/(2a + 3nd - d) = 3 Therefore, (S_(3n))/(S_(2n)- S_n )= 3 Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 10 Maths Chapter 5 Arithmetic Progression Q 17 \| Page 58
2 Proper fractions with linear factors Firstly we describe how to calculate partial fractions for proper fractions where the denominator may be written as a product of linear factors. The steps are as follows: $\phantom{\rule{2em}{0ex}}\bullet \phantom{\rule{1em}{0ex}}$ Factorise the denominator. $\phantom{\rule{2em}{0ex}}\bullet \phantom{\rule{1em}{0ex}}$ Each factor will produce a partial fraction. A factor such as $3x+2$ will produce a partial fraction of the form $\frac{A}{3x+2}$ where $A$ is an unknown constant. In general a linear factor $ax+b$ will produce a partial fraction $\frac{A}{ax+b}$ . The unknown constants for each partial fraction may be different and so we will call them $A$ , $B$ , $C$ and so on. $\phantom{\rule{2em}{0ex}}\bullet \phantom{\rule{1em}{0ex}}$ Evaluate the unknown constants by equating coefficients or using specific values of $x$ . The sum of the partial fractions is identical to the original algebraic fraction for all values of $x$ . Key Point 14 A linear factor $ax+b$ in the denominator gives rise to a single partial fraction of the form $\frac{A}{ax+b}$ The steps involved in expressing a proper fraction as partial fractions are illustrated in the following Example. Example 41 Express $\frac{7x+10}{2{x}^{2}+5x+3}$ in terms of partial fractions. Solution Note that this fraction is proper. The denominator is factorised to give $\left(2x+3\right)\left(x+1\right)$ . Each of the linear factors produces a partial fraction. The factor $2x+3$ produces a partial fraction of the form $\frac{A}{2x+3}$ and the factor $x+1$ produces a partial fraction $\frac{B}{x+1}$ , where $A$ and $B$ are constants which we need to find. We write $\phantom{\rule{2em}{0ex}}\frac{7x+10}{\left(2x+3\right)\left(x+1\right)}=\frac{A}{2x+3}+\frac{B}{x+1}$ By multiplying both sides by $\left(2x+3\right)\left(x+1\right)$ we obtain $\phantom{\rule{2em}{0ex}}7x+10=A\left(x+1\right)+B\left(2x+3\right)$ …() We may now let $x$ take any value we choose . By an appropriate choice we can simplify the right-hand side. Let $x=-1$ because this choice eliminates $A$ . We find $\begin{array}{rcll}7\left(-1\right)+10& =& A\left(0\right)+B\left(-2+3\right)& \text{}\\ 3& =& B\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$ so that the constant $B$ must equal 3. The constant $A$ can be found either by substituting some other value for $x$ or alternatively by ‘equating coefficients’. Observe that, by rearranging the right-hand side, Equation () can be written as $\phantom{\rule{2em}{0ex}}7x+10=\left(A+2B\right)x+\left(A+3B\right)$ Comparing the coefficients of $x$ on both sides we see that $7=A+2B$ . We already know $B=3$ and so $\begin{array}{rcll}7& =& A+2\left(3\right)& \text{}\\ & =& A+6\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$ from which $A=1$ . We can therefore write $\phantom{\rule{2em}{0ex}}\frac{7x+10}{2{x}^{2}+5x+3}=\frac{1}{2x+3}+\frac{3}{x+1}$ We have succeeded in expressing the given fraction as the sum of its partial fractions. The result can always be checked by adding the fractions on the right. Express $\frac{9-4x}{3{x}^{2}-x-2}$ in partial fractions. First factorise the denominator: $\left(3x+2\right)\left(x-1\right)$ Because there are two linear factors we write $\phantom{\rule{2em}{0ex}}\frac{9-4x}{3{x}^{2}-x-2}=\frac{A}{3x+2}+\frac{B}{x-1}$ Multiply both sides by $\left(3x+2\right)\left(x-1\right)$ to obtain the equation from which to find $A$ and $B$ : $9-4x=A\left(x-1\right)+B\left(3x+2\right)$ Substitute an appropriate value for $x$ to obtain $B$ : Substitute $x=1$ and get $B=1$ Equating coefficients of $x$ to obtain the value of $A$ : $-4=A+3B$ , $A=-7$ since $B=1$ Finally, write down the partial fractions: $\frac{-7}{3x+2}+\frac{1}{x-1}$ Exercises 1. Find the partial fractions of 1. $\frac{5x-1}{\left(x+1\right)\left(x-2\right)}$ , 2. $\frac{7x+25}{\left(x+4\right)\left(x+3\right)}$ , 3. $\frac{11x+1}{\left(x-1\right)\left(2x+1\right)}$ . Check by adding the partial fractions together again. 2. Express each of the following as the sum of partial fractions: 1. $\frac{3}{\left(x+1\right)\left(x+2\right)}$ , 2. $\frac{5}{{x}^{2}+7x+12}$ , 3. $\frac{-3}{\left(2x+1\right)\left(x-3\right)}$ , 1(a) $\frac{2}{x+1}+\frac{3}{x-2}$ , 1(b) $\frac{3}{x+4}+\frac{4}{x+3}$ 1(c) $\frac{4}{x-1}+\frac{3}{2x+1}$ , 2(a) $\frac{3}{x+1}-\frac{3}{x+2}$ , 2(b) $\frac{5}{x+3}-\frac{5}{x+4}$ , 2(c) $\frac{6}{7\left(2x+1\right)}-\frac{3}{7\left(x-3\right)}$ .
We had a very good set of solutions sent in from Bomere Heath Primary School. Here are a couple of examples but all of their contributions are well worth looking at here: BomereSchoolSolutions.doc or BomereSchoolSolutions.pdf Shauna and Bethany: First we found all the combinations for adding and subtracting four consecutive numbers. Then we found out the answer for them. We did this two times then we looked for patterns such as: 1)+++ =2 middle numbers added together then x2. 2)--- =last number x2 in the negatives. 3)+-- =-4. 4)++- =first number times 2. 5)-++ =3rd number x2. 6)--+ =0. 7)-+- =-2. 8)+-+ = 2nd number x2. 9)= All even. and that’s what we found out:) From Archie and Jodie: Firstly we wrote down four consecutive numbers, in our case 3, 4, 5, 6 then we found all the diffrent combinations eg: 3+4+5+6, 3-4-5-6, 3+4+5-6, 3+4-5-6 After that we found out all the answers for them. Next we did the same process but with 22, 23, 24, 25 and found out all the combinations eg: 22+23-24+25, 22-23+24+25, 22-23-24+25, 22-23+24-25 Then we some patterns like with + + + if you add the two middle numbers together then double it you get the answer. And with - + - the answer is always -2. And with - - + the answer is always 0. And with + - + you double the second number to get the answer. And with + - - the answer is always -4. And with - + + you double the third number to get the answer. And with + + - you double the first number to get the answer. And with - - - you double the end number to get the answer. We hope you like our solution and hope it helps you if you do this challenge. We had the following from St. Paul's Primary School: 5+6+7+8=26 5-6-7-8=-16 5+6+7-8=10 5-6-7+8=0 5-6+7-8=-2 5+6-7+8=12 5+6-7-8=-4 5-6+7+8=14 In each answer we noticed that there are pairs of answers which belong together. For example, 26 and -16, 12 and -2. The reason for this is that they have the same unit number. We also noticed that the difference between these pairs is 10, except for the pair 26 and -16, which is 42. All the positive answers are over 10. If there are two subtractions signs the answer is a negative number. If there are two add signs the answer will be a positive number. In 2016 we had a long Word document sent in from year 5 at Richardson Dees Primary School. Here is a very small part as a taster, but you can read the very worthwhile document here.doc . 1. 4+5+6+7= 22 2. -4-5-6-7= -22 3. 4-5+6+7= 12 - double the 3rd number 4. 4+5-6+7= 10 - double the 2nd number 5. 4+5+6-7= 8 - double the 1st number 6. -4+5+6+7= 14 - double the 4th number Thank you for all the solutions that were sent in, it was really good to read about all the things you have found out. Keep up investigating Maths!
# “Prove” 3 = 0. Can You Spot The Mistake? Hey, this is Presh Talwalkar. Here’s a portion of an email I received. Hi I’m Lucas from Brazil, and I love your videos. Here’s a problem from a book with math problems that I could not solve and the book doesn’t say the answer, it just said “This one we leave to the students!” I translated the problem for you, and i’m also sending you a photo of the original problem and the book link from where I got the problem I was so grateful to Lucas, I replied right away. “Thanks for writing and for translating too, I appreciate the original photo” So what was the problem? It was a false proof that 3 is equal to 0 Here’s how it starts. Let x be a solution of x squared plus x plus 1 equals 0 Since x is not equal to 0 we can divide both sides by x Let’s simplify this to the following form we have x plus 1, plus 1 over x is equal to 0 Now, from the first equation we have x plus 1 is equal to negative x squared We’re going to substitute that into the second equation so that first x plus 1 becomes negative x squared, then we have plus 1 over x is equal to 0 We’ll now rearrange this equation to get 1 over x is equal to x squared This means 1 is equal to x cubed so x is equal to 1 We now take x equals 1 and substitute it into our original equation x squared plus x plus 1 equals 0 We get 1 squared plus 1 plus 1 is equal to 0 which means 3 is equal to 0 This is clearly a nonsensical result But every single step seemed to be correct So where is the mistake in this false proof that 3 is equal to 0? Can you figure it out? Give this problem a try and when you’re ready keep watching the video for the explanation. So let’s take a look at this false proof Which step introduces the mistake from which the ultimate result of three equals zero originates? That would be between the second and third steps where we substitute x plus 1 is equal to negative x squared The problem with this step is that it creates an extraneous solution x equals 1, which is not a solution to the original equation, x squared plus x plus 1 equals 0 So let me explain that in a little bit more detail We start out where x is a solution of x squared plus x plus 1 equals 0, so far so good This has two solutions, which we can get from the quadratic formula We have negative 1 plus i times the square root of 3 all over 2, and we have negative 1 minus i times the square root of 3 all over 2 Now x is not equal to 0 so it is valid for us to divide both sides by x Now we have a new equation x plus 1 plus 1 over x is equal to 0 This equation again has the same two solutions So far so good Now, we’re going to substitute x plus 1 is equal to negative x squared We get the equation negative x squared plus 1 over x is equal to 0. Now what happens when you solve this equation? Well you end up with another solution so you have the same two solutions as before but then you end up with the new solution of x equals 1 Notice if you substitute x equals 1 into this equation you end up with negative 1 plus 1 equals 0 So x equals 1 is a solution to this equation But it was not a solution to the previous two equations So the lesson is that when we’ve substituted x plus 1 is equal to negative x squared, we’ve got an equation that has another solution, and this is not a solution to the original equation x squared plus x plus 1 is equal to 0 So you have to be careful if a step introduces an extraneous solution, Or else you could end up with a nonsensical result like three is equal to zero Did you identify where the mistake originated? Thanks for watching this video please subscribe to my channel I make videos on math you can catch me on my blog Mind Your Decisions. If you like this video you can check out my books which are linked in the video description and you can support me on Patreon. If you have a topic suggestion or a puzzle you can email me presh mindyourdecisions com And you can catch me on social media either at mindyourdecisions or @preshtalwalkar .
# How do you solve the system of equations: 5x - 2y = 23 ; 5x + 2y = 17? Aug 14, 2015 The solutions are color(blue)(x=4,y=-3/2 #### Explanation: $5 x - \textcolor{b l u e}{2 y} = 23$.........equation $\left(1\right)$ $5 x + \textcolor{b l u e}{2 y} = 17$.......equation $\left(2\right)$ Solving through elimination. Adding the equations results in elimination of color(blue)(2y: $5 x - \cancel{\textcolor{b l u e}{2 y}} = 23$ $5 x + \cancel{\textcolor{b l u e}{2 y}} = 17$ $10 x = 40$ color(blue)(x=4 Now finding $y$ from equation $1$ $5 x - 2 y = 23$ $5 x - 23 = 2 y$ $20 - 23 = 2 y$ $- 3 = 2 y$ color(blue)(y=-3/2
Home > Maths > Number > Ratio Maths Ratio Print # Division in a given ratio Question Dave and Lisa win £500 between them. They agree to divide the money in the ratio 2:3. How much does each person receive? The ratio 2:3 tells us that for every 2 parts Dave receives, Lisa will receive 3 parts. There are 5 parts in total. £500 represents 5 parts. Therefore, £100 represents 1 part. Dave receives 2 parts: 2 × £100 = £200 Lisa receives 3 parts: 3 × £100 = £300 ## Order It's important to notice what order the parts of the ratio are written in. 2:3 is not the same as 3:2. In the example, the ratio of Dave's money to Lisa's was 2:3. If we swap the order to 3:2 then Dave would get more than Lisa. To keep it the same as in the example we could say that the ratio of Lisa's money to Dave's would be 3:2 Question Q1. Amelia and Shabana win a sum of money, which they agree to share in the ratio 5:3. If Amelia receives £150, how much will Shabana receive? Q2. A necklace is made using gold and silver beads in the ratio 3:2. If there are 80 beads in the necklace: a) How many are gold? b) How many are silver? A1. Amelia receives 5 parts, which is equivalent to £150. Therefore, £150 represents 5 parts. £150 ÷ 5 represents one part. So one part is £30 Shabana receives 3 parts: 3 × £30 = £90. A2. Gold : Silver = 3:2, so there are 5 parts altogether. 80 ÷ 5 = 16, so 1 part represents 16 beads. a) Gold = 3 × 16 = 48 beads b) Silver = 2 × 16 = 32 beads BBC iD
# Vectors & Scalars: Operations, Resolution of Vectors, Laws of Vectors with Solved Examples 0 Save Vectors and Scalars are used to represent the physical quantities so that they can be used for calculations. Every physical quantity is classified as scalar or vector. Using the quantities in their scalar or vector form simplifies the calculation. In this article, we will learn about how to identify the vector and scalar quantities, the rules of mathematical operations for them, Unit Vectors, Resolution of vectors, Rectangular components of a vector, Analytical method, Parallelogram law and Triangular Law in detail with Solved Examples. So let’s start Physical Quantity: The quantities which can be measured or defined are known as physical quantities in the form of vectors and scalars. Some examples are— mass, amount of substance, current, velocity, acceleration, force, volume, density, pressure, temperature, time, etc. The physical quantities can be divided into the following two groups: 1. Scalar quantity 2. Vector quantity ## Vectors & Scalars: What is a Scalar Quantity? • The quantity having only magnitude and no direction is called a scalar quantity. • Some scalar quantities are mass, volume, time, density, distance, speed, electric potential, energy, electric charge, electric current, etc. For example, A man is 65 kg. Here the mass of the man is 65 kg which needs only magnitude (number) to describe it which is equal to 65 kg. The addition, subtraction and multiplication of scalar quantities follow the simple rules of Algebra because they have only magnitudes. ## Vectors & Scalars: What is a Vector Quantity? • The quantity having both magnitudes as well as direction is called a vector quantity. • Some vector quantities are Velocity, acceleration, force, displacement, momentum, electric field, weight, etc. • For example, A car is moving with 20 m/s velocity towards the east. Here magnitude is 20 m/s and the direction is east. So the velocity of the car is a vector quantity. • A vector quantity is represented as A So, \|A\|or A is the magnitude of the vector and an arrow is a direction. • The magnitude of the vector is equal to the length of the line from head to tail. Characteristics of vector quantity: 1. They have both magnitudes as well as direction. 2. They do not follow the simple rules of algebra like a scalar. 3. They change if either the magnitude or the direction changes or both changes. ### Types of Vectors The various types of vectors are listed below: 1) Unit vector • The vector having magnitude equal to 1 is called the unit vector. • The unit vector of A along the direction of is ᠎A and it is given by:᠎᠎᠎ $$\hat{A} ={\vec{A} \over A}= {\vec{A} \over \|\vec{A}\|}$$ • The unit vectors along the x, y and z-axis is i, j and k respectively. 2) Equal vector The two vectors A and Bare equal vectors if both magnitudes, as well as direction, are the same for both. $$\vec{A}=\vec{B} \\ \|\vec{A}\|=\|\vec{B}\|$$ 3) Zero vector • The vector having magnitude equal to zero is called the null vector. It is generally represented by O. • A point is generally taken as a null vector. $$\|\vec{A}\|=0$$ 4) Parallel vector • The two vectors A and B are parallel vectors (A \|\| B) if they have the same direction but their magnitude may or may not be the same. • If the vectors are in opposite directions then they are called antiparallel vectors. 5) Coplanar vector • If the two or more vectors lie in the same plane or they are parallel to the same plane then the vectors are said to be coplanar vectors. • If they are not in the same plane then they are said to be non – planar vectors. ## Magnitude or Length of a Vector Let A=x i+y j+z k be a A vector. The magnitude or the length of is given by: The magnitude or length = \|A\| = x2+y2+z2 Q) Find the unit vector along in the given vector A=4 i+3 j Ans. A=4 i+3 j The magnitude of vector = \|A\| = 42+32 = 25=5 $$\text {Unit vector} (\hat{A})= {\vec{A} \over{\|{\vec{A}\|}}} = {4\over 5} \hat{i} + {3\over 5} \hat{j}$$ ## Resolution of Vector We have a vector F where the magnitude of the vector is F and the angle with horizontal is θ. The vector has two components- 1. Vertical component = F Sinθ 2. Horizontal component = F Cosθ ## Types of Coordinate System Coordinate systems are essential for representing any quantity on paper in mathematical form. Coordinate system is nothing but a set of reference points that define the boundaries of the system we need to study. These reference points are universal in nature, and can be used to state any physical quantity at a particular instant of time or over a period of time. We mostly use two types of coordinate systems. They are as given below: ### Cartesian Coordinate System The Cartesian Coordinate system or the Rectangular System is one of the most basic coordinate systems. We all are quite familiar with it from our Mathematics Textbook. It consists of two axes, X-axis and Y-axis, mutually perpendicular to each other, intersecting at a point called origin. The X-axis is given by the equation y=0 and the Y-axis is given by the equation x=0. The x-coordinate of a point is known as Abscissa. The y-coordinate of a point is known as Ordinate. ### Polar Coordinate System The polar coordinate system or the Radial coordinate system consists of angle and radius. The angle with the polar axis is measured counterclockwise from the axis to the line. The point will have a unique distance from the origin (r). The relationship between polar and cartesian coordinates is given by: x=rcos and y=rsin r= x2+y2and tan=(y/x) Hope this article helped you to understand Vectors and Scalars completely. Do try our Testbook App to get more practice to clear all your doubts. It’s absolutely free! ## Vectors & Scalars FAQs Q.1 What are scalar quantities? Ans.1 The quantity having only magnitude and no direction is called a scalar quantity. Q.2 What are vector quantities? Ans.2 The quantity having both magnitudes as well as direction is called a vector quantity. Q.3 What is the parallelogram of vectors? Ans.3 According to this law, the two non-zero vectors are the two adjacent sides of a parallelogram then the resultant of both the vectors is given by the diagonal of the parallelogram passing through the point of intersection of the two vectors. Q.4 What is the triangle of vectors? Ans.4 According to this law, if two vectors are represented by two sides of a triangle in magnitude and direction were taken in the same order then the third side of that triangle represents in magnitude and direction of the resultant vector. Q.5 What is the resolution of the vector? Ans.5 We have a vector F where the magnitude of the vector is F and the angle with horizontal is θ. The vector has two components- Vertical component = F Sinθ Horizontal component = F Cosθ
AP Statistics Curriculum 2007 StudentsT (Difference between revisions) Revision as of 04:52, 4 February 2008 (view source)IvoDinov (Talk \| contribs) (→Approach I (exact!))← Older edit Revision as of 04:52, 4 February 2008 (view source)IvoDinov (Talk \| contribs) (→Approach III (Approximate))Newer edit → Line 74: Line 74: ====Approach III (Approximate)==== ====Approach III (Approximate)==== - : $X \sim B(n=12, p=0.44)$ The sample proportion is $\hat{p} = X/n \approx N [m = p = 0.44; (p(1-p)/n)1/2=0.1433]$ Thus, $P(X=6) = P(\hat{p}=0.5) \approx P(p_1 \leq \hat{p} \leq p_2)$, where $p_1=0.5-1/24$ and $p_2=0.5+1/24$. Standardize $Z = (p-0.44)/0.1433$ to get: + : $X \sim B(n=12, p=0.44)$ The sample proportion is $\hat{p} = X/n \approx N [m = p = 0.44; (p(1-p)/n)1/2=0.1433]$ Thus, $P(X=6) = P(\hat{p}=0.5) \approx P(p_1 \leq \hat{p} \leq p_2)$, where $p_1=0.5-1/24$ and $p_2=0.5+1/24$. Standardize $Z = (p-0.44)/0.1433$ to get - $P(X=6) \approx P(p_1 \leq \hat{p} \leq p_2) = P(Z_1 \leq Z \leq Z_2) = 0.211.$ + :$P(X=6) \approx P(p_1 \leq \hat{p} \leq p_2) = P(Z_1 \leq Z \leq Z_2) = 0.211.$ General Advance-Placement (AP) Statistics Curriculum - Student's T Distribution Very frequently in practice we do now know the population variance and therefore need to estimate it using the sample-variance. This requires us to introduce the T-distribution, which is a one-parameter distribution connecting $Cauchy=T_{(df=1)} \longrightarrow T_{(df)}\longrightarrow N(0,1)=T_{(df=\infty)}$. Student's T Distribution The Student's t-distribution arises in the problem of estimating the mean of a normally distributed population when the sample size is small and the population variance is unknown. It is the basis of the popular Student's t-tests for the statistical significance of the difference between two sample means, and for confidence intervals for the difference between two population means. Suppose X1, ..., Xn are independent random variables that are Normally distributed with expected value μ and variance σ2. Let $\overline{X}_n = {X_1+X_2+\cdots+X_n \over n}$ be the sample mean, and ${S_n}^2=\frac{1}{n-1}\sum_{i=1}^n\left(X_i-\overline{X}_n\right)^2$ be the sample variance. We already discussed the following statistic: $Z=\frac{\overline{X}_n-\mu}{\sigma/\sqrt{n}}$ is normally distributed with mean 0 and variance 1, since the sample mean $\scriptstyle \overline{X}_n$ is normally distributed with mean μ and standard deviation $\scriptstyle\sigma/\sqrt{n}$. Gosset studied a related quantity under the pseudonym Student), $T=\frac{\overline{X}_n-\mu}{S_n / \sqrt{n}},$ which differs from Z in that the (unknown) population standard deviation $\scriptstyle \sigma$ is replaced by the sample standard deviation Sn. Technically, $\scriptstyle(n-1)S_n^2/\sigma^2$ has a Chi-square distribution $\scriptstyle\chi_{n-1}^2$ distribution. Gosset's work showed that T has a specific probability density function, which approaches Normal(0,1) as the degree of freedom (df=sample-size -1) increases. Example Suppose a researcher wants to examine CD4 counts for HIV(+) patients seen at his clinic. She randomly selects a sample of n = 25 HIV(+) patients and measures their CD4 levels (cells/uL). Suppose she obtains the following results and we are interested in calculating a 95% (α = 0.025) confidence interval for μ: Variable N N* Mean SE of Mean StDev Minimum Q1 Median Q3 Maximum CD4 25 0 321.4 14.8 73.8 208.0 261.5 325.0 394.0 449.0 What do we know from the background information? $\overline{y}= 321.4$ s = 73.8 SE = 14.8 n = 25 $CI(\alpha)=CI(0.025): \overline{y} \pm t_{\alpha\over 2} {1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}.$ $321.4 \pm t_{(24, 0.025)}{73.8\over \sqrt{25}}$ $321.4 \pm 2.064\times 14.8$ [290.85,351.95] CI Interpretation Still, does this CI (290.85, 351.95) mean anything to us? Consider the following information: The U.S. Government classification of AIDS has three official categories of CD4 counts – asymptomatic = greater than or equal to 500 cells/uL • AIDS related complex (ARC) = 200-499 cells/uL • AIDS = less than 200 cells/uL • Now how can we interpret our CI? SOCR CI Experiments The SOCR Confidence Interval Experiment provides empirical evidence that the definition and the construction protocol for Confidence intervals are consistent. Activities • A biologist obtained body weights of male reindeer from a herd during the seasonal round-up. He measured the weight of a random sample of 102 reindeer in the herd, and found the sample mean and standard deviation to be 54.78 kg and 8.83 kg, respectively. Suppose these data come from a normal distribution. Calculate a 99% confidence interval. • Suppose the proportion of blood type O in the population is 0.44. If we take a random sample of 12 subjects and make a note of their blood types what is the probability that exactly 6 subjects have type O blood type in the sample? Approach I (exact!) P(X = 6) = ? Where $X\sim B(12, 0.44)$ $P(X=6)={12\choose 6}p^6(1-p)^{6}=\frac{12!}{6!(6)!}0.44^6 0.56^6=0.2068$, using SOCR Binomial interactive GUI or calculator. Approach II (Approximate) $X \sim B(n=12, p=0.44)$ $X (approx.) \sim N [\mu = n p = 5.28; \sigma=(np(1-p))1/2=1.7]$. $P(X=6) \approx P(Z_1\leq Z \leq Z_2)$, where $Z = {{X-5.28} \over {1.7}}$ and X1 = 5.5, X2 = 6.5. So, $P(X=6)\approx P(Z_1 \leq Z \leq Z_2)=0.211.$ Approach III (Approximate) $X \sim B(n=12, p=0.44)$ The sample proportion is $\hat{p} = X/n \approx N [m = p = 0.44; (p(1-p)/n)1/2=0.1433]$ Thus, $P(X=6) = P(\hat{p}=0.5) \approx P(p_1 \leq \hat{p} \leq p_2)$, where p1 = 0.5 − 1 / 24 and p2 = 0.5 + 1 / 24. Standardize Z = (p − 0.44) / 0.1433 to get $P(X=6) \approx P(p_1 \leq \hat{p} \leq p_2) = P(Z_1 \leq Z \leq Z_2) = 0.211.$
# 2023 AMC 10B Problems/Problem 9 (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) ## Problem The numbers $16$ and $25$ are a pair of consecutive positive squares whose difference is $9$. How many pairs of consecutive positive perfect squares have a difference of less than or equal to $2023$? $\text{(A)}\ 674 \qquad \text{(B)}\ 1011 \qquad \text{(C)}\ 1010 \qquad \text{(D)}\ 2019 \qquad \text{(E)}\ 2017$ ## Solution 1 Let x be the square root of the smaller of the two perfect squares. Then, $(x+1)^2 - x^2 =x^2+2x+1-x^2 = 2x+1 \le 2023$. Thus, $x \le 1011$. So there are $\boxed{\text{(B)}1011}$ numbers that satisfy the equation. ~andliu766 A very similar solution offered by ~darrenn.cp and ~DarkPheonix has been combined with Solution 1. Minor corrections by ~milquetoast Note from ~milquetoast: Alternatively, you can let $x$ be the square root of the larger number, but if you do that, keep in mind that $x=1$ must be rejected, since $(x-1)$ cannot be $0$. ## Solution 2 The smallest number that can be expressed as the difference of a pair of consecutive positive squares is $3$, which is $2^2-1^2$. The largest number that can be expressed as the difference of a pair of consecutive positive squares that is less than or equal to $2023$ is $2023$, which is $1012^2-1011^2$. These numbers are in the form $(x+1)^2-x^2$, which is just $2x+1$. These numbers are just the odd numbers from 3 to 2023, so there are $[(2023-3)/2]+1=1011$ such numbers. The answer is $\boxed{\text{(B)}1011}$. ~Aopsthedude ~Math-X ## Video Solution ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
The square source of 121 is expressed as √121 in the radical type and together (121)½ or (121)0.5 in the exponent form. The square root of 121 is 11. It is the positive solution that the equation x2 = 121. The number 121 is a perfect square. You are watching: What is the square root of 121 Square root of 121: 11Square source of 121 in exponential form: (121)½ or (121)0.5Square root of 121 in radical form: √121 1 What Is the Square source of 121? 2 Is Square root of 121 Rational or Irrational? 3 How to find the Square source of 121? 4 FAQs on Square source of 121 We understand that addition has one inverse procedure in subtraction and also multiplication has actually an inverse procedure in the division. Similarly, finding the square root is an inverse operation of squaring. The square source of 121 is the number that gets multiplied to chin to give the number 121. A reasonable number is a number that can be to express in the type of p/q, where p and q are integers and also q is no equal to 0. We already found that 121 = 11. The number 11 is a reasonable number. So, the square root of 121 is a rational number. ## How to uncover the Square source of 121? We will comment on two techniques of detect the square source of 121 Prime FactorizationLong division ### Square root of 121 By prime Factorization Prime administer is a way of to express a number together a product of its prime factors. The prime factorization of 121 is 121 = 11 × 11 = 112. To discover the square source of 121, we take one number from every pair of the exact same numbers and also we multiply them. 121 = 11 × 11 121 = 11 ### Square source of 121 By Long Division The value of the square source of 121 by long department method consists of the complying with steps: Step 1: starting from the right, we will pair increase the number by placing a bar over them.Step 2: find a number which, once multiplied come itself, gives the product much less than or same to 1. So, the number is 1. Putting the divisor as 1, we obtain the quotient together 1 and the remainder 0Step 3: double the divisor and enter it through a empty on its right. Assumption: v the largest feasible digit to to fill the blank which will additionally become the new digit in the quotient, together that once the new divisor is multiplied to the brand-new quotient the product is much less than or equal to the dividend. Divide and also write the remainder. See more: The Second Number In An Ordered Pair, What Is The Second Number Of An Ordered Pair Explore Square roots utilizing illustrations and also interactive examples Important Notes: The square root is the inverse procedure of squaring.We can uncover the square root of 121 using prime factorization, recurring subtraction, and also the long department method.The square root of a number is both negative and optimistic for the exact same numerical worth i.e., the square source of 121 will it is in 11. Think Tank: We know that (-11) × (-11) =121. So, deserve to we say that -11 is a square root of 121?Can you identify a quadratic equation who roots room 121 and -121?
Perfect Practice Makes Perfect # Sequences ## Tuesday’s Twister #12 – Sequences May 13th, 2015 by John Lehet “Lost time is never found again.” – Ben Franklin When working with students in elementary school, I like to talk about sequences or number patterns. I often present them with the start of a sequence, say the first three numbers. I then ask the class to give me the next term (or number) in the sequence. Letting everyone in the class mull it over for a bit, I ask for the pattern (the code essentially) and the next few numbers. When I first did this, I was amazed! I had a sequence in mind, but the students kept giving different, yet very viable sequences. Not only that, they could justify their answers by supplying the “code” for the sequence. I quickly realized that with just three numbers at the start of a sequence, the possibilities were plentiful. It was quite the challenge for the students also. So, I started doing this as a “break the ice” activity with classrooms. We would use the same three numbers again and again, and see how many different (yet viable) sequences could be made. When they give me a sequence and its corresponding justification, I would say “great, let’s find another” and erase it leaving only the first three “seed numbers” that I had originally written. I would then ask them to give me another different sequence and the whole thing would start over again. Each time I do this, I am still amazed at how many different sequences they come up with and how challenging they find it. Here’s the start of a sequence for you to try … ## Here’s three numbers that start a sequence: 2, 3, 5, . . . What do you think the next number can be? How about the number after that? Remember, it has to follow a pattern, so you can easily find each successive number by applying the pattern. I’ve given this pattern to numerous classrooms. Each time, they came up with a variety of different answers, all of which make a valid sequence. For the next number, classes have given 7, some have given 8 while others have given 14. In all, from all of the classes, I have received 9 different sequences that these three numbers can generate (and there’s even more!). My challenge to you is to find as many different sequences that can be generated using the first three numbers 2, 3, 5. Remember don’t just come up with the sequence, but identify the pattern that it follows. Challenge yourself to see if you can find nine different sequences. Good luck, have fun and challenge others, the more the merrier!
# Video: Describing Relationships and Extending Terms in Arithmetic Sequences Kathryn Kingham Through examples, describe relationships in arithmetic sequences. Then extend the sequences to 𝑛 positions, with 𝑛th term formulae being a simple multiple of 𝑛 (e.g., 6𝑛, 51𝑛, etc.). Use tables to clearly see the relationships in sequences. 08:41 ### Video Transcript Describing Relationships and Extending terms in Arithmetic Sequences Here is an example of an arithmetic sequence. A sequence is any ordered list of numbers. An arithmetic sequence is a sequence found by adding the same number to the previous term. For example, you add nine to twenty-eight and that equals thirty-seven; you add nine to thirty-seven and that equals forty-six. Each of these numbers are terms in the sequence. Someone might ask you find the next term in this arithmetic sequence. When they say find the next term, we know that they’re looking for the value that comes after sixty-four here. And because we’re also given the information that it’s an arithmetic sequence, we also know that we will be adding to find the next value. Each step in this sequence has been adding nine to the previous term. This lets us know that whatever sixty-four plus nine is, that’s our next term The next term in this sequence is seventy-three. Here’s an example of a simple arithmetic sequence. We’re adding two each time When we’re dealing with sequences, each of the terms also has a position. You can see the labels here. The two is in first position. The four is in second position. Six is in third position, and then fourth and fifth, and so on. Positions are really important. Let’s take a closer look at what is happening here with these positions. Position one has a value of two. Position two has a value of four. To get from position one to position two, we need to add two. And to go from position two to position three, we need to add two. There is a relationship here between the position of a term and the term’s value. This column shows us that to solve for position one, we multiply one times two; to solve for position two, the value, we take two and we multiply it by two; to solve for position three, we multiply three by two to get six. In the operation the first number we’re looking at is the position number. The two is the number that we’re adding to each term. Some of you might be wondering, “Well why would I take the time to figure out the multiplication when I could just add two to six? Six plus two is eight, that’s pretty simple.” And that’s true. If I wanted to find the next number in this sequence, I would probably just say ten plus two is twelve; the sixth position is twelve. But if the question said something like what is the eightieth position in this sequence, you don’t want to add two eighty times. Now we’re going to add the number eighty, the position eighty, to our table. We would follow the same operation. In the eightieth position of this sequence is the number one hundred and sixty. Finding that by multiplying eighty times two is significantly faster than trying to add two each time eighty times. You might also come across the question: what is an algebraic expression for finding terms in this sequence? In that case we don’t know what position we’re looking for. We’re looking to write an expression that can be used for all positions. When dealing with positions and sequences, we usually use the letter 𝑛 to represent an unknown position. We could find the value of a term in position 𝑛 by multiplying the position by two, 𝑛 times two. We could say 𝑛 times two for our expression, or simply two 𝑛. We can plug in any position number here and find the term that would be in that position. We could plug in position one hundred, position seven, position fifteen, it doesn’t matter. This expression works for solving this sequence. Here’s another example find the next term in this sequence: fifty-one, one hundred and two, one hundred and fifty-three. What’s next? The first question we should ask is what is being added to each term. Last time that was really easy because it was two. If you don’t immediately recognise what’s being added, here’s what you can do. You can take the term in the second position and subtract the term in the first position: one hundred and two minus fifty-one is fifty-one. You could also subtract one hundred and two from fifty-three or the position two number from the position three number. Both of these equal fifty-one. Fifty-one is what is being added each time. To find the next term, we need to take the third term, one hundred and fifty-three, and add fifty-one to that. The next term in this sequence is two hundred and four. Here’s what a table for this sequence would look like. The operation here would be to take the position and multiply it by fifty-one because fifty-one is what we’re adding here. Fifty-one 𝑛 would be the expression we could use to take any position and find its value. Here is our last example find the seventieth term in the following sequence: six, twelve, eighteen, twenty-four. We’re trying to answer the question what is being added to each term, and we know that here each term is six more than the previous term. But I’m not just trying to find out what are we adding to each term. I need to find out what expression can I use to find the seventieth term. I noticed the pattern is that you take the position and you multiply it by six. So in order for me to find the seventieth term, I need to multiply that by six. When I do that I have a solution of four hundred and twenty. I also know that I can find any term in this sequence by taking the position 𝑛 and multiplying it by six; that expression is important. That’s all for this video. Now you can go try some sequences on your own.
# Statistics Fundamentals ### Statistics Fundamentals Introduction This page simply lists various identities used in statistics and provides basic explanatory notes and links to pages with more detailed information There are two primary branches of statistics : descriptive statistics and inferential statistics . Descriptive Statistics is the the branch of statistics which relates to collecting, summarising and presenting data sets. Inferential Statistics is the branch of statistics which analyses sample data to arrive at conclusions about a population An example of descriptive statistics is the average age of people who voted for a party in a election. An example of inferential statistics is taking a sample of 1000 english mechanical engineers to enable calculation of the average salary of mechanical engineers in england. The primary identities of statistics are Population : All members of a group under consideration. Sample : The part of the population selected for analysis such as to obtain information about the population Parameter: A numerical measure describing a characteristic of a population. example mean μ. Statistic:A numerical measure describing a characteristic of a sample. example: arithmetic mean xm Variable:A characteristic of an item that is analysed using statistical methods. example length, income The above identities relates to samples and populations. Parameters of populations as identified below including mean μ, variance σ 2, and standard deviation σ are true values. These true values are ideal assuming that the whole population (N) can be measured or the sample number (n) is infinite. The expectation E(X) is the sum of all the products formed by multiplying each event in a probability. E(X) = x i p i. distribution by its corresponding probability. Notes Name Symbol Description Link Population The total of the variables under study. e.g. the population of the UK : The population of the world: The total number of fish in a lake. Samples And Populations Population Size N The total number in the population. e.g. The population of the UK. The upper case N is also used for the number of elementary events for a random variable Population mean μ The sum of the value of each item of the population divided by the number of items in the population. (The mean is synonymous with the Expection E(X).) Population variance σ 2 The sum of [square (the value of each item in the population - the value of the population mean )] divided n (number in the population). This is used to determine the standard deviation. Population Standard Deviation σ The square root of the variance. This is a measure of the spread of the population. If every item in a population has a very similar dimension then the standard deviation is small. Sample a random collection of units from the population. Collect to establish information on the population as a whole. Sample size n number of items in the sample Sample mean xm the sum of the value of each item of the sample divided by the sample size Discrete variable A variable which is counted. Examples include population, number of coins, number of fish. Continuous variable A variable which is measured . Examples include length of a bar, height of a person, distance to a planet. Sample variance sx 2 The sum of [square (the value of each item in the sample - the value of the sample mean )]divided by (n-1) . n = number of items in the sample. This is used to determine the standard deviation. Sample Standard Deviation σ The square Root of the variance. This is a measure of the spread of the sample. If every item in a sample has a very similar dimension then the standard deviation is small. Sample median m The sample median is the middle value (in the case of an odd-sized sample), or average of the two middle values (in the case of an even-size sample), when the values in a sample are arranged in ascending order.- Not used on this website Sample mode The value in a sample which occurs most often. A set of numbers with one mode is called a uni-modal set and a set with two modes is called a bimodal. Not used on this website Range The difference between the largest and smallest value in a set of values for a variable. Quartiles If a set of numbers e.g. a sample is arranged in order of magnitude then the values which divide the set into four equal parts are called quartiles. The lower quartile is the value of the item at (n+1)/4. The upper quartile value is the value of the item at (3(n+1)/4 and the value of the middle quartile is the same as the median. Not used on this website Deciles If a set of numbers is arranged in order of magnitude then the deciles are the values of the numbers which divide the set into ten groups. Not used on this website Percentiles If a set of numbers is arranged in order of magnitude then the percentiles are the values of the numbers which divide the set into 100 groups. Not used on this website Probability p A numerical measure of how likely it is that some event will occur.1 or 100% is the probability of certainty. .0 or 0% is the probability that the event will definitely not occur Probability Hypothesis A statistical hypothesis is an assumption about the distribution of a random variable. Example: The mean height of children in a school is 1,2m is a hypothesis which may be accepted or rejected. Hypothesis Null Hypothesis Ho The null hypothesis, H0 represents a theory that has been put forward, because it is considered to be true or because it is to be used as a basis but has not been proved. Example Ho :The mean height of children in a school is 1,2m Alternative Hypothesis H1 Is the alternative hypothesis which results on rejection of the null hypothesis: Example.H1 : the height of children in a school is not 1,2m Permutation n P(r) A permutation is an arrangement of (r) things with the order being important. abcd is different to bcda. Permutations / Combinations Combination n C(r) A combination is an arrangement of (r)things with the order being important. abcd is a combination the letter order is not important Expectation E(X) Synonymous with mean this is the sum of the products of each value in a distribution and its respective probability Expectation Uniform Distribution A distribution in which every possible value of the variable has the same probability is called a rectangular or uniform distribution. Discrete Distributions Binomial Distribution This distribution relates to the number of times a success occurs in n independent trials. Poisson Distribution The Poisson distribution is simply a limiting case of the binomial distribution with p -> 0, and n - > infinity such that the mean is m = np which approaches a finite value. Typically, a Poisson random variable is a count of the number of events that occur in a certain time interval or spatial area. Hypergeometric Distribution Where the binomial probability function involves sampling with replacement i.e. each trial is in dependent the hypergeometric distribution involves sampling without replacement. The trials are therefore not independent. Normal Distribution The normal probability distribution is the continuous distribution which is most representative of events that occur in the natural world subject to countless variables. It is mostly used for continuous variables and discrete variable with large samples. Normal Distribution t-test When sample sizes are small, and the standard deviation of the population is unknown it is normal to use the distribution of the t statistic. t distribution- Chi-Squared test The Chi-squared test is a tool which enables determination how much a sample distribution can deviate from a population if the hypothesis of equivalence is true. Chi-Test F- test When comparing two samples it is often necessary to test the validity that the samples are from the same distribution. The F ratio test is used for this purpose F-test
## HIGHEST COMMON FACTOR WITH VARIABLES WORKSHEET On the webpage, "Highest common factor with variables worksheet" we are going to see some practice questions. (1) Find the highest common factor of 7 x²yz⁴ , 21x²y⁵z³ (2) Find the highest common factor of x²y, x³y , x²y² (3) Find the highest common factor of 25 b c⁴ d³ , 35 b² c⁵ , 45 c³ d (4) Find the highest common factor of 35 x⁵ y³ z⁴ , 49 x² y z³ , 14 x y² z² (5) Find the highest common factor of 35 n²m, 21 m² n (6) Find the highest common factor of 66 y x, 30 x² y (7) Find the highest common factor of 60 y, 56 x² (8) Find the highest common factor of 80 x³, 30 x²y ## Highest common factor with variables worksheet - answers To find the highest common factor of the given numbers or for algebraic expressions we have to follow the steps: Step 1: List the prime factors of each of the given number. For algebraic expression we have to find factors of them. Step 2: List the common factors of the given numbers or common factors. Step 3: Multiply those common factors. Question 1 : Find the highest common factor of 7 x² y z⁴ , 21 x²y⁵ z³ Solution : Step 1 : Prime factors of 7 x² y z⁴ = 7 x x² x y z Prime factors of 7 x² y z⁴ = 7 3 x x y x y⁴ x Step 2 : Common factors that we find in both terms are 7,x², y and z³ Step 3 : Multiply the common factors 7,x², y and z³ Hence, the highest common factor is 7 x² y z³ Let us see the next example on "Highest common factor with variables worksheet". Question 2 : Find the highest common factor of x² y, x³ y , x² y² Solution: Step 1 : Prime factors of x² y = x² x Prime factors of x³ y = x² x x x Prime factors of x² y² = x² x x y Step 2 : The common factors that we find from given terms are x² and y Step 3 : Multiply the common factors x² and y Hence, the highest common factor is x² y Let us see the next example on "Highest common factor with variables worksheet". Question 3 : Find the highest common factor of 25 b c⁴ d³ , 35 b² c⁵ , 45 c³ d Solution: Step 1 : Prime factors of 25 b c⁴ d³ = 5 x 5 x b x c³ x d x d x d Prime factors of 35 b² c⁵ = 7 x 5 x b x b x c³ x c² Prime factors of 45 c³ d = 3 x 3 x 5 x c³ x d Step 2 : The common factors that we find from given terms are 5 and c³ Step 3 : Multiply the common factors 5 and c³ Hence, the highest common factor is 5 c³ Let us see the next example on "Highest common factor with variables worksheet". Question 4 : Find the highest common factor of 35 x⁵ y³ z⁴ , 49 x² y z³ , 14 x y² z² Solution: Step 1 : Prime factors of 35 x⁵ Y³z⁴ x 5 x x x x y² x x z² x z² Prime factors of 49 x² y z³ x 7 x x x x x x z² x z Prime factors of 14 x y² z² x 7 x x x y xx The common factors that we find from given terms are 7, x, y and z² Step 3 : Multiply the common factors 7, x, y and z² Hence, the highest common factor is 7 x y z² Let us see the next example on "Highest common factor with variables worksheet". Question 5 : Find the highest common factor of 35 n²m, 21 m² n Solution: Step 1 : Prime factors of 35 n²m x 5 x n x n m Prime factors of 21 m² n x 3 x m x m x n Step 2 : The common factors are 7, m and n Step 3 : Multiply the common factors 7, m and n Hence, the highest common factor is 7 m n Let us see the next example on "Highest common factor with variables worksheet". Question 6 : Find the highest common factor of 66 y x, 30 x² y Solution: Step 1 : Prime factors of 66 y x = 2 x 3 x 11 x y x x Prime factors of 30 x² y = 3 x 2 x 5 x x x x x y Step 2 : The common factors are 2, 3, x and y Step 3 : Multiply the common factors 2, 3, x and y Hence, the highest common factor is 6xy Let us see the next example on "Highest common factor with variables worksheet". Question 7 : Find the highest common factor of 60 y, 56 x² Solution: Step 1 : Prime factors of 60 y = 2 xx 3 x 5 x y Prime factors of 56 x² = 2 x 2 x 2 x 7 x x x x Step 2 : The common factors are 2 and 2 Step 3 : Multiply the common factors 2 and 2 Hence, the highest common factor is 4 Let us see the next example on "Highest common factor with variables worksheet". Question 8 : Find the highest common factor of 80 x³, 30 x²y Solution: Step 1 : Prime factors of 80 x³ = 2 x 2 x 2 x 2 xx x² x x Prime factors of 30 x²y = 2 x 3 x 5 x x² x y Step 2 : The common factors are 2, 5 and x² Step 3 : Multiply the common factors 2, 5 and x² Hence, the highest common factor is 10 x² After having gone through the stuff given above, we hope that the students would have understood "Highest common factor with variables worksheet". Apart from the stuff, "highest common factor with variables worksheet", if you need any other stuff in math, please use our google custom search here. 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The Bahaí calendar is written of 19 months, each v 19 days. For this reason the number of days in a year in this solar calendar is 19 × 19 = 361. Finding the square source is the inverse procedure of squaring the number. In this mini-lesson permit us discover to calculation the square root of 361 and also to refer the square root of 361 in the most basic radical form. You are watching: What is the square root of 361 Square source of 361361 = 19Square that 361: 3612 = 130,321 1 What Is the Square root of 361? 2 Is Square source of 361 rational or Irrational? 3 How to find the Square root of 361? 4 Thinking the end of The Box! 5 Important note on Square source of 361 6 FAQs top top Square source of 361 ## What Is the Square source of 361? + 19 × +19 = 361 and - 19 × - 19 = 361361 = ± 19 ## Is Square source of 361 reasonable or Irrational? The square root of 361 is a perfect square number. Thus it is a whole number, which can be expressed together a rational number of the kind p/q. The square source of 361 is a rational number. ## How to find the Square source of 361? The square source of or any type of number deserve to be calculation in plenty of ways. To cite a few: element factorization method, repeated subtraction technique and the long division method. ### Square source of 361 by repetitive Subtraction Method Any perfect square number is the sum of consecutive odd numbers. Because 361 is a perfect square, it is the amount of continually odd numbers. Thus by repetitive subtraction we can verify the square source of 361 is 19. 361 - 1 = 360360 - 3 = 357357 - 5 = 352352 - 7 = 345345 - 9 = 336336 - 11 = 325325 - 13 = 312312 - 15 = 297297- 17 = 280280 - 19 = 261261 - 21 = 240240 - 23 = 217217 - 25 = 202202 - 27 = 175175 - 29 = 146146 - 31 = 115115 - 33 = 8282 - 35 = 3737- 37 = 0 We have done the recurring subtraction 19 times. Thus 361 = 19. ### Square root of 361 through the Long department Method Let"s see exactly how to discover the square source of 361 through the long department method. Right here are the desirable steps to it is in followed. See more: What Type Of Star Is Altarf Mean, Star Facts: Al Tarf Write 361 as 3 61. Division 3 through 1 and also get the remaider as 2. Lug the pair the 61 down. We have actually 2 61 to be divided now.Multiply the quotient by 2 and have 2x as the brand-new divisor. Find a number in the place of x such that 2x × x gives 61 or much less than that. We discover 29 × 9 is 261. We attain the remainder as 0. Thus, 361 = 19. Explore Square roots making use of illustrations and also interactive examples Think Tank Do you know that the sum of very first 19 odd number (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37) = 361? You can uncover this sum without yes, really addition. Can you try this out with any other perfect square, together well and also check for yourself? Important Notes
# Finding the next terms of a geometric sequence with whole numbers Online Quiz Following quiz provides Multiple Choice Questions (MCQs) related to Finding the next terms of a geometric sequence with whole numbers. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz. Q 1 - The first three terms of a geometric sequence are 162, 54, and 18. Find the next two terms of this sequence. ### Explanation Step 1: The geometric sequence given is 162, 54, 18… The common ratio is $\frac{54}{162} = \frac{18}{54} = \frac{1}{3}$ Step 2: The next two terms of the sequence are − $\frac{18}{3} = 6;\:\frac{6}{3} = 2$. So the terms are 6 and 2 Q 2 - The first three terms of a geometric sequence are 6, -24, and 96. Find the next two terms of this sequence. ### Explanation Step 1: The geometric sequence given is 6, −24, 96… The common ratio is $\frac{−24}{6} = \frac{96}{−24} = −4$ Step 2: The next two terms of the sequence are − 96(−4) = −384; −384(−4) = 1536. So the terms are −384 and 1536 Q 3 - The first three terms of a geometric sequence are 8, 40, and 200. Find the next two terms of this sequence. ### Explanation Step 1: The geometric sequence given is 8, 40, 200… The common ratio is $\frac{40}{8} = \frac{200}{40} = 5$ Step 2: The next two terms of the sequence are − 200 × 5 = 1000; 1000 × 5 = 5000. So the terms are 1000 and 5000. Q 4 - The first three terms of a geometric sequence are 8, -4, and 2. Find the next two terms of this sequence. ### Explanation Step 1: The geometric sequence given is 8, −4, 2… The common ratio is $\frac{−4}{8} = \frac{2}{−4} = \frac{−1}{2}$ Step 2: The next two terms of the sequence are − $2 \times \frac{−1}{2} = −1;\: −1 \times \frac{−1}{2} = \frac{1}{2}$. So the terms are −1 and $\frac{1}{2}$ Q 5 - The first three terms of a geometric sequence are 3, 27, and 243. Find the next two terms of this sequence. ### Explanation Step 1: The geometric sequence given is 3, 27, 243… The common ratio is $\frac{27}{3} = \frac{243}{27} = 9$ Step 2: The next two terms of the sequence are − 243 × 9 = 2187; 2187 × 9 = 19683. So the terms are 2187 and 19683 Q 6 - The first three terms of a geometric sequence are 3, 24, and 192. Find the next two terms of this sequence. ### Explanation Step 1: The geometric sequence given is 3, 24, 192… The common ratio is $\frac{24}{3} = \frac{192}{24} = 8$ Step 2: The next two terms of the sequence are − 192 × 8 = 1536; 1536 × 8 = 12288. So the terms are 1536 and 12288 Q 7 - The first three terms of a geometric sequence are 4, 16, and 64. Find the next two terms of this sequence. ### Explanation Step 1: The geometric sequence given is 4, 16, 64… The common ratio is $\frac{16}{4} = \frac{64}{16} = 4$ Step 2: The next two terms of the sequence are − 64 × 4 = 256; 256 × 4 = 1024. So the terms are 256 and 1024 Q 8 - The first three terms of a geometric sequence are 2, -10, and 50. Find the next two terms of this sequence. ### Explanation Step 1: The geometric sequence given is 2, −10, 50… The common ratio is $\frac{−10}{2} = \frac{50}{−10} = −5$ Step 2: The next two terms of the sequence are − 50(−5) = −250; −250 × −5 = 1250. So the terms are −250 and 1250 Q 9 - The first three terms of a geometric sequence are −6, 18, and −54. Find the next two terms of this sequence. ### Explanation Step 1: The geometric sequence given is −6, 18, −54… The common ratio is $\frac{18}{−6} = \frac{−54}{18} = −3$ Step 2: The next two terms of the sequence are − −54(−3) = 162; 162 (−3) = 486. So the terms are 162 and −486 Q 10 - The first three terms of a geometric sequence are 8, -32, and 128. Find the next two terms of this sequence. ### Explanation Step 1: The geometric sequence given is 8, −32, 128… The common ratio is $\frac{−32}{8} = \frac{128}{−32} = −4$ Step 2: The next two terms of the sequence are − 128(−4) = −512; −512(−4) = 2048. So the terms are −512 and 2048 finding_next_terms_of_geometric_sequence_with_whole_numbers.htm
# Problems on Quadratic Equation We will solve different types of problems on quadratic equation using quadratic formula and by method of completing the squares. We know the general form of the quadratic equation i.e., ax$$^{2}$$ + bx + c = 0, that will help us to find the nature of the roots and formation of the quadratic equation whose roots are given. 1. Solve the quadratic equation 3x$$^{2}$$ + 6x + 2 = 0 using quadratic formula. Solution: The given quadratic equation is 3x$$^{2}$$ + 6x + 2 = 0. Now comparing the given quadratic equation with the general form of the quadratic equation ax$$^{2}$$ + bx + c = 0 we get, a = 3, b = 6 and c = 2 Therefore, x = $$\frac{- b ± \sqrt{b^{2} - 4ac}}{2a}$$ ⇒ x = $$\frac{- 6 ± \sqrt{6^{2} - 4(3)(2)}}{2(3)}$$ ⇒ x = $$\frac{- 6 ± \sqrt{36 - 24}}{6}$$ ⇒ x = $$\frac{- 6 ± \sqrt{12}}{6}$$ ⇒ x = $$\frac{- 6 ± 2\sqrt{3}}{6}$$ ⇒ x = $$\frac{- 3 ± \sqrt{3}}{3}$$ Hence, the given quadratic equation has two and only two roots. The roots are $$\frac{- 3 - \sqrt{3}}{3}$$ and $$\frac{- 3 - \sqrt{3}}{3}$$. 2. Solve the equation 2x$$^{2}$$ - 5x + 2 = 0 by the method of completing the squares. Solutions: The given quadratic equation is 2x$$^{2}$$ - 5x + 2 = 0 Now dividing both sides by 2 we get, x$$^{2}$$ - $$\frac{5}{2}$$x + 1 = 0 ⇒ x$$^{2}$$ - $$\frac{5}{2}$$x = -1 Now adding $$(\frac{1}{2} \times \frac{-5}{2})$$ = $$\frac{25}{16}$$ on both the sides, we get ⇒ x$$^{2}$$ - $$\frac{5}{2}$$x + $$\frac{25}{16}$$ = -1 + $$\frac{25}{16}$$ ⇒ $$(x - \frac{5}{4})^{2}$$ = $$\frac{9}{16}$$ ⇒ $$(x - \frac{5}{4})^{2}$$ = ($$\frac{3}{4}$$)$$^{2}$$ ⇒ x - $$\frac{5}{4}$$ = ± $$\frac{3}{4}$$ ⇒ x = $$\frac{5}{4}$$ ± $$\frac{3}{4}$$ ⇒ x = $$\frac{5}{4}$$ - $$\frac{3}{4}$$ and $$\frac{5}{4}$$ + $$\frac{3}{4}$$ ⇒ x = $$\frac{2}{4}$$ and $$\frac{8}{4}$$ ⇒ x = $$\frac{1}{2}$$ and 2 Therefore, the roots of the given equation are $$\frac{1}{2}$$ and 2. 3. Discuss the nature of the roots of the quadratic equation 4x$$^{2}$$ - 4√3 + 3 = 0. Solution: The given quadratic equation is 4x$$^{2}$$ - 4√3 + 3 = 0 Here the coefficients are real. The discriminant D = b$$^{2}$$ - 4ac = (-4√3 )$$^{2}$$ - 4 4 3 = 48 - 48 = 0 Hence the roots of the given equation are real and equal. 4. The coefficient of x in the equation x$$^{2}$$ + px + q = 0 was taken as 17 in place of 13 and thus its roots were found to be -2 and -15. Find the roots of the original equation. Solution: According to the problem -2 and -15 are the roots of the equation x$$^{2}$$ + 17x + q = 0. Therefore, the product of the roots = (-2)(-15) = $$\frac{q}{1}$$ ⇒ q = 30. Hence, the original equation is x$$^{2}$$ – 13x + 30 = 0 ⇒ (x + 10)(x + 3) = 0 ⇒ x = -3, -10 Therefore, the roots of the original equation are -3 and -10. ` ### New! Comments Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question. Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
The Law of Cosines Home > Lessons > Law of Cosines Search \| Updated February 9th, 2018 Introduction Here are the sections within this lesson: Equations: The Law of Cosines Triangles are a fundamental geometrical shape. Triangles appear within several disciplines, some of which are architecture, engineering, astronomy, and chemistry. This is why mathematicians have studied them and consequently have several relations to enumerate their sides and angles. Here is one such relation. These are the equations collectively called The Law of Cosines. As it will be thoroughly explained within the next section, these equations are useful for two cases: 1) when two sides and the angle between them are known within a triangle and 2) when all three sides of a triangle are known. If either case is known, then the law of cosines is helpful. If you are not familiar with how to label a triangle or you have never worked with sine, cosine, or tangent, please read our trigonometry basics section before moving on to the next sections within this page. Calculating the Length of a Side Let's take a look at some given information within the following diagram. The law of cosines is a helpful tool for this situation because we know a triangle's angle and the sides that are next to it, as emphasized via the colored graphic below. Now that we know the lawe of cosines can be used for the problem, we now need to label the triangle's angles. The triangle can be arbitrarily labeled with the letters 'A,' 'B,' and 'C.' We will progress with the following arrangement of labeled angles. Now we can start labeling the sides. The side opposite angle-A, our 35-degree angle, is side-a. Likewise, side-b is 375, side-c is 400. Using this set of information, we have to determine which law of cosines equation is best to use. Since we know angle-A, we should find the formula that has angle-A in it. Using this equation, we need to plug in the given information within it. A calculator can be used to simplify the right side of the equation. To cancel the square on the right side of the equation, we must employ its inverse function, which is the square root. We will take the square root of both sides of the equation. Again, with the use of a calculator, we can find side-a by actually using the square root function of a calculator. It has been shown that the law of cosines is useful for calculating the side of a triangle when we know two sides and the angle between those sides. The side we calculate will always be the side opposite the given angle Use this video and quiz to reinforce the lesson. Calculating the Measure of an Angle Here is a given situation involving a triangle. The given information above is one where we know all three sides of a triangle. Let's label the angles with 'A,' 'B,' and 'C.' Labeling the angles is arbitrary; but, for the sake of continuing along with the given information, we will go with this arrangement of angles. Recall, we look to the sides opposite sides an angle and label them accordingly. For instance, side-c is opposite angle-C. The graphic below demonstrates this labeling of sides. It is now clear that side-a is 44, side-b is 29, and side-c is 50. Our next challenge is determine what we can calculate. The only variables that are unknown are angles. So, we must choose which angle to solve for. We can solve for any of them, making our choice an another arbitrary choice. Let's pick angle-C to find. There are three law of cosine equations to use. Let's review them. Of the three equations, only one of them contains angle-C. Since we have chosen to solve for angle-C, we will use that equation. Let's plug in the given information: side-a is 44, side-b is 29, and side-c is 50. It is now time to start cleaning up the equation so we can solve for angle-C. Let's square the numbers and multiply the factors in front of the cosine function. Doing so will yield this: The steps that we have taken so far have been basic. It has involved basic the substitution of values, squaring some numbers, and multiplying factors. Now, we have to tackle some algebra. Commonly, students make an error at this step. Some students think these two numbers can be subtracted. However, these numbers cannot be combined because they are not like terms. Do not make the mistake of combining these values. Instead of combining unlike terms, we have to cancel the 2777 by subtracting that value from both sides of the equation. This is the result of subtracting 2777 from both sides of the equation. Our next task is to cancel the coefficient. The -2552 must be canceled by dividing that value from both sides of the equation. Insead of dividing the right side, let's leave the value as a fraction so that we can avoid rounding a decimal value. It's better to round the value once at the conclusion of our calculations. Doing so will allow us to retain the most accurate answer possible. We have nearly solved our equation for angle-C. However, we have the cosine of angle-C. The next step must involve canceling the cosine function. The only way to cancel the cosine function is to use the inverse cosine function, like so. This is what our equation becomes after we cancel the cosine function with its inverse cosine function. All that remains now is the use of a calculator. We have to place this information into a calculator to get the final value for angle-C. The steps above show how we can arrive at an angle when we are given three sides of a triangle. Use this video and quiz to help you reinforce the lesson. Choosing the Correct Law There are two laws, The Law of Cosines and the Law of Sines (see the Related Lessons section for the lesson). This table will help determine which law to use given the circumstances that are present. Let it be known to use the Law of Sines we do have to know three pieces of information: 1) we either know two angles and a side or 2) two sides and an angle. When considering the first case, knowing two angles of a triangle means we can find the third angle. If we know a side, we automatically know the angle opposite that side because we know all the angles of the triangle. For the second case, we must know an angle that is opposite one of the two given sides. Having an angle and a side opposite it is the indicator we must use the Law of Sines. Instructional Videos View these videos to help you learn how to use the Law of Cosines. Quizmasters This quizmaster will test your skills. Use it to see if you can calculate with The Law of Cosines. Activities This activity is related to the lessons above. ctivity: Law of Cosines Related Lessons We would like to invite you to learn from our other advanced trigonometry lesson.
# Difference between revisions of "2016 AMC 10A Problems/Problem 22" ## Problem For some positive integer $n$, the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3$. How many positive integer divisors does the number $81n^4$ have? $\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425$ ## Solution 1 Since the prime factorization of $110$ is $2 \cdot 5 \cdot 11$, we have that the number is equal to $2 \cdot 5 \cdot 11 \cdot n^3$. This has $2 \cdot 2 \cdot 2=8$ factors when $n=1$. This needs a multiple of 11 factors, which we can achieve by setting $n=2^3$, so we have $2^{10} \cdot 5 \cdot 11$ has $44$ factors. To achieve the desired $110$ factors, we need the number of factors to also be divisible by $5$, so we can set $n=2^3 \cdot 5$, so $2^{10} \cdot 5^4 \cdot 11$ has $110$ factors. Therefore, $n=2^3 \cdot 5$. In order to find the number of factors of $81n^4$, we raise this to the fourth power and multiply it by $81$, and find the factors of that number. We have $3^4 \cdot 2^{12} \cdot 5^4$, and this has $5 \cdot 13 \cdot 5=\boxed{\textbf{(D) }325}$ factors. ## Solution 2 $110n^3$ clearly has at least three distinct prime factors, namely 2, 5, and 11. The number of factors of $p_1^{n_1}\cdots p_k^{n_k}$ is $(n_1+1)\cdots(n_k+1)$ when the $p$'s are distinct primes. This tells us that none of these factors can be 1. The number of factors is given as 110. The only way to write 110 as a product of at least three factors without $1$s is $2\cdot 5\cdot 11$. We conclude that $110n^3$ has only the three prime factors 2, 5, and 11 and that the multiplicities are 1, 4, and 10 in some order. I.e., there are six different possible values of $n$ all of the form $n=p_1\cdot p_2^3$. $81n^4$ thus has prime factorization $81n^4=3^4\cdot p_1^4\cdot p_2^{12}$ and a factor count of $5\cdot5\cdot13=\boxed{\textbf{(D) }325}$ ~savannahsolver ## Video Solution ~ pi_is_3.14 2016 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 23 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions 2016 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
# Recurrence Relation Posted By on April 29, 2016 In this chapter, we will discuss how recursive techniques can derive sequences and be used for solving counting problems. The procedure for finding the terms of a sequence in a recursive manner is called recurrence relation. We study the theory of linear recurrence relations and their solutions. Finally, we introduce generating functions for solving recurrence relations. ## Definition A recurrence relation is an equation that recursively defines a sequence where the next term is a function of the previous terms (Expressing Fn as some combination of Fi with i<n). Example − Fibonacci series: Fn = Fn−1 + Fn−2, Tower of Hanoi: Fn = 2Fn−1 + 1 ## Linear Recurrence Relations A linear recurrence equation of degree k is a recurrence equation which is in the format xn = A1 xn−1 + A2 xn−1 + A3 xn−1 +… Ak xn−k (An is a constant and Ak ≠ 0) on a sequence of numbers as a first-degree polynomial. These are some examples of linear recurrence equations − Recurrence relations Initial values Solutions Fn = Fn−1 + Fn−2 a1 = a2 = 1 Fibonacci number Fn = Fn−1 + Fn−2 a1 = 1, a2 = 3 Lucas number Fn = Fn−2 + Fn−3 a1 = a2 = a3 = 1 Padovan sequence Fn = 2Fn−1 + Fn−2 a1 = 0, a2 = 1 Pell number ### How to solve linear recurrence relation Suppose, a two ordered linear recurrence relation is: Fn = AFn−1 + BFn−2 where A and B are real numbers. The characteristic equation for the above recurrence relation is − x2 − Ax − B = 0 Three cases may occur while finding the roots − Case 1 − If this equation factors as (x − x1)(x − x1) = 0 and it produces two distinct real roots x1 and x2, then Fn = ax1n+ bx2n is the solution. [Here, a and b are constants] Case 2 − If this equation factors as (x − x1)2 = 0 and it produces single real root x1, then Fn = a x1n + bn x1n is the solution. Case 3 − If the equation produces two distinct real roots x1 and x2 in polar form x1 = r ∠ θ and x2 = r ∠(− θ), then Fn = rn (a cos(nθ) + b sin(nθ)) is the solution. ### Problem 1 Solve the recurrence relation Fn = 5Fn−1 − 6Fn−2 where F0 = 1 and F1 = 4 Solution The characteristic equation of the recurrence relation is − x2 − 5x + 6 = 0, So, (x − 3) (x − 2) = 0 Hence, the roots are − x1 = 3 and x2 = 2 The roots are real and distinct. So, this is in the form of case 1 Hence, the solution is − Fn = ax1n + bx2n Here, Fn = a3n + b2n (As x1 = 3 and x2 = 2) Therefore, 1 = F0 = a30 + b20 = a+b 4 = F1 = a31 + b21 = 3a+2b Solving these two equations, we get a = 2 and b = −1 Hence, the final solution is − Fn = 2.3n + (−1) . 2n = 2.3n − 2n ### Problem 2 Solve the recurrence relation Fn = 10Fn−1 − 25Fn−2 where F0 = 3 and F1 = 17 Solution The characteristic equation of the recurrence relation is − x2 − 10x −25 = 0, So, (x − 5)2 = 0 Hence, there is single real root x1 = 5 As there is single real valued root, this is in the form of case 2 Hence, the solution is − Fn = ax1n + bnx1n 3 = F0 = a.50 + b.0.50 = a 17 = F1 = a.51 + b.1.51 = 5a+5b Solving these two equations, we get a = 3 and b = 2/5 Hence, the final solution is − Fn = 3.5n + (2/5) .n.2n ### Problem 3 Solve the recurrence relation Fn = 2Fn−1 − 2Fn−2 where F0 = 1 and F1 = 3 Solution The characteristic equation of the recurrence relation is − x2 −2x −2 = 0 Hence, the roots are − x1 = 1 + i andx2 = 1 − i In polar form, x1 = r ∠ θ andx2 = r ∠(− θ), where r = √2 and θ = π / 4 The roots are imaginary. So, this is in the form of case 3. Hence, the solution is − Fn = (√2 )n (a cos(n. π / 4) + b sin(n. π / 4)) 1 = F0 = (√2 )0 (a cos(0. π / 4) + b sin(0. π / 4) ) = a 3 = F1 = (√2 )1 (a cos(1. π / 4) + b sin(1. π / 4) ) = √2 ( a/√2 + b/√2) Solving these two equations we get a = 1 and b = 2 Hence, the final solution is − Fn = (√2 )n (cos(n. π / 4) + 2 sin(n. π / 4)) ## Particular Solutions A recurrence relation is called non-homogeneous if it is in the form Fn = AFn−1 + BFn−2 + F(n)where F(n) ≠ 0 The solution (an) of a non-homogeneous recurrence relation has two parts. First part is the solution (ah) of the associated homogeneous recurrence relation and the second part is the particular solution (at). So, an = ah + at. Let F(n) = cxn and x1 and x2 are the roots of the characteristic equation − x2 = Ax + B which is the characteristic equation of the associated homogeneous recurrence relation − • If x ≠ x1 and x ≠ x2, then at = Axn • If x = x1, x ≠ x2, then at = Anxn • If x = x1 = x2, then at = An2xn ### Problem Solve the recurrence relation Fn = 3Fn−1 + 10Fn−2 + 7.5n where F0 = 4 and F1 = 3 Solution The characteristic equation is − x2 − 3x −10 = 0 Or, (x − 5)(x + 2) = 0 Or, x1= 5 and x2 = −2 Since, x = x1 and x ≠ x2, the solution is − at = Anxn = An5n After putting the solution into the non-homogeneous relation, we get − An5n = 3A(n − 1)5n−1 + 10A(n − 2)5n−2 + 7.5n Dividing both sides by 5n−2, we get − An52 = 3A(n − 1)5 + 10A(n − 2)50 + 7.52 Or, 25An = 15An − 15A + 10An − 20A + 175 Or, 35A = 175 Or, A = 5 So, Fn = n5n+1 Hence, the solution is − Fn = n5n+1 + 6.(−2)n −2.5n ## Generating Functions Generating Functions represents sequences where each term of a sequence is expressed as a coefficient of a variable x in a formal power series. Mathematically, for an infinite sequence, say a0, a1, a2,…………, ak,………, the generating function will be − Gx=a0+a1x+a2x2+.........+akxk+.........=k=0akxkGx=a0+a1x+a2x2+………+akxk+………=∑k=0∞akxk ### Some Areas of Application Generating functions can be used for the following purposes − • For solving a variety of counting problems. For example, the number of ways to make change for a Rs. 100 note with the notes of denominations Rs.1, Rs.2, Rs.5, Rs.10, Rs.20 and Rs.50 • For solving recurrence relations • For proving some of the combinatorial identities • For finding asymptotic formulae for terms of sequences ### Problem 1 What are the generating functions for the sequences {ak} with ak= 2 and = 3k? Solution When ak = 2, generating function, G(x)=k=02xk=2+2x+2x2+2x3+.........G(x)=∑k=0∞2xk=2+2x+2×2+2×3+……… When ak=3k,G(x)=k=03kxk=0+3x+6x2+9x3+......ak=3k,G(x)=∑k=0∞3kxk=0+3x+6×2+9×3+…… ### Problem 2 What is the generating function of the infinite series; 1, 1, 1, 1, ……….? Solution Here, ak = 1, for 0 ≤k≤ ∞. Hence, G(x)=1+x+x2+x3+........=1(1x)G(x)=1+x+x2+x3+……..=1(1−x) ### Some Useful Generating Functions • For ak=ak,G(x)=k=0akxk=1+ax+a2x2+.........=1(1ax)ak=ak,G(x)=∑k=0∞akxk=1+ax+a2x2+………=1(1−ax) • For ak=(k+1),G(x)=k=0(k+1)xk=1+2x+3x2.........=1(1x)2ak=(k+1),G(x)=∑k=0∞(k+1)xk=1+2x+3×2………=1(1−x)2 • For ak=cnk,G(x)=k=0cnkxk=1+cn1x+cn2x2+.........+x2=(1+x)nak=ckn,G(x)=∑k=0∞cknxk=1+c1nx+c2nx2+………+x2=(1+x)n • For ak=1k!,G(x)=k=0xkk!=1+x+x22!+x33!.........=exak=1k!,G(x)=∑k=0∞xkk!=1+x+x22!+x33!………=ex #### Posted by Akash Kurup Founder and C.E.O, World4Engineers Educationist and Entrepreneur by passion. Orator and blogger by hobby Website: http://world4engineers.com
Title Beginning Algebra Learning Objectives After completing this tutorial, you should be able to: Read a bar graph. Read a line graph. Read a double line graph. Draw and read a Venn diagram. Introduction In this tutorial we will be reading graphs. Graphs can be used to visually represent the relationship of data. It can help organize and show people statistics, which can be good for some and not so good for others, depending on what the statistics show. Organizing data graphically can come in handy in fields like business, sports, teaching, politics, advertising, etc.. Let's start looking at some graphs. Tutorial Bar Graph A bar graph can be used to give a visual representation of the relationship of data that has been collected. It is made up of a vertical and a horizontal axis and bars that can run vertically or horizontally. Vertical Bar Graph If the bars are vertical, match the top of the bar with the vertical axis found at the side of the overall graph to find the information the bar associates with on the vertical axis. You will find what the bar associates with on the horizontal axis at the base of the bar. The bar graph below has vertical bars: The horizontal axis represents years and the vertical axis represents profit in thousands of dollars. The first bar on the left associates with the year 1999 AND the profit of \$20,000. The red line shows how the top of the bar lines up with 20 on the vertical axis. The second bar from the left associates with the year 2000 and the profit of \$30,000. The blue line shows how the top of the bar lines up with 30 on the vertical axis. Horizontal Bar Graph If the bars are horizontal, match the right end of the bar with the horizontal axis found at the bottom of the overall graph to find the information the bar associates with on the horizontal axis. You will find what the the bar associates with on the vertical axis at the left end of the bar. The bar graph below has horizontal bars: (Note that this graph shows the same information the above graph does, just with horizontal bars instead of vertical bars.) The vertical axis represents years and the horizontal axis represents profit in thousands of dollars. The first bar on the bottom associates with the year 1999 AND the profit of \$20,000. The red line shows how the right end of the bar lines up with 20 on the horizontal axis. The second bar from the bottom associates with the year 2000 and the profit of \$30,000. The blue line shows how the right end of the bar lines up with 30 on the horizontal axis. Example 1: The bar graph below shows the number of students in a math class that received the grades shown. Use this graph to answer questions 1a - 1d. 1a. Find the number of students who received an A. 1b. Find the number of students who received an F. 1c. Find the number of students who passed the course (D or higher). 1d. Which grade did the most students receive? The bar that associates with the grade A is the first bar on the left. The top of that bar matches with 6 on the vertical axis. 6 students received an A. The bar that associates with the grade F is the fifth bar from the left. The top of that bar matches with 2 on the vertical axis. 2 students received an F. 1c. Find the number of students who passed the course (D or higher). (return to bar graph) We will have to do a little calculating here. We will need to find the number of students that received an A, B C, and D and then ad them together. The bar that associates with the grade A is the first bar on the left. The top of that bar matches with 6 on the vertical axis. The bar that associates with the grade B is the second bar from the left. The top of that bar matches with 16 on the vertical axis. The bar that associates with the grade C is the third bar from the left. The top of that bar matches with 12 on the vertical axis. The bar that associates with the grade D is the fourth bar from the left. The top of that bar matches with 4 on the vertical axis. 6 + 16 + 12 + 4 = 38 students passed the course. It looks like more students received a B than any other single grade. Example 2: The bar graph below shows the number of civilians holding various federal government jobs. Use the graph to answer questions 2a - 2d. 2a. About how many civilians work for Congress? 2b. About how many civilians work for the State Department? 2c. About how many civilians work for the armed forces (Navy, Air Force, and Army)? 2d. Which federal government job listed has the most civilian workers? The bar that associates with Congress is the fourth bar up. The right of that bar lines up a little to the left of 50 on the horizontal axis. Note how the question asks ABOUT how many. In some cases, if it does not directly line up with a number that is marked you may need to approximate. This is very close to and less than 50. A good approximation is 25. About 25,000 civilians work for Congress. The bar that associates with the State Department is the sixth bar up. The right of that bar lines up with 50 on the horizontal axis. About 50,000 civilians work for the State Department. 2c. About how many civilians work for the armed forces (Navy, Air Force, and Army)? (return to bar graph) We will have to do a little calculating on this one. We will need to find the number of civilians that work for each branch of the armed services and then add them up. The bar that associates with the Navy is the third bar up. The right of that bar ends between 300 and 350 on the horizontal axis. 310 is a good approximation for this number. The bar that associates with the Air Force is the second bar up. The right of that bar ends between 200 and 250 on the horizontal axis. 210 is a good approximation for this number. The bar that associates with the Army is the first bar from the bottom. The right of that bar ends just under 350 on the horizontal axis. 340 is a good approximation for this number. About 310,000 + 210,000 + 340,000 = 860,000 civilians work for the State Department. 2d. Which federal government job listed has the most civilian workers? (return to bar graph) It looks like the Army has the most civilian workers. Line Graph A line graph is another way to give a visual representation of the relationship of data that has been collected. It is made up of a vertical and horizontal axis and a series of points that are connected by a line. Each point on the line matches up with a corresponding vertical axis and horizontal axis value on the graph. In some cases, you are giving a value from the horizontal axis and you need to find its corresponding value from the vertical axis. You find the point on the line that matches the given value from the horizontal axis and then match it up with its corresponding vertical axis value to find the value you are looking for. You would do the same type of process if you were given a vertical axis value and needed to find a horizontal axis value. The graph below is a line graph: (Note that this graph shows the same information the above graphs under vertical and horizontal graphs do, just with a line instead of bars.) The horizontal axis represents years and the vertical axis represents profit in thousands of dollars. The first point on the left associates with the year 1999 AND the profit of \$20,000. The red line shows how it lines up with 20 on the vertical axis and 1999 on the horizontal axis. The second point from the left associates with the year 2000 and the profit of \$30,000. The blue line shows how it lines up with 30 on the vertical axis and 2000 on the horizontal axis. Example 3: The line graph below shows the distance traveled of a vacationer going 70 mph down I-40 from 0 to 4 hours. Use the graph to answer questions 3a - 3b. 3a. How far has the vacationer traveled at 3 hours? 3b. How long does it take the vacationer to travel 140 miles? 3a. How far has the vacationer traveled at 3 hours? (return to line graph) The point that matches with 3 on the horizontal axis also matches with 210 on the vertical axis. The vacationer has traveled 210 miles. 3b. How long does it take the vacationer to travel 140 miles? (return to line graph) The point that matches with 140 on the vertical axis also matches with 2 on the horizontal axis. It takes the vacationer 2 hours to travel 140 miles. Example 4: The line graph below shows the profit a local candy company made over the months of September through December of last year. Use the graph to answer questions 4a - 4c. 4a. About how much was the profit in the month of October? 4b. Which month had the lowest profit? 4c. What is the difference between the profits of November and December? The point that matches with October on the horizontal axis also matches between 20 and 25 on the vertical axis. It looks to be about 23. The profit for the month of October is about \$23,000. It looks like September had the lowest profit. 4c. What is the difference between the profits of November and December? (return to line graph) The point that matches with November on the horizontal axis also matches with 15 on the vertical axis. The point that matches with December on the horizontal axis also matches with 20 on the vertical axis. The difference between the profits of November and December would be 20,000 - 15,000 = \$5,000. Double Line Graph A double line graph is another way to give a visual representation of the relationship of data that has been collected. It is similar to the line graph mentioned above. The difference is there are two lines of data instead of one. It is made up of a vertical and horizontal axis and two series of points each one connected by a line. The legend will show which line represents what set of points. Most times a solid line and a dashed line are used. But varying colors can also distinguish the two lines apart. Each point on each line matches up with a corresponding vertical axis and horizontal axis value on the graph. In some cases, you are giving a value from the horizontal axis and you need to find its corresponding value from the vertical axis. You find the point on the line that matches the given value from the horizontal axis and then match it up with its corresponding vertical axis value to find the value you are looking for. You would do the same type of process if you were given a vertical axis value and needed to find a horizontal axis value. The graph below is a double line graph: The horizontal axis represents the year and the vertical axis represents profit in thousands of dollars. The legend towards the top of the graph indicates which line represents which product. The solid line corresponds with Product A and the dashed line goes with Product B. The first point on the solid line on the left associates with the year 1995 AND the profit of \$30,000. The second point on the solid line from the left associates with the year 1996 AND the profit of \$40,000. The third point on the solid line from the left associates with the year 1997 AND the profit of \$40,000. The fourth point on the solid line from the left associates with the year 1998 AND the profit of \$30,000. The fifth point on the solid line from the left associates with the year 1999 AND the profit of \$60,000. The first point on the dashed line on the left associates with the year 1995 AND the profit of \$20,000. The second point on the dashed line from the left associates with the year 1996 AND the profit of \$20,000. The third point on the dashed line from the left associates with the year 1997 AND the profit of \$15,000. The fourth point on the dashed line from the left associates with the year 1998 AND the profit of \$40,000. The fifth point on the dashed line from the left associates with the year 1999 AND the profit of \$50,000. Example 5: The double line graph below shows the total enrollment of students in a local college from 1990 - 1995, broken down into part-time and full-time students. Use the graph to answer questions 5a - 5c. 5a. What was the full-time enrollment in 1992? 5b. For what year shown on the graph did the number of part-time students exceed the previous year’s number of part-time students by the greatest number? 5c. What was the total enrollment from 1993 to 1995? 5a. What was the full-time enrollment in 1992? (return to double line graph) Since we are looking for full-time students, are we going to look at the solid or dashed line? According to the legend, we need to look at the dashed line. The point that is on the dashed line and matches with 1992 on the horizontal line also matches with 200 on the vertical line. There were 200 full-time students enrolled in 1992. 5b. For what year shown on the graph did the number of part-time students exceed the previous year’s number of part-time students by the greatest number? (return to double line graph) Since we are looking for part-time students, are we going to look at the solid or dashed line? According to the legend, we need to look at the solid line. When looking at the graph, we are only interested in a rise in the number of part-time students. From 1990 to 1991, the number of part-time students went up 100 to 150. From 1991 to 1992, it went down from 150 to 50. From 1992 to 1993, there was increase from 50 to 250. From 1993 to 1994, there was another increase, this time from 250 to 300. The last years, 1994 - 1995, it held steady at 300. So, what year exceeded the previous number of part-time students by the greatest number? Looks like 1993. There were 200 more part-time students in 1993 than there were in 1992. 5c. What was the total enrollment from 1993 to 1995? (return to double line graph) Let’s break this down into part-time and full-time students. Looking at the dashed line to see the number of full-time students we get 250 + 400 + 500 = 1150. Looking at the solid line to see the number of part-time students we get 250 + 300 + 300 = 850. Putting those together, we have 1150 + 850 = 2000 students who were enrolled from 1993 to 1995. Venn Diagrams Venn diagrams are a visual way of organizing information. It can be very helpful when you have a problem to solve that categorizes or shows relationships between things. A common use for Venn diagrams is analyzing the results of a survey. For example, you may have a survey of students asking them which classes they like and perhaps you listed math and english. The student could check 0, 1, or 2 of these choices. You would strategically place the results in a Venn diagram. If they only choose math, then they would go in a particular region of the diagram that shows that, if they picked both, they would go into the area of the diagram that depicts that, etc. Of course there are other uses for the Venn diagram, that is one of the more common ones. The graph below is a Venn diagram: This diagram represents the results of a survey of people who were asked if they liked Coke or Pepsi. They could choose only Coke, only Pepsi, both, or neither. Note that a lot of times you do not see the letter U or the roman numerals on a Venn Diagram (just the box and the circles), I use them as references so you know what area of the diagram I'm talking about in the lesson. The rectangle box represents the universal set U. The universal set is the set of all elements considered in a problem. In this example, the universal set are all the people who took the survey. The circles represent the categories or subsets involving the universal set. In this example, the two categories or choices on the survey were Coke and Pepsi. When you draw a Venn diagram, you want to overlap the circles in case there are some that pick both categories. We need to make sure we accurately place those people and do not count them more than one time. The roman numerals are called region numbers. Region I represents everyone who selected ONLY Coke which was 575 people. Region II is where the two circles intersect or overlap. It represents everyone who selected BOTH Coke and Pepsi which was 100 people. Region III represents everyone who selected ONLY Pepsi which was 225 people. Region IV is inside the rectangle, but outside the circles. It represents everyone who selected NEITHER Coke nor Pepsi which was 15 people. Example 6: A teacher took a survey on pets in her class of 40 students. 12 students said they had a cat. 9 students said they had a dog. 2 said they had both a cat and a dog. How many students picked neither? How many students had only a cat? How many students had only a dog? 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: The bar graph below shows the profit a cd store made over the months of September through December of last year. Use the graph to answer questions 1a - 1c. 1a. About how much was the profit in September? (answer/discussion to 1a - 1c) 1b. Which month had the highest profit? (answer/discussion to 1a - 1c) 1c. What is the difference between the profits of October and November? (answer/discussion to 1a - 1c) Practice Problems 2a - 2c: The line graph below shows last week's high temperatures in Fahrenheit. Use the graph to answer questions 2a - 2c. 2a. How much was Thursday's high temperature? (answer/discussion to 2a - 2c) 2b. Which day had the lowest high temperature? (answer/discussion to 2a - 2c) 2c. What temperature occurred the most? (answer/discussion to 2a - 2c) Practice Problems 3a - 3c: The double line graph below shows the total enrollment of people who work out at a local gym from 1998 - 2002, broken down into males and females. Use the graph to answer questions 3a - 3c. 3a. What was the ratio between male gym members and female gym members in 2000? (answer/discussion to 3a - 3c) 3b. For what year shown on the graph did male gym membership not change from the year before? (answer/discussion to 3a - 3c) 3c. What was the total enrollment of the gym from 1998 to 2000? (answer/discussion to 3a - 3c) Practice Problems 4a - 4c: A group of students were asked if they liked rock or country music. The results were as follows: 27 said they liked rock, 20 said they liked country, 5 liked both, and 3 liked neither. 4a. How many students chose only rock? (answer/discussion to 4a - 4c) 4b. How many students chose only country? (answer/discussion to 4a - 4c) 4c. How many students were surveyed? (answer/discussion to 4a - 4c) 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 July 25, 2011 by Kim Seward.
# When to Use the Law of Sines and Cosines You probably learned the basic trigonometric functions in school – sine, cosine, and tangent. But do you remember when to use each one? Checkout this video: ## Introduction In trigonometry, there are a few different ways to find missing sides or angles in a triangle. Two of the most common methods are using the law of sines and the law of cosines. But when should you use each method? In this article, we’ll explore when it’s appropriate to use the law of sines and cosines so that you can solve any triangle problem that comes your way. ## What is the law of sines? In mathematics, the law of sines, sine rule, sine formula, or sine relation is an equality relating the lengths of certain sides of a triangle to the sines of its angles. According to the law, a/sin(A) = b/sin(B) = c/sin(C) where a, b and c are the lengths of the sides of the triangle and A, B and C are the angles opposite those respective sides. The law can be applied to any triangle, whether it is right angled or not. It can also be extended to non-right angled triangles if one replaces sin(A), sin(B), sin(C) by the so-called directional cosines which are in fact the ratios of certain sides in a corresponding reference triangle. ## What is the law of cosines? The law of cosines is used to find the side lengths or angles of a triangle when we know some or all of the other side lengths or angles. This can be really useful in many different geometry and trigonometry problems. The law of cosines is also sometimes called the cosine rule. ## How to use the law of sines The law of sines and cosines are two related formulas that are often used together to solve problems. They are both ways of finding missing sides or angles in a triangle, when you know some of the other sides or angles. To use the law of sines, you need to know: – The length of one side of the triangle – The angle opposite that side – The other two angles in the triangle To use the law of cosines, you need to know: – The lengths of two sides of the triangle – The angle between those two sides – Either: --the length of the third side --or one of the other two angles in the triangle ## How to use the law of cosines There are two main trigonometric functions- Sine and Cosine. These functions help us to calculate the missing sides and angles of a triangle when we only know some of the information. The law of sines and cosines are both very useful, but it can be tricky to know when to use each one. In this article, we’ll go over when to use the law of cosines so that you can tackle any trigonometry problem with confidence! The law of cosines states that: c^2 = a^2 + b^2 – 2ab * cos(C) This means that if you know the lengths of two sides of a triangle, as well as the angle between them (angle C in the equation), you can use the law of cosines to find the length of the third side. This is particularly useful when you’re working with a oblique triangle- a triangle where none of the angles are 90 degrees. To use the law of cosines, simply plug in the known values for a, b, and C into the equation. Then, solve for c. This will give you the missing side length that you were looking for! ## Applications of the law of sines The law of sines is typically used in two scenarios: when you know two angles and the length of one side of a triangle (AAS), or when you know two sides and the angle between them (SAS). We will explore these scenarios in greater depth below. AAS: When you know two angles and the length of one side of a triangle, you can use the law of sines to find the remaining lengths. This is helpful when you are trying to find the height of an object, for example. To use the law of sines in this scenario, you will set up your equation as follows: a/sin(A) = b/sin(B) = c/sin(C) In this equation, a is the known side, B and C are the known angles, and b and c are the unknown sides. You can solve for either unknown side using this equation. SAS: When you know two sides and the angle between them, you can use the law of cosines to find all three missing parts. This is helpful when you are trying to find an object that is hidden from view. To use the law of cosines in this scenario, you will set up your equation as follows: a^2 = b^2 + c^2 – 2bc*cos(A) ## Applications of the law of cosines There are many applications for the law of cosines, including navigation, astronomy, and solving triangles. In navigation, the law of cosines can be used to find the distance between two points on a sphere, such as two ships at sea. In astronomy, the law of cosines can be used to find the distance to a star or other celestial objects. Finally, the law of cosines can be used to solve any triangle, whether it is right angled or not. ## Advantages of using the law of sines There are many different situations in which you might need to use the law of sines or cosines. For example, you might need to find the height of an object when you only know the angles of the triangle formed by the object and the ground, or you might need to find the length of a side of a triangle when you only know the lengths of two other sides. In general, you can use either the law of sines or cosines whenever you known two angles and one side (or two sides and one angle) of a triangle, but it’s sometimes more convenient to use one over the other. One advantage of using the law of sines is that it doesn’t require any information about angles that are not adjacent to the side you’re trying to find. So, if you know angle A and side a, and you want to find angle B, you can just use the law of sines without having to know angle C. This can be helpful if angle C is difficult to measure or if you don’t have all of the information about the triangle. Another advantage of using the law of sines is that it sometimes yields simpler calculations than using the law of cosines. This can be particularly helpful if you’re working with large numbers or decimals. However, there are also some advantages to using the law of cosines. One is that it can be used in cases where one angle is obtuse (greater than 90 degrees). The law of sines can only be used with acute angles (angles less than 90 degrees). Another advantage is that it sometimes yields more accurate results than using the law of sines, especially when working with small numbers or decimals. Ultimately, which method you use is up to you – both have their advantages and disadvantages. ## Advantages of using the law of cosines There are many mathematical formulas that can be used to solve certain types of problems. Some formulas are more useful than others in specific situations. The law of cosines is a good example of a formula that has specific advantages and disadvantages. The law of cosines is a mathematical formula that can be used to solve certain problems involving triangles. The formula is derived from the Pythagorean theorem, which states that the sum of the squares of the sides of a right triangle is equal to the square of the hypotenuse. The law of cosines can be used to find the length of a side of a triangle when two other sides and the angle between them are known. This is sometimes more accurate than using the Pythagorean theorem alone. In addition, the law of cosines can be used to find angles when two sides and one angle are known. This can be very useful in certain situations, such as when working with surveying data. One disadvantage of using the law of cosines is that it can be more difficult to remember than some other formulas. In addition, the law of cosines only works with triangles, so it cannot be used to solve problems involving other shapes. ## Conclusion We hope you enjoyed this lesson on the law of sines and cosines. Now that you know when and how to use these formulas, you’ll be able to solve many types of problems involving triangles. Be sure to practice what you’ve learned so that you can master these concepts. Scroll to Top
# All About blocks will build a 12 by 12 room 0 “Building blocks are a timeless toy that have captivated the imaginations of children for generations. But did you know that these simple, colorful pieces can also be used to create functional and practical structures? In this article, we will explore the world of All About blocks and their latest project: a 12 by 12 room. From the innovative design to the endless possibilities for customization, let’s discover how these blocks are revolutionizing the way we build and play.” Table of Contents ## How many blocks will build a 12 by 12 room To determine how many blocks will be needed to build a 12 by 12 room, we must first consider the type of blocks that will be used and the dimensions of the blocks themselves. In this scenario, we will assume that standard concrete blocks with dimensions of 8 inches by 8 inches by 16 inches will be used. To calculate the number of blocks needed, we must first determine the perimeter and area of the room. The perimeter of a 12 by 12 room would be 12 + 12 + 12 + 12 = 48 feet long. This means that there would be 48 feet of wall to be built. Next, we need to determine the height of the wall. Assuming a standard ceiling height of 8 feet, we can calculate that the total height of the walls would be 8 feet. To calculate the total number of blocks, we first need to convert the dimensions of the blocks to feet. 8 inches is equal to 0.666 feet and 16 inches is equal to 1.333 feet. This means that each block has a face area of 0.666 feet by 1.333 feet. To determine the number of blocks needed, we can divide the total wall area (48 feet x 8 feet = 384 square feet) by the area of one block (0.666 feet x 1.333 feet = 0.888 square feet). This results in approximately 433 blocks needed to build the 12 by 12 room. However, this calculation does not account for any windows or doors in the room. If there are openings, we will need to calculate the area of these openings and subtract them from the total wall area before determining the number of blocks needed. In conclusion, to build a 12 by 12 room using standard concrete blocks with dimensions of 8 inches by 8 inches by 16 inches, approximately 433 blocks would be needed without any openings. It is always recommended to account for some extra blocks for any potential variations or mistakes during construction. ## How to calculate number of concrete blocks do you need. Building projects often require the use of concrete blocks, also known as concrete masonry units (CMUs). These blocks are commonly used in walls, foundations, and other structural elements in construction. If you are planning a project that involves concrete blocks, it is important to accurately calculate the number of blocks you will need to complete the project. Here’s a guide on how to calculate the number of concrete blocks you need for your project. 1. Determine the Dimension of the Blocks: The first step is to find out the size of the concrete blocks you will be using. Typically, concrete blocks come in standard sizes of 8x8x16 inches or 6x8x16 inches. Make sure to measure the length, width, and height of the blocks in inches. 2. Measure the Dimensions of the Project: Measure the length and height of the wall or foundation where you will be using the concrete blocks. Multiply these two dimensions to get the total area in square inches. 3. Allow for Mortar Joints: Mortar joints are the spaces between the blocks that are filled with mortar to bond them together. These joints typically add 3/8 of an inch to the overall width of the blocks. So, when calculating the number of blocks, you need to take into account the mortar joints as well. 4. Find the Number of Blocks per Square Foot: To determine the number of blocks you need per square foot, divide the total area in square inches by the area of one block (length x width). This will give you the number of blocks needed per square foot. 5. Calculate the Total Number of Blocks: Now, multiply the number of blocks you need per square foot by the total square footage of the project. This will give you the total number of blocks you need for the project. Make sure to round up to the nearest whole number. See also All About bags of concrete do I need for a 10×20 slab 6. Add Extra Blocks for Wastage: It is always recommended to add extra blocks to account for wastage due to breakage or cutting. A good rule of thumb is to add 5-10% extra blocks to your total estimate. 7. Consider Different Sizes and Shapes: If your project involves different sizes or shapes of blocks, you will need to repeat the above steps for each type and then add the total number of blocks together. In summary, to calculate the number of concrete blocks you need, you need to know the block dimensions, the project dimensions, and factor in mortar joints and wastage. By following these steps, you can accurately estimate the number of blocks required for your project. It is always recommended to double-check your calculations to avoid any shortage or excess of blocks during construction. ## How many blocks will build a 12 by 12 room To determine how many blocks will be needed to build a 12 by 12 room, we must first consider the type of blocks that will be used and the dimensions of the blocks themselves. In this scenario, we will assume that standard concrete blocks with dimensions of 8 inches by 8 inches by 16 inches will be used. To calculate the number of blocks needed, we must first determine the perimeter and area of the room. The perimeter of a 12 by 12 room would be 12 + 12 + 12 + 12 = 48 feet long. This means that there would be 48 feet of wall to be built. Next, we need to determine the height of the wall. Assuming a standard ceiling height of 8 feet, we can calculate that the total height of the walls would be 8 feet. To calculate the total number of blocks, we first need to convert the dimensions of the blocks to feet. 8 inches is equal to 0.666 feet and 16 inches is equal to 1.333 feet. This means that each block has a face area of 0.666 feet by 1.333 feet. To determine the number of blocks needed, we can divide the total wall area (48 feet x 8 feet = 384 square feet) by the area of one block (0.666 feet x 1.333 feet = 0.888 square feet). This results in approximately 433 blocks needed to build the 12 by 12 room. However, this calculation does not account for any windows or doors in the room. If there are openings, we will need to calculate the area of these openings and subtract them from the total wall area before determining the number of blocks needed. In conclusion, to build a 12 by 12 room using standard concrete blocks with dimensions of 8 inches by 8 inches by 16 inches, approximately 433 blocks would be needed without any openings. It is always recommended to account for some extra blocks for any potential variations or mistakes during construction. ## How many blocks can build 2 rooms When constructing any type of structure, one key element is the materials used. In the case of building rooms, one common building material is blocks. But how many blocks are needed to build two rooms? As a civil engineer, I have the knowledge and expertise to provide an accurate estimate. First, it’s important to determine the size of the rooms being constructed. The size can vary depending on the intended use of the rooms, but for this estimation, let’s assume that each room is a standard size of 12 feet by 12 feet. This gives a total area of 144 square feet for each room. Next, the type and size of blocks being used must be considered. There are various types of blocks, such as concrete blocks, cinder blocks, and clay bricks. For the purpose of this estimation, let’s use standard concrete blocks that are 8 inches by 8 inches by 16 inches in size. To calculate the number of blocks needed, we first convert the square footage into square inches by multiplying 144 square feet by 144 (12 inches x 12 inches). This gives a total of 20,736 square inches for each room. Now, we need to factor in the size of the blocks. A standard 8 inch by 8 inch by 16 inch concrete block has a volume of 0.89 cubic feet. This is equivalent to 119,808 cubic inches. Dividing the total square inches of each room (20,736) by the volume of one block (119,808), we get an estimated 0.17 blocks per square inch. Multiplying this by the total square footage of two rooms (288 square feet) gives us an estimated 49 blocks needed for both rooms. However, it’s always recommended to account for waste or extra blocks for any potential future repairs. Therefore, it’s safe to increase this number by 15-20%. This gives a total of 56-59 blocks needed to build two rooms. See also All About sand do you mix with cement Additionally, if there are any openings in the walls for windows or doors, those areas will require more blocks. The exact number will depend on the size of the openings, but it’s important to factor this into the estimation as well. In conclusion, using the above estimation, approximately 56-59 standard concrete blocks are needed to build two standard sized rooms. However, it’s important to keep in mind that this is just an estimation and the actual number may vary depending on various factors such as the type of blocks used and any additional openings in the walls. A thorough calculation should be done by a qualified professional before beginning any construction project. ## How many blocks can build 3 rooms The number of blocks needed to build 3 rooms can vary depending on the size and design of the rooms, as well as the specific type and size of blocks being used. However, as a general estimate, we can provide a rough calculation to give an idea of the approximate number of blocks that may be required. Firstly, it is important to determine the dimensions of the rooms in question. For this example, we will assume that each room is 4 meters by 5 meters, giving a total of 20 square meters for each room. Therefore, for 3 rooms, the total area would be 60 square meters. Next, we need to decide on the thickness of walls. For this, we will use the standard practice of using 8-inch concrete blocks, which equates to 0.2 meters. We also need to consider the mortar joint thickness, which is typically 10 mm or 0.01 meters. Now, we can calculate the number of blocks required for the walls of the 3 rooms. Taking into account the wall thickness and mortar joints, the area covered by one block would be 0.22 square meters (0.2 x 0.2 + 0.01). Therefore, the total number of blocks needed for the walls of the 3 rooms would be 272 (60 ÷ 0.22). However, this calculation does not include the blocks needed for the floors, ceilings, and doors. Additionally, some areas may require double walls or extra support walls, which would increase the number of blocks needed. Therefore, it is always advisable to add an extra 5-10% to the estimated number of blocks for unforeseen requirements. Moreover, the above calculation is based on using standard 8-inch concrete blocks. If you plan to use larger or smaller blocks, the number of blocks required would differ. It is always recommended to consult with a structural engineer or get a professional estimate to determine the accurate number of blocks required for your specific project. In conclusion, to build 3 rooms, approximately 272 blocks would be needed for the walls, with additional blocks for floors, ceilings, and doors. However, it is best to get a professional estimate to ensure accurate calculations and avoid any disruptions or delays during the construction process. ## How many blocks can build 4 rooms The number of blocks needed to build 4 rooms depends on various factors such as the size of the rooms, the type of blocks used, and the design of the rooms. Let’s assume we are using standard concrete blocks measuring 8 inches by 8 inches by 16 inches and each room has a standard size of 10 feet by 12 feet. Firstly, we need to determine the area of each room. Since we are using a standard size of 10 feet by 12 feet, the area of each room would be 120 square feet (10 x 12 = 120). Next, we need to calculate the volume of each block in cubic feet. One concrete block measures 8 inches by 8 inches by 16 inches, which is equivalent to 0.444 cubic feet (0.067 cubic yards). To determine the number of blocks needed for each room, we divide the area of the room by the volume of one block. Thus, for each room, we would need approximately 270 blocks (120/0.444 = 270). Therefore, to build 4 rooms, we would need approximately 1080 blocks (270 x 4 = 1080). However, this is just a rough estimate. The number of blocks required may vary depending on the design of the rooms, including the number of windows, doors, and other openings that require additional blocks for support. Additionally, the thickness of the walls may also affect the number of blocks needed. In conclusion, building 4 rooms using standard concrete blocks would require approximately 1080 blocks. As a civil engineer, it is essential to consider various factors and make precise calculations to ensure the right amount of materials is used for construction projects. These calculations not only help in cost estimation but also ensure structural stability and durability. See also Introduction of Well Foundation ## How many blocks can build 5 rooms The number of blocks required to build 5 rooms varies depending on the size of the rooms and the construction techniques used. However, we can make an estimate based on standard building practices. Assuming the rooms are of average size, measuring 4 meters by 4 meters (13 feet by 13 feet), and the walls are constructed using standard concrete blocks with a size of 20 cm by 20 cm by 40 cm (8 inches by 8 inches by 16 inches), we can calculate the number of blocks needed for each room. A standard room has an area of 16 square meters (172 square feet). To calculate the perimeter of the room, we need to add up the lengths of all four walls, which in this case is 4+4+4+4 = 16 meters (52 feet). Since one standard block has a length of 20 cm (0.2 meters), we would need 80 blocks to construct one wall. Since there are four walls in the room, we would require a total of 320 blocks for one room. Therefore, for 5 rooms, the total number of blocks required would be 320 x 5 = 1600 blocks. However, this estimate does not take into account additional factors that can affect the quantity of blocks needed, such as door and window openings, corners, and partition walls. It is always advisable to consult a structural engineer or a construction expert for a more accurate estimation based on the specific design of the rooms. Furthermore, the number of blocks needed also depends on the structural design and construction method used. For example, if the walls are built using reinforced concrete rather than just concrete blocks, fewer blocks would be required. In conclusion, the number of blocks needed to build 5 rooms can vary, but as per our estimate, it would require approximately 1600 blocks. It is always best to consult a professional for an accurate estimation based on the specific project requirements. ## How many blocks can build 6 rooms As a civil engineer, one of my tasks is to determine the necessary materials and quantities for a construction project. If the goal is to build 6 rooms, the number of blocks needed will depend on various factors such as the size and design of the rooms, the type of block used, and the construction method. Generally, the unit of measurement for blocks in construction projects is in square feet, and the number of blocks needed for a room can be calculated using its floor area. The average size of a room can be around 150-200 square feet, but this can vary depending on the intended use and design. Let’s assume that the size of each room is 150 square feet, and the desired block size is 8 inches by 8 inches with a thickness of 4 inches. This type of block is commonly used for residential projects, and its coverage area is 64 square inches or 0.444 square feet. If we divide the room’s floor area of 150 square feet by 0.444 square feet, we get approximately 338 blocks per room. So for 6 rooms, we would need a total of 2,028 blocks (338 blocks per room x 6 rooms). However, it is essential to consider the waste or breakage of blocks during the construction process, so we usually add around 5-10% to the estimated quantity. Adding 10% to the total, we would need 2,231 blocks to build 6 rooms. It is also important to note that this calculation is just an estimate based on the assumptions made. The actual number of blocks needed can vary depending on the factors mentioned earlier. For instance, if the rooms are larger or if a different block size is used, the quantity will change. Additionally, we also need to consider other materials such as mortar, reinforcement, and labor costs, which will affect the overall project cost. In conclusion, the number of blocks needed to build 6 rooms can vary depending on different factors. However, as a rough estimate, using 8x8x4 inch blocks with a 10% allowance for waste, we would need around 2,231 blocks to construct 6 rooms. As a civil engineer, it is my responsibility to ensure that the correct calculations and preparations are made to ensure a successful construction project. ## Conclusion In conclusion, All About blocks are an innovative and efficient solution for building a 12 by 12 room. These blocks are easy to use, cost-effective, and environmentally friendly. With their interlocking design, they provide a sturdy and durable structure that can withstand various weather conditions. Whether it is for residential or commercial purposes, All About blocks offer endless possibilities for customizing the 12 by 12 room to meet individual needs. From creating extra living space to building storage units, these blocks are a versatile choice for any construction project. So, if you are planning to build a 12 by 12 room, All About blocks should definitely be considered as the go-to option for a hassle-free and successful build.
# The Binomial Expansion - PowerPoint PPT Presentation 1 / 23 The Binomial Expansion. Introduction. You first met the Binomial Expansion in C2 In this chapter you will have a brief reminder of expanding for positive integer powers We will also look at how to multiply out a bracket with a fractional or negative power I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. The Binomial Expansion Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - #### Presentation Transcript The Binomial Expansion ### Introduction • You first met the Binomial Expansion in C2 • In this chapter you will have a brief reminder of expanding for positive integer powers • We will also look at how to multiply out a bracket with a fractional or negative power • We will also use partial fractions to allow the expansion of more complicated expressions Teachings for Exercise 3A ### The Binomial Expansion You need to be able to expand expressions of the form (1 + x)n where n is any real number Find: Always start by writing out the general form Sub in: n = 4 x = x Work out each term separately and simplify Every term after this one will contain a (0) so can be ignored  The expansion is finite and exact 3A ### The Binomial Expansion You need to be able to expand expressions of the form (1 + x)n where n is any real number Find: Always start by writing out the general form Sub in: n = 3 x = -2x Work out each term separately and simplify It is VERY important to put brackets around the x parts Every term after this one will contain a (0) so can be ignored  The expansion is finite and exact 3A ### The Binomial Expansion You need to be able to expand expressions of the form (1 + x)n where n is any real number Find: Rewrite this as a power of x first Write out the general form (it is very unlikely you will have to go beyond the first 4 terms) Sub in: n = -1 x = x Work out each term separately and simplify • With a negative power you will not get a (0) term • The expansion is infinite • It can be used as an approximation for the original term 3A ### The Binomial Expansion You need to be able to expand expressions of the form (1 + x)n where n is any real number Find: Rewrite this as a power of x first Write out the general form (it is very unlikely you will have to go beyond the first 4 terms) Sub in: n = 1/2 x = -3x Work out each term separately and simplify  You should use your calculator carefully • With a fractional power you will not get a (0) term • The expansion is infinite • It can be used as an approximation for the original term 3A ### The Binomial Expansion You need to be able to expand expressions of the form (1 + x)n where n is any real number and state the values of x for which it is valid… Find the Binomial expansion of: Write out the general form Sub in: n = 1/3 x = -x Work out each term separately and simplify Imagine we substitute x = 2 into the expansion The values fluctuate (easier to see as decimals)  The result is that the sequence will not converge and hence for x = 2, the expansion is not valid 3A ### The Binomial Expansion You need to be able to expand expressions of the form (1 + x)n where n is any real number and state the values of x for which it is valid… Find the Binomial expansion of: Write out the general form Sub in: n = 1/3 x = -x Work out each term separately and simplify Imagine we substitute x = 0.5 into the expansion The values continuously get smaller  This means the sequence will converge (like an infinite series) and hence for x = 0.5, the sequence IS valid… 27 3A ### The Binomial Expansion You need to be able to expand expressions of the form (1 + x)n where n is any real number and state the values of x for which it is valid… Find the Binomial expansion of: Write out the general form Sub in: n = 1/3 x = -x Work out each term separately and simplify How do we work out for what set of values x is valid? The reason an expansion diverges or converges is down to the x term… If the term is bigger than 1 or less than -1, squaring/cubing etc will accelerate the size of the term, diverging the sequence If the term is between 1 and -1, squaring and cubing cause the terms to become increasingly small, to the sum of the sequence will converge, and be valid Write using Modulus The expansion is valid when the modulus value of x is less than 1 3A ### The Binomial Expansion You need to be able to expand expressions of the form (1 + x)n where n is any real number and state the values of x for which it is valid… Find the Binomial expansion of: Write out the general form: Sub in: n = -2 x = 4x Work out each term separately and simplify The ‘x’ term is 4x… Divide by 4 3A ### The Binomial Expansion You need to be able to expand expressions of the form (1 + x)n where n is any real number and by using x = 0.01, find an estimate for √2 Find the Binomial expansion of: Write out the general form: Sub in: n = 1/2 x = -2x Work out each term separately and simplify 3A ### The Binomial Expansion You need to be able to expand expressions of the form (1 + x)n where n is any real number and by using x = 0.01, find an estimate for √2 Find the Binomial expansion of: x = 0.01 Rewrite left using a fraction Square root top and bottom separately Multiply by 10 Divide by 7 3A Teachings for Exercise 3B ### The Binomial Expansion You can use the expansion for (1 + x)n to expand (a + bx)n by taking out a as a factor Find the first 4 terms in the Binomial expansion of: Take a factor 4 out of the brackets Both parts in the square brackets are to the power 1/2 You can work out the part outside the bracket Write out the general form: Sub in: n = 1/2 x = x/4 Work out each term carefully and simplify it Remember we had a 2 outside the bracket  Multiply each term by 2 Multiply by 4 3B ### The Binomial Expansion You can use the expansion for (1 + x)n to expand (a + bx)n by taking out a as a factor Find the first 4 terms in the Binomial expansion of: Take a factor 2 out of the brackets Both parts in the square brackets are to the power -2 You can work out the part outside the bracket Write out the general form: Sub in: n = -2 x = 3x/2 Work out each term carefully and simplify it Remember we had a 1/4 outside the bracket  Divide each term by 4 Multiply by 2, divide by 3 3B Teachings for Exercise 3C ### The Binomial Expansion You can use Partial fractions to simplify the expansions of more difficult expressions Find the expansion of: up to and including the term in x3 Express as Partial Fractions Cross-multiply and combine The numerators must be equal If x = 2 If x = -1 Express the original fraction as Partial Fractions, using A and B 3C ### The Binomial Expansion You can use Partial fractions to simplify the expansions of more difficult expressions Find the expansion of: up to and including the term in x3 Both fractions can be rewritten Expand each term separately Write out the general form: Sub in: x = x n = -1 Work out each term carefully Remember that this expansion is to be multiplied by 3 3C ### The Binomial Expansion You can use Partial fractions to simplify the expansions of more difficult expressions Find the expansion of: up to and including the term in x3 Both fractions can be rewritten Expand each term separately Take a factor 2 out of the brackets (and keep the current 2 separate…) Both parts in the square brackets are raised to -1 Work out 2-1 This is actually now cancelled by the 2 outside the square bracket! 3C ### The Binomial Expansion You can use Partial fractions to simplify the expansions of more difficult expressions Find the expansion of: up to and including the term in x3 Both fractions can be rewritten Expand each term separately Write out the general form: Sub in: x = -x/2 n = -1 Work out each term carefully 3C ### The Binomial Expansion You can use Partial fractions to simplify the expansions of more difficult expressions Find the expansion of: up to and including the term in x3 Both fractions can be rewritten Replace each bracket with its expansion Subtract the second from the first (be wary of double negatives in some questions) 3C ### Summary • We have been reminded of the Binomial Expansion • We have seen that when the power is a positive integer, the expansion is finite and exact • With negative or fractional powers, the expansion is infinite • We have seen how to decide what set of x-values the expansion is valid for • We have also used partial fractions to break up more complex expansions
## M2 - Row operations as matrix multiplication #### Example 1 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_1 \to R_1 + -4R_2$$. 2. Give a $$4 \times 4$$ matrix $$N$$ that may be used to perform the row operation $$R_1 \to -5R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_1 \to R_1 + -4R_2$$ and then $$R_1 \to -5R_1$$ to $$A$$ (note the order). 1. $$C= \left[\begin{array}{cccc} 1 & -4 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$N= \left[\begin{array}{cccc} -5 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$NCA$$ #### Example 2 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$M$$ that may be used to perform the row operation $$R_3 \to R_3 + 4R_2$$. 2. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_1 \to -4R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_1 \to -4R_1$$ and then $$R_3 \to R_3 + 4R_2$$ to $$A$$ (note the order). 1. $$M= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 4 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$C= \left[\begin{array}{cccc} -4 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$MCA$$ #### Example 3 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_2 \to R_2 + -4R_4$$. 2. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_4 \leftrightarrow R_2$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_2 \to R_2 + -4R_4$$ and then $$R_4 \leftrightarrow R_2$$ to $$A$$ (note the order). 1. $$P= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$Q= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right]$$ 3. $$QPA$$ #### Example 4 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_1 \to R_1 + 5R_4$$. 2. Give a $$4 \times 4$$ matrix $$B$$ that may be used to perform the row operation $$R_2 \to -5R_2$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_2 \to -5R_2$$ and then $$R_1 \to R_1 + 5R_4$$ to $$A$$ (note the order). 1. $$P= \left[\begin{array}{cccc} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$B= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & -5 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$PBA$$ #### Example 5 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_3 \to R_3 + 4R_1$$. 2. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_1 \to 5R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_3 \to R_3 + 4R_1$$ and then $$R_1 \to 5R_1$$ to $$A$$ (note the order). 1. $$C= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 4 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$Q= \left[\begin{array}{cccc} 5 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$QCA$$ #### Example 6 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$N$$ that may be used to perform the row operation $$R_3 \to R_3 + -4R_2$$. 2. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_1 \to 5R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_1 \to 5R_1$$ and then $$R_3 \to R_3 + -4R_2$$ to $$A$$ (note the order). 1. $$N= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & -4 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$C= \left[\begin{array}{cccc} 5 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$NCA$$ #### Example 7 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_4 \leftrightarrow R_1$$. 2. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_2 \to 5R_2$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_4 \leftrightarrow R_1$$ and then $$R_2 \to 5R_2$$ to $$A$$ (note the order). 1. $$C= \left[\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right]$$ 2. $$Q= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$QCA$$ #### Example 8 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_4 \leftrightarrow R_1$$. 2. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_4 \to 2R_4$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_4 \leftrightarrow R_1$$ and then $$R_4 \to 2R_4$$ to $$A$$ (note the order). 1. $$Q= \left[\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right]$$ 2. $$C= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{array}\right]$$ 3. $$CQA$$ #### Example 9 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_2 \to R_2 + 4R_4$$. 2. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_4 \to -5R_4$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_4 \to -5R_4$$ and then $$R_2 \to R_2 + 4R_4$$ to $$A$$ (note the order). 1. $$Q= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$C= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -5 \end{array}\right]$$ 3. $$QCA$$ #### Example 10 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_1 \to R_1 + 3R_2$$. 2. Give a $$4 \times 4$$ matrix $$B$$ that may be used to perform the row operation $$R_3 \to 3R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_3 \to 3R_3$$ and then $$R_1 \to R_1 + 3R_2$$ to $$A$$ (note the order). 1. $$P= \left[\begin{array}{cccc} 1 & 3 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$B= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$PBA$$ #### Example 11 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$B$$ that may be used to perform the row operation $$R_1 \to 4R_1$$. 2. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_2 \leftrightarrow R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_1 \to 4R_1$$ and then $$R_2 \leftrightarrow R_1$$ to $$A$$ (note the order). 1. $$B= \left[\begin{array}{cccc} 4 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$C= \left[\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$CBA$$ #### Example 12 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$B$$ that may be used to perform the row operation $$R_4 \to R_4 + -3R_2$$. 2. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_4 \to 4R_4$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_4 \to R_4 + -3R_2$$ and then $$R_4 \to 4R_4$$ to $$A$$ (note the order). 1. $$B= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -3 & 0 & 1 \end{array}\right]$$ 2. $$P= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 4 \end{array}\right]$$ 3. $$PBA$$ #### Example 13 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$B$$ that may be used to perform the row operation $$R_1 \to R_1 + -5R_3$$. 2. Give a $$4 \times 4$$ matrix $$M$$ that may be used to perform the row operation $$R_1 \to -3R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_1 \to -3R_1$$ and then $$R_1 \to R_1 + -5R_3$$ to $$A$$ (note the order). 1. $$B= \left[\begin{array}{cccc} 1 & 0 & -5 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$M= \left[\begin{array}{cccc} -3 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$BMA$$ #### Example 14 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_3 \leftrightarrow R_4$$. 2. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_3 \to 4R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_3 \leftrightarrow R_4$$ and then $$R_3 \to 4R_3$$ to $$A$$ (note the order). 1. $$Q= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}\right]$$ 2. $$C= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$CQA$$ #### Example 15 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$M$$ that may be used to perform the row operation $$R_3 \to R_3 + -4R_2$$. 2. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_3 \to 5R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_3 \to 5R_3$$ and then $$R_3 \to R_3 + -4R_2$$ to $$A$$ (note the order). 1. $$M= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & -4 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$P= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$MPA$$ #### Example 16 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_4 \leftrightarrow R_3$$. 2. Give a $$4 \times 4$$ matrix $$N$$ that may be used to perform the row operation $$R_2 \to R_2 + -4R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_2 \to R_2 + -4R_1$$ and then $$R_4 \leftrightarrow R_3$$ to $$A$$ (note the order). 1. $$Q= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}\right]$$ 2. $$N= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ -4 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$QNA$$ #### Example 17 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_4 \leftrightarrow R_3$$. 2. Give a $$4 \times 4$$ matrix $$B$$ that may be used to perform the row operation $$R_4 \to R_4 + -2R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_4 \leftrightarrow R_3$$ and then $$R_4 \to R_4 + -2R_3$$ to $$A$$ (note the order). 1. $$P= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}\right]$$ 2. $$B= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & -2 & 1 \end{array}\right]$$ 3. $$BPA$$ #### Example 18 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_4 \to R_4 + 2R_3$$. 2. Give a $$4 \times 4$$ matrix $$N$$ that may be used to perform the row operation $$R_2 \leftrightarrow R_4$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_4 \to R_4 + 2R_3$$ and then $$R_2 \leftrightarrow R_4$$ to $$A$$ (note the order). 1. $$Q= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 2 & 1 \end{array}\right]$$ 2. $$N= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right]$$ 3. $$NQA$$ #### Example 19 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$M$$ that may be used to perform the row operation $$R_2 \to R_2 + 4R_3$$. 2. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_3 \to 5R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_2 \to R_2 + 4R_3$$ and then $$R_3 \to 5R_3$$ to $$A$$ (note the order). 1. $$M= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 4 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$Q= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$QMA$$ #### Example 20 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_4 \to R_4 + 4R_1$$. 2. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_4 \leftrightarrow R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_4 \to R_4 + 4R_1$$ and then $$R_4 \leftrightarrow R_1$$ to $$A$$ (note the order). 1. $$P= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 4 & 0 & 0 & 1 \end{array}\right]$$ 2. $$Q= \left[\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right]$$ 3. $$QPA$$ #### Example 21 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$M$$ that may be used to perform the row operation $$R_1 \to R_1 + 5R_2$$. 2. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_2 \leftrightarrow R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_2 \leftrightarrow R_1$$ and then $$R_1 \to R_1 + 5R_2$$ to $$A$$ (note the order). 1. $$M= \left[\begin{array}{cccc} 1 & 5 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$Q= \left[\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$MQA$$ #### Example 22 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_1 \to 2R_1$$. 2. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_3 \leftrightarrow R_2$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_1 \to 2R_1$$ and then $$R_3 \leftrightarrow R_2$$ to $$A$$ (note the order). 1. $$Q= \left[\begin{array}{cccc} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$P= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$PQA$$ #### Example 23 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_4 \to 2R_4$$. 2. Give a $$4 \times 4$$ matrix $$N$$ that may be used to perform the row operation $$R_4 \leftrightarrow R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_4 \to 2R_4$$ and then $$R_4 \leftrightarrow R_1$$ to $$A$$ (note the order). 1. $$Q= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{array}\right]$$ 2. $$N= \left[\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right]$$ 3. $$NQA$$ #### Example 24 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$N$$ that may be used to perform the row operation $$R_2 \to R_2 + 3R_3$$. 2. Give a $$4 \times 4$$ matrix $$B$$ that may be used to perform the row operation $$R_2 \leftrightarrow R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_2 \to R_2 + 3R_3$$ and then $$R_2 \leftrightarrow R_1$$ to $$A$$ (note the order). 1. $$N= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$B= \left[\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$BNA$$ #### Example 25 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_3 \to R_3 + 4R_2$$. 2. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_4 \to 4R_4$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_4 \to 4R_4$$ and then $$R_3 \to R_3 + 4R_2$$ to $$A$$ (note the order). 1. $$P= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 4 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$Q= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 4 \end{array}\right]$$ 3. $$PQA$$ #### Example 26 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_3 \to -3R_3$$. 2. Give a $$4 \times 4$$ matrix $$N$$ that may be used to perform the row operation $$R_4 \leftrightarrow R_2$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_3 \to -3R_3$$ and then $$R_4 \leftrightarrow R_2$$ to $$A$$ (note the order). 1. $$C= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -3 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$N= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right]$$ 3. $$NCA$$ #### Example 27 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$M$$ that may be used to perform the row operation $$R_2 \leftrightarrow R_1$$. 2. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_2 \to R_2 + -4R_4$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_2 \to R_2 + -4R_4$$ and then $$R_2 \leftrightarrow R_1$$ to $$A$$ (note the order). 1. $$M= \left[\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$C= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$MCA$$ #### Example 28 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$N$$ that may be used to perform the row operation $$R_4 \to R_4 + -3R_3$$. 2. Give a $$4 \times 4$$ matrix $$M$$ that may be used to perform the row operation $$R_1 \to 4R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_1 \to 4R_1$$ and then $$R_4 \to R_4 + -3R_3$$ to $$A$$ (note the order). 1. $$N= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & -3 & 1 \end{array}\right]$$ 2. $$M= \left[\begin{array}{cccc} 4 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$NMA$$ #### Example 29 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$B$$ that may be used to perform the row operation $$R_1 \to 5R_1$$. 2. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_4 \to R_4 + -5R_2$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_4 \to R_4 + -5R_2$$ and then $$R_1 \to 5R_1$$ to $$A$$ (note the order). 1. $$B= \left[\begin{array}{cccc} 5 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$C= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -5 & 0 & 1 \end{array}\right]$$ 3. $$BCA$$ #### Example 30 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_4 \to R_4 + -2R_3$$. 2. Give a $$4 \times 4$$ matrix $$M$$ that may be used to perform the row operation $$R_2 \to 3R_2$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_4 \to R_4 + -2R_3$$ and then $$R_2 \to 3R_2$$ to $$A$$ (note the order). 1. $$Q= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & -2 & 1 \end{array}\right]$$ 2. $$M= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$MQA$$ #### Example 31 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_3 \leftrightarrow R_1$$. 2. Give a $$4 \times 4$$ matrix $$M$$ that may be used to perform the row operation $$R_1 \to -3R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_3 \leftrightarrow R_1$$ and then $$R_1 \to -3R_1$$ to $$A$$ (note the order). 1. $$C= \left[\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$M= \left[\begin{array}{cccc} -3 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$MCA$$ #### Example 32 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_4 \to R_4 + -3R_1$$. 2. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_3 \to 5R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_3 \to 5R_3$$ and then $$R_4 \to R_4 + -3R_1$$ to $$A$$ (note the order). 1. $$P= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -3 & 0 & 0 & 1 \end{array}\right]$$ 2. $$C= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$PCA$$ #### Example 33 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_3 \leftrightarrow R_1$$. 2. Give a $$4 \times 4$$ matrix $$B$$ that may be used to perform the row operation $$R_4 \to 5R_4$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_4 \to 5R_4$$ and then $$R_3 \leftrightarrow R_1$$ to $$A$$ (note the order). 1. $$Q= \left[\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$B= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 5 \end{array}\right]$$ 3. $$QBA$$ #### Example 34 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_1 \to 5R_1$$. 2. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_2 \to R_2 + -5R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_1 \to 5R_1$$ and then $$R_2 \to R_2 + -5R_3$$ to $$A$$ (note the order). 1. $$Q= \left[\begin{array}{cccc} 5 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$P= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & -5 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$PQA$$ #### Example 35 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$B$$ that may be used to perform the row operation $$R_2 \to R_2 + 5R_3$$. 2. Give a $$4 \times 4$$ matrix $$N$$ that may be used to perform the row operation $$R_4 \leftrightarrow R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_2 \to R_2 + 5R_3$$ and then $$R_4 \leftrightarrow R_1$$ to $$A$$ (note the order). 1. $$B= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$N= \left[\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right]$$ 3. $$NBA$$ #### Example 36 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_4 \leftrightarrow R_1$$. 2. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_3 \to 4R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_4 \leftrightarrow R_1$$ and then $$R_3 \to 4R_3$$ to $$A$$ (note the order). 1. $$C= \left[\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right]$$ 2. $$Q= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$QCA$$ #### Example 37 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_1 \leftrightarrow R_4$$. 2. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_4 \to -3R_4$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_4 \to -3R_4$$ and then $$R_1 \leftrightarrow R_4$$ to $$A$$ (note the order). 1. $$C= \left[\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right]$$ 2. $$P= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -3 \end{array}\right]$$ 3. $$CPA$$ #### Example 38 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$B$$ that may be used to perform the row operation $$R_1 \to -2R_1$$. 2. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_4 \leftrightarrow R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_1 \to -2R_1$$ and then $$R_4 \leftrightarrow R_3$$ to $$A$$ (note the order). 1. $$B= \left[\begin{array}{cccc} -2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$C= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}\right]$$ 3. $$CBA$$ #### Example 39 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_4 \to R_4 + -2R_3$$. 2. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_1 \to 2R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_1 \to 2R_1$$ and then $$R_4 \to R_4 + -2R_3$$ to $$A$$ (note the order). 1. $$C= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & -2 & 1 \end{array}\right]$$ 2. $$Q= \left[\begin{array}{cccc} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$CQA$$ #### Example 40 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_1 \to R_1 + 3R_2$$. 2. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_3 \to 5R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_1 \to R_1 + 3R_2$$ and then $$R_3 \to 5R_3$$ to $$A$$ (note the order). 1. $$Q= \left[\begin{array}{cccc} 1 & 3 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$P= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$PQA$$ #### Example 41 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_3 \to R_3 + -4R_4$$. 2. Give a $$4 \times 4$$ matrix $$M$$ that may be used to perform the row operation $$R_2 \leftrightarrow R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_3 \to R_3 + -4R_4$$ and then $$R_2 \leftrightarrow R_3$$ to $$A$$ (note the order). 1. $$Q= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -4 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$M= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$MQA$$ #### Example 42 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$M$$ that may be used to perform the row operation $$R_2 \leftrightarrow R_3$$. 2. Give a $$4 \times 4$$ matrix $$N$$ that may be used to perform the row operation $$R_2 \to 3R_2$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_2 \to 3R_2$$ and then $$R_2 \leftrightarrow R_3$$ to $$A$$ (note the order). 1. $$M= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$N= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$MNA$$ #### Example 43 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$C$$ that may be used to perform the row operation $$R_3 \leftrightarrow R_1$$. 2. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_4 \to R_4 + -3R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_3 \leftrightarrow R_1$$ and then $$R_4 \to R_4 + -3R_3$$ to $$A$$ (note the order). 1. $$C= \left[\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$Q= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & -3 & 1 \end{array}\right]$$ 3. $$QCA$$ #### Example 44 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$B$$ that may be used to perform the row operation $$R_3 \to -4R_3$$. 2. Give a $$4 \times 4$$ matrix $$M$$ that may be used to perform the row operation $$R_2 \leftrightarrow R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_3 \to -4R_3$$ and then $$R_2 \leftrightarrow R_3$$ to $$A$$ (note the order). 1. $$B= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -4 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$M= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$MBA$$ #### Example 45 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$M$$ that may be used to perform the row operation $$R_2 \to R_2 + -2R_4$$. 2. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_3 \leftrightarrow R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_2 \to R_2 + -2R_4$$ and then $$R_3 \leftrightarrow R_1$$ to $$A$$ (note the order). 1. $$M= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$P= \left[\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$PMA$$ #### Example 46 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$B$$ that may be used to perform the row operation $$R_4 \to R_4 + 3R_3$$. 2. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_4 \to 3R_4$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_4 \to 3R_4$$ and then $$R_4 \to R_4 + 3R_3$$ to $$A$$ (note the order). 1. $$B= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 3 & 1 \end{array}\right]$$ 2. $$Q= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 3 \end{array}\right]$$ 3. $$BQA$$ #### Example 47 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_2 \to R_2 + -4R_4$$. 2. Give a $$4 \times 4$$ matrix $$B$$ that may be used to perform the row operation $$R_2 \leftrightarrow R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_2 \leftrightarrow R_3$$ and then $$R_2 \to R_2 + -4R_4$$ to $$A$$ (note the order). 1. $$P= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$B= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$PBA$$ #### Example 48 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$N$$ that may be used to perform the row operation $$R_4 \leftrightarrow R_1$$. 2. Give a $$4 \times 4$$ matrix $$B$$ that may be used to perform the row operation $$R_2 \to R_2 + -2R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_2 \to R_2 + -2R_3$$ and then $$R_4 \leftrightarrow R_1$$ to $$A$$ (note the order). 1. $$N= \left[\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right]$$ 2. $$B= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$NBA$$ #### Example 49 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$Q$$ that may be used to perform the row operation $$R_1 \to R_1 + 4R_4$$. 2. Give a $$4 \times 4$$ matrix $$N$$ that may be used to perform the row operation $$R_2 \leftrightarrow R_1$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_1 \to R_1 + 4R_4$$ and then $$R_2 \leftrightarrow R_1$$ to $$A$$ (note the order). 1. $$Q= \left[\begin{array}{cccc} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$N= \left[\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$NQA$$ #### Example 50 π Let $$A$$ be a $$4 \times 4$$ matrix. 1. Give a $$4 \times 4$$ matrix $$P$$ that may be used to perform the row operation $$R_3 \leftrightarrow R_1$$. 2. Give a $$4 \times 4$$ matrix $$M$$ that may be used to perform the row operation $$R_3 \to 3R_3$$. 3. Use matrix multiplication to describe the matrix obtained by applying $$R_3 \leftrightarrow R_1$$ and then $$R_3 \to 3R_3$$ to $$A$$ (note the order). 1. $$P= \left[\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 2. $$M= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ 3. $$MPA$$
# Digital Roots Help Check Addition and Multiplication (classes 9-29 thru 10-5) The digital root of a number is the number obtained by adding all the digits, then adding the digits of that number, and then continuing until a single-digit number is reached. Digital roots were known to the Roman bishop Hippolytos as early as the third century. It was employed by Twelfth-century Hindu mathematicians as a method of checking answers to multiplication, division, addition and subtraction. For example, the digital root of 65,536 is 7, because 6 + 5 + 5 + 3 + 6 = 25 and 2 + 5 = 7. This addition of the digits in a number is further simplified by first discarding or casting away any digits whose sum is 9. The remainder is set down, in each case as the digital root. For example, the digital root of 972,632 is 2 because we cast out the nines or numbers who's sum is nine like the 7 + 2 and the 6 + 3. The remainder of 2 is the digital root. When casting out nines, if there is a remainder of zero, the digital root is 9. The digital root of 927 is 9 because we cast out the 9 and the 7 + 2 leaving a remainder of zero. USING DIGITAL ROOTS TO VERIFY ADDITION (this is for 1st through 5th graders) Suppose we are adding two numbers: 345 + 789. · The Digital Root of 789 is 6. · The Digital Root of 345 is 3 · Now add 3 + 6 = 9 · This implies that the Digital Root of our final answer obtained after adding 345 + 789 will be 9. · Let’s verify that 345 + 789 = 1134 · The Digital Root of 1134=1+1+3+4=9 This equals the sum obtained at by adding the digital roots of the original two numbers above. USING DIGITAL ROOTS TO VERIFY MULTIPLICATION (this is for 3rd through 5th graders) Suppose we are multiplying two numbers: 12 x 8 · The Digital Root of 12 is 3. · The Digital Root of 8 is 8. · Now multiply 3x8=24; whose digital root is 6. · This implies that the Digital Root of our final answer obtained after multiplying 12 by 8 will be 6. · Let’s verify that 12x8=96 · The Digital Root of 96 is 9+6=15 and 1+5=6 This equals the product obtained at by multiplying the digital roots of the original two numbers above. Fill in the multiplication table in the pdf file. However, instead of filling each box in with the products of the two numbers on the horizontal and vertical axis, use the DIGITAL ROOTS. *now, look for magnificent patterns in the digital roots. USE THE WORKSHEETS IN THE PDF FILE TO PRACTICE USING DIGITAL ROOTS TO VERIFY ADDITION AND MULTIPLICATION AttachmentSize Digital_Roots.pdf262.49 KB
Select Page # Year 4 ### We can divide a two-digit number or a three-digit number by a one-digit number using efficient written methods #### What we are learning: Once we know our tables, long division is not as difficult as it sometimes looks. If the answer to the division is within our tables knowledge we should know the answer, e.g. 81 ÷ 9 is 9 or 56 ÷ 8 is 7. If the calculation takes us beyond the range of our tables knowledge we need an efficient written method to work out the answer. For example, if we have 96 ÷ 4 this is well beyond 10 x 4 = 40 or even 12 x 4 = 48, which we know. We can set the division out like this. We know we are trying to find out how many times 4 will ‘go into’ 96. The first stage is to ask ourselves how many times 4 will go into 96 by using multiples of 10. We know that 4 x 10 = 40, 4 x 20 = 80 and 4 x 30 = 120. Since 120 is too big we know that 4 goes into 96 twenty times, which is 80. We set it out like this: Note that the 2 is in the tens column and represents the 20 that we have multiplied by 4 to get 80. We can now take the 80 away from 96 to see how many we have left to divide like this: We can now ask how many times 4 will go into 16. We know from our tables that 4 x 4 = 16. We put the 4 at the top to make the ‘quotient’ (answer), insert the 16 underneath (as we now that 4 x 4 = 16) and we take this away to see if there is a remainder. Now we can see that 4 (the divisor) goes into 96 (the dividend) exactly 24 times (the quotient) and there is no remainder. We can extend this method to divide a three-digit number by a one-digit number. For example, if we have 767 ÷ 3 we set it out the same way: This time we start by dividing with multiples of 100. We know that 3 x 100 = 300, 3 x 200 = 600 and 3 x 300 = 900, so 3 will go into 767 200 times. We put the 200 into the hundreds column like this and take the 600 away from our dividend. Having divided with multiples of 100 we can now use multiples of 10. We know that 3 x 50 is 150 which works and that3 x 60 is 180 which is too big. We insert the 50 into the tens column and take the 150 away from the dividend that is left like this: Finally we can divide the 17 by 3 and complete the division to leave a remainder of 2 The answer (quotient) is therefore 255 remainder 2 #### Activities you can do at home: First try dividing two-digit numbers by a one-digit number using the examples on the activity sheet provided. Once your child is confident with these, move on to dividing three-digit numbers by a one-digit number. #### Good questions to ask and discuss: Can you explain what the following are: divisor, dividend, quotient, and remainder? Pick a number in the division calculation on the page that you have finished together and ask Where does this number come from? What part does it play in the calculation? This will get your child to explain part of the process to you to ensure s/he understands all parts of it. #### If your child: Can’t explain what s/he is doing and gets confused part way through the process Work on a simpler calculation to re-establish the method. Go back to the beginning of the calculation and talk through each stage together. Long division is one of the calculations with the most stages in it, so it is not surprising that children get confused sometimes. By talking it through you will quickly see whether they understand what they are doing. ##### ACTIVITY SHEETS Activity sheet PDF Activity sheet answers PDF ##### Extension Activity Please use this activity when you think your child understands the unit of work. It will deepen and extend your child’s understanding of this unit.
# 35 9.1 Practice B Geometry Answers ## Introduction Welcome to our comprehensive guide on 9.1 Practice B Geometry answers! In this article, we will provide you with step-by-step solutions and explanations for the problems found in the 9.1 Practice B worksheet. Whether you're a student looking for help with your homework or a teacher seeking additional resources for your students, you've come to the right place. Let's dive in and explore the answers to these geometry problems. ### Problem 1: Finding the Area of a Triangle To start off, let's take a look at problem 1. In this problem, you are given the lengths of two sides of a triangle and the measure of the included angle. The task is to find the area of the triangle. To solve this problem, you can use the formula for the area of a triangle: A = 1/2 * base * height. In this case, the base is one of the given sides, and the height can be found using trigonometry. By plugging in the given values and performing the necessary calculations, you will find the area of the triangle. ### Problem 2: Applying the Pythagorean Theorem The second problem involves the use of the Pythagorean Theorem. You are given the lengths of two sides of a right triangle and need to find the length of the third side. The Pythagorean Theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. By applying this theorem and rearranging the equation, you can solve for the missing side length. ### Problem 3: Calculating the Volume of a Cylinder In problem 3, you are tasked with finding the volume of a cylinder. The formula for the volume of a cylinder is V = π * r^2 * h, where r is the radius of the base and h is the height of the cylinder. By plugging in the given values and performing the necessary calculations, you will determine the volume of the cylinder. ### Problem 4: Solving for Unknown Angles Next, let's move on to problem 4, which involves solving for unknown angles in a triangle. You are given the measures of two angles and need to find the measure of the third angle. Remember that the sum of the measures of the angles in a triangle is always 180 degrees. By subtracting the measures of the known angles from 180, you can find the measure of the unknown angle. ### Problem 5: Finding the Surface Area of a Prism Problem 5 focuses on finding the surface area of a prism. The formula for the surface area of a prism is SA = 2 * base area + lateral area. The base area can be found by calculating the area of the base shape, and the lateral area is the sum of the areas of the rectangular faces. By plugging in the given values and performing the necessary calculations, you will determine the surface area of the prism. ### Problem 6: Applying the Angle Bisector Theorem The sixth problem involves the use of the Angle Bisector Theorem. You are given a triangle with an angle bisector and need to find the lengths of the segments it divides the opposite side into. The Angle Bisector Theorem states that in a triangle, an angle bisector divides the opposite side into segments that are proportional to the lengths of the other two sides. By setting up a proportion and solving for the unknown lengths, you can find the desired segment lengths. ### Problem 7: Determining Similarity of Triangles In problem 7, you are asked to determine the similarity of two triangles. Two triangles are considered similar if their corresponding angles are congruent and their corresponding sides are proportional. By comparing the angles and side lengths of the given triangles, you can determine if they are similar or not. ### Problem 8: Calculating the Circumference of a Circle Problem 8 involves finding the circumference of a circle. The formula for the circumference of a circle is C = 2 * π * r, where r is the radius of the circle. By plugging in the given radius and performing the necessary calculations, you will determine the circumference of the circle. ### Problem 9: Applying the Law of Cosines The ninth problem requires the use of the Law of Cosines. You are given the lengths of two sides and the measure of the included angle in a triangle and need to find the length of the third side. The Law of Cosines states that in a triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of their lengths and the cosine of the included angle. By applying this theorem and rearranging the equation, you can solve for the missing side length. ### Problem 10: Finding the Surface Area of a Cone Finally, let's tackle problem 10, which involves finding the surface area of a cone. The formula for the surface area of a cone is SA = π * r * (r + l), where r is the radius of the base and l is the slant height of the cone. By plugging in the given values and performing the necessary calculations, you will determine the surface area of the cone. ## Conclusion That concludes our guide on 9.1 Practice B Geometry answers. We hope that this article has provided you with the solutions and explanations you were looking for. Remember, practice is key when it comes to mastering geometry, so don't hesitate to work through additional problems and seek further assistance if needed. Good luck with your studies!
# A Step by Step Guide To Solve Simultaneous Equation Sometimes, a pair of simultaneous equations may contain one linear and one quadratic. In order to solve this type of equation, we begin with the linear equation. In the linear equation, the substitution method is used, whereby one of the unknown is made the subject of the formula. The value got is thereafter re-substituted in the other equation which is quadratic, in two unknowns to obtain a simpler quadratic equation in one unknown. The quadratic equation in one unknown is then solved using any method, to obtain one or more values. The value(s) are then substituted again in any of the original two equations to obtain the one or two values for the other unknown. Having been familiar with the required steps for solving simultaneous, linear, and quadratic equation, let’s solve some examples Solve the equation x+y=5—(1) x^2-2y^2=1—(2) Make x the subject of the equation in equation 1 x=5-y—(3) Substitute x = 5y into equation 2 (5-y)^2-y^2=1 Letâ€™s expand the bracket first 5×5-yx5-yx5-yx-y-2y^2=1 25-5y-5y+y^2-2y^2=1 25=10y+y^2-2y^2=1 25=10y-y^2= 1 How? y^2-2y^2=-? y^2 1-2 = -1 Take 25 to the right hand side -10y-y^2=1-25 -10y-y^2=-24 In order to rearrange the equation, multiply each term by -1 -1x-10y-1x-y^2=-1x-24s +10y+y^2=+24 y^2+10y-24=1—(4) Solve by factorization i.ey^2 x-24=-24y^2 Middle term is 10y two factors whose sum gives 10y and product gives -24y^2 Are +12y and-2y We replace 10y with the two factors ie+12y and-2y in equation 4 (y^2+12y)-(2y-24)=0 y(y^ +12)-2(y+12)=0 The factors are (y+12)(y-2)=0 y+12=0 and y-2=0 y=-12 or y=2 Substitute the two values of y into equation 3 to obtain the two values of x x=5-y When y=-12 x=5-(-12)=5+12=17 When y = 2 x=5-2=3 Therefore, y = -12, x = 17 and y = 2, x = 3 ### Example 2: Solve the equation x=3y=2 and x^2+2y^2=3 Simultaneously Solution x=3y=2—(1) x^2+2y^2=3—(2) Make the subject of the equation in equation 1 x-3y=2 x^ =2+3y—(3) Substitute x=2+3y into equation 2 (2+3y)^2+2y^2=3 Let expand the bracket 2×2+2x3y+3yx2+3yx3y+2y^2=3 4+6y+6y+9y^2+2y^2=3 4+12y+11y^2=3 11y^2+12y=-1 11y^2+12y+1=0 Now we have a quadratic equation let solve it by factorization (11y^2+11y)+(y+1)=0 11(y^ +1)+(y+1)=0 (y^ +1)(11y+1)=0 y+1=0 and 11y+1=0 y=-1 and 11y/11=(-1)/11 y=(-1)/11 Substitute y= -1 and y=(-1)/11 into equation 3 x=2+2y When y = -1 x=2+3x-1 x=2-3=-1 When y = -1 x= -1 When y=(-1)/11 x=2+3x (-1)/11 x=2+((-3)/11) x=2+2 (-3)/11 x=11×2/11-3=22/11-3= 19/11 x=1 8/11 x=-1,y=-1 and y=(-1)/(-1),x=1 8/11 ### Example 3: Solve the equation x+2y=2 and x^2+2xy=8 Solution x+2y= 2—(1) x^2+2xy=8—(2) Make x the subject in equation 1 x+2y= 2 x=2-2y—(3) Substitute equation 3 into equation 2 (2-2y)^2+2(2-2y)y= 8 Let expand the bracket 2×2-2yx2-2yx2-2yx-2y 4-4y-4y+4y^2 4-8y+4y^2 Again, 2(2-2y)y 2(2y-2y^2 )=4y-4y^2 4-8y+4y^2+4y-4y^2=8 4-8y+4y+4y^2-4y^2=8 4-4y=8 N/B: 4y^2-4y^2=0 -4y=8-4 -4y/(-4y)=8/(-4) y=-1 Substitute y = -1 into equation 3 x=2-2y x=2-2(-1) x=2-(-1) x=2+2=4 Therefore, x=4,y=1 ### Example 4: Solve the equation2x+y=5 and x^2+y^2=25, simultaneously Solution 2x+y=5—(1) x^2+y^2=25—(2) Make y the subject in equation 1 2x^ +y^ =5 y=5-2x—(3) Substitute equation (3) into equation 2 x^2+(5-2x)^2 = 25 x^2+25-20x+4x^2 = 25 x^2+4x^2+25-20x=25 5x^2-20x=25-25 5x^2-20x=0 Divide through by 5x^2/5-20x/5=0 x^2-4x=0 x^ (x-4)=0 That means x =0 and x-4=0, x = 4 Therefore, x =0 or x= 4 Substitute the values of x into equation 3 y=5-2x When x =0 y=5-2(0)=5-0 y=5 When x = 4 y=5-2×4 y=5-8=-3 Therefore, the solution is (0,5) and (4, -3) From what you have learnt in this topic, you can see it with me that simultaneous linear and the quadratic equation is not difficult to understand if you follow the procedure required. I am optimistic that you can tackle any problem involving one-linear one quadratic simultaneous equation.
# Inscribed Angle Theorems Proof \| Inscribed Angle Theorem Formula ### What is Inscribed Angle Theorems? In Geometry, the inscribed angle is formed in the interior of a circle with two secant lines intersecting on the circle. Or this could be taken as the angle subtended at a specific point on the circle by two other points. In brief, an inscribed angle is defined by two chords of the circle sharing an endpoint. The inscribed angle theorem is related to the measure of an inscribed angle to the central angle subtending over the same arc. The theorem states that an inscribed angle θ in the circle is half of the central angle i.e. 2θ subtends over the same arc on the circle. This is the reason why the angle ## Inscribed Angle Theorem The use inscribed angle is pretty common when you study geometry during your early schools or colleges. One special case of the Theorem is the Thales’ Theorem that states that the angle subtended by a diameter would always be 90-degree that is a right angle. If you are taking the consequence of the theorem then the sum of opposite angles of a cyclic quadrilateral would result in 180-degrees. Conversely, any Quadrilateral for which this condition is true can be inscribed within a circle. This theorem would act as the baseline for many other theorems as frequently used by students with respect to the circle. It will allow the intersection of two chords within a circle and the products of the length of their pieces is almost equal. According to the Inscribed Angle Theorems in the mathematics, the inscribed angle (A) is half the central angle (C)Circle Theorems. It can also be written as A = C / 2 mathematically or C = 2A. this hardly counts in terms of logic but easy to remember. So, you choose your best way to memorize the theorem in your own style. ## Inscribed Angle Theorem Formula Circles are used everywhere around and inscribed angles are the special angles that sits within a circle on the vertex, on the circumference of the circle. This is true that every inscribed angle shares a special relationship with the intercepted arc. The vertex is defined as the common endpoint of two sides of the angle and the sides are named as the chord of a circle. The chord is the line segment who endpoints also sit on the circumference of a circle. One of the endpoints is named as the vertex and another endpoint sits across the circle. The arc formed by the inscribed angle is named as the intercepted arc formed between two chords of angle and intersected by the chords too. The intercepted angle and the intercepted arc always share a special relationship. Here, in the figure below, you can see how to define the inscribed angle, vertex, and the chord in the Geometry. Here, in the example, the intercepted arc is measured as 48-degree and the inscribed angle would be the half of the intercepted arc i.e. 24-degrees.
# matrix multiplication This article gives an overview of the various ways to perform matrix multiplication. ## Ordinary matrix product By far the most important way to multiply matrices is the usual matrix multiplication. It is defined between two matrices only if the number of columns of the first matrix is the same as the number of rows of the second matrix. If A is an m-by-n matrix and B is an n-by-p matrix, then their product is an m-by-p matrix denoted by AB (or sometimes A Â· B). The product is given by for each pair i and j with 1 ≤ im and 1 ≤ jp. The algebraic system of "matrix units" summarises the abstract properties of this kind of multiplication. ### Calculating directly from the definition The picture to the left shows how to calculate the (1,2) element and the (3,3) element of AB if A is a 4Ã—2 matrix, and B is a 2Ã—3 matrix. Elements from each matrix are paired off in the direction of the arrows; each pair is multiplied and the products are added. The location of the resulting number in AB corresponds to the row and column that were considered. ### The coefficients-vectors method This matrix multiplication can also be considered from a slightly different viewpoint : it adds vectors together after being multiplied by different coefficients. If A and B are matrices given by: and then For example: The rows in the matrix on the left are the list of coefficients. The matrix on the right is the list of vectors. In the example, the first row is [1 0 2], and thus we take 1 times the first vector, 0 times the second vector, and 2 times the third vector. The equation can be simplified further by using outer products: The terms of this sum are matrices of the same shape, each describing the effect of one column of A and one row of B on the result. The columns of A can be seen as a coordinate system of the transform, i.e. given a vector x we have where are coordinates along the "axes". The terms are like , except that contains the ith coordinate for each column vector of B, each of which is transformed independently in parallel. The example revisited: The vectors and have been transformed to and in parallel. One could also transform them one by one with the same steps: ### Vector-lists method The ordinary matrix product can be thought of as a dot product of a column-list of vectors and a row-list of vectors. If A and B are matrices given by: and where A1 is the vector of all elements of the form a1,x A2 is the vector of all elements of the form a2,x etc, and B1 is the vector of all elements of the form bx,1 B2 is the vector of all elements of the form bx,2 etc, then ### Properties Matrix multiplication is not commutative (that is, ABBA), except in special cases. It is easy to see why: you cannot expect to switch the proportions with the vectors and get the same result. It is also easy to see how the order of the factors determines the result when one knows that the number of columns in the proportions matrix has to be the same as the number of rows in the vectors matrix: they have to represent the same number of vectors. Although matrix multiplication is not commutative, the determinants of AB and BA are always equal (if A and B are square matrices of the same size). See the article on determinants for an explanation. This notion of multiplication is important because if A and B are interpreted as linear transformations (which is almost universally done), then the matrix product AB corresponds to the composition of the two linear transformations, with B being applied first. Additionally, all notions of matrix multiplication described here share a set of common properties described below. ### Algorithms The complexity of matrix multiplication, if carried out naively, is O(nÂ³), but more efficient algorithms do exist. Strassen's algorithm, devised by Volker Strassen in 1969 and often referred to as "fast matrix multiplication", is based on a clever way of multiplying two 2 × 2 matrices which requires only 7 multiplications (instead of the usual 8). Applying this trick recursively gives an algorithm with a cost of . In practice, though, it is rarely used since it is awkward to implement and it lacks numerical stability. The constant factor implied in the big O notation is about 4.695. The algorithm with the lowest known exponent, which was presented by Don Coppersmith and Shmuel Winograd in 1990, has an asymptotic complexity of O(n2.376). It is similar to Strassen's algorithm: a clever way is devised for multiplying two k × k matrices with less than kÂ³ multiplications, and this technique is applied recursively. It improves on the constant factor in Strassen's algorithm, reducing it to 4.537. However, the constant term implied in the O(n2.376) result is so large that the Coppersmith–Winograd algorithm is only worthwhile for matrices that are too big to handle on present-day computers. Since any algorithm for multiplying two n × n matrices has to process all 2 × nÂ² entries, there is an asymptotic lower bound of Ω(nÂ²) operations. Raz (2002) proves a lower bound of for bounded coefficient arithmetic circuits over the real or complex numbers. Cohn et al. (2003, 2005) put methods such as the Strassen and Coppersmith–Winograd algorithms in an entirely different, group-theoretic context. They show that if families of wreath products of Abelian with symmetric groups satisfying certain conditions exists, matrix multiplication algorithms with essential quadratic complexity exist. Most researchers believe that this is indeed the case (Robinson, 2005). ## Scalar multiplication The scalar multiplication of a matrix A = (aij) and a scalar r gives a product rA of the same size as A. The entries of rA are given by If we are concerned with matrices over a ring, then the above multiplication is sometimes called the left multiplication while the right multiplication is defined to be When the underlying ring is commutative, for example, the real or complex number field, the two multiplications are the same. However, if the ring is not commutative, such as the quaternions, they may be different. For example For two matrices of the same dimensions, we have the Hadamard product, also known as the entrywise product and the Schur product. It can be generalized to hold not only for matrices but also for operators. The Hadamard product of two m-by-n matrices A and B, denoted by AB, is an m-by-n matrix given by (AB)ij = aijbij. For instance . Note that the Hadamard product is a submatrix of the Kronecker product (see below). The Hadamard product is studied by matrix theorists, and it appears in lossy compression algorithms such as JPEG, but it is virtually untouched by linear algebraists. It is discussed in (Horn & Johnson, 1994, Ch. 5). ## Kronecker product Main article: Kronecker product. For any two arbitrary matrices A and B, we have the direct product or Kronecker product A B defined as Note that if A is m-by-n and B is p-by-r then A B is an mp-by-nr matrix. Again this multiplication is not commutative. For example . If A and B represent linear transformations V1W1 and V2W2, respectively, then A B represents the tensor product of the two maps, V1 V2W1 W2. ## Common properties All three notions of matrix multiplication are associative: and distributive: and . and compatible with scalar multiplication: Note that these three separate couples of expressions will be equal to each other only if the multiplication and addition on the scalar field are commutative, i.e. the scalar field is a commutative ring. See Scalar multiplication above for a counter-example such as the scalar field of quaternions. ## Frobenius inner product The Frobenius inner product, sometimes denoted is the component-wise inner product of two matrices as though they are vectors. In other words, the sum of the entries of the Hadamard product. That is, This inner product induces the Frobenius norm. ## References • Henry Cohn, Robert Kleinberg, Balazs Szegedy, and Chris Umans. Group-theoretic Algorithms for Matrix Multiplication. arXiv:math.GR/0511460. Proceedings of the 46th Annual Symposium on Foundations of Computer Science, 23-25 October 2005, Pittsburgh, PA, IEEE Computer Society, pp. 379–388. • Henry Cohn, Chris Umans. A Group-theoretic Approach to Fast Matrix Multiplication. arXiv:math.GR/0307321. Proceedings of the 44th Annual IEEE Symposium on Foundations of Computer Science, 11-14 October 2003, Cambridge, MA, IEEE Computer Society, pp. 438–449. • Coppersmith, D., Winograd S., Matrix multiplication via arithmetic progressions, J. Symbolic Comput. 9, p. 251-280, 1990. • Horn, Roger; Johnson, Charles: "Topics in Matrix Analysis", Cambridge, 1994. • R. Raz. On the complexity of matrix product. In Proceedings of the thirty-fourth annual ACM symposium on Theory of computing. ACM Press, 2002. • Robinson, Sara, Toward an Optimal Algorithm for Matrix Multiplication, SIAM News 38(9), November 2005. PDF • Strassen, Volker, Gaussian Elimination is not Optimal, Numer. Math. 13, p. 354-356, 1969. matrix (plural matrices) is a rectangular table of elements (or entries), which may be numbers or, more generally, any abstract quantities that can be added and multiplied. In mathematics, a matrix unit is an idealisation of the concept of a matrix, with a focus on the algebraic properties of matrix multiplication. The topic is comparatively obscure within linear algebra, because it entirely ignores the numeric properties of matrices; it is mostly spatial vector, or simply vector, is a concept characterized by a magnitude and a direction. A vector can be thought of as an arrow in Euclidean space, drawn from an initial point A pointing to a terminal point B. coefficient is a constant multiplicative factor of a certain object. For example, the coefficient in 9x2 is 9. The object can be such things as a variable, a vector, a function, etc. Outer product typically refers to the tensor product or to operations with similar cardinality such as exterior product. The cardinality of these operations is that of cartesian products. The name contrasts with the inner product, which is the product in the opposite order. dot product, also known as the scalar product, is an operation which takes two vectors over the real numbers R and returns a real-valued scalar quantity. It is the standard inner product of the Euclidean space. In linear algebra, a column vector is an m × 1 matrix, i.e. a matrix consisting of a single column of elements. The transpose of a column vector is a row vector and vice versa. In linear algebra, a row vector is a 1 × n matrix, that is, a matrix consisting of a single row: The transpose of a row vector is a column vector. Commutativity is a widely used mathematical term that refers to the ability to change the order of something without changing the end result. It is a fundamental property in most branches of mathematics and many proofs depend on it. In algebra, a determinant is a function depending on n that associates a scalar, det(A), to every nÃ—n square matrix A. The fundamental geometric meaning of a determinant is as the scale factor for volume when A In algebra, a determinant is a function depending on n that associates a scalar, det(A), to every nÃ—n square matrix A. The fundamental geometric meaning of a determinant is as the scale factor for volume when A In mathematics, a linear map (also called a linear transformation or linear operator) is a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. As a branch of the theory of computation in computer science, computational complexity theory investigates the problems related to the amounts of resources required for the execution of algorithms (e.g. In computational complexity theory, big O notation is often used to describe how the size of the input data affects an algorithm's usage of computational resources (usually running time or memory). In the mathematical discipline of linear algebra, the Strassen algorithm, named after Volker Strassen, is an algorithm used for matrix multiplication. It is asymptotically faster than the standard matrix multiplication algorithm, but slower than the fastest known algorithm, and is Volker Strassen is a German mathematician. He received in 2003, with three others, the Paris Kanellakis Award of the ACM, for the Solovay-Strassen primality test. In 1971 Strassen published a paper together with Arnold SchÃ¶nhage on asymptotically-fast integer multiplication; 19th century - 20th century - 21st century 1930s 1940s 1950s - 1960s - 1970s 1980s 1990s 1966 1967 1968 - 1969 - 1970 1971 1972 Also: :1969 (number) : In the mathematical subfield of numerical analysis, numerical stability is a desirable property of numerical algorithms. The precise definition of stability depends on the context, but it is related to the accuracy of the algorithm. In computational complexity theory, big O notation is often used to describe how the size of the input data affects an algorithm's usage of computational resources (usually running time or memory). Don Coppersmith is a cryptographer and mathematician. He was involved in the design of the Data Encryption Standard block cipher at IBM, particularly the design of the S-boxes, strengthening them against differential cryptanalysis. Shmuel Winograd is a computer scientist, noted for his work on fast algorithms for arithmetic, and in particular for the algorithm known as the Coppersmith-Winograd algorithm. From 1970-1974 and 1980-1994 he was the director of the Mathematical Science Department at IBM. 20th century - 21st century 1960s 1970s 1980s - 1990s - 2000s 2010s 2020s 1987 1988 1989 - 1990 - 1991 1992 1993 Year 1990 (MCMXC) was a common year starting on Monday (link displays the 1990 Gregorian calendar). Group theory is the mathematical study of symmetry, as embodied in the structures known as groups. These are sets with a closed binary operation satisfying the following three properties: 1. The operation must be associative. 2. There must be an identity element. In mathematics, the wreath product of group theory is a specialized product of two groups, based on a semidirect product. Wreath products are an important tool in the classification of permutation groups and also provide a way of constructing interesting examples of groups. In mathematics, a ring is an algebraic structure in which addition and multiplication are defined and have properties listed below. The branch of abstract algebra which studies rings is called ring theory. Commutativity is a widely used mathematical term that refers to the ability to change the order of something without changing the end result. It is a fundamental property in most branches of mathematics and many proofs depend on it. quaternions are a non-commutative extension of complex numbers. They were first described by the Irish mathematician, Sir William Rowan Hamilton, in 1843 and applied to mechanics in three-dimensional space.
# Finding Z – scores & Normal Distribution Using the Standard Normal Distribution Week 9 Chapter’s 5.1, 5.2, 5.3. ## Presentation on theme: "Finding Z – scores & Normal Distribution Using the Standard Normal Distribution Week 9 Chapter’s 5.1, 5.2, 5.3."— Presentation transcript: Finding Z – scores & Normal Distribution Using the Standard Normal Distribution Week 9 Chapter’s 5.1, 5.2, 5.3 Normal Distribution Normal Distribution - is a very important statistical data distribution pattern occurring in many natural phenomena, such as height, blood pressure, grades, IQ, baby birth weights, etc. Normal Curve - when graphing the normal distribution as a histogram, it will create a bell- shaped curve known as a normal curve. It is based on Probability! You’ll see! Normal Distribution Curve: What is this curve all about? The shape of the curve is bell-shaped The graph falls off evenly on either side of the mean. (symmetrical) 50% of the distribution lies on the left of the mean 50% lies to the right of the mean. (above) The spread of the normal distribution is controlled by the standard deviation. The mean and the median are the same in a normal distribution. (and even the mode) Features of Standard Normal Curve Mean is the center 68% of the area is within one S.D. 95% of area is within two S.D.’s 99% of area is within 3 S.D.’s As each tail increases/decreases, the graph approaches zero (y axis), but never equals zero on each end. For each of these problems we will need pull-out table IV in the back of text What is a Z – Score? Z-score’s allow us a method of converting, proportionally, a study sample to the whole population. Z-Score’s are the exact number of standard deviations that the given value is away from the mean of a NORMAL CURVE. Table IV always solves for the area to the left of the Z-Score! Finding the area to the left of a Z (Ex. 1) – Find the area under the standard normal curve that lies to the left of Z=1.34. Finding the area to the right of a Z (Ex. 2) - Find the area under the standard normal curve that lies to the right of Z = -1.07. Finding the area in-between two Z’s (Ex. 3) - Find the area under the standard normal curve that lies between Z=-2.04 and Z=1.25. Formula: x = data value u = population mean Practice examples: For each of the following examples, Look for the words "normally distributed" in a question before using Table IV to solve them. Don’t forget - Table IV always solves for the area to the left of the Z-Score! Finding Probabilities The shaded area under the curve is equal to the probability of the specific event occurring. Ex (4) - A shoe manufacturer collected data regarding men's shoe sizes and found that the distribution of sizes exactly fits a normal curve. If the mean shoe size is 11 and the standard deviation is 1.5. (a)What is the probability of randomly selecting a man with a shoe size smaller than 9.5? (b)If I surveyed 40 men, how many would be expected to wear smaller than 9.5? How did we get that answer: -1.00 is a Z-score (# of S.D.’s from the mean) that refers to the area to the left of that position. Find it in Table IV. -1.00 =.1587 We want the area to the left of that curve, so, this is the answer. Table IV gives us the answer for area to the left of the curve. (b).1587(40) = 6.3 = 6 This is how many SD’s from the mean Ex (5) – Gas mileage of vehicles follows a normal curve. A Ford Escape claims to get 25 mpg highway, with a standard deviation of 1.6 mpg. A Ford Escape is selected at random. (a) What is the probability that it will get more than 28 mpg? (b) If I sampled 250 Ford Escapes, how many would I expect to get more than 28 mpg? How did we get that answer: 1.875 is a Z-score (# of S.D.’s from the mean) that refers to the area to the left of that position. 1.875 =.9696 We want the area to the right of that curve, thus 1-.9696 =.0304 Ex (6) – This past week gas prices followed a normal distribution curve and averaged \$3.73 per gallon, with a standard deviation of 3 cents. What percentage of gas stations charge between \$ 3.68 and \$ 3.77? Ex (7) – This week gas prices followed a normal distribution curve and averaged \$3.71 per gallon, with a standard deviation of 3 cents. (a)What percent of stations charge at least \$3.77? (b)What percent will charge less than \$3.71? (c)What percent will charge less than \$3.69? (d)What percent will charge in-between \$3.67and \$3.75 per gallon? (e)If I sampled 30 gas stations, how many would charge between \$3.67and \$3.75 per gallon? NOW – Going back from probabilities to Z-Scores: Chapter 5.3 – Finding Z-scores from probabilities – Transforming a Z-score to an X-value Look up.9406 on Table 4 What Z-score corresponds to this area? Finding Z-score’s of area to the left (Ex. 8) (a)Find the Z-score so that the area to the left is 10.75% (b) Find the Z-score that represents the 75 th percentile? (c) Find the Z-score so that the area to the left is.88 (d) Find the Z-score so that the area to the left is.9880 Transforming a Z-score to an x-value Try: 90 th percentile Look for the three ingredients to solve for x: Population mean, standard deviation, and you will need the Z-score that corresponds to the given percent (or probability) (Ex. 9) – The national average on the math portion of the SAT is a 510 with a standard deviation of 130. SAT scores follow a normal curve. (a) What score represents the 90 th percentile? (b) What score will place you at the 35 th percentile? Finding Z-score’s of area to the left Finding Z-score’s of area to the Right (Ex. 10) – Find the Z score so that the area under the standard normal curve to the right is.7881 Find the Z score of area to the Right (Ex. 11) – A batch of Northern Pike at a local fish hatchery has a mean length of a 8 inches just as they are released to the wild. Their lengths are normally distributed with a standard deviation of 1.25 inches. What is the shortest length that could still be considered part of the top 15% of lengths? Find the Z score of area in-between two Z’s (Ex. 12) – Find the Z score that divides the middle 90% of the area under the standard normal curve. Critical Two-tailed Z value: - Used to find the remaining percent on the outside of the area under the curve. -90% is equal to 1-.90=.10 -.10/2 =.05 = 5% M&M Packaging Ex (13) – A bag of M&M’s contains 40 candies with a Standard deviation of 3 candies. The packaging machine is considered un-calibrated if it packages bags outside of 80%, centered about the mean. What interval must the candies be between for sale? The Central Limit Theorem: As the sample sizes increases, the sampling distribution becomes more accurate in representation of the entire population. Thus, As additional observations are added to the sample, the difference of the Sample mean and the population mean approaches ZERO. ( No difference) Download ppt "Finding Z – scores & Normal Distribution Using the Standard Normal Distribution Week 9 Chapter’s 5.1, 5.2, 5.3." Similar presentations
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 2.3: Limits $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ In the previous two sections we computed some quantities of interest (slope, velocity) by seeing that some expression "goes to'' or "approaches'' or "gets really close to'' a particular value. In the examples we saw, this idea may have been clear enough, but it is too fuzzy to rely on in more difficult circumstances. In this section we will see how to make the idea more precise. There is an important feature of the examples we have seen. Consider again the formula ${-19.6\Delta x-4.9\Delta x^2\over \Delta x}.$ We wanted to know what happens to this fraction as "$$\Delta x$$ goes to zero.'' Because we were able to simplify the fraction, it was easy to see the answer, but it was not quite as simple as "substituting zero for $$\Delta x$$,'' as that would give ${-19.6\cdot 0 - 4.9\cdot 0\over 0},$ which is meaningless. The quantity we are really interested in does not make sense "at zero,'' and this is why the answer to the original problem (finding a velocity or a slope) was not immediately obvious. In other words, we are generally going to want to figure out what a quantity "approaches'' in situations where we can't merely plug in a value. If you would like to think about a hard example (which we will analyze later) consider what happens to $$(\sin x)/x$$ as $$x$$ approaches zero. Example $$\PageIndex{1}$$ Does $$\sqrt{x}$$ approach 1.41 as $$x$$ approaches 2? Solution In this case it is possible to compute the actual value $$\sqrt{2}$$ to a high precision to answer the question. But since in general we won't be able to do that, let's not. We might start by computing $$\sqrt{x}$$ for values of $$x$$ close to 2, as we did in the previous sections. Here are some values: $$\sqrt{2.05} = 1.431782106$$, $$\sqrt{2.04} = 1.428285686$$, $$\sqrt{2.03} = 1.424780685$$, $$\sqrt{2.02} = 1.421267040$$, $$\sqrt{2.01} = 1.417744688$$, $$\sqrt{2.005} = 1.415980226$$, $$\sqrt{2.004} = 1.415627070$$, $$\sqrt{2.003} = 1.415273825$$, $$\sqrt{2.002} = 1.414920492$$, $$\sqrt{2.001} = 1.414567072$$. So it looks at least possible that indeed these values "approach'' 1.41---already $$\sqrt{2.001}$$ is quite close. If we continue this process, however, at some point we will appear to "stall.'' In fact, $$\sqrt{2}=1.414213562\ldots$$, so we will never even get as far as 1.4142, no matter how long we continue the sequence. So in a fuzzy, everyday sort of sense, it is true that $$\sqrt{x}$$ "gets close to'' 1.41, but it does not "approach'' 1.41 in the sense we want. To compute an exact slope or an exact velocity, what we want to know is that a given quantity becomes "arbitrarily close'' to a fixed value, meaning that the first quantity can be made "as close as we like'' to the fixed value. Consider again the quantities ${-19.6\Delta x-4.9\Delta x^2\over \Delta x}=-19.6-4.9\Delta x.$ These two quantities are equal as long as $$\Delta x$$ is not zero; if $$\Delta x$$ is zero, the left hand quantity is meaningless, while the right hand one is $$-19.6$$. Can we say more than we did before about why the right hand side "approaches'' $$-19.6$$, in the desired sense? Can we really make it "as close as we want'' to $$-19.6$$? Let's try a test case. Can we make $$-19.6-4.9\Delta x$$ within one millionth $$0.000001$$ of $$-19.6$$? The values within a millionth of $$-19.6$$ are those in the interval $$(-19.600001,-19.599999)$$. As $$\Delta x$$ approaches zero, does $$-19.6-4.9\Delta x$$ eventually reside inside this interval? If $$\Delta x$$ is positive, this would require that $$-19.6-4.9\Delta x> -19.600001$$. This is something we can manipulate with a little algebra: \eqalign{-19.6-4.9\Delta x&> -19.600001\cr -4.9\Delta x&>-0.000001\cr \Delta x& < -0.000001/-4.9\cr \Delta x& < 0.0000002040816327\ldots\cr } \nonumber Thus, we can say with certainty that if $$\Delta x$$ is positive and less than $$0.0000002$$, then $$\Delta x < 0.0000002040816327\ldots$$ and so $$-19.6-4.9\Delta x > -19.600001$$. We could do a similar calculation if $$\Delta x$$ is negative. So now we know that we can make $$-19.6-4.9\Delta x$$ within one millionth of $$-19.6$$. But can we make it "as close as we want''? In this case, it is quite simple to see that the answer is yes, by modifying the calculation we've just done. It may be helpful to think of this as a game. I claim that I can make $$-19.6-4.9\Delta x$$ as close as you desire to $$-19.6$$ by making $$\Delta x$$ "close enough'' to zero. So the game is: you give me a number, like $$10^{-6}$$, and I have to come up with a number representing how close $$\Delta x$$ must be to zero to guarantee that $$-19.6-4.9\Delta x$$ is at least as close to $$-19.6$$ as you have requested. Now if we actually play this game, I could redo the calculation above for each new number you provide. What I'd like to do is somehow see that I will always succeed, and even more, I'd like to have a simple strategy so that I don't have to do all that algebra every time. A strategy in this case would be a formula that gives me a correct answer no matter what you specify. So suppose the number you give me is $$\epsilon$$. How close does $$\Delta x$$ have to be to zero to guarantee that $$-19.6-4.9\Delta x$$ is in $$(-19.6-\epsilon, -19.6+\epsilon)$$? If $$\Delta x$$ is positive, we need: \eqalign{-19.6-4.9\Delta x&> -19.6-\epsilon\cr -4.9\Delta x&>-\epsilon\cr \Delta x& < -\epsilon/-4.9\cr \Delta x& < \epsilon/4.9\cr } \nonumber So if I pick any number $$\delta$$ that is less than $$\epsilon/4.9$$, the algebra tells me that whenever $$\Delta x < \delta$$ then $$\Delta x < \epsilon/4.9$$ and so $$-19.6-4.9\Delta x$$ is within $$\epsilon$$ of $$-19.6$$. (This is exactly what I did in the example: I picked $$\delta = 0.0000002 < 0.0000002040816327\ldots$$.) A similar calculation again works for negative $$\Delta x$$. The important fact is that this is now a completely general result---it shows that I can always win, no matter what "move'' you make. Now we can codify this by giving a precise definition to replace the fuzzy, "gets closer and closer'' language we have used so far. Henceforward, we will say something like "the limit of $$(-19.6\Delta x-4.9\Delta x^2)/\Delta x$$ as $$\Delta x$$ goes to zero is $$-19.6$$,'' and abbreviate this mouthful as $\lim_{\Delta x\to 0} {-19.6\Delta x-4.9\Delta x^2\over \Delta x} = -19.6. \nonumber$ Here is the actual, official definition of "limit''. Definition $$\PageIndex{2}$$: Limits Suppose $$f$$ is a function. We say that $$\displaystyle \lim_{x\to a}f(x)=L$$ if for every $$\epsilon>0$$ there is a $$\delta > 0$$ so that whenever $$0 < \|x-a\| < \delta$$, $$\|f(x)-L\| < \epsilon$$. The $$\epsilon$$ and $$\delta$$ here play exactly the role they did in the preceding discussion. The definition says, in a very precise way, that $$f(x)$$ can be made as close as desired to $$L$$ (that's the $$\|f(x)-L\| < \epsilon$$ part) by making $$x$$ close enough to $$a$$ (the $$0 < \|x-a\| < \delta$$ part). Note that we specifically make no mention of what must happen if $$x=a$$, that is, if $$\|x-a\|=0$$. This is because in the cases we are most interested in, substituting $$a$$ for $$x$$ doesn't even make sense. Make sure you are not confused by the names of important quantities. The generic definition talks about $$f(x)$$, but the function and the variable might have other names. In the discussion above, the function we analyzed was ${-19.6\Delta x-4.9\Delta x^2\over \Delta x}. \nonumber$ and the variable of the limit was not $$x$$ but $$\Delta x$$. The $$x$$ was the variable of the original function; when we were trying to compute a slope or a velocity, $$x$$ was essentially a fixed quantity, telling us at what point we wanted the slope. (In the velocity problem, it was literally a fixed quantity, as we focused on the time 2.) The quantity $$a$$ of the definition in all the examples was zero: we were always interested in what happened as $$\Delta x$$ became very close to zero. Armed with a precise definition, we can now prove that certain quantities behave in a particular way. The bad news is that even proofs for simple quantities can be quite tedious and complicated; the good news is that we rarely need to do such proofs, because most expressions act the way you would expect, and this can be proved once and for all. Example $$\PageIndex{3}$$ Let's show carefully that $$\displaystyle \lim_{x\to 2} x+4 = 6$$. Solution This is not something we "need'' to prove, since it is "obviously'' true. But if we couldn't prove it using our official definition there would be something very wrong with the definition. As is often the case in mathematical proofs, it helps to work backwards. We want to end up showing that under certain circumstances $$x+4$$ is close to 6; precisely, we want to show that $$\|x+4-6\| < \epsilon$$, or $$\|x-2\| < \epsilon$$. Under what circumstances? We want this to be true whenever $$0 < \|x-2\| < \delta$$. So the question becomes: can we choose a value for $$\delta$$ that guarantees that $$0 < \|x-2\| < \delta$$ implies $$\|x-2\| < \epsilon$$? Of course: no matter what $$\epsilon$$ is, $$\delta=\epsilon$$ works. So it turns out to be very easy to prove something "obvious,'' which is nice. It doesn't take long before things get trickier, however. Example $$\PageIndex{4}$$ It seems clear that $$\displaystyle \lim_{x\to 2} x^2=4$$. Let's try to prove it. Solution We will want to be able to show that $$\|x^2-4\| < \epsilon$$ whenever $$0 < \|x-2\| < \delta$$, by choosing $$\delta$$ carefully. Is there any connection between $$\|x-2\|$$ and $$\|x^2-4\|$$? Yes, and it's not hard to spot, but it is not so simple as the previous example. We can write $$\|x^2-4\|=\|(x+2)(x-2)\|$$. Now when $$\|x-2\|$$ is small, part of $$\|(x+2)(x-2)\|$$ is small, namely $$(x-2)$$. What about $$(x+2)$$? If $$x$$ is close to 2, $$(x+2)$$ certainly can't be too big, but we need to somehow be precise about it. Let's recall the "game'' version of what is going on here. You get to pick an $$\epsilon$$ and I have to pick a $$\delta$$ that makes things work out. Presumably it is the really tiny values of $$\epsilon$$ I need to worry about, but I have to be prepared for anything, even an apparently "bad'' move like $$\epsilon=1000$$. I expect that $$\epsilon$$ is going to be small, and that the corresponding $$\delta$$ will be small, certainly less than 1. If $$\delta\le 1$$ then $$\|x+2\| < 5$$ when $$\|x-2\| < \delta$$ (because if $$x$$ is within 1 of 2, then $$x$$ is between 1 and 3 and $$x+2$$ is between 3 and 5). So then I'd be trying to show that $\|(x+2)(x-2)\| < 5\|x-2\| < \epsilon. \nonumber$ So now how can I pick $$\delta$$ so that $$\|x-2\| < \delta$$ implies $$5\|x-2\| < \epsilon$$? This is easy: use $$\delta=\epsilon/5$$, so $$5\|x-2\| < 5(\epsilon/5) = \epsilon$$. But what if the $$\epsilon$$ you choose is not small? If you choose $$\epsilon=1000$$, should I pick $$\delta=200$$? No, to keep things "sane'' I will never pick a $$\delta$$ bigger than 1. Here's the final "game strategy:'' When you pick a value for $$\epsilon$$ I will pick $$\delta=\epsilon/5$$ or $$\delta=1$$, whichever is smaller. Now when $$\|x-2\| < \delta$$, I know both that $$\|x+2\| < 5$$ and that $$\|x-2\| < \epsilon/5$$. Thus $\|(x+2)(x-2)\| < 5(\epsilon/5) = \epsilon. \nonumber$ This has been a long discussion, but most of it was explanation and scratch work. If this were written down as a proof, it would be quite short, like this: Proof that $$\displaystyle \lim_{x\to 2}x^2=4$$. Given any $$\epsilon$$, pick $$\delta=\epsilon/5$$ or $$\delta=1$$, whichever is smaller. Then when $$\|x-2\| < \delta$$, $$\|x+2\| < 5$$ and $$\|x-2\| < \epsilon/5$$. Hence $\|x^2-4\|=\|(x+2)(x-2)\| < 5(\epsilon/5) = \epsilon \nonumber.$ It probably seems obvious that $$\displaystyle \lim_{x\to2}x^2=4$$, and it is worth examining more closely why it seems obvious. If we write $$x^2=x\cdot x$$, and ask what happens when $$x$$ approaches 2, we might say something like, "Well, the first $$x$$ approaches 2, and the second $$x$$ approaches 2, so the product must approach $$2\cdot2$$.'' In fact this is pretty much right on the money, except for that word "must.'' Is it really true that if $$x$$ approaches $$a$$ and $$y$$ approaches $$b$$ then $$xy$$ approaches $$ab$$? It is, but it is not really obvious, since $$x$$ and $$y$$ might be quite complicated. The good news is that we can see that this is true once and for all, and then we don't have to worry about it ever again. When we say that $$x$$ might be "complicated'' we really mean that in practice it might be a function. Here is then what we want to know: Theorem $$\PageIndex{5}$$ Suppose $$\displaystyle \lim_{x\to a} f(x)=L$$ and $$\displaystyle\lim_{x\to a}g(x)=M$$. Then $\lim_{x\to a} f(x)g(x) = LM.$ Proof We have to use the official definition of limit to make sense of this. So given any $$\epsilon$$ we need to find a $$\delta$$ so that $$0 < \|x-a\| < \delta$$ implies $$\|f(x)g(x)-LM\| < \epsilon$$. What do we have to work with? We know that we can make $$f(x)$$ close to $$L$$ and $$g(x)$$ close to $$M$$, and we have to somehow connect these facts to make $$f(x)g(x)$$ close to $$LM$$. We use, as is so often the case, a little algebraic trick: \eqalign{\|f(x)g(x)-LM\|&= \|f(x)g(x)-f(x)M+f(x)M-LM\|\cr &=\|f(x)(g(x)-M)+(f(x)-L)M\|\cr &\le \|f(x)(g(x)-M)\|+\|(f(x)-L)M\|\cr &=\|f(x)\|\|g(x)-M\|+\|f(x)-L\|\|M\|.\cr} \nonumber This is all straightforward except perhaps for the "$$\le$$'. That is an example of the triangle inequality , which says that if $$a$$ and $$b$$ are any real numbers then $$\|a+b\|\le \|a\|+\|b\|$$. If you look at a few examples, using positive and negative numbers in various combinations for $$a$$ and $$b$$, you should quickly understand why this is true; we will not prove it formally. Since $$\displaystyle \lim_{x\to a}f(x) =L$$, there is a value $$\delta_1$$ so that $$0 < \|x-a\| < \delta_1$$ implies $$\|f(x)-L\| < \|\epsilon/(2M)\|$$, This means that $$0 < \|x-a\| < \delta_1$$ implies $$\|f(x)-L\|\|M\| < \epsilon/2$$. You can see where this is going: if we can make $$\|f(x)\|\|g(x)-M\| < \epsilon/2$$ also, then we'll be done. We can make $$\|g(x)-M\|$$ smaller than any fixed number by making $$x$$ close enough to $$a$$; unfortunately, $$\epsilon/(2f(x))$$ is not a fixed number, since $$x$$ is a variable. Here we need another little trick, just like the one we used in analyzing $$x^2$$. We can find a $$\delta_2$$ so that $$\|x-a\| < \delta_2$$ implies that $$\|f(x)-L\| < 1$$, meaning that $$L-1 < f(x) < L+1$$. This means that $$\|f(x)\| < N$$, where $$N$$ is either $$\|L-1\|$$ or $$\|L+1\|$$, depending on whether $$L$$ is negative or positive. The important point is that $$N$$ doesn't depend on $$x$$. Finally, we know that there is a $$\delta_3$$ so that $$0 < \|x-a\| < \delta_3$$ implies $$\|g(x)-M\| < \epsilon/(2N)$$. Now we're ready to put everything together. Let $$\delta$$ be the smallest of $$\delta_1$$, $$\delta_2$$, and $$\delta_3$$. Then $$\|x-a\| < \delta$$ implies that $$\|f(x)-L\| < \|\epsilon/(2M)\|$$, $$\|f(x)\| < N$$, and $$\|g(x)-M\| < \epsilon/(2N)$$. Then \eqalign{\|f(x)g(x)-LM\|&\le\|f(x)\|\|g(x)-M\|+\|f(x)-L\|\|M\|\cr & < N{\epsilon\over 2N}+\left\|{\epsilon\over 2M}\right\|\|M\|\cr &={\epsilon\over 2}+{\epsilon\over 2}=\epsilon.\cr} \nonumber This is just what we needed, so by the official definition, $$\displaystyle \lim_{x\to a}f(x)g(x)=LM$$. $$\square$$ A handful of such theorems give us the tools to compute many limits without explicitly working with the definition of limit. Theorem $$\PageIndex{6}$$ Suppose that $$\displaystyle \lim_{x\to a}f(x)=L$$ and $$\displaystyle \lim_{x\to a}g(x)=M$$ and $$k$$ is some constant. Then \eqalign{ &\lim_{x\to a} kf(x) = k\lim_{x\to a}f(x)=kL\cr &\lim_{x\to a} (f(x)+g(x)) = \lim_{x\to a}f(x)+\lim_{x\to a}g(x)=L+M\cr &\lim_{x\to a} (f(x)-g(x)) = \lim_{x\to a}f(x)-\lim_{x\to a}g(x)=L-M\cr &\lim_{x\to a} (f(x)g(x)) = \lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x)=LM\cr &\lim_{x\to a} {f(x)\over g(x)} = {\lim_{x\to a}f(x)\over\lim_{x\to a}g(x)}={L\over M},\hbox{ if $$M$$ is not 0}\cr } Roughly speaking, these rules say that to compute the limit of an algebraic expression, it is enough to compute the limits of the "innermost bits'' and then combine these limits. This often means that it is possible to simply plug in a value for the variable, since $$\displaystyle \lim_{x\to a} x =a$$. Example $$\PageIndex{7}$$ Compute $$\displaystyle \lim_{x\to 1}{x^2-3x+5\over x-2}$$. Solution If we apply the theorem in all its gory detail, we get \eqalign{ \lim_{x\to 1}{x^2-3x+5\over x-2}&= {\lim_{x\to 1}(x^2-3x+5)\over \lim_{x\to1}(x-2)}\cr &={(\lim_{x\to 1}x^2)-(\lim_{x\to1}3x)+(\lim_{x\to1}5)\over (\lim_{x\to1}x)-(\lim_{x\to1}2)}\cr &={(\lim_{x\to 1}x)^2-3(\lim_{x\to1}x)+5\over (\lim_{x\to1}x)-2}\cr &={1^2-3\cdot1+5\over 1-2}\cr &={1-3+5\over -1} = -3\cr } \nonumber It is worth commenting on the trivial limit $$\displaystyle \lim_{x\to1}5$$. From one point of view this might seem meaningless, as the number 5 can't "approach'' any value, since it is simply a fixed number. But 5 can, and should, be interpreted here as the function that has value 5 everywhere, $$f(x)=5$$, with graph a horizontal line. From this point of view it makes sense to ask what happens to the height of the function as $$x$$ approaches 1. Of course, as we've already seen, we're primarily interested in limits that aren't so easy, namely, limits in which a denominator approaches zero. There are a handful of algebraic tricks that work on many of these limits. Example $$\PageIndex{8}$$ Compute $$\displaystyle \lim_{x\to1}{x^2+2x-3\over x-1}$$. Solution We cannot simply plug in $$x=1$$ because that makes the denominator zero. However: \eqalign{ \lim_{x\to1}{x^2+2x-3\over x-1}&=\lim_{x\to1}{(x-1)(x+3)\over x-1}\cr &=\lim_{x\to1}(x+3)=4\cr } While theorem $$\PageIndex{6}$$ is very helpful, we need a bit more to work easily with limits. Since the theorem applies when some limits are already known, we need to know the behavior of some functions that cannot themselves be constructed from the simple arithmetic operations of the theorem, such as $$\sqrt{x}$$. Also, there is one other extraordinarily useful way to put functions together: composition. If $$f(x)$$ and $$g(x)$$ are functions, we can form two functions by composition: $$f(g(x))$$ and $$g(f(x))$$. For example, if $$f(x)=\sqrt{x}$$ and $$g(x)=x^2+5$$, then $$f(g(x))=\sqrt{x^2+5}$$ and $$g(f(x))=(\sqrt{x})^2+5=x+5$$. Here is a companion to theorem $$\PageIndex{6}$$ for composition: Theorem $$\PageIndex{9}$$ Suppose that $$\displaystyle \lim_{x\to a}g(x)=L$$ and $$\displaystyle \lim_{x\to L}f(x)=f(L)$$. Then $\lim_{x\to a} f(g(x)) = f(L).$ Note the special form of the condition on $$f$$: it is not enough to know that $$\displaystyle \lim_{x\to L}f(x) = M$$, though it is a bit tricky to see why. Many of the most familiar functions do have this property, and this theorem can therefore be applied. For example: Theorem $$\PageIndex{10}$$ Suppose that $$n$$ is a positive integer. Then $\lim_{x\to a}\root n\of{x} = \root n\of{a},$ provided that $$a$$ is positive if $$n$$ is even. This theorem is not too difficult to prove from the definition of limit. Another of the most common algebraic tricks was used in section 2.1. Here's another example: Example $$\PageIndex{11}$$ Compute $$\displaystyle \lim_{x\to-1} {\sqrt{x+5}-2\over x+1}$$. Solution \eqalign{\lim_{x\to-1} {\sqrt{x+5}-2\over x+1}&= \lim_{x\to-1} {\sqrt{x+5}-2\over x+1}{\sqrt{x+5}+2\over \sqrt{x+5}+2}\cr &=\lim_{x\to-1} {x+5-4\over (x+1)(\sqrt{x+5}+2)}\cr &=\lim_{x\to-1} {x+1\over (x+1)(\sqrt{x+5}+2)}\cr &=\lim_{x\to-1} {1\over \sqrt{x+5}+2}={1\over4}\cr} At the very last step we have used theorems 2.3.9 and $$\PageIndex{10}$$. Occasionally we will need a slightly modified version of the limit definition. Consider the function $$f(x)=\sqrt{1-x^2}$$, the upper half of the unit circle. What can we say about $$\displaystyle \lim_{x\to 1}f(x)$$? It is apparent from the graph of this familiar function that as $$x$$ gets close to 1 from the left, the value of $$f(x)$$ gets close to zero. It does not even make sense to ask what happens as $$x$$ approaches 1 from the right, since $$f(x)$$ is not defined there. The definition of the limit, however, demands that $$f(1+\Delta x)$$ be close to $$f(1)$$ whether $$\Delta x$$ is positive or negative. Sometimes the limit of a function exists from one side or the other (or both) even though the limit does not exist. Since it is useful to be able to talk about this situation, we introduce the concept of one sided limit: Definition $$\PageIndex{12}$$: One Sided Limits Suppose that $$f(x)$$ is a function. We say that $$\displaystyle \lim_{x\to a^-}f(x)=L$$ if for every $$\epsilon>0$$ there is a $$\delta > 0$$ so that whenever $$0 < a-x < \delta$$, $$\|f(x)-L\| < \epsilon$$. We say that $$\lim_{x\to a^+}f(x)=L$$ if for every $$\epsilon>0$$ there is a $$\delta > 0$$ so that whenever $$0 < x-a < \delta$$, $$\|f(x)-L\| < \epsilon$$. Usually $$\displaystyle \lim_{x\to a^-}f(x)$$ is read "the limit of $$f(x)$$ from the left'' and $$\displaystyle \lim_{x\to a^+}f(x)$$ is read "the limit of $$f(x)$$ from the right''. Example $$\PageIndex{13}$$ Discuss $$\displaystyle \lim_{x\to 0}{x\over\|x\|}$$, $$\displaystyle \lim_{x\to 0^-}{x\over\|x\|}$$, and $$\displaystyle \lim_{x\to 0^+}{x\over\|x\|}$$. Solution The function $$f(x)=x/\|x\|$$ is undefined at 0; when $$x>0$$, $$\|x\|=x$$ and so $$f(x)=1$$; when $$x < 0$$, $$\|x\|=-x$$ and $$f(x)=-1$$. Thus $$\displaystyle \lim_{x\to 0^-}{x\over\|x\|}=\lim_{x\to 0^-}-1=-1$$ while $$\displaystyle \lim_{x\to 0^+}{x\over\|x\|}=\lim_{x\to 0^+}1=1$$. The limit of $$f(x)$$ must be equal to both the left and right limits; since they are different, the limit $$\displaystyle \lim_{x\to 0}{x\over\|x\|}$$ does not exist. ## Contributors • Integrated by Justin Marshall.
# How do you find the missing side b given 5in 15in using the pythagorean theorem? Feb 7, 2017 See the entire solution process below: #### Explanation: The Pythagorean Theorem states: ${a}^{2} + {b}^{2} = {c}^{2}$ Where: $a$ and $b$ are sides of the triangle and $c$ is the hypotenuse. In this problem we will be solving for $b$ after substituting for $a$ and $c$. By definition the hypotenuse is always the largest value therefore we will substitute $15$ for $c$ and $5$ for $a$ giving: ${5}^{2} + {b}^{2} = {15}^{2}$ $25 + {b}^{2} = 225$ $25 - \textcolor{red}{25} + {b}^{2} = 225 - \textcolor{red}{25}$ $0 + {b}^{2} = 200$ ${b}^{2} = 200$ $\sqrt{{b}^{2}} = \sqrt{200}$ $b = \sqrt{200}$ in or $b = 14.14$ in rounded to the nearest hundredth.
# How do you find the slope and intercepts of x-2y=8? Dec 8, 2016 $\text{slope " =1/2," intercepts are } - 4 , 8$ #### Explanation: The equation of a line in $\textcolor{b l u e}{\text{slope-intercept form}}$ is. $\textcolor{red}{\overline{\underline{\| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = m x + b} \textcolor{w h i t e}{\frac{2}{2}} \|}}}$ where m represents the slope and b, the y-intercept. $\text{Rearrange " x-2y=8" into this form}$ Leave - 2y on the left side and subtract x from both sides. $\cancel{x} \cancel{- x} - 2 y = - x + 8$ $\Rightarrow - 2 y = - x + 8$ divide ALL terms on both sides by - 2 $\frac{\cancel{- 2} y}{\cancel{- 2}} = \frac{- 1}{- 2} x + \frac{8}{- 2}$ $\Rightarrow y = \frac{1}{2} x - 4 \leftarrow \text{ now in above form}$ $\Rightarrow \text{slope " =1/2" and y-intercept } = - 4$ To find the x-intercept, substitute y = 0 into the equation and solve for x. $\Rightarrow x - \left(2 \times 0\right) = 8 \Rightarrow x = 8 \text{ is the x-intercept}$
# triangle in circle If we set it to 50% it will create a circle. First off, a definition: Inscribed Angle: an angle made from points sitting on the circle's circumference. Solution to Problem : Let B and N be the two points of tangency of the circle (see figure below). See circumcenter of a triangle for more about this. Finally, rectangle and circle? Circumscribe a Circle on a Triangle. So cropping is quick, highly secured and consumes less bandwidth. The problem appears to be simple, but it is actually very difficult! Find the radius of the circle. This is very similar to the construction of an inscribed hexagon, except we use every other vertex instead of all six. Inscribe an equilateral triangle in the circle using the given point as a vertex. The equilateral triangle represents unity, recovery and service in response to the physical, mental and spiritual parts of the disease of alcoholism. And what that does for us is it tells us that triangle ACB is a right triangle. Plug the area into the triangle formula. This is a right triangle, and the diameter is its hypotenuse. 1. Let a be the length of the sides, A - the area of the triangle, p the perimeter, R - the radius of the circumscribed circle, r - the radius of the inscribed circle, h - the altitude (height) from any side.. For example, if you found the area of the circle to be 50.24 cm, your formula will look like this: =. Examples: Input: R = 4 Output: 20.784 Explanation: Area of equilateral triangle inscribed in a circle of radius R will be 20.784, whereas side of the triangle … That'd be a total of 6. One side of the triangle is 18cm. The circumcircle always passes through all three vertices of a triangle. This is the largest equilateral that will fit in the circle, with each vertex touching the circle. A Euclidean construction. Find length of triangle as well as radius of inscribed circle 1 Circle with inscribed right triangle with inscribed circle with inscribed right triangle with …: When a triangle is inserted in a circle in such a way that one of the side of the triangle is diameter of the circle then the triangle is right triangle. Goal: 5L 6E. It can be any line passing through the center of the circle and touching the sides of it. cropping is much Faster, since we are not uploading your images to our server. This strategy encourages students to reflect on their learning and process information presented in the lesson. circle with vertical fill 9678: 25ce : bullseye 9679: 25cf : black circle 9680: 25d0 : circle with left half black 9681: 25d1 : circle with right half black 9682: 25d2 : circle with lower half black 9683: 25d3 : circle with upper half black 9684: 25d4 : circle with upper right quadrant black 9685: 25d5 : circle … So we can just apply the Pythagorean theorem here. Therefore, the area of a triangle equals the half of the rectangular area, To circumscribe a triangle, all you need to do is find the […] Points A, B and C are the centres of three circles, each one of which touches the other two. How to Circumscribe a Circle on a Triangle using just a compass and a straightedge. Use the mouse to drag a point along the perimeter of the circle. Triangle-Square-Circle. We are given the following triangle with sides equal to 50 cm, 35 cm and 40 cm. 15 squared plus 8 squared-- let me do this in magenta-- is going to be the length of side AB squared. Since you want the area of each figure to be the same, use the area you previously calculated for the circle. This code for Calculate Area for Rectangle, Square, Triangle, Circle in C# language. Equilateral Triangle: How to construct an equilateral triangle, inscribed in a given circle? #circle { background: lightblue; border-radius: 50%; width: 100px; height: 100px; } A CSS Circle Triangles. We want to find area of circle inscribed in this triangle. Now draw a diameter to it. Area of a triangle, the radius of the circumscribed circle and the radius of the inscribed circle: Rectangular in the figure below is composed of two pairs of congruent right triangles formed by the given oblique triangle. The length of a leg of an isosceles right triangle is #5sqrt2# units. All of them should be constructed simultaneously. As you drag it, the triangle and angles are redisplayed. Answer To Impossible Viral Problem – Triangle In A Circle (Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks). In laymen’s terms, any triangle can fit into some circle with all its corners touching the circle. You can receive one more (hidden) V-star if there are several possible objects that satisfy the statement of a problem. Similar to other closing strategies, it asks students to pick out important pieces of information and to question anything they don’t completely understand. For the right triangle in the above example, the circumscribed circle is simple to draw; its center can be found by measuring a distance of 2.5 units from A along $$\overline{AB}$$. Start with the construction of a hexagon inscribed in a circle with the same size as the given circle… Geometry calculator for solving the inscribed circle radius of a scalene triangle given the length of side c and angles A, B and C. We have one relation among semi-perimeter of triangle and the radius of circle inscribed in such a triangle which is: Because the larger triangle with sides 15, x, and 25 has a base as the diameter of the circle, it is a right triangle and the angle opposite the diameter must be 90. Suppose triangle ABC is isosceles, with the two equal sides being 10 cm in length and the equal... What is the basic formula for finding the area of an isosceles triangle? This free triangle calculator computes the edges, angles, area, height, perimeter, median, as well as other values of a triangle. Circumscribe: To draw on the outside of, just touching the corner points but never crossing.. Steps: Construct the perpendicular bisector of one side of triangle; Construct the perpendicular bisector of another side Click hereto get an answer to your question ️ ABC is an isosceles triangle, in which AB = AC , circumscribed about a circle. How to construct (draw) an equilateral triangle inscribed in a given circle with a compass and straightedge or ruler. The lengths of AM and BC are equal to 6 and 18 cm respectively. We then have three right triangles. Equilateral triangle formulas. So far i'm able to draw triangle inside circle but i try to move little dot its get out of the circle. In Euclidean geometry, any three points, when non-collinear, determine a unique triangle and simultaneously, a unique plane (i.e. This video shows how to inscribe a circle in a triangle using a compass and straight edge. 4. Show that BC is bisected at the point of contact ? View a scaled diagram of the resulting triangle, or explore many other math calculators, as well as hundreds of other calculators addressing finance, health, fitness, and more. Triangles are a little trickier. We have to set the borders on the element to match a triangle. Geometry calculator for solving the inscribed circle radius of a right triangle given the length of sides a, b and c. Note that the center of the circle can be inside or outside of the triangle. Every triangle can be circumscribed by a circle, meaning that one circle — and only one — goes through all three vertices (corners) of any triangle. However, there are several steps involved in making each of them, such as choosing if you want the shape filled in or outlined. Making a circle, square, or triangle in Adobe Photoshop is a pretty straight-forward process. The center of the circle is not given. Let’s solve the problem in the following steps: The Triangle appears in Masonry in two forms, the Right Triangle, i.e., that Triangle which has one of its angles a right angle, ninety degrees, or the one-fourth part of a Circle, and the Equilateral Triangle, i.e., that Triangle which has all its sides equal, each to the other, and, … triangle AOB and triangle AOC are congruent right triangles. Given an integer R which denotes the radius of a circle, the task is to find the area of an equilateral triangle inscribed in this circle.. A triangle is a polygon with three edges and three vertices.It is one of the basic shapes in geometry.A triangle with vertices A, B, and C is denoted .. In the figure below, triangle ABC is tangent to the circle of center O at two points. Coins on a Plate. Thus, the Pythagorean theorem can be used to find the length of x. x 2 + 15 2 = 25 2 Rather than do the calculations, notice that the triangle is a 3-4-5 triangle (multiplied by 5). That'd be a total of 6. Some interesting things about angles and circles. Find the length of one side of the triangle if the radius of circumscribing circle is 9cm. We need a different procedure for acute and obtuse triangles, since for an acute triangle the center of the circumscribed circle will be inside the triangle, and it will be outside for an obtuse triangle. Connect the points to form a triangle. Videos This video gives a review of the following circle theorems: arrow theorem, bow theorem, cyclic quadrilateral, semi-circle, radius-tangent theorem, alternate segment theorem, chord center theorem, dual tangent theorem. Its center is at the point where all the perpendicular bisectors of the triangle's sides meet. The area of circle = So, if we can find the radius of circle, we can find its area. Circle Theorems. Prove that the perimeter of the triangle ABC is equal to the diameter of the largest circle. Doing the same with the triangle and rectangle, and it's again easy to get two intersections per triangle side and impossible to get more. crop a circle in the image, is an online tool, used to crop round circle in your images. A triangle inside a circle represents the Sobriety Circle and Triangle Symbol used by the Alcoholics Anonymous group. Display the angles in the triangle. The area of a triangle inscribed in a circle is 42.23cm2. Play around with a triangle and circle, and it's easy to get two intersections per triangle side, but seems impossible to get more. To prove this first draw the figure of a circle. Follow the steps below for how to create and edit each shape. 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Courses Courses for Kids Free study material Offline Centres More Store The area of the circle ${{x}^{2}}+{{y}^{2}}=16$ exterior to the parabola ${{y}^{2}}=6x$isa). $\dfrac{4}{3}\left( 4\pi -\sqrt{3} \right)$b). $\dfrac{4}{3}\left( 8\pi -\sqrt{3} \right)$c). $\dfrac{4}{3}\left( 4\pi +\sqrt{3} \right)$d). $\dfrac{4}{3}\left( 8\pi +\sqrt{3} \right)$ Last updated date: 17th Sep 2024 Total views: 80.4k Views today: 1.80k Verified 80.4k+ views Hint: Substitute ${{y}^{2}}=6x$ in the equation of circle ${{x}^{2}}+{{y}^{2}}=16$ and find the value of x and find the points which intersects parabola with circle. Thus, to find the required area, subtract the area of the circle with the area of the parabola which is exterior to the circle. Complete step-by-step solution: First, we will draw the corresponding figure, draw the circle ${{x}^{2}}+{{y}^{2}}=16$, where center is (0, 0). Radius of the circle = $r=\sqrt{16}=4$ And draw the parabola, ${{y}^{2}}=6x$. Now, we need to find the intersection A and B i.e., the intersection points of the parabola with the circle. We know, ${{y}^{2}}=6x$, put the value of ${{y}^{2}}$in the equation of circle, we get \begin{align} & \Rightarrow {{x}^{2}}+{{y}^{2}}=16 \\ & \Rightarrow {{x}^{2}}+6x=16 \\ & \Rightarrow {{x}^{2}}+6x-16=0-(1) \\ \end{align} Equation (1) represents a quadratic equation, similar to the general equation $a{{x}^{2}}+bx+c=0$. By comparing them we get, $\therefore a=1,b=6,c=-16$ Substitute the values in $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, we get \begin{align} & =\dfrac{-6\pm \sqrt{{{6}^{2}}-4\times 1\times \left( -16 \right)}}{2}=\dfrac{-6\pm \sqrt{36+64}}{2}=\dfrac{-6\pm \sqrt{100}}{2} \\ & =\dfrac{-6\pm 10}{2} \\ \end{align} $\therefore$We get the roots as $\left( \dfrac{-6+10}{2} \right)$ and $\left( \dfrac{-6-10}{2} \right)$. That is the roots are 2 and -8. So, we have x = 2 and x = -8, substitute them in the equation of parabola, we get \begin{align} & x=2\Rightarrow {{y}^{2}}=6x=6\times 2=12 \\ & y=\pm \sqrt{12}=\pm 2\sqrt{3} \\ & x=-8\Rightarrow {{y}^{2}}=6\times \left( -8 \right)=-32 \\ \end{align} Since, the square cannot be negative, x = -8 is not possible. $\therefore$Point A is $\left( 2,2\sqrt{3} \right)$ and $B\left( 2,-2\sqrt{3} \right)$. Now we will find the required area, from figure we see that Required area = Area of circle – Area OACB Now, we need to find the area of the circle. Area of circle$=4\times$Area of 1st quadrant $=4\times \int\limits_{0}^{4}{ydx}$ We have to get the value of y from the equation of the circle \begin{align} & {{x}^{2}}+{{y}^{2}}=16 \\ & \Rightarrow {{y}^{2}}=16-{{x}^{2}} \\ & \therefore y=\pm \sqrt{16-{{x}^{2}}} \\ \end{align} Since, we need to find the area of the $1^{st}$ quadrant, we take y positive. $\therefore y=\sqrt{16-{{x}^{2}}}-(1)$ $\therefore$Area of the circle$=4\times \int\limits_{0}^{4}{\sqrt{16-{{x}^{2}}}dx}$ $=4\int\limits_{0}^{4}{\sqrt{{{4}^{2}}-{{x}^{2}}}dx}$ We know the formula, $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+c-(2)$ Replace a by 4, we get $\therefore$Area of circle $=4\left[ \dfrac{x}{2}\sqrt{{{4}^{2}}-{{x}^{2}}}+\dfrac{{{4}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{4} \right]_{0}^{4}$ Area of circle $=4\left[ \dfrac{x}{2}\sqrt{16-{{x}^{2}}}+8{{\sin }^{-1}}\dfrac{x}{4} \right]_{0}^{4}$ Applying the limits, we get \begin{align} & =4\left[ \left( \dfrac{4}{2}\sqrt{16-{{4}^{2}}}+8{{\sin }^{-1}}\dfrac{4}{4} \right)-\left( \dfrac{0}{2}\sqrt{{{16}^{2}}-{{0}^{2}}}+8{{\sin }^{-1}}\dfrac{0}{4} \right) \right] \\ & =4\left[ 0+8{{\sin }^{-1}}1-0-8{{\sin }^{-1}}0 \right] \\ & =4\times \left[ 0+8\times \dfrac{\pi }{2}-0-0 \right]=4\times 8\times \dfrac{\pi }{2}=16\pi -(3) \\ \end{align} Now, we need to find the area of OACB. OACB is symmetric about the y-axis. $\therefore$area OACB$=2\times$area OAC. So, we get $\therefore$area OACB = 2(area OAD + area ADC)……..(A) Now we will find all the areas separately. Area ACD As x-coordinate of point A is ‘2’ and x-coordinate of point C is ‘4’, so Area ACD$=\int\limits_{2}^{4}{ydx}$ As we are taking $1^{st}$ quadrant $y=\sqrt{16-{{x}^{2}}}$ $\therefore$ area ACD$=\int\limits_{2}^{4}{\sqrt{16-{{x}^{2}}}dx}=\int\limits_{2}^{4}{\sqrt{{{4}^{2}}-{{x}^{2}}}dx}$ Use the formula mentioned in equation (2), we get Area ACD$=\left[ \dfrac{x}{2}\sqrt{{{4}^{2}}-{{x}^{2}}}+\dfrac{{{4}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{4} \right]_{2}^{4}$ Applying the limits, we get \begin{align} & =\left[ \left( \dfrac{4}{2}\sqrt{{{4}^{2}}-{{4}^{2}}}+8{{\sin }^{-1}}\dfrac{4}{4} \right)-\left( \dfrac{2}{2}\sqrt{{{4}^{2}}-{{2}^{2}}}+\dfrac{{{4}^{2}}}{2}{{\sin }^{-1}}\dfrac{2}{4} \right) \right] \\ & =0+8{{\sin }^{-1}}1-2\sqrt{3}-8\times {{\sin }^{-1}}\dfrac{1}{2} \\ & =8\times \dfrac{\pi }{2}-2\sqrt{3}-8\times \dfrac{\pi }{6} \\ & =4\pi -\dfrac{4\pi }{3}-2\sqrt{3} \\ & =\dfrac{12\pi -4\pi }{3}-2\sqrt{3} \\ \end{align} Area ACD$=\dfrac{8\pi }{3}-2\sqrt{3}$-(4) Now we will find the Area CAD$=\int\limits_{0}^{2}{ydx}$ We have to find y from the equation of parabola. \begin{align} & {{y}^{2}}=6x\Rightarrow y=\pm \sqrt{6x} \\ & \therefore y=\sqrt{6x} \\ \end{align} Therefore, the area OAD will become, $=\int\limits_{0}^{2}{\sqrt{6x}dx}$ Taking out the constant, we get $=\sqrt{6}\int\limits_{0}^{2}{\sqrt{x}.dx}=\sqrt{6}\int\limits_{0}^{2}{{{x}^{\dfrac{1}{2}}}dx}$ On integrating, we get \begin{align} & =\sqrt{6}\left[ \dfrac{{{x}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1} \right]_{0}^{2} \\ & =\sqrt{6}{{\left[ \dfrac{{{x}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right]}^{2}} \\ & =\dfrac{2\sqrt{6}}{3}\left[ {{x}^{\dfrac{3}{2}}} \right]_{0}^{2} \\ \end{align} On Applying the limits and solving, we get \begin{align} & =\dfrac{2\sqrt{6}}{3}\times \left[ {{2}^{\dfrac{3}{2}}}-{{0}^{\dfrac{3}{2}}} \right] \\ & =\dfrac{2\sqrt{6}}{3}\times {{2}^{\dfrac{3}{2}}} \\ & =\dfrac{2\sqrt{2}\times \sqrt{3}}{3}\times 2\sqrt{2} \\ & =\dfrac{4\times 2}{\sqrt{3}} \\ \end{align} Therefore, the area of OAD wil be$=\dfrac{8}{\sqrt{3}}-(5)$ Now consider equation (A), we get $\therefore$area OACB = 2(area OAD + area ADC)……..(A) Substituting values from equation (4) and (5), we get $\therefore$Area OACB $=2\left( \dfrac{8}{\sqrt{3}}+\dfrac{8\pi }{3}-2\sqrt{3} \right)$ On solving this we get $\therefore$Area OACB $=2\left( \dfrac{8}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}+\dfrac{8\pi }{3}-2\sqrt{3} \right)=2\left( \dfrac{8\sqrt{3}}{3}+\dfrac{8\pi }{3}-2\sqrt{3} \right)$ Taking the LCM, we get $\therefore$Area OACB $=2\left( \dfrac{8\sqrt{3}+8\pi -2\sqrt{3}\times 3}{3} \right)$ $\therefore$Area OACB $=2\left( \dfrac{2\sqrt{3}+8\pi }{3} \right)$ $\therefore$Required area = Area of circle – area OACB Required area $=16\pi -2\left( \dfrac{2\sqrt{3}+8\pi }{3} \right)$ Required area $=\left( \dfrac{16\pi \times 3-4\sqrt{3}-16\pi }{3} \right)$ Required area $=\left( \dfrac{32\pi -4\sqrt{3}}{3} \right)$ $\therefore$ Required area is $\dfrac{4}{3}\left( 8\pi -\sqrt{3} \right)$ square units. Hence, the correct option is (c). Note: We know the formula for finding the area of circle $=\pi {{r}^{2}}$ , we can find the radius from the equation given to us, i.e., ${{x}^{2}}+{{y}^{2}}=16$. Hence, from this, we get the radius as 4. Now apply this radius directly in the formula for finding the area of a circle as, Area of circle $=\pi \times {{4}^{2}}=16\pi$. Hence, you can also find the area using this simple formula.
# Using Mathematical Terms ## Using Mathematical Terms Welcome to the fascinating world of mathematical terms! Anyone, who is approaching Middle School, needs to prepare to solve increasingly more and difficult mathematical problem-solving and reasoning questions. These questions often use high level mathematical terms. In this text, we'll explore crucial terms like sum, term, product, factor, quotient, and coefficient. These terms are the building blocks of mathematics, and understanding them is key to mastering many mathematical concepts. ## Understanding Mathematical Terms It is important to have a deep understanding of each of the mathematical terms to understand what mathematical operation needs to be completed. Therefore, you will find a detailed explanation of sum, term, product, factor, quotient, and coefficient below. ### Sum – Definition and Application The sum is the result of adding two or more numbers together. In a written task such as ‘The sum of two numbers is 15. If one number is 8, what is the other number?’ it is important to understand the role the word ‘sum’ plays in this problem. When we see the word sum, it doesn’t always mean find the sum, but rather it is a clue as to what needs to be done to solve the problem. Sometimes you might need to use the inverse operation, which for sum would be subtraction. ### Term– Definition and Application In mathematics, a term is a single mathematical expression. It can be a number, a variable, or numbers and variables multiplied together. In mathematics, especially in algebra, term refers to a single mathematical expression. It can be a number, a variable, or the product of numbers and variables. In a written task such as, "Identify the terms in the expression 3x + 4y - 5," it's important to recognize that terms are separated by plus or minus signs. Here, 3x, 4y, and -5 are distinct terms. Understanding this helps you to break down expressions into simpler parts for further operations or simplifications. ### Product – Definition and Application The product is the result of multiplying two or more numbers. The term product refers to the result of multiplying two or more numbers or expressions. In a task like, "Find the product of 7 and 7," you must recognize that finding the product means performing multiplication. This term is crucial in understanding how to combine numbers and variables in algebraic expressions. Sometimes, the question might suggest the product is a given value, in which case you might need to use the inverse operation, division, to solve the problem. ### Factor– Definition and Application A factor is a number that divides into another number without leaving a remainder. A factor is a number or expression that divides another number or expression evenly, with no remainder. In problems like, "List all the factors of 24," you need to identify numbers that can perfectly divide 24 (such as 1, 2, 3, 4, 6, 8, 12, 24). Understanding factors is key in simplifying fractions, finding the greatest common factors, and solving division-related problems. It’s about recognizing the building blocks that make up a number. Factors are important, and they are applied in Prime Factorization, Least Common Multiples, and Finding the Greatest Common Factor. ### Quotient– Definition and Application The quotient is the result of dividing one number by another. The quotient is the result of division. In a task like, "Divide 15 by 3 to find the quotient," you must recognize that the quotient is the answer to a division problem. This concept is essential in understanding how division works and in solving problems that require dividing numbers. It's important to note that the quotient helps in breaking down larger numbers into smaller, more manageable parts. ### Coefficient – Definition and Application A coefficient is a number used to multiply a variable. In algebra, a coefficient is a number used to multiply a variable. For example, in the expression 4x + 3, 4 is the coefficient of x. When you encounter problems like, "Identify the coefficient in the term 7y," you must be able to identify that the coefficient is the numerical part of the term that is attached to the variable. Understanding coefficients is crucial in algebra, as it helps in simplifying expressions and solving equations. It’s about recognizing how numbers and variables are linked in algebraic expressions. ## Mathematical Terms – Application Test your understanding of mathematical terms with these questions. What is the sum of 5 and 8? Identify the terms in 2x + 7. What is the product of 4 and 6? Is 2 a factor of 14? What is the quotient of 10 divided by 2? ## Mathematical Terms Summary Key Learnings from this Text: • Understanding mathematical terms like sum, term, product, factor, quotient, and the coefficient is essential. • These terms are used in various mathematical operations and expressions. • Recognizing and applying these terms correctly enhances your mathematical comprehension and problem-solving skills. • Here is a table of the terms and definitions for you to reference or make a copy of for your use. Term Definition Sum The result of adding two or more numbers together. Term A single mathematical expression, which can be a number, variable, or both combined. Product The result of multiplying two or more numbers. Factor A number that divides another number exactly, without leaving a remainder. Quotient The result of dividing one number by another. Coefficient A number that multiplies a variable in an algebraic expression. Keep practicing these terms in different mathematical contexts to strengthen your understanding. Remember, these terms are not just numbers or expressions; they help structure the world of mathematics! Now you should be ready to explore Using Variables to Represent Numbers in Expressions and Equivalent Expressions ## Mathematical Terms – Frequently Asked Questions What exactly is a sum? How is a term used in mathematics? What does 'product' mean in math? How do you define a factor? What is a quotient? What role does a coefficient play in an expression? Can a term consist of only a variable? Is the sum always larger than the numbers added? Are factors always smaller than the number they divide? Can coefficients be negative? ## Using Mathematical Terms exercise Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the learning text Using Mathematical Terms . • ### Match the terms and definitions. Hints • Sum is used for addition • Product is used for multiplication • Quotient is used for division • A factor is a whole number that can be multiplied by another whole number to produce a given product • A coefficient is the number in front of a variable in a mathematical expression • A term is an expression that has numbers and/or variables Solution Here is the completed table. • ### What is the correct mathematical term? Hints The product is the result of multiplying two or more numbers. A coefficient is a number used to multiply a variable. The sum of 4 and 6 is 10. Solution Sum refers to the result of adding two or more numbers together. It's the total obtained when you combine quantities. For example, in the addition problem 3 + 5, the sum is 8. • ### Identify the coefficient in the term 7y. Hints A coefficient is a number used to multiply a variable. So the coefficient here is 5. What is y multiplied by? Solution A coefficient is a number used to multiply a variable, so the coefficient is 7. • ### Identify the product in the equation. Hints The product is the result of multiplying two factors. The product of 2 x 3 is 6. Solution The product is the result of multiplying two factors. • ### The sum of two numbers is 20. If one number is 9, what is the other number? Hints What number do we add to 9 to get 20? Or, what is 20 - 9? How many numbers do we count up from 9 to 20? Solution If we count up from 9 to 20, we count 11. 9 + 11 = 20, so the answer is 11. • ### Look at the bolded items. Match each term to the example. Hints Sum- the result of adding two or more numbers The quotient is found by dividing two numbers. You multiply two factors to find the product, or answer, of a multiplication equation. Solution The sum is the result of adding 2 or more numbers. • 8 + 6 = 14 A term is a single mathematical expression. • 3 + a A product is the result of multiplying 2 or more numbers. • 15 x 26 = 390 A factor is a number that divides another evenly. • 15 x 26 = 390 A quotient is the result of dividing one number by another. • 28 $\div$ 2 = 14 A coefficient multiplies a variable in an expression. • 5x + 8 = 2 Rating Ø 5.0 / 1 ratings The authors sofatutor Team Using Mathematical Terms CCSS.MATH.CONTENT.6.EE.A.2.B
# Recursively defined sequences ## Sequences and Series ### 2.2.3 Recursively defined sequences Now that we know we can interchange limits and algebraic operations, we can compute the limits of many sequences. One such class are recursively defined sequences, that is, sequences where the next number in the sequence is computed using a formula from a fixed number of preceding elements in the sequence. Example 2.2.8: Let{xn}be defined byx1:=2 and xn+1:=xn−x 2 n−2 2xn . We must first find out if this sequence is well-defined; we must show we never divide by zero. Then we must find out if the sequence converges. Only then can we attempt to find the limit. So let us prove xn exists and xn>0 for all n(so the sequence is well-defined and bounded below). Let us show this by . We know thatx1=2>0. For the induction step, suppose xn>0. Then xn+1=xn−x 2 n−2 2xn = 2x2nx2n+2 2xn = x2n+2 2xn . It is always true thatx2 n+2>0, and asxn>0, then x 2 n+2 2xn >0 and hencexn+1>0. Next let us show that the sequence is monotone decreasing. If we show thatx2n−2≥0 for alln, thenxn+1≤xnfor alln. Obviouslyx12−2=4−2=2>0. For an arbitrarynwe have x2n+1−2= x2 n+2 2xn 2 −2=x 4 n+4x2n+4−8x2n 4x2 n = x4 n−4x2n+4 4x2 n = x2 n−22 4x2 n . Since any square is nonnegative,xn2+1−2≥0 for alln. Therefore,{xn}is monotone decreasing and bounded (xn>0 for alln), and so the limit exists. It remains to find the limit. Write 2xnxn+1=x2n+2. Since{xn+1}is the 1-tail of{xn}, it converges to the same limit. Let us definex:=limxn. Take the limit of both sides to obtain 2x2=x2+2, orx2=2. Asxn>0 for allnwe getx≥0, and thereforex=√2. You may have seen the above sequence before. It is theNewton’s method for finding the square root of 2. This method comes up very often in practice and converges very rapidly. We used the fact thatx21−2>0, although it was not strictly needed to show convergence by considering a tail of the sequence. In fact the sequence converges as long asx16=0, although with a negativex1we would arrive atx=√2. By replacing the 2 in the numerator we obtain the square root of any positive number. These statements are left as an exercise. You should, however, be careful. Before taking any limits, you must make sure the sequence converges. Let us see an example. Example 2.2.9: Supposex1:=1 andxn+1:=x2n+xn. If we blindly assumed that the limit exists (call itx), then we would get the equationx=x2+x, from which we might concludex=0. However, it is not hard to show that{xn}is unbounded and therefore does not converge. The thing to notice in this example is that the method still works, but it depends on the initial valuex1. If we setx1:=0, then the sequence converges and the limit really is 0. An entire branch of mathematics, called dynamics, deals precisely with these issues. See . ### Some convergence tests It is not always necessary to go back to the definition of convergence to prove that a sequence is convergent. We first give a simple convergence test. The main idea is that{xn}converges toxif and only if{\|xn−x\|}converges to zero. Proposition 2.2.10. Let{xn}be a sequence. Suppose there is an xRand a convergent sequence {an}such that lim n→∞an=0 and \|xn−x\| ≤an for all n. Then{xn}converges andlimxn=x. Proof. Letε >0 be given. Note thatan≥0 for alln. Find anM∈Nsuch that for alln≥M we havean=\|an−0\|<ε. Then, for alln≥Mwe have \|xn−x\| ≤an<ε. As the proposition shows, to study when a sequence has a limit is the same as studying when another sequence goes to zero. In general, it may be hard to decide if a sequence converges, but for certain sequences there exist easy to apply tests that tell us if the sequence converges or not. Let us see one such test. First let us compute the limit of a very specific sequence. Proposition 2.2.11. Let c>0. (i) If c<1, then lim n→∞c n=0. (ii) If c>1, then{cn}is unbounded. Proof. First supposec<1. Asc>0, thencn>0 for allnNby . Asc<1, thencn+1<cn for alln. So we have a decreasing sequence that is bounded below. Hence, it is convergent. Let L:=limcn. The 1-tail{cn+1}also converges toL. Taking the limit of both sides ofcn+1=c·cn, we obtainL=cL, or(1c)L=0. It follows thatL=0 asc6=1. Now supposec>1. Suppose for contradiction that the sequence is bounded above byB>0, that is,cnBfor alln N. Then for alln, 1 c n = 1 cn ≥ 1 B >0. If we look at the above proposition, we note that the ratio of the(n+1)th term and thenth term isc. We generalize this simple result to a larger class of sequences. The following lemma will come up again once we get to series. Lemma 2.2.12(Ratio test for sequences). Let{xn}be a sequence such that xn6=0for all n and such that the limit L:= lim n→∞ \|xn+1\| \|xn\| exists. (i) If L<1, then{xn}converges andlimxn=0. (ii) If L>1, then{xn}is unbounded (hence diverges). IfLexists, butL=1, the lemma says nothing. We cannot make any conclusion based on that information alone. For example, the sequence {1/n}converges to zero, butL=1. The constant sequence{1}converges to 1, not zero, andL=1. The sequence{(1)n}does not converge at all, andL=1 as well. Finally, the sequence{n}is unbounded, yet againL=1. The statement may be strengthened, see exercises and . Proof. SupposeL<1. As \|xn+1\| \|xn\| ≥0 for alln, thenL≥0. Pickrsuch thatL<r<1. We wish to compare the sequence to the sequencern. The idea is that while the sequence is not going to be less thanLeventually, it will eventually be less thanr, which is still less than 1. The intuitive idea of the proof is illustrated in . 1 L r Figure 2.4:Proof of ratio test in picture. The short lines represent the ratios \|xn+1\| \|xn\| approachingL. Asr−L>0, there exists anM∈Nsuch that for alln≥Mwe have \|xn+1\| \|xn\| −L <r−L Therefore, fornM, \|xn+1\| \|xn\| −L<r−L or \|xn+1\| \|xn\| <r. Forn>M(that is fornM+1) write \|xn\|=\|xM\|\|xM+1\| \|xM\| \|xM+2\| \|xM+1\|··· \|xn\| \|xn−1\| <\|xM\|rr···r=\|xM\|r n−M = (\|x M\|r−M)rn. The sequence{rn}converges to zero and hence\|xM\|r−Mrnconverges to zero. By , theM-tail of{xn}converges to zero and therefore{xn}converges to zero. Now supposeL>1. Pickrsuch that 1<r<L. AsLr>0, there exists anMNsuch that for alln≥M \|xn+1\| \|xn\| −L <L−r. Therefore, \|xn+1\| \|xn\| >r. Again forn>M, write \|xn\|=\|xM\|\|xM+1\| \|xM\| \|xM+2\| \|xM+1\|··· \|xn\| \|xn−1\| >\|xM\|rr···r=\|xM\|r n−M = (\|x M\|r−M)rn. The sequence{rn}is unbounded (sincer>1), and therefore{xn}cannot be bounded (if\|xn\| ≤B for alln, thenrn< B \|xM\|r M for alln, which is impossible). Consequently,{xn}cannot converge. Example 2.2.13: A simple application of the above lemma is to prove lim n→∞ 2n n! =0. Proof: Compute 2n+1/(n+1)! 2n/n! = 2n+1 2n n! (n+1)!= 2 n+1. It is not hard to see that{n+12 }converges to zero. The conclusion follows by the lemma. Example 2.2.14: A more complicated (and useful) application of the ratio test is to prove lim n→∞n 1/n=1. Proof: Letε>0 be given. Consider the sequence(1+n ε)n . Compute (n+1)/(1+ε)n+1 n/(1+ε)n = n+1 n 1 1+ε. The limit of n+1n =1+1n asn∞is 1, and so lim n→∞ (n+1)/(1+ε)n+1 n/(1+ε)n = 1 1+ε <1. Therefore, n (1+ε)n converges to 0. In particular, there exists anNsuch that forn≥N, we have n (1+ε)n <1, or n<(1+ε) n, orn1/n<1+ε. Asn1, thenn1/n1, and so 0n1/n1<ε. Consequently, limn1/n=1. ### Exercises Exercise2.2.1: Prove . Hint: Use constant sequences and . Exercise2.2.2: Prove part of . Exercise2.2.3: Prove that if{xn}is a convergent sequence, k∈N, then lim n→∞x k n= lim n→∞xn k . Hint: Use . Exercise2.2.4: Suppose x1:=12 and xn+1:=x2n. Show that{xn}converges and findlimxn. Hint: You cannot divide by zero! Exercise2.2.5: Let xn:=n−cosn(n). Use the to show that{xn}converges and find the limit. Exercise2.2.6: Let xn:=n12 and yn:=1n. Define zn:=yxnn and wn:=yxnn. Do{zn}and{wn}converge? What are the limits? Can you apply ? Why or why not? Exercise2.2.7: True or false, prove or find a counterexample. If{xn}is a sequence such that{xn2}converges, then{xn}converges. Exercise2.2.8: Show that lim n→∞ n2 2n =0. Exercise2.2.9: Suppose{xn}is a sequence and suppose for some x∈R, the limit L:= lim n→∞ \|xn+1−x\| \|xn−x\| exists and L<1. Show that{xn}converges to x. Exercise2.2.10(Challenging): Let{xn}be a convergent sequence such that xn≥0and k∈N. Then lim n→∞x 1/k n = lim n→∞xn 1/k . Hint: Find an expression q such that x1n/k−x1/k xn−x = 1 q. Exercise2.2.11: Let r>0. Show that starting with any x16=0, the sequence defined by xn+1:=xn−x 2 n−r 2xn converges to√r if x1>0and−√r if x1<0. Exercise2.2.12: a) Suppose{an}is a bounded sequence and{bn}is a sequence converging to 0. Show that{anbn}converges to 0. b) Find an example where{an}is unbounded,{bn}converges to 0, and{anbn}is not convergent. Exercise2.2.13(Easy): Prove the following stronger version of , the ratio test. Suppose{xn} is a sequence such that xn6=0for all n. a) Prove that if there exists an r<1and MNsuch that for all n≥M we have \|xn+1\| \|xn\| ≤r, then{xn}converges to0. b) Prove that if there exists an r>1and MNsuch that for all n≥M we have \|xn+1\| \|xn\| ≥r, then{xn}is unbounded. Exercise2.2.14: Suppose x1:=c and xn+1:=xn2+xn. Show that{xn}converges if and only if−1≤c≤0, in which case it converges to 0. Exercise2.2.15: Prove lim n→∞(n 2+1)1/n =1. Exercise2.2.16: Prove that (n!)1/n is unbounded. Hint: Show thatCn ### Limit superior, limit inferior, and Bolzano–Weierstrass Outline Related documents
# How many sides are there in a Heptagon? 7 sides A. ### How many sides are there in a septagon? 7 • #### What is the name of a 9 sided shape? In geometry, a nonagon /ˈn?n?g?n/ (or enneagon /ˈ?niː?g?n/) is a nine-sided polygon or 9-gon. The name "nonagon" is a prefix hybrid formation, from Latin (nonus, "ninth" + gonon), used equivalently, attested already in the 16th century in French nonogone and in English from the 17th century. • #### What is a shape that has 9 sides? 2D Shapes Triangle - 3 SidesSquare - 4 Sides Pentagon - 5 SidesHexagon - 6 sides Heptagon - 7 SidesOctagon - 8 Sides Nonagon - 9 SidesDecagon - 10 Sides More • #### How many triangles are in a Heptagon? A Heptagon, also known as a septagon, has 7 sides. The number of triangles that can be drawn from 1 vertex in any polygon with n sides, is n-2. Thus you can draw 5 triangle from a single vertex of a Heptagon. B. ### What is the angle of Heptagon? To find the measure of the interior angles, we know that the sum of all the angles is 900 degrees (from above) And there are seven angles So, the measure of the interior angle of a regular heptagon is about 128.57 degrees. • #### Do all quadrilaterals have 360 degrees? The Quadrilateral Sum Conjecture tells us the sum of the angles in any convex quadrilateral is 360 degrees. Remember that a polygon is convex if each of its interior angles is less that 180 degree. • #### Is a hexagon regular or irregular? A hexagon is an example of a polygon, or a shape with many sides. A regular hexagon has six sides that are all congruent, or equal in measurement. A regular hexagon is convex, meaning that the points of the hexagon all point outward. All of the angles of a regular hexagon are congruent and measure 120 degrees. • #### What is the sum of the exterior angles of a Heptagon? To find the measure of the interior angles, we know that the sum of all the angles is 900 degrees (from above) And there are seven angles So, the measure of the interior angle of a regular heptagon is about 128.57 degrees. C. ### What is the shape of a Heptagon? A polygon is an enclosed two-dimensional shape with straight sides. This includes triangles, quadrilaterals, pentagons, etc but not circles or ovals. A heptagon is a seven-sided polygon. Along with seven sides, a heptagon has seven vertices and angles. • #### Are all 4 sided shapes are parallelograms? A parallelogram is a four-sided shape with opposite sides that are parallel. Because of the parallel lines, the opposites sides are equal in length. There are some unique properties of the angles inside parallelograms. First, the opposite angles are equal. • #### What is the shape of a hexagon? Shapes for kids: Hexagons. The hexagon is a six sided polygon. Each side must be closed and the shape not only needs six sides to qualify as a hexagon, but it must have six angles. • #### Is a hexagon regular or irregular? A hexagon is an example of a polygon, or a shape with many sides. A regular hexagon has six sides that are all congruent, or equal in measurement. A regular hexagon is convex, meaning that the points of the hexagon all point outward. All of the angles of a regular hexagon are congruent and measure 120 degrees. Updated: 2nd October 2019
Вы находитесь на странице: 1из 29 # Trigonometric Equation ## Theory Notes - Trigonometric Equation 1. TRIGONOMETRIC EQUATIONS An equation involving one or more trigonometrical ratios of unknown angle is called trigonometric equation e.g. cos2x 4 sinx = 1. It is to be noted that a trigonometrical identity is satisfied for every value of the unknown angle whereas, trigonometric equation is satisfied only for some values (finite or infinite in number) of unknown angle. e.g. sin2x + cos2x = 1 is a trigonometrical identity as it is satisfied for every value of x R. ## 2. SOLUTION OFATRIGONOMETRIC EQUATION A value of the unknown angle which satisfies the given equation is called a solution of the equation 1 e.g. / 6 is a solution of sin = . 2 3. GENERAL SOLUTION Since trigonometrical functions are periodic functions, solutions of trigonometric equations can be .in generalized with the help of the periodicity of the trigonometrical functions. The solution consisting of all possible solutions of a trigonometric equation is called its general solution. ps We use the following formulae for solving the trigonometric equations: (n I) LM OP te N 2 2 Q yS FG IJ ud ## tan = tan = n + where H , 2 2 K , n I . St LM OP,n I . sin2 = sin2 = n , where 0, N 2Q cos = cos = n , where LM0, OP,n I . 2 2 N 2Q tan = tan = n , where LM0, IJ ,n I . N 2K 2 2 Note: (For these type of equations students must use these solution to get correct answers) sin = 0 = n, cos = 0 = (2n + 1) , 2 tan = 0 = n, sin = 1 = (4n + 1) 2 sin = 1 = (4n 1) 2 Page 1 of 29 www.StudySteps.in Trigonometric Equation cos = 1 = 2n cos = 1 = (2n + 1) ## sin = sin and cos = cos = 2n + Note: Everywhere in this chapter n is taken as an integer, if not stated otherwise. The general solution should be given unless the solution is required in a specified interval or range. DRILL EXERCISE - 1 ## Solve the following trigonometric equation 1 3 1. tan + cot = 2 2. sin = and cos = 2 2 3. cos2 sin 1 4 =0 4. .in sin sin 5 sin 3 , where 0 ps 1 te 4 yS ud ## (a) Solution of equations by factorising.Consider the equation ; (2 sin x - cos x) (1 + cos x) = sin2 x. St ## (b) Solutions of equations reducible to quadratic equations. Consider the equation ; 3 cos2 x - 10 cos x + 3 = 0 ## (c) Solving equations by introducing an Auxilliary argument. Consider the equation ; sin x cos x 2 and 3 cos x sin x 2 . (d) Solving equations by Transforming a sum of Trigonometric functions into a product. Consider the example ; sin 5 x + sin 2 x - sin 4x = 0 (e) Solving equations by transforming a product of trigonometric functions into a sum. Consider the equation ; sin 5x . cos 3x = sin 6x . cos 2x. (f) Solving equations by a change of variable : (i) b g Equations of the form P sin x cos x,sin x.cos x 0 , where P(x, z) is a polynominal , can be solved by the change.cos x sin x = t 1 2sin x.cos x = t2 . Consider the equation ; sin x + cos x = 1 + sin x . cos x . Page 2 of 29 www.StudySteps.in Trigonometric Equation (ii) Equation of the form of a .sin x + b . cos x + d = 0, where a ; b and d are real numberes and a , b 0 can be solved by changing sin x and cos x into their corresponding tangent of half the angle.Consider the equation 3cos x + 4 sin x = 5. (iii) Many equations can be solved by introducing a new variable e.g. the equation FG 1 IJ 0 by substituting, H 2K sin4 2x + cos4 2x = sin 2x. cos 2x changes to 2( y 1) y sin 2x . cos 2x = y. (g) Solving equations with the use of the Boundness of the functions sin x and cos x. Consider FGx IJ FG x IJ H K H K the equation ; sin x cos 2 sin x 1 sin 2 cos x .cos x 0 . 4 4 ## 5. SOME IMPORTANT POINTS TO REMEMBER While solving a trigonometric equation, squaring the equation at any step should be avoided as far as possible. If squaring is necessary, check the solution for extraneous values. .in Never cancel terms containing unknown terms on the two sides, which are in product. It may cause loss of genuine solution. ps The answer should not contain such values of angles, which make any of the terms undefined. Domain should not be changed. If it is changed, necessary corrections must be incorporated. te Check that the denominator is not zero at any stage while solving equations. yS Some times you may find that your answers differ from those in the package in their notations. This may be due to the different methods of solving the same problem. Whenever you ud come across such situation, you must check their authenticity. This will ensure that your While solving trigonometric equations you may get same set of solution repeated in your St answer. It is necessary for you to exclude these repetitions, e.g. n + , (n I) forms a 2 k part of , k I the second part of the second set of solution (you can check by 5 10 k putting k = 5 m + 2 (mI). Hence the final answer is ,kI. 5 10 Some times the two solution set consist partly of common values. In all such cases the common part must be presented only once. Now we present some illustrations for solving the different forms of trigonometric equations. Which will highlight the importance of above mentioned points. Page 3 of 29 www.StudySteps.in Trigonometric Equation DRILL EXERCISE - 2 ## Solve the folloiwng trigonometric equations 1. cos sin cos 2 sin 2 2. sin 5x + sin 3x = sin 4x 13 3. sin2x + cos4x =0 4. 4 sin4x + 12 cos2 x = 7 16 1 x 5. (1 tan )(1 sin 2) 1 tan 6. sin x + sin 2x + sin 3 x = cot . 2 2 2 7. 2cos 2x = 3.2 cos x 4 . 8. sin6 2x + cos6 2x = 7/16. 9. sin2x + sin22x = 1 Illustration 1: Solve: 7cos2 + 3sin2 = 4 Solution: Given 7cos2 + 3sin2 = 4 .in ps or, 7cos2 + 3 (1 cos2) = 4 or, 4cos2 = 1 te 2 2 1 yS cos cos 2 2 3 ud n 3 n I Illustration 2: St ## Solve: 3tan ( - 150) = tan ( + 150) Solution: Given, 3tan ( - 150) = tan ( + 150) tan 150 3 or, tan 15 0 1 tan 15 0 tan 15 4 0 tan 15 tan 15 2 or, 0 0 (By componendo and dividendo) sin 15 0 15 0 2 sin 15 15 or 0 0 or, 2 sin2 = 2 or sin2 = 1 = sin 2 n 2 = n + (-1)n = + (- 1)n , n I 2 2 4 Page 4 of 29 www.StudySteps.in Trigonometric Equation Illustration 3: 1 Solve: cos cos2 cos3 = 4 Solution: 4cos cos2 cos3 = 1 or, (2cos3 cos) 2cos2 = 1 or, (cos4 + cos2) 2cos2 - 1 = 0 or, 2cos4 cos2 + 2cos22 - 1 = 0 or, 2cos4 cos2 + cos4 = 0 or, cos4 [2cos2 + 1] = 0 If cos4 = 0, 4 = (2n + 1) = (2n + 1) 2 8 If 2cos2 + 1 = 0 1 2 2 or, cos2 = = cos 2 2m 2 3 3 = m .in 3 Hence, = (2n + 1) or m where n, m I ps 8 3 Illustration 4: te ## Solve: tanx + tan2x + tan3x = 0. yS Solution: tanx + tan2x + tan3x = 0 ud ## or, tanx + tan2x + tan (x + 2x) = 0 tan x tan 2 x or, tanx + tan2x + 1 tan x tan 2x 0 St 1 or, (tanx + tan2x) 1 1 tan x tan 2x 0 If tanx + tan2x = 0, tanx = tan2x or, tanx = tan(2x) x = n + (2x) or, 3x = n n x= 3 1 If 1 + = 0 then, 1 tanx tan2x = 1 1 tan x tan 2 x 2 tan x or, tanx tan2x = 2 or, tanx 2 1 tan 2 x or, tan2x = 1 tan2x 1 or, 2tan2x = 1 or, tan2x = 2 1 1 1 or, tanx = or x = m tan 2 2 Page 5 of 29 www.StudySteps.in Trigonometric Equation x = m + n 1 Hence x = or m tan 1 , n , m I 3 2 Illustration 5 : Solve: 2sin2x 5sinx cosx 8cos2x = 2. Solution: In such problems we divide both sides by cos2x. This converts the given equation in a quadratic equation in tanx, which can be easily solved. Clearly, cosx 0 For if cosx = 0, then 2sin2x = 2 sin2x = 1 which is impossible. Given equation is 2sin2x 5sinx cosx 8cos2x = 2 or, 2tan2x 5tanx 8 = 2sec2x [dividing both sides by cos2x] or, 2tan2x 5tanx 8 + 2 (1 + tan2x) = 0 or, 4 tan2x 5tanx 6 = 0 or, 4tan2x 8tanx + 3tanx 6 = 0 or, 4tanx (tanx 2) + 3 (tanx 2) = 0 .in or, (tanx 2) (4tanx + 3) = 0 either tanx 2 = 0 ps tanx = 2 = tan (suppose) x = n + = n + tan12 te 3 or, 4tanx + 3 = 0 tanx = = tan (suppose) 4 yS 3 x = n + = m + tan1 . where n , m I 4 ud ## 6. SOLVING SIMULTANEOUS EQUATIONS Here we will discuss problems related to the solution of two equations satisfied simultaneously. We St ## may divide the problems in two categories. (i) Two equations in one unknown (ii) Two equations in two unknowns. Illustration 6: Find all values of lying between 0 and 2, satisfying the following equations, rsin = 3 and r + 4sin = 2 ( 3 + 1) Solution: Given equations are, rsin = 3 ... (i) and r + 4sin = 2( 3 + 1) ... (ii) To find the value of , we must eliminate r. 3 Now, from (i), r = sin Page 6 of 29 www.StudySteps.in Trigonometric Equation Substituting the value of r in (ii), we get, 3 4 sin 2 ( 3 1) sin or, 4sin2 2 3 sin 2sin + 3 = 0 or, 2sin (2sin 3 ) 1 (2sin 3 ) = 0 or, (2sin 3 ) (2sin 1) = 0 3 If 2sin 3 = 0, sin = = sin = n + (1)n 2 3 3 1 If 2sin 1 = 0, sin = = sin = n + (1)n 2 6 6 2 5 Values of lying between 0 and 2 are , , , 6 3 3 6 Illustration 7: Find the smallest positive values of x and y satisfying .in xy= , and cotx + coty = 2 4 Solution: ps Given xy= ... (i) 4 cotx + coty = 2 ... (ii) te ## From (ii), sin(x + y) = 2sinx.siny = cos (x y) cos(x + y) yS = cos cos (x + y) 4 ud 1 sin (x + y) + cos (x + y) = cos 4 2 St 1 1 1 sin (x + y) + cos (x + y) = 2 2 2 cos (x + y ) = cos 4 3 x+y = 2n 4 3 x + y = 2n . (iii) 3 4 7 for n = 0, x + y = (sin ce x , y 0) . (iv) 12 5 From (i) and (iv), x = , y 12 6 5 Hence least positive values of x and y are and respectively.. 12 6 Page 7 of 29 www.StudySteps.in Trigonometric Equation DRILL EXERCISE - 3 Solve the following simultaneous equations 1 3 1 2 1. sin x sin y = and cos x cos y = 2. yx= and cos (x) cos (y) = 4 4 4 2 1 1 3. cos (x y) = and cos (x + y) = 4. sin x + cos y = 1 and x + y = 2 2 3 1 5. tan x + tan y = 2 and cos x cos y = 2 7. TRIGONOMETRIC INEQUATIONS To solve trigonometric inequation of the type f(x) a, or f(x) a where f(x) is some trigonometric ratio we take following steps. (i) Draw the graph of f(x) in a interval length equal to fundamental period of f(x). (ii) Draw the line y = a. (iii) Take the portion of the graph for which inequation is satisfied. .in (iv) To generalise add pn (n I) and take union over set of integers , where p is fundamental period of f(x). ps Illustration 8: Find the solution set of the inequation sin x > 1/2. te Solution: When sinx = 1/2, the two values of x between 0 and 2 are /6 and 5/6. yS y ud y = 1/2 /6 5/6 x St ## From, the graph of y = sin x, it is obvous that, between 0 and 2 5 sinx > 1/2 for x . 6 6 Hence sin x > 1/ 2 2n + /6 < x < 2n+ 5/6 5 The required solution set = n 2 n , 2 n I 6 6 Page 8 of 29 www.StudySteps.in Trigonometric Equation Illustration 9: 1 Find the solution set of < tan x 1. 3 Solution: 1 < tan x 1 3 6 < x O 6 4 2 4 2 The required solution set will be n , n nI 6 4 DRILL EXERCISE - 4 .in Find the solution set for following trigonometric inequation 1 Sin Sin + cos < 1 ps 1. 2. 2 x te 1 3. log2 sin < 1 4. sin x < 2 2 yS ud ## 8. PROBLEMS BASED ON BOUNDARY CONDITIONS If the problem involves only one equation consisting of more than one variable or equation involves St variable of different natures then the boundary conditions of trigonometric functions is generally used. It must be noted that \|sinx\| 1 ; \| cosx \| 1; \|sec x\| 1; \| cosec x \| 1; \|tan x\| 0; \|cot x\| 0 Illustration 9: Solve sin2x + cos2 y = 2sec2z Solution: L.H.S. = sin2x + cos2y 2 R.H.S. = 2sec2z 2 Hence L.H.S. = R.H.S. only when sin2x = 1, cos2 y = 1, sec2 z = 1 cos2x = 0, sin2 y = 0, cos2 z = 1 cos x = 0, sin y = 0, sin z = 0 x = (2m + 1) , y = n and z = t where m, n, t are integers. 2 Page 9 of 29 www.StudySteps.in Trigonometric Equation Illustration 10: x Solve: 2 cos 2 sin 2 x x 2 x 2 , 0 x / 2 2 Solution: In this problem, terms on the two sides of the equation are different in nature, L.H.S. is in trigonometric form, whereas R.H.S. is in algebraic form. Hence, we will use boundary conditions. x L.H.S. = 2 cos2 sin 2 x 2 = (1 + cosx) sin2x < 2 (since 1 + cosx < 2 and sin2 x 1) 1 R.H.S. = x2 + 2 x2 Hence L.H.S. is never equal to R. H. S. Therefore, the given equation has no solution. DRILL EXERCISE - 5 x2 x 1. sin4 x cos7 x = 1. 2. Solve: 2 cos 2 2 x 2 x . .in 6 ## 3. 2cos x 2cos x = 4 4. sin2 x + cos2 y = 2 cosec2 z ps te 5. 2 sec2 x 1 = sin 3y yS ud St Page 10 of 29 www.StudySteps.in Trigonometric Equation Drill Exercise - 1 5 1. = m , nI 2. = 2n + , nI 4 6 2 5 3. = n + (1)n , n I 4. 0, , , , & 6 6 3 3 6 5 5. , 6. = n ,nI 3 3 3 Drill Exercise - 2 2 m n 1. 2 n or , m, n I 2. x= , x = 2m , n, m I 3 6 4 3 n 3. x= 4. x = 2n .in 2 6 4 FG IJ 2 n 5. H n or m K , m, n I 6. x , n I ps 4 7 7 n x , n I te 7. x n, n I 8. 4 12 yS 9. (2n + 1) , nI 6 ud Drill Exercise - 3 St 1. {( (6n + 6k 1)/6, (6n 6k 1)/6), ( (6n + 6k + 1)/6, (6n 6k + 1)/6}, (n, k I). ## 2. {n, (4n + 1)/4, ((4n 1)/4, n)} (n I) 3. {( (6k + 6n 1)/6, (2k 2n 1)/2) , ( (2k + 2n 1)/2, (6k 6n 1)/6}, (k, n I). 1 5 4. ((1) sin n 2 3 n, (1) n sin 1 2 3 n , (n I). 12 12 ## 5. {( (4m + 1)/4, (8n 4m + 1)/4)}, (n, m I) Page 11 of 29 www.StudySteps.in Trigonometric Equation Drill Exercise - 4 7 11 1. 2n 6 , 2n 6 2. 2n , 2n 2 nI nI ## 3. (4n, (12n + 1)/3 ( (12n + 1)/3, 2 + 4n), (n I) 5 n 4. 2n, 2n 6 2n 6 , (2n 1) 5. R , n I n I 4 Drill Exercise - 5 1. x 2n , n I , x n , nI 2. x=0 2 3. 4. x = (2n + 1) 2 .in , y = m, z = (2t + 1) , (n, m, t I) 2 ps 5. x = n, y = (4m + 1) , (n, m I) te 3 yS Example 1 : ud Solution: St ## Here sinx + cosx =2 2 sinx cosx = 2 sin2x (1) 1 1 or 2 ( 2 sin x 2 cos x ) 2 sin 2 x or sin x sin 2 x 4 or, 2x = n + (1)n x 4 Taking n even, n = 2m, m I, 2x = 2m + x + 4 x = 2m + where m 4 Taking n odd, n = 2m + 1, mI 2x = (2m + 1) x 4 Page 12 of 29 www.StudySteps.in Trigonometric Equation 2m 1 3x = (2m + 1) or x 4 3 12 1 1 3 Thus, x = 2m or 2m , where ml. 4 3 4 Example 2 : Find the general solution of the equation [cos (x/4) 2sinx] sin x + [1 + sin (x/4) 2 cosx] cosx = 0 Solution: From the given equation x x sin x cos 4 sin 4 cos x 2 (sin x cos x ) cos x 0 2 2 or, sin (x + x/4) 2 + cosx = 0 or, sin(5x/4) + cosx 2 = 0 (1) .in Which is possible only if sin(5x/4) and cos x both are equal to their maximum value 1. Thus (1) is satisfied if sin (5x/4) = 1 (2) ps and cos x = 1 (3) te ## From (2), sin (5x/4) = 1 yS 5x/4 = 2n + /2 or x = 2 (4n + 1) /5, where nI (4) ud ## and from (3), cos x = 1 x = 2m, where m I (5) St Now the value of x satisfy both (2) and (3) are the value of x which are common in (4) and (5). Thus we require the integral values of m and n, such that 2(4n + 1) /5 = 2m m = (4n + 1)/5 (5) Since both m and n are integers, therefore we can take only those integral values of n for which 4n + 1 is divisible by 5. Obviously n = 1 is the least positive integral value of n for which 4n + 1 is divisible by 5. Thus we can take n = 1 + 5 l, l I . From (4), x = (2/5) [4(1 + 5l ) + 1] = 2(4l + 1) , where lI Example 3 : Solve the equation cos2 [ (sin x 2 cos x )] tan ( x tan x ) 1 2 2 2 4 4 Solution: Given cos2 [ (sin x 2 cos x )] tan ( x tan x ) 1 2 2 2 4 4 Page 13 of 29 www.StudySteps.in Trigonometric Equation or, sin 2 { (sin x 2 cos 2 x )} tan 2 ( x tan 2 x ) 0 4 4 It is possible only when sin 2 { (sin x 2 cos 2 x )} 0 (1) 4 and tan 2 x tan 2 x 0 (2) 4 from equation (1) sin2 (sin x 2 cos x ) = 0 2 (sin x + 2 cos 2 x ) = n, n I 4 or, sinx + 2 cos2x = 4n \|sinx + cos2x \| \|sin x\| + \|cosx\|2 1 + < 4 The equation has no solution for n 0 we consider n = 0 sinx + cos2x = 0 .in ps i.e., sin2x sinx 1 = 0 or, (sinx 2 ) ( 2 sinx + 1) = 0 te sinx 2 yS 1 sinx = x = 2k /4, . k I 2 Also these values of x satisfy the equation (2), therefore the general solution of given equation is ud x = 2k , kI. St Example 4 : Find the general solution of the equation sin4x + cos4x = sinx cosx Solution: The given equation can be rewritten as 4sin4x + 4cos4x = 4sinx cosx or, (1 cos2x)2 + (1 + cos2x)2 = 2sin2x or, 2(1 + cos22x) = 2sin 2x 1 + cos22x = sin 2x or, 1 + 1 sin22x = sin2x sin22x + sin2x = 2 This relation is possible if and only if sin2x = 1 (4n 1) or, 2x = 2n + x n ( n I) 2 4 4 Page 14 of 29 www.StudySteps.in Trigonometric Equation Example 5 : tan + tan( + (/3)) + tan( + (2/3)) = 3 Solution: From the given equation, tan tan( / 3) tan tan(2 / 3) tan + 1 tan tan( / 3) 1 tan tan(2 / 3) 3 tan 3 tan 3 or, tan 3 1 3 tan 1 3 tan ## tan (1 3 tan 2 ) (tan 3).(1 3 tan ) (tan 3) (1 3 tan ) or, =3 (1 3 tan ) (1 3 tan ) 3 tan tan 3 or, 3 3 1 3 tan 2 or, tan3 = 1 = tan(/4) or, 3 = n + (/4) or, = (4n + 1) (/12), where n I Example 6 : .in ps 1 1 Solve x and y : 4 sin x 3 cos y 11 , sinx 5.16 2.3 cos y 2 te Solution: yS 1 Let, 4sinx = , 3 cos y ud ## Then the equation becomes + = 11 (1) 52 2 = 2 St (2) On solving we get 12 = 2, 5 If = 2, 4 sin x 2; 2 2 sin x 2 ; 2sinx = 1; 1 sin x = 2 12 12 If = , then 4sinx = which is impossible as 4sinx > 0 5 5 When = 2, we get = 11 2 = 9 1 3 cos y 9 32 Page 15 of 29 www.StudySteps.in Trigonometric Equation 1 1 2; cosy = cos y 2 1 1 Thus we have sinx = , cos y 2 2 x = n + (1)n and y = 2m , where m, n I 6 3 Example 7 : Find all values of for which the equation sin4x + cos4x + sin2x + = 0 has atleast one solution. Also find the general solution of the equation for that . Solution: Here (sin2x + cos2x)2 2sin2x. cos2x + sin2x + = 0 1 2 or, 1 sin 2 x sin 2 x 0 2 or, sin22x 2sin2x 2(1 + ) = 0 sin2x = 2 4 8(1 ) 2 1 2 3 .in (1) ps 3 But sin 2x is real; so 2 + 3 0, i.e., te 2 Also, 1 sin2x 1 yS 11 2 3 1 2 3 1 ud As 1 + 3 So 1 + 2 3 = 1 = St 2 Also 1 1 2 3 1 0 2 3 2 3 1 , 2 2 ## from (1), sin 2x = 1 2 3 3 1 where , 2 2 2x = n + (1)n, nI and sin = 1 2 3 x= n 2 (1) n sin 1 1 2 3 Page 16 of 29 www.StudySteps.in Trigonometric Equation 3 1 where n and , 2 2 Example 8 : Solve for and : tan + tan = 4 tan3 + tan3 = 2 Solution: Here, tan + tan = 4 or, (tan 2) + (tan 2) = 0 Putting tan = x + 2 and tan = y + 2, then x + y = 0 (1) again, tan3 + tan3 = 2 ## 3 tan tan 3 3 tan tan 3 or, =2 1 3 tan 2 1 3 tan 2 or, .in 3tan 3tan3 9tan tan2 + 3tan3tan2 + 3tan tan3 9tan2.tan + 3tan2. tan3 = 2(1 3tan2 3tan2 + 9tan2 tan2 ps or, 3.4 {(tan + tan)3 3tan.tan (tan + tan)} 9.4 tan tan + 3.4 tan2 tan2 = 2 6{(tan + tan)2 2tan. tan} + 18 tan2. tan2 te yS ## = 2 96 + 12 tan. tan+ 18 tan2. tan2 or, 6tan2.tan2 + 36 tan tan 42 = 0 ud tan. tan == 7, 1 St ## or, (x + 2) (x + 2) = 7, 1 (using (1)) or, 4 x2 = 7, 1 ; x2 = 11, 3 Now, x2 = 11 x = 11 tan = 2 11 = n + tan1 (2 11 ) where nI. x2 = 3 x = 3 tan = 2 3 = n + tan1 (2 3 ) where n I Also, tan = 4 tan = 2 11 , 2 3 = m + tan1 ( 2 11 ) or m + tan1 (2 3 ), where m I. Example 9 : Page 17 of 29 www.StudySteps.in Trigonometric Equation ( 3 sin 2 x cos 2 x 2 ) For x ( , ), solve the equation ( 3 sin x cos x ) =4 Solution: The given equation is ( 3 sin 2 x cos 2 x 2 ) or, ( 3 sin x cos x ) 4 ## ( 3 sin 2 x cos 2 x 2 3 sin x cos x ) or, 2 sin x 6 =4 2 sin x 6 or, 2 sin x 6 4 Hence, 2 sin x 2 6 or, sin x 1 or, x + 2n 6 6 2 or, x = 2n 2 6 .in ps As x ( , ) , te 2 x= and x = are the solutions of the given equation 3 3 yS Example 10 : ud ## Solve for x and y: 12 sinx + 5 cosx = 2y2 8y + 21 St Solution: 12sinx + 5cosx = 2y2 8y + 21 12 5 12 2 5 2 sin x cos x 2( y 2 4 y 4) 13 13 13 5 13cos(x ) = 2(y 2)2 + 13, where cos = 13 Clearly, LHS 13 because the greatest value of cos (x ) is 1 Also RHS 13 because the least value of RHS is 13 the equation can hold if the value of each side = 13 Thus cos(x ) = 1 and y = 2 x = 2n and y = 2 x = 2n + and y = 2 x = 2n + cos1 and y = 2, where nI. 13 Page 18 of 29 www.StudySteps.in Trigonometric Equation Example 11 : 2 3 Find the solution set of the system of equations, x + y = and cos x + cos y = , where x 3 2 and y are real. Solution: 2 x + y = 3 , cos x + cos y = 23 2 cos x 2 y . cos x 2 y = 23 2 cos . cos 3 xy 2 = 23 [x+y= 3 ] 2 ## cos x 2 y = 23 which is not possible . Example 12 : Find the values of x, 0 x 2 , such that sin x + sin 2x + sin 3x = cos x + cos 2x + cos 2x. .in ps Solution: The given equation can be written as te ## (sin x + sin 3x) + sin 2x = (cos x + cos 3x) + cos 2x 2 sin 2x cos x + sin 2x = 2 cos 2x cos x + cos 2x yS ## sin 2x (2 cos x + 1) cos 2x (2 cos x + 1) = 0 (sin 2x cos 2x) (2 cos x + 1) = 0 That is, either sin 2x cos 2x = 0 or 2 cos x + 1 = 0. In former case ud tan 2x = 1 2x = n (n I) 4 St (4n 1) 5 9 13 x x , , , 8 8 8 8 8 1 2 4 If 2 cos x + 1 = 0, then cos x = , that is x or 2 3 3 Example 13 : If 32 tan 8 2 cos 2 3cos and 3cos 2 1 , then find the general value of . Solution : 1 Given 3cos 2 1 or cos 2 3 1 1 1 cos 2 31 Now, tan 2 .........(i) 1 cos 2 1 1 2 3 Page 19 of 29 www.StudySteps.in Trigonometric Equation ## Now 32 tan 8 2 cos 2 3cos 4 1 or 2 cos 3cos 32 2 2 or, 2 cos 2 3cos 2 0 2 or 2 cos 2 4 cos cos 2 0 or, (cos 2) (2 cos 1) 0 1 2 or 2cos 1 0 [ cos 2] or, cos c os 2 3 2 2n , where n 0, 1, 2,......... 3 Example 14 : Solve the equation e cos x = ecos x + 4 Solution : Given equation is ecos x = ecos x + 4 1 or, z 4 0 , where ecosx = z (suppose) .in z or, z2 4z 1 = 0 ps 4 16 4.1(1) z 2 te z 2 5 yS z 2 5 ecos x = 2 5 ud cos x = log e 2 5 1 and is imposible St no solution Example 15 : For what value of k the equation sin x + cos(k + x) + cos(k x) = 2 has real solutions ? Solution : Given equation is sin x + cos(k + x) + cos(k x) = 2 or sin x + 2 cos k . cos x = 2 or 2 cos x . cos x + sin x = 2 This equation is of the form a cos x + b sin x = c Here a = 2 cos k, b = 1 and c = 2 Since for real solutions, \| c \| a 2 b 2 \| 2 \| 1 4 cos 2 k or 2 1 4 cos 2 k 3 1 1 cos 2 k sin 2 k sin 2 k 0 4 4 4 1 1 1 1 or sin k sin k 0 sin k 2 2 2 2 Page 20 of 29 www.StudySteps.in Trigonometric Equation n k n 6 6 .in ps te yS ud St Page 21 of 29 www.StudySteps.in Trigonometric Equation ## SOLVED OBJECTIVE EXAMPLES Example 1 : 1 If sinx + cosx = y y , x [0, ] , then (a) x = /4 (b) x = 2 (c) x = (d) x = 3/4 6 Solution: 1 y 2 and \|sinx + cosx\| y 2 1 Hence y + 2 and sinx + cosx = 2 , which is possible for y = 1, x = /4. y Hence (a) is the correct answer. .in ps Example 2 : If x2 4x + 5 siny = 0, y[0, 2), then te (a) x = 1, y = 0 (b) x = 1, y = /2 yS (c) x = 2, y = 0 (d) x = 2, y = /2 Solution: ud (x 2)2 + 1 = sin y x = 2, sin y = 1 x = 2, y = /2 St ## Hence (d) is the correct answer Example 3 : The set of all x in (, ) satisfying \|4sinx 1\| < 5 is given by 3 3 (a) , (b) , 10 10 10 10 3 (c) , (d) none of these 10 10 Solution: We have \|4sinx1\| < 5 5 < 4sinx 1 < 5 5 1 5 1 4 sin x 4 Page 22 of 29 www.StudySteps.in Trigonometric Equation sin sin x cos 10 5 sin sin x sin 10 2 5 3 sin sin x sin 10 10 3 x , 10 10 Hence (a) is the correct answer. Example 4 : The values of x between 0 and 2, which satisfy the equation sin x 8 cos 2 x 1 are in A.P. with common difference (a)/4 (b)/8 .in (c) 3/8 (d) 5/8 Solution: ps We have sin x 8 cos 2 x 1 1 te sinx \|cosx\| = 2 2 yS Case I when cosx > 0 ud 1 In this case sinx cosx = 2 2 St 1 3 9 13 sin2x = + 2x , , , 2 4 4 4 4 3 9 13 x= , , , 8 8 8 8 3 As x lies between 0 and 2 and cos x > 0, x = , 8 8 CaseII 1 When cosx < 0 . In this case sinx \|cosx\| = 2 2 1 1 sinx cosx = or sin 2 x 2 2 2 5 7 13 15 5 7 x= , , , x= , as cosx < 0 8 8 8 8 8 8 Page 23 of 29 www.StudySteps.in Trigonometric Equation 3 5 7 Thus the values of x satisfying the given equation which lie between 0 and 2 are , , , 8 8 8 8 These are in A.P. with common difference 4 Hence (a) is the correct answer. Example 5 : The number of points inside or on the circle x2 + y2 = 4 satisfying tan4x + cot4 x + 1 = 3 sin2y is (a) one (b) two (c) four (d) infinite Solution: tan4x + cot4x + 1 = (tan2x cot2x)2 + 3 3 3sin2y 3 tan2 x = cot2x, sin2 y = 1 tanx = 1, siny = 1 x = /4, 3/4, But x 4 2 x 2 x = /4 only 2 ## siny = 1 y = /2, 3/2, .. But y2 4 y = /2 only .in ps So four solutions are possible, te ## Hence (c) is the correct answer. yS Example 6 : The equation 3sin2x + 10 cos x 6 = 0 is satisfied if ( n I ) ud ## (a) x = n + cos1(1/3) (b) x = n cos1(1/3) St n (c) x = 2n cos1 (1/3) (d) x = cos1 (1/3) 2 Solution: The given equation is equivalent to 3(1 cos2x) + 10 cos x 6 = 0 3cos2x 10 cosx + 3 = 0 (3cosx 1) (cos x 3) = 0 Therefore cos x = 1/3 (because cos x 3). Hence x = 2n cos1 (1/3), n I Hence (c) is the correct answer. Example 7 : The equation 2 sin x cos 2 x 2 sin x sin 2 x cos 2 x sin 2 x has a root for which 2 2 (a) sin 2x = 1 (b) sin 2x = 1 (c) cos x = 1/2 (d) cos 2x = 1/2 Page 24 of 29 www.StudySteps.in Trigonometric Equation Solution: The given equation can be written as x 2 sin (cos 2 x sin 2 x ) cos 2 x sin 2 x 2 x x or 2 sin cos 2 x cos 2 x 2 sin 1 cos 2 x 0 2 2 Hence cos 2x = 0 or sin (x/2) = 1/2. That is , 2x = n + /2 or x/2 = k + (1)k (n, k I). In other words, 6 n x or x = 2k + (1)k 2 4 3 n If x = , then sin 2x = 1, and if x = 2k + (1)k , 2 4 3 1 2 1 .in cos x = cos and cos 2x = cos 3 2 3 2 Hence (d) is the correct answer. ps Example 8 : te A solution of the equation (1 tan) (1 + tan) sec2 + 2 tan 2 = 0 where lies in the interval yS ## (/2, /2) is given by ud (a) = 0 (b) 3 St (c) = (d) 4 6 Solution: (1 tan) (1 + tan) sec2 + 2 tan 0 2 (1 tan4) + 2 tan 2 = 0 2t = t2 1 ## (where t = tan2 ) By inspection (or by graph) we find y = 2t and y = t2 1 intersect in t = 3 tan2 = 3 tan = 3 8 y = 2t = /3 as shown in the figure 1 y = t21 0 3 t Hence (b) is the correct answer. 1 Page 25 of 29 www.StudySteps.in Trigonometric Equation Example 9 : Number of solutions of the equation tanx + secx = 2 cosx lying in the interval [0, 2] is (a) 0 (b) 1 (c) 2 (d) 3 Solution: The given equation can be written as 1 sin x = 2cosx 1 + sin x = 2 cos2 x = 2 (1 sin2x) cos x 2sin2x + sinx 1 = 0 (1 + sinx) (2 sinx 1) = 0 sin x = 1 or 1/2 3 1 Now sinx = 1 x = for which the given equation is not meaningful and x = 2 2 5 x= or 6 6 The required number of solutions are 2. .in Hence (c) is the correct answer. ps Example 10 : The number of solutions of the equation 2cos = \|sinx\| , , is te 2 2 yS ## (a) zero (b) 2 (c) 4 (d) more than four ud Solution: We have 2cos = \|sin x\| St ## It is true only for \|sin x\| = 1 sin x = 1 So, x = 2n /2 x [2, 2] 3 3 x= , , , 2 2 2 2 Then the no. of solutions = 4. Example 11 : For what and only what values of lying between 0 and is the inequality, sin cos3 > sin3 cos valid ? (a) 0 , (b) 0 , 4 2 (c) , (d) none of these 4 2 Solution: Page 26 of 29 www.StudySteps.in Trigonometric Equation We have sin cos3 > sin3 cos sin cos (cos2 sin2 ) > 0 sin cos (1 tan2 ) > 0 ( sin > 0 for 0 < < ) cos (1 tan ) > 0 2 ## cos > 0 and 1 tan2 > 0 cos < 0 and 1 tan2 < 0 3 0 , or , 4 4 Example 12 : tan x = tan x if : 2 k 1 (a) x 2 k 1 , k (b) x k , 2 2 2 k 1 2 k 1 (c) x k , (d) x , k , k N 2 2 Solution: .in R.H.S. 0 for all x, the given condition is true for those values of x which lie in the I or III quadrant and the values of x given by A abd B satisfy these conditions. ps Example 13 : te 1 The number of solutions of the equation, cot x = cot x + (0 x 2) is : sin x yS (a) 0 (b) 1 (c) 2 (d) 3 ud Solution: St If cos x > 0 1 then (impossible) sin x Now if cot x < 0 1 then cot x = cot x + sin x 2 cos x 1 =0 sin x 1 cos x = 2 2 x = 2 n , n I and 0 x 2 3 2 4 then x= , 3 3 Page 27 of 29 www.StudySteps.in Trigonometric Equation Example 14 : The general solution of, sin x 3 sin 2x + sin 3x = cos x 3 cos 2x + cos 3x is n (a) n + (b) 8 2 8 n 3 (c) ( 1)n (d) 2 n + cos 1 2 8 2 Solution: sin x - 3 sin 2x + sin 3x - cos x - 3 cos 2x + cos 3x 2 sin 2x cos x - 3 sin 2x = 2 cos 2x cos x - 3 cos 2x (2 cos x - 3) sin 2x = cos 2x (2 cos x - 3) 2 cos x - 3 0 tan 2x = 1 n x = 2 8 Example 15 : The equation, cos 2x + a sin x = 2a 7 possesses a solution if : (a) a < 2 (c) a > 6 (b) 2 a 6 .in (d) a is any integer ps Solution: te k = a (say) k sin x + (1 - 2 sin 2 x) = 2 k - 7 yS 2 sin2 x - k sin x + 2 (k - 4) = 0 k k2 16 k 64 k (k 8) ud sin x = 4 = 4 = 12 (k - 4) , 2 k4 sin x 2 sin x = St 2 k4 1 2 1 2k6 Page 28 of 29 www.StudySteps.in Trigonometric Equation .in ps te yS ud St Page 29 of 29 www.StudySteps.in
13 teachers like this lesson Print Lesson ## Objective #### Big Idea The main idea of this lesson is that students compare dividing radicals by hand without rationalizing and realize why rationalizing came about and how it works. ## Warm Up 20 minutes This Warm up is intended to take about 20 minutes for the students to complete and for me to review with the class. This lesson is based from the History of Mathematics and the story of rationalizing. When students enter the classroom: • I have a square root table posted at the front of the room. • We discuss the perfect squares as I post them on the board. I ask, what is the value of square root of two? • Square root of one is one • Square root of four is two • So square root of 2 and square root of 3 are between one and two • Looking at the square root table, square root of two is approximately 1.414 that we will use for our first exercise today. I tell students that the calculator is a relatively recent invention. Before its arrival students had to calculate the fraction "1 divided by the square root of two" by hand. So, that is their first task today, to divide 1 by the square root of two. I tell them to approximate the square root of two with the rational number 1.414. 1. Students write 1 divided by 1.414 on their individual white boards and begin working. This problem is very difficult for students. They begin struggling of what to do with the decimal. It is interesting to watch as I walk around the room and try to prompt them with questioning. 2. Finally some students realize to move the decimal in the divisor of 1.414 three places, and then have to figure out to move the decimal after 1 three places. If they finally get to the problem they need, the student is having to divide 1000 by 1414. 3. Next, they must add zeros after the 1000.000 and begin dividing into 10000 first which is the tenths place. I did not have very many students get the correct answer, but it does help them to appreciate the complexity of the algorithm for dividing radicals (without a calculator). I have one sample of student work in the resources (Note: not a correct answer). I do not reveal the answer before we move on to work problems two and three. I do let the students know that to get a common denominator, you multiply by 4/4 or 3/3 or 2/2 because this is equivalent to multiplying an expression by one, using the Identity Property. I introduce rationalizing on top of this foundation. I begin our work on Problem 2 by saying, "Mathematicians began rationalizing to change the form of the fraction so that it is easier to divide by hand. This was important before the invention of the calculator." Then I ask, "What radical can we multiply by to multiply by a one?" With some assistance, the students agree on square root of two divided by square root of two. So now I ask the students to divide the square root of two divided by 2. So students write out 1.414 divided by 2, which is much easier for them. Most students were successful at finding that the fraction was equal to .707. This Warm Up activity takes time, but it helps students remember why to rationalize the denominator when it has a radical. It is not mathematically incorrect to leave a radical in the denominator. But, there are operations where it is helpful to have the number written in this form. Since these operations were once common, the practice of rationalizing the denominator was standardized, although it is less necessary these days. ## Independent Practice 30 minutes After the Warm Up activity, for Independent Practice I give my students a Rationalizing Denominators Worksheet. There are twenty-four problems on this practice, so it will take the students about 30 minutes to complete. This worksheet also helps my students to review their perfect squares and the writing of ratios. It is intended to reinforce the discussion of rationalizing the denominators of fractions to simplify radical expressions. This always seems to cause the students difficulty, so I am hoping the history lesson helps them remember the not only the procedure, but why we are rationalizing. I want the students to recognize that the form is being changed, but not the number value. After the students have completed the practice, I have them self-grade their own papers using a pen or colored pencil. No erasers are allowed at this time. Then I have them hand the worksheet in so that I can check their progress. ## Exit Slip 10 minutes This Exit slip only takes about 10 minutes for the students to complete. I use it as a quick formative assessment to check student understanding on being able to not only rationalize the denominator, but explain the reasoning behind it. I show a student sample of the exit slip in the resource section. This student rationalized the fraction correctly, and expressed reasoning of why to rationalize. Even though most of the students rationalized correctly, several of the students skip over the parts that I ask them to write about the math. Purposely planning for writing and discussing math to occur on a daily basis helps students to know it is expected every day.
Thursday, September 21, 2023 # What Is Initial Value In Math ## New York State Common Core Math Grade 8 Module 6 Lesson 2 Initial Value of a Linear Function Lesson 2 Summary When a linear function is given by the equation of a line of the form y = mx +c, the rate of change is m and initial value is b. Both are easy to identify. The rate of change of a linear function is the slope of the line it represents. It is the change in the values of y per a one unit increase in the values of x. • A positive rate of change indicates that a linear function is increasing. • A negative rate of change indicates that a linear function is decreasing. • Given two lines each with positive slope, the function represented by the steeper line has a greater rate of change. The initial value of a linear function is the value of the y-variable when the x value is zero. In 2008, a collector of sports memorabilia purchased 5 specific baseball cards as an investment. Let y represent the cards resale value and x represent the number of years since purchase. Each of the cards’ resale values after 0, 1, 2, 3, and 4 years could be modeled by linear equations as follows: Card A: y = 5 – 0.7x Card B: y = 4 + 2.6x Card C: y = 10 + 0.9x Card D: y = 10 – 1.1x Card E: y = 8 + 0.25x 1. Which card are decreasing in value each year? How can you tell? 2. Which card had the greatest initial values at purchase ? 3. Which card is increasing in value the fastest from year to year? How can you tell? 4. If you were to graph the equations of the resale values of Card B and Card C, which card’s graph line would be steeper? Explain. ## Existence And Uniqueness Of Solutions The PicardâLindelÃ¶f theorem guarantees a unique solution on some interval containing t0 if f is continuous on a region containing t0 and y0 and satisfies the Lipschitz condition on the variable y.The proof of this theorem proceeds by reformulating the problem as an equivalent integral equation. The integral can be considered an operator which maps one function into another, such that the solution is a fixed point of the operator. The Banach fixed point theorem is then invoked to show that there exists a unique fixed point, which is the solution of the initial value problem. An older proof of the PicardâLindelÃ¶f theorem constructs a sequence of functions which converge to the solution of the integral equation, and thus, the solution of the initial value problem. Such a construction is sometimes called “Picard’s method” or “the method of successive approximations”. This version is essentially a special case of the Banach fixed point theorem. Hiroshi Okamura obtained a necessary and sufficient condition for the solution of an initial value problem to be unique. This condition has to do with the existence of a Lyapunov function for the system. A simple example is to solve y ## Introduction To Differential Equations #### Part 2: Initial value problems Our mathematical model describing the spread of the rumor consists of two parts. The first is the differential equation dS/dt = k S S = 2 This second condition is called an initial condition — the value of dependent variable at the first value of the independent variable under consideration. A differential equation together with an initial condition is called an initial value problem. The implied “problem” is to find a function that satisfies both conditions. We need to discuss the term “solution.” A solution to a differential equation is a function that satisfies the relation for all values of the independent variable under consideration. For example, Y = exp is a solution of the differential equation dY/dt = Yt The function Y = 3 exp is another solution to this differential equation. A solution to an initial value problem is a solution to the differential equation that also satisfies the initial condition. So, Y = exp is a solution to the initial value problem dY/dt = YtY = 1 but, Y = 3 exp is not. One can find a symbolic description for the solution of the initial value problem dS/dt = k S tS = 2 Here is such a description: For now, we will not discuss how you could find such a solution. Rather, we will just verify that it works. • Use your computer algebra system to check that the given function S is indeed a solution to the initial value problem • dS/dt = k S tS = 2 Function Recommended Reading: Who Are Paris Jackson’s Biological Parents ## What Is An Initial Condition An initial condition is a starting point; Specifically, it gives dependent variable values for a certain independent variable. It allows you to zoom in on a specific solution. In general, an initial condition can be any starting point. For example, you might want to define an initial pressure or a starting balance in a bank account. In statistics, its a nuisance parameter in unit root testing . In calculus, the term usually refers to the starting condition for finding the particular solution for a differential equation. ## Laplace Transforms Of Derivatives In the rest of this chapter well use the Laplace transform to solve initial value problems for constant coefficient second order equations. To do this, we must know how the Laplace transform of \ is related to the Laplace transform of \. The next theorem answers this question. ##### Theorem 8.3.1 Suppose \ is continuous on \\) and of exponential order \, and \ is piecewise continuous on \.\) Then \ and \ have Laplace transforms for \ and Proof We know from Theorem 8.1.6 that \\) is defined for \. We first consider the case where \ is continuous on \\). Integration by parts yields \ &= e^f-f+s\int^T_0 e^f\,dt \label \end \] for any \. Since \ is of exponential order \, \=0\) and the integral in on the right side of Equation \ref converges as \ if \. Therefore which proves Equation \ref. Suppose \ and \ is only piecewise continuous on \, with discontinuities at \. For convenience, let \ and \. Integrating by parts yields Summing both sides of this equation from \ to \ and noting that yields Equation \ref, so Equation \ref follows as before. ##### Example 8.3.1: Recommended Reading: Formal Charge Of Cf4 ## Use In Differential Equations The initial condition in a differential equation is usually what is happening when the initial time is at zero . For example, lets say you have some function g, you might be given the following initial condition: • g = 40 • g = 32 An initial condition leads to a particular solution; If you dont have an initial value, youll get a general solution. Watch the video for two examples: A second order differential equation with an initial condition. initial value problem For example, the differential equation needs a general solution of a function or series of functions :dydx = 19×2 + 10But if an initial condition is specified, then you must find a particular solution . For example:dydx19x2 + 10; y = 5.Finding a particular solution for a differential equation requires one more stepsimple substitutionafter youve found the general solution. ## What Is It Initial Value the x-axis the y-axis Step-by-step explanation: 1. What are the coordinates of the point that is plotted by moving 4 right from the origin and 2 down? 2 .Match each pair of coordinate points to their location on the coordinate plane. x-value is 6 and y-value is -1 x-value is y-value is 0 x-value is 3 and y-value is 5 x-value is -4 and y-value is 5 x-value is -2 and y-value is -7 3. Which of the following statements abut the coordinate plane is true? The origin has an x-value of 0 and any y-value. The origin has an x-value of 0 and any y-value. Positive y-values are above the origin. Positive y-values are above the origin. Negative y-values are to the right of the origin. Negative y-values are to the right of the origin. Negative x-values are to the right of the origin. Negative x-values are to the right of the origin. 4. What is the horizontal or vertical distance between the points and ? 5. Select the inequality symbol that should be used to represent the following situation. To ride the roller coaster, you must be at least 64 inches tall. 6. At least 83 students attended the play. Which inequality represents this situation? You May Like: What Does K Mean In Physics ## On Definition Of Solution Of Initial Value Problem For Fractional Differential Equation Of Variable Order • Department of Mathematics, China University of Mining and Technology Beijing, Beijing 100083, China • 2. School of Science, Shandong Jiaotong University, Jinan, 250357, China • Received: 27 December 2020Accepted: 13 April 2021 22 April 2021 • MSC : 26A33 • We propose a new definition of continuous approximate solution to initial value problem for differential equations involving variable order Caputo fractional derivative based on the classical definition of solution of integer order differential equation. Some examples are presented to illustrate these theoretical results. Keywords: Citation: Shuqin Zhang, Jie Wang, Lei Hu. On definition of solution of initial value problem for fractional differential equation of variable order. AIMS Mathematics, 2021, 6: 6845-6867. doi: 10.3934/math.2021401 ## Students Are Also Searching For Grade 8 Math #4.2b, Slope, Rate of change and Initial value b • hair would be dry and brittle without the presence of • during which changes of state do atoms overcome the attractive forces between them? • which of the following statements about factors of production is the most accurate? If you have more homework to do you can use the search bar to find the answer to other homework: 40 have done it today and 15 in the last hour.
To Go Types of Numbers # Properties of Exponents ## A Little Bit About Zero 0 raised to any positive exponent is 0. This makes sense, because if you multiply one or more copies of 0 together, you'll just get 0. Turns out it's hard for 0 to become anything other than 0. Even if he really applies himself. Any nonzero number raised to the 0 power is 1. Think about it this way: 24 = 16 23 = 8 22 = 4 21 = 2 As the exponent drops by 1, the answer is divided in half. If we drop the exponent by 1 once more and divide the answer in half again, we get 2 = 1. We can't believe how many times you just dropped that exponent. Can't you be more careful? 0 is a troublesome number. We want 0 raised to a power to be zero, but we also want any number raised to the 0 power to be 1. There's no way to win! This means that 00 is undefined. If it's not too late, don't think about this too hard. It'll make your head hurt. ### Multiplication Sample: 25 × 27 This means that you need to multiply 5 copies of 2 together, and then multiply that result by 7 copies of 2 multiplied together. That's a total of 12 copies of 2. So 25 × 27 = 212. Why so many copies of 2? What are you, passing them out at a meeting? If we have the same base with two different exponents and we're multiplying these numbers, as in the above example, the exponents get added together. In symbols, if a, b and c are real numbers, ab × ac = a(b+c). This makes sense so long as you remember what exponents are abbreviating. ### Negative Exponents So far, we have only looked at exponents that are positive integers. Let's try to figure out what a number would be when raised to a negative exponent. Suppose we want to understand what 3-1 means. Well, let's use what we know about multiplication of exponents. We know that, because we add exponents during multiplication, 31 × 3-1 would be 31+(-1) = 3 = 1. This tells us that 3-1 is the multiplicative inverse, or reciprocal, of 3! So . Did you follow that? If not, double back and read this paragraph again until it sinks in. It won't kill you. Now what happens if we take bigger powers? 5-7, for example. In this case, we'll look at 57 × 5-7 = 57+-7 = 5 = 1. So 5-7 is the same as 1/57. Are you loving this stuff as much as we think you are? ### Division Sample: 25 ÷ 22 This means , so we're just canceling out two of our 2's. Buh-bye, guys. You shall be missed. By reducing, our fraction equals 23. In general, ab ÷ ac = a(b - c), because we start out with b copies of a, divide out c copies, and are left with b - c copies. Note that a is not equal to 0. Notice that if b > c, you are left with a positive exponent. But if b < c, you have a negative exponent. Which shouldn't stress you out any, as you now know what to do with them. Sample: 42 ÷ 44 This translates to: . See what we did there on the end? Always look for ways that an expression can be further simplified. Sample: 63 ÷ 67 means 3 copies of 6 divided by 7 copies of 6: Cancel out 3 copies of 6 from the top and bottom of the fraction to get . Be careful: In order to use the properties above, the base of the exponents has to be the same. We can't combine, for example, 43 and 52. 3 copies of 4 and 2 copies of 5 multiply to give you 3 copies of 4 and 2 copies of 5, and that's unfortunately as nice as it gets with exponent notation. Which isn't very nice. Hear that, Santa? ### Exponentiation Sample: (25)3 means (2 × 2 × 2 × 2 × 2)3. You can't just add the 5 and the 3 together in this instance, because what we're actually being asked to do is take 3 copies of (2 × 2 × 2 × 2 × 2), or 15 copies of 2 multiplied together. Looks a little like we're going to be multiplying exponents here. In fact, it looks a lot like that. (25)3 = 215 So, in general, (ab)c = ab × c. ### Raising Products to a Power Sample: (6 × 7)3 = (6 × 7)(6 × 7)(6 × 7) = 63 × 73. In general, if a and b are real numbers and c is a whole number, (a × b)c = ac × bc. ### Raising Quotients to a Power Sample: If a and b are real numbers and c is a whole number, . Next Page: Prime Factorization Previous Page: Exponents and Powers - Whole Numbers Calculator
# 2.1 - Place Value The student will a) read, write, and identify the place and value of each digit in a three-digit numeral, with and without models; b) identify the number that is 10 more, 10 less, 100 more, and 100 less than a given number up to 999; c) compare and order whole numbers between 0 and 999; and d) round two-digit numbers to the nearest ten ### BIG IDEAS • So that I can understand the value of each number in order to compare them • So that I can round numbers and get a close number when the exact number is not needed • So that I can understand that numbers are based on a simple pattern of tens where each place has ten times the value of the place to its right • So that I can understand that reading, writing and identifying place value is essential to many other concepts in math ### UNDERSTANDING THE STANDARD • The number system is based on a simple pattern of tens where each place has ten times the value of the place to its right. • Numbers are written to show how many hundreds, tens, and ones are in the number. • Opportunities to experience the relationships among hundreds, tens, and ones through hands-on experiences with manipulatives are essential to developing the ten-to-one place value concept of our number system and to understanding the value of each digit in a three-digit number. This structure is helpful when comparing and ordering numbers. • Manipulatives that can be physically connected and separated into groups of tens and leftover ones (e.g., snap cubes, beans on craft sticks, pennies in cups, bundle of sticks, beads on pipe cleaners, etc.) should be used. • Ten-to-one trading activities with manipulatives on place value mats provide experiences for developing the understanding of the places in the base-10 system. • Models that clearly illustrate the relationships among ones, tens, and hundreds, are physically proportional (e.g., the tens piece is ten times larger than the ones piece). • Flexibility in thinking about numbers is critical (e.g., 84 is equivalent to 8 tens and 4 ones, or 7 tens and 14 ones, or 5 tens and 34 ones, etc.). This flexibility builds background understanding for the ideas used when regrouping. When subtracting 18 from 174, a student may choose to regroup and think of 174 as 1 hundred, 6 tens, and 14 ones. • Hundreds charts can serve as helpful tools as students develop an understanding of 10 more, 10 less, 100 more and 100 less. • Rounding a number to the nearest ten means determining which two tens the number lies between and then which ten the number is closest to (e.g., 48 is between 40 and 50 and rounded to the nearest ten is 50, because 48 is closer to 50 than it is to 40). • Rounding is an estimation strategy that is often used to assess the reasonableness of a solution or to give an estimate of an amount. • Vertical and horizontal number lines are useful tools for developing the concept of rounding. Rounding to the nearest ten using a number line is done as follows: • Locate the number on the number line. • Identify the two closest tens the number comes between. • Determine the closest ten. • If the number in the ones place is 5 (halfway between the two tens), round the number to the higher ten. • Mathematical symbols (>, <) used to compare two unequal numbers are called inequality symbols. ### ESSENTIALS The student will use problem solving, mathematical communication, mathematical reasoning, connections, and representations to • Demonstrate understanding of the ten-to-one relationships among ones, tens, and hundreds, using manipulatives. (a) • Write numerals, using a model or pictorial representation (i.e., a picture of base-10 blocks). (a) • Read three-digit numbers when shown a numeral, a model of the number, or a pictorial representation of the number. (a) • Identify and write the place (ones, tens, hundreds) of each digit in a three-digit numeral. (a) • Determine the value of each digit in a three-digit numeral (e.g., in 352, the 5 represents 5 tens and its value is 50). (a) • Use models to represent numbers in multiple ways, according to place value (e.g., 256 can be 1 hundred, 14 tens, and 16 ones, 25 tens and 6 ones, etc.). (a) • Use place value understanding to identify the number that is 10 more, 10 less, 100 more, or 100 less than a given number, up to 999. (b) • Compare two numbers between 0 and 999 represented with concrete objects, pictorially or symbolically, using the symbols (>, <, or =) and the words greater than, less than or equal to. (c) • Order three whole numbers between 0 and 999 represented with concrete objects, pictorially, or symbolically from least to greatest and greatest to least. (c) • Round two-digit numbers to the nearest ten. (d) ### KEY VOCABULARY Updated: Aug 22, 2018
## Volume Solution3 In this page volume solution3 we are going to see solution of each questions of the worksheet volume worksheet1. Question 7: A rectangular sheet of metal foil with dimension 66 cm x 12 cm is rolled to form a cylinder of height 12 cm. Find the volume of the cylinder. Solution: Length of metal sheet = 66 cm Width of sheet = 12 cm Now the metal sheet is in the form of cylinder. Here height of the cylinder is also 12 cm. Now 66 cm is to be considered as circumference of the base of the cylinder. circumference of the base of the cylinder = 66 cm 2 Π r = 66 2 x (22/7) x r = 66 r = 66 x (1/2) x (7/22) r = 3 x (1/2) x (7) r = 21/2 r = 10.5 cm Volume of the right circular cylinder = Π r² h = (22/7) x (10.5)² x 12 = (22/7) x (10.5) x(10.5) x 12 = (22) x (1.5) x (10.5) x 12 = 4158 cm³ Volume of the the right circular cylinder = 4158 cm³ Question 8: A lead pencil is in the shape of right circular cylinder. The pencil is 28 cm long and its radius is 3 mm. If the lead is of radius 1 mm,then find the volume of the wood used in the pencil. Solution: Height of the pencil = 28 cm Outer radius (R) = 3 mm Inner radius (r) = 1 mm Now we have to change the units from mm to cm. For that we have to divide it by 10. R = 3/10 = 0.3 cm r = 1/10 = 0.1 cm To find the volume of the wood used to make that pencil,simply we can find the volume of the pencil.Pencil is in the shape of hollow cylinder Volume of hollow cylinder = Π h (R² - r²) = (22/7) x 28 x (0.3² - 0.1²) = (22) x 4 x (0.09 - 0.01) = (22) x 4 x (0.08) = 7.04 cm³ Volume of wood used in the pencil = 7.04 cm³ volume solution3 volume solution3
## A Visual Guide to Voronoi Graph & Delaunay Triangulation Voronoi diagrams help us find the closest points from an arbitrary point. Let’s start by displaying a set of points on a plane. The Voronoi diagram draws boundaries between each of those points. Around each point, regions will develop called the Voronoi regions. Let’s see how we would draw those boundaries. Each boundary should be in the middle of two points. Let’s begin by drawing a boundary between point 1 and point 2. The boundary should be directly in the middle separating the two points. Let’s move to creating a boundary between points 2 and 3. Wherever the points intersect, we’ll remove that line segment. We’ll create the boundary between point 2 and 6 next. We also have to create a boundary between point 1 and point 6. This will require trimming on a couple of different borders. The final product looks like the following. Let’s move on and create a border between 2 and 7. Next, we’ll create a border between points 6 and 7. Next, we’ll create a border between points 6 and 8. Next, we’ll create a border between points 7 and 8. Next, we’ll create a border between points 5 and 7. Next, we’ll create a border between points 7 and 9. Let’s draw a border between points 5 and 9. A couple of more points to go. We’ll draw the boundary for points 3 and 5 next. Next, we’ll create a border between points 3 and 4. We’ll draw a boundary between points 4 and 5. From the looks of the diagram, there are a couple of boundaries that were not accounted for, the boundary between point 1 and 3 and the boundary between point 8 and 9. Let’s quickly adjust those and complete the Voronoi diagram. Once we have the regions defined, we can draw the lines that make up the Delaunay triangulation. The simple rule of thumb is, if the points share a border, draw a line connecting them. Point 1 shares a boundary with points 2, 3, and 6, so we’ll draw lines connecting those points. Although it may look like the line crosses two boundary lines when going from point 1 to 3, we can quickly modify that line to show where it will cross the border only once. Point 2 shares boundaries with unconnected points 3, 6, and 7; it’s already connected to point 1. Point 3 shares boundaries with unconnected points 4, 5, and 7. Point 4 shares a boundary with the unconnected point 5. Point 5 shares boundaries with unconnected points 7 and 9. Point 6 shares boundaries with unconnected points 7 and 8. Point 7 shares boundaries with unconnected points 8 and 9. Point 8 shares a boundary with the unconnected point 9. Point 9 is already connected to all other points that share its boundary, so this completes the construction of the Delaunay triangles. If we knew the weight of each edge, we could construct a minimum spanning tree after the creation of the Delaunay triangles. How would we find the Voronoi diagram for a set of points in a rectilinear metric? Let’s start with a small set of points on a grid. The distance from p1 to p2 is 4 units regardless of the path that you take. The middle point going through either route is 2 units away. We’ll start by drawing a line connecting those two points. To separate the points equally, we’ll draw lines protruding from each endpoint that maximizes the area of each point. Next, we’ll examine the boundary for points 1 and 3. This will create a border similar to the one separating points 1 and 2. Again, regardless of the path that you take, to get to the midpoint between those two points, it will require 2 units of travel. We’ll connect those two points and we’ll shift the bottom vertical line. Next, we must draw a boundary between points 2 and 3. The midpoint is 1 unit away, so a horizontal line will be drawn separating those two points. A border needs to be constructed between points 2 and 4. The distance to point 4 from point 2 is 6 units. The midpoint of the trajectory is 3 units from points 2 and 4. We will draw a line connecting those two points and we’ll draw vertical lines from the endpoints. There is still the boundary between points 3 and 4 that needs to be constructed. Since the distance between points 3 and 4 is the same as the distance between points 2 and 4, a similar approach is taken to construct the boundary between them. The midpoint is 3 units away, so we’ll draw a line that connects the midpoints from the top and from the bottom. The right-bottom, vertical line will be moved to the new endpoint. We can again construct the Delaunay triangles by connecting the points that share a boundary. Point 1 shares boundaries with points 2 and 3. Point 2 shares boundaries with unconnected points 3 and 4. Point 3 shares a boundary with the unconnected point 4. All points are now connected to their immediate neighbors. This completes the Delaunay triangulation. Since we know the distance to each point, we can construct a minimum spanning tree. There are a couple of variations of the minimum spanning tree for this problem, but we’ll choose one. The total length of this MST is 12. If you liked what you read, check out my book, An Illustrative Introduction to Algorithms.
# How do you solve using the completing the square method x^2 + 2x - 63 = 0? Mar 8, 2016 $\left(x + 9\right) \left(x - 7\right)$ #### Explanation: To solve the equation x^2+2x−63=0, using the completing the square method, one has to complete the square using the variable $x$. We know the identity ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$, hence we should halve the coefficient of $x$ and square it and then add and subtract the 'square' in the trinomial. Here coefficient of $x$ is 2 and hence squareof half of it is $1$. Hence x^2+2x−63=0 can be written as x^2+2x+1-1−63=0 or $\left({x}^{2} + 2 \times x \times 1 + {1}^{2}\right) - 64$ or ${\left(x + 1\right)}^{2} - {8}^{2}$ Now using the identity ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, this becomes $\left(\left(x + 1\right) + 8\right) \times \left(\left(x + 1\right) - 8\right)$ or $\left(x + 9\right) \left(x - 7\right)$
top of page Search # Lessons learned from the 2023 Algebra 1 STAAR The 2023 Algebra 1 STAAR introduced statewide online testing and several new item types. Using a modified version of the statewide item analysis report, I examined the readiness standards that had less than 50% mastery. Each standard has both an analysis of the items themselves to infer what made them so difficult and instructional implications for educators to ensure a more successful 2024 STAAR test. Â Standard # of items % mastery A.11B 2 21 A.7C 1 30 A.8A 1 30 A.2A 2 33.5 A.9D 2 34.5 A.2I 1 39 A.5A 2 42.5 A.9C 2 43.5 A.6A 2 45.5 A.9C 2 49.5 Â Access the slide deck here. ### A.11B - 21% overall mastery simplify numeric and algebraic expressions using the laws of exponents, including integral and rational exponents #10 - 38% correct #35 - 4% full credit, 38% partial credit, 58% no credit ###### Analysis • Fractions add a layer of complexity to dividing exponents or finding the power of a power • The drag and drop response required you to use the same number twice (not all drag and drops allow using a response more than once) ###### Instructional Implications • Utilize fractions in when asking students to multiply or divide by exponents • Bundle multiple laws of exponents into a single problem (e.g., radicals and division) • Have students practice rewriting dividing exponents as multiplying by negative exponents (e.g., x^4/x^2 = x^4 (x^-2) ) Â ### A.7C - 30% overall mastery determine the effects on the graph of the parent function f(x) = x2 when f(x) is replaced by af(x), f(x) + d, f(x - c), f(bx) for specific values of a, b, c, and d #38 - 30% full credit, 35% partial credit, 35% no credit ###### Analysis • Drag and drop removed certainty for partial guesses • Students had to analyze both a vertical and horizontal transformation simultaneously • The horizontal transformation (- 3) actually moves the vertex to the right ###### Instructional Implications • Bundle problems so that students multiple transformations together • Have students practice graphing the child function (correctly with exponent) to analyze the vertex Â Watch the full walkthrough of all 50 items on the 2023 Algebra 1 STAAR below. Â A.8A - 30% overall mastery solve quadratic equations having real solutions by factoring, taking square roots, completing the square, and applying the quadratic formula #41 - 30% correct ###### Analysis • Distribution of answer selections (30%, 28%, 27%, 14%) indicative of guessing • Multiple strategies could have been used to solve the problem ###### Instructional Implications • Focus on correctly applying quadratic formula • Emphasize problems that require simplifying a radical • Have students solve the same problem using at least two different strategies Â ### A.2A - 33.5% overall mastery determine the domain and range of a linear function in mathematical problems; determine reasonable domain and range values for real-world situations, both continuous and discrete; and represent domain and range using inequalities #7 - 33% full credit, 22% partial credit, 44% no credit #44 - 34% correct ###### Analysis • Responses were required in verbal form (rather than as inequalities) • Students had to differentiate between open and closed circles • Problems required analysis of both domain and range simultaneously ###### Instructional Implications • Bundle problems so that students assess both domain and range together • Have students draw rays that satisfy requirements of both domain and range • Provide error analysis opportunities for students to disprove why graphs do NOT represent a given domain and range Â ### A.9D - 34.5% overall mastery graph exponential functions that model growth and decay and identify key features, including y- intercept and asymptote, in mathematical and real-world problems #9 - 14% full credit, 59% partial credit, 27% no credit #42 - 55% correct ###### Analysis • Both problems asked students to identify features of exponential graphs (i.e., asymptote, y-intercept) • Common error for asymptote was most likely labeling x-axis as x = 0 ###### Instructional Implications • When working with standard form (y = ab ), also have students identify a as the y-intercept and explain why • When given an exponential function in equation or graph form, have students create a real-world situation to describe it, including naming the y-intercept Â Â ### A.2I - 39% overall mastery write systems of two linear equations given a table of values, a graph, and a verbal description #37 - 39% correct ###### Analysis • Distribution of answer selections (16%, 21%, 24%, 39%) indicative of guessing • Solving the problem required • Finding slope of each line • Substituting an ordered pair and solving for b for each line • Transforming each resulting slope-intercept form equation into standard form ###### Instructional Implications • Focus on applying correct sequence of formulas • Practice building stamina and how to check their answers with substitution or by graphing discrete points along with the systems of equations Â ### A.5A - 42.5% overall mastery solve linear equations in one variable, including those for which the application of the distributive property is necessary and for which variables are included on both sides #4 - 22% correct #29 - 53% correct ###### Analysis • Distribution of answer selections for #4 (27%, 21%, 30%, 22%) indicative of guessing • This type of problem (linear equations and distributive property) has never been asked without explicitly providing the equation ###### Instructional Implications • Find and utilize verbal descriptions of linear equations (using the distributive property) that do not explicitly provide the equation • Have students work in pairs to create their own problems (w/o explicit equations) and exchange with peers to solve Â ### A.9C - 43.5% overall mastery write exponential functions in the form f(x) = ab^x (where b is a rational number) to describe problems arising from mathematical and real-world situations, including growth and decay #14 - 37% correct #36 - 50% correct ###### Analysis • Students were not provided a growth factor for one problem • Growth factor of 4% had to be translated to 1.04 ###### Instructional Implications • Have students practice finding growth/decay factor when given initial and first-year amount • Provide rationale for translating a growth factor to 1 + decimal and a decay factor to 1 - decimal • Give students percentage (e.g., decay rate of 22%) and practice converting to decay factor (e.g., 0.78) Â ### A.6A - 45.5% overall mastery determine the domain and range of quadratic functions and represent the domain and range using inequalities #11 - 47% correct #32 - 44% correct ###### Analysis • Students needed to graph at least one problem to correctly identify range • Improper fractions in the table made visual sequencing challenging ###### Instructional Implications • Reinforce difference between domain and range and that domain is rarely limited in quadratic equations • Practice identifying range from a table by sketching a rough parabola with given points • Practice plotting discrete points in a graphing calculator to visually determine range Â ### A.5C - 49.5% overall mastery solve systems of two linear equations with two variables for mathematical and real-world problems #23 - 68% correct #40 - 31% correct ###### Analysis • Distribution of answer selections for #40 (16%, 26%, 27%, 31%) indicative of guessing • A system of equations question has never been asked using a table ###### Instructional Implications • Have students practice taking systems of equations questions (verbal descriptions and equations) and translating them to tables • Reinforce the qualities that define a system of equations (two or more equations that share variables) to better identify them in real-world situations
# The cost of preparing the walls of a room 12 m Question: The cost of preparing the walls of a room 12 m long at the rate of Rs 1.35 per square metre is Rs 340.20 and the cost of matting the floor at 85 paise per square metre is Rs 91.80. Find the height of the room. Solution: The cost of preparing 4 walls of a room whose length is $12 \mathrm{~m}$ is $\mathrm{Rs} 340.20$ at a rate of $\mathrm{Rs} 1.35 / \mathrm{m}^{2}$. $\therefore$ Area of the four walls of the room $=\frac{\text { total cost }}{\text { rate }}=\frac{\mathrm{Rs} 340.20}{\mathrm{Rs} 1.35}=252 \mathrm{~m}^{2}$ Also, the cost of matting the floor at 85 paise $/ \mathrm{m}^{2}$ is Rs $91.80$. $\therefore$ Area of the floor $=\frac{\text { total cost }}{\text { rate }}=\frac{\mathrm{Rs} 91.80}{\mathrm{Rs} 0.85}=108 \mathrm{~m}^{2}$ Hence, breadth of the room $=\frac{\text { area of the floor }}{\text { length }}=\frac{108}{12}=9 \mathrm{~m}$ Suppose that the height of the room is $\mathrm{h} \mathrm{m} .$ Then, we have : Area of four walls $=2 \times($ length $\times$ height $+$ breadth $\times$ height $)$ $\Rightarrow 252=2 \times(12 \times \mathrm{h}+9 \times \mathrm{h})$ $\Rightarrow 252=2 \times(21 \mathrm{~h})$ $\Rightarrow 21 \mathrm{~h}=\frac{252}{2}=126$ $\Rightarrow \mathrm{h}=\frac{126}{21}=6 \mathrm{~m}$ $\therefore$ The height of the room is $6 \mathrm{~m}$.
This page's content is no longer actively maintained, but the material has been kept on-line for historical purposes. The page may contain broken links or outdated information, and parts may not function in current web browsers. ## Introduction to Statistics #### I - Definitions 1. mean: the average of a set of numbers median: The number found in the middle when looking at the set of numbers from smallest to largest mode: most commonly occuring value in a set of numbers 2. Example for mean, median, and mode: Sample data set: 5° 10° 10° 7° 3° mean = 5 + 10 + 10 + 7 + 3 = 35, 35/5 = 7 median = 3° 5° 7° 10° 10° mode = 10° 1. Variance: a measure of how data points differ from the mean 2. Data Set 1: 3, 5, 7, 10, 10 Data Set 2: 7, 7, 7, 7, 7 - Data Set 1: mean = 7, median = 7 - Data Set 2: mean = 7, median = 7 But we know that the two data sets are not identical! The variance shows how they are different. 3. Formula for variance: `S2 = (1/(N-1) × (the sum of (each data point - mean)2)` Formula applied to data set 1: S2data set 1 = ( 1/(5-1) ) × ( (3-7)2 + (5-7)2 + (7-7)2 + (10-7)2 + (10-7)2 ) note: N = number of data points S2data set 1 = 1/4 × { (-4)2 + (-2)2 + (0)2 + (3)2 + (3)2 } S2data set 1 = 1/4 × ( 16 + 4 + 0 + 9 + 9 ) S2 = 1/4 × 38 S2 = 38/4 S2 = 9.5 for data set 1 Formula applied to data set 2: S2data set 2 = (1/4) × (0 + 0 + 0 + 0 + 0) S2 = 1/4(0) S2 = 0 for data set 2 4. Standard Deviation is "S," the square root of the variance: `S = Square root of [(1/(N-1)) × (sum of (each data point - mean)2)]` • measure of the difference from the mean. Large S means the data is spread widely around the mean. • units are the same as the data itself 5. S for data set 1 above is 3.08. S for data set 2 above is 0.
kidzsearch.com > wiki Explore:web images videos games # Set theory A Venn diagram illustrating the intersection of two sets. Set theory is the study of sets in mathematics. Sets are collections of objects. We refer to these objects as "elements" or "members" of the set. To write a set, one wraps the numbers in {curly brackets}, and separates them with commas. For example. the set {1, 2, 3} holds 1, 2, and 3. Sets are also often referred to using capital roman letters such as $A$, $B$, $C$.[1] There are three methods one can use to describe a set: Description method, Roster method (tabular form) and Rule method (set builder form). ## History Set theory was created around 1874 by Georg Cantor.[2] It had to be made better because collections of objects can cause problems if we work with them without explaining them better. Russell's paradox was one of the problems. Think about the set of all sets that are not members of themselves. If it were inside itself, then the rule which defines it would mean that it is not inside itself. But, if it were not inside itself, then the rule which defines it would mean that it is inside itself. This was a serious problem, and it meant that the old set theory was broken. It was improved by people including Zermelo and Bertrand Russell.[3] ## Theory Set theory begins by giving some examples of things that are sets. Then it gives rules in which you can make other sets from the already known sets. Collections of objects that are not sets are called (proper) classes.[3] It is possible to do mathematics using only sets, rather than classes, so that the problems that classes cause in mathematics do not occur. • Example: An object o and a set A. If o is a member (or element) of A, we write oA.[1] Since sets are objects, the membership relation can relate sets as well. A binary relation between two sets is the subset relation, also called set inclusion. If all the members of set A are also members of set B, then A is a subset of B, marking AB. For example, {1,2} is a subset of {1,2,3}, but {1,4} is not. From this example, it is clear that a set is a subset of itself. In cases where one wishes to not to have this, the term proper subset is meant not to have this possibility. The self-considering objects in set theory was considered too, with some example numbers being 1={1}, 2={1, 2}, 3={1, 2, 3} and so on. ## References 1. Bagaria, Joan (2020). Zalta, Edward N.. ed. The Stanford Encyclopedia of Philosophy (Spring 2020 ed.). Metaphysics Research Lab, Stanford University.
These are my notes on quadratic models in precalculus. When a mathematical model leads to a quadratic function, the properties of the model. We can use the quadratic function to determine the maximum or minimum value of the function. The fact that the graph of a quadratic function has a maximum or minimum value enables us to answer questions involving optimization or finding the maximum or minimum values involving quadratic functions. Solving Applied Problems In economics, revenue is defined as the amount of money received from the sale of an item and is equal to the unit selling price of the item times the number of units that were sold. $$R=xp$$ In economics, the Law of Demand states that p and x are related. As one increases, the other decreases. The equation that relates p and x is called the demand equation. Example 1 The marketing department at Texas Instruments has found that, when certain calculators are sold at a certain price, the number of calculators sold is given by the demand equation: $$x=21000-150p$$ 1. Express the revenue as a function of price 2. What unit price should be used to maximize revenue 3. If this price is charged, what is the maximum revenue 4. How many units are sold at this price 5. Graph Solution: 1. The revenue is $$R=xp$$ where $$x=21000-150p$$. $$(21000-150p)p= -150p^2+21000p$$ 2. The function is a quadratic function with a=-150, b=21000, and c=0. Because a<0, the vertex is the highest point on the parabola. The revenue is therefore a maximum when the price is: $$p=\frac{-b}{2a} = -\frac{21000}{2(-159)} = \frac{21000}{-300} = 70.00$$ 3. The maximum revenue is: $$R(70) = -150(70)^2 + 21000(70) = 735000$$ 4. The number of calculators sold is given by the demand equation: $$x=21000-150p$$ At a price of $$p=70$$, $$x=21000-150(70) = 10500$$ Example 2 Maximizing the Area Enclosed by a Fence A farmer has 2000 yards of fence to enclose a rectangular field. What are the dimensions of the rectangle that encloses the most area? Solution: The available fence represents the perimeter of the rectangle. If x is the length and w is the width, then: $$2x+2w=2000$$ The area of the rectangle is: $$a=xw$$ To express area in terms of a single variable, we solve the equation for w and substitute the result in $$a=xw$$. Then area involves only the variable x. You could also solve the equation for x and express area in terms of w alone. $$2x + 2w = 2000$$ $$2w = 2000-2x$$ $$w = \frac{2000-2x}{2} = (1000-x)$$ So, the area is: $$a = xw = x(1000-x) = -x^2 + 1000x$$ Now, area is a quadratic function of x. $$a(x)=-x^2+1000x$$ When you graph this, you see a<0, so the vertex is a maximum point on the graph. The maximum value occurs at: $$x=-\frac{b}{2a} = -\frac{1000}{2(-1)} = 500$$ The maximum value of a is: $$a(-\frac{b}{2a}) = a(500) = -500^2 + 1000(500) = -250000 + 500000 = 250000$$ The largest rectangle that can be enclosed by 2000 yards of fence has an area of 250000 square yards. Its dimensions are 500 by 500 yards. Example 3 Analyzing the motion of a projectile A projectile is fired from a cliff 500 feet above the water at an inclination of 45 degrees to the horizontal, with a muzzle velocity of 400 feet per second. In physics, it is established that the height of the projectile above the water is given by: $$h(x)=\frac{-32x^2}{(400)^2} + x + 500$$ X is the horizontal distance of the projectile from the base of the cliff. 1. Find the maximum height of the projectile 2. How far from the base of the cliff will the projectile strike the water? Solution: 1. The height of the projectile is given by a quadratic function. $$h(x)=\frac{-32x^2}{(400)^2}+x+500=\frac{-1}{5000}x^2+x+500$$ We are looking for the maximum value of h. Since a<0, the maximum value is obtained at the vertex. $$x=-\frac{b}{2a}=-\frac{1}{2(-1/5000)}=\frac{5000}{2}=2500$$ The maximum height of the projectile is: $$h(2500)=\frac{-1}{5000}(2500)^2+2500+500=-1250+2500+500=1750ft$$ 2. The projectile will strike the water when the height is zero. To find the distance x traveled, we need to solve the equation: $$h(x)=\frac{-1}{5000}x^2+x+500=0$$ We find the descriminant first. $$b^2-4ac=1^2-4(\frac{-1}{5000})(500)=1.4$$ Then: $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-1 \pm \sqrt{1.4}}{2(-1/5000)}=5458ft$$
# Area of a Triangle ## Formula Area of triangle = (1/2) × "base" × "height" = 1/2 × b × h. # Area of a triangle: To get the area of the triangle, we first choose one of the sides to be the base (b). Then we draw a perpendicular line segment from a vertex of the triangle to the base. This is the height (h) of the triangle. The area of a triangle is equal to half the product of the base and the height. Area of triangle = (1/2) × "base" × "height" = 1/2 × b × h All the congruent triangles are equal in the area but the triangles equal in the area need not be congruent. ## Areas of different types of triangles: Consider an acute and an obtuse triangle. Area of each triangle = (1/2) × "base" × "height" = 1/2 × b × h ## Example Find the area of the following triangle: Area of triangle = 1/2bh = 1/2 × QR × PS = 1/2 × 4cm × 2cm = 4 cm2 ## Example Find the area of the following triangle: Area of triangle = 1/2bh = 1/2 × MN × LO = 1/2 × 3cm × 2cm = 3 cm2 ## Example Find BC, if the area of the triangle ABC is 36 cm2 and the height AD is 3 cm. Height = 3 cm, Area = 36 cm2 Area of the triangle ABC =1/2bh 36 = 1/2 × b × 3 b = (36 xx 2)/3 = 24 cm So, BC = 24 cm. ## Example In ∆PQR, PR = 8 cm, QR = 4 cm and PL = 5 cm. Find: (i) the area of the ∆PQR (ii) QM. (i) QR = base = 4 cm, PL = height = 5 cm Area of the triangle PQR =1/2bh = 1/2 xx 4 xx 5 = 10 cm2 (ii) PR = base = 8 cm, QM = height = ?, Area = 10 cm2 Area of triangle = 1/2 xx b xx h 10 = 1/2 xx 8 xx h h = 10/4 = 5/2 = 2.5. QM = 2.5 cm If you would like to contribute notes or other learning material, please submit them using the button below. ### Shaalaa.com Area of Triangle [00:13:58] S 0%
Question # If A=begin{pmatrix}8 & 0 4 & -2 3&6 end{pmatrix} text{ and } B=begin{pmatrix}2 & -2 4 & 2 -5&1 end{pmatrix} ,then find the matrix X such that 2A+3X=5B Matrices If $$A=\begin{pmatrix}8 & 0 \\ 4 & -2 \\ 3&6 \end{pmatrix} \text{ and } B=\begin{pmatrix}2 & -2 \\ 4 & 2 \\ -5&1 \end{pmatrix}$$ ,then find the matrix X such that $$2A+3X=5B$$ 2021-02-03 Step 1 We have to find the matrix X such that $$2A+3X=5B$$ where matrices are given as: $$A=\begin{pmatrix}8 & 0 \\ 4 & -2 \\ 3&6 \end{pmatrix} \text{ and } B=\begin{pmatrix}2 & -2 \\ 4 & 2 \\ -5&1 \end{pmatrix}$$ Here we have to use operation of matrices, in operation of matrices we focus on corresponding elements of the matrix. Example: $$2\begin{pmatrix}a & b \\c & d \end{pmatrix}=\begin{pmatrix}2a & 2b \\ 2c & 2d \end{pmatrix} , \frac{1}{2}\begin{pmatrix}a & b \\c & d \end{pmatrix}=\begin{pmatrix}\frac{a}{2} & \frac{b}{2} \\\frac{c}{2} & \frac{d}{2} \end{pmatrix}$$ $$\begin{pmatrix}a & b \\c & d \end{pmatrix}+\begin{pmatrix}x & y \\ z & w \end{pmatrix} = \begin{pmatrix}a+x & b+y \\ c+z & d+w \end{pmatrix}$$ Step 2 Putting given value of matrices into the expression, we get $$2A+3X=5B$$ $$2\begin{pmatrix}8 & 0 \\ 4 & -2 \\ 3&6 \end{pmatrix}+3X=5\begin{pmatrix}2 & -2 \\ 4 & 2 \\ -5&1 \end{pmatrix}$$ $$\begin{pmatrix}2\times8 & 2\times0 \\ 2\times4 & 2(-2) \\ 2\times3&2\times6 \end{pmatrix}+3X=\begin{pmatrix}5\times2 & 5(-2) \\ 5\times4 & 5\times2 \\ 5(-5)&5\times1 \end{pmatrix}$$ $$\begin{pmatrix}16 & 0 \\ 8 & -4 \\ 6&12 \end{pmatrix}+3X=\begin{pmatrix}10 & -10 \\ 20 & 10 \\ -25&5 \end{pmatrix}$$ $$3X=\begin{pmatrix}10 & -10 \\ 20 & 10 \\ -25&5 \end{pmatrix}-\begin{pmatrix}16 & 0 \\ 8 & -4 \\ 6&12 \end{pmatrix}$$ $$=\begin{pmatrix}10-16 & -10-0 \\ 20-8 & 10+4 \\ -25-6&5-12 \end{pmatrix}$$ $$3X=\begin{pmatrix}-6 & -10 \\ 12 & 14 \\ -31&-7 \end{pmatrix}$$ $$X=\frac{1}{3} \begin{pmatrix}-6 & -10 \\ 12 & 14 \\ -31&-7 \end{pmatrix}$$ $$X= \begin{pmatrix}\frac{-6}{3} & \frac{-10}{3} \\ \frac{12}{3} & \frac{14}{3} \\ \frac{-31}{3}&\frac{-7}{3} \end{pmatrix}$$ $$=\begin{pmatrix}-2 & \frac{-10}{3} \\ 4 & \frac{14}{3} \\ \frac{-31}{3}&\frac{-7}{3} \end{pmatrix}$$ Hence, $$X=\begin{pmatrix}-2 & \frac{-10}{3} \\ 4 & \frac{14}{3} \\ \frac{-31}{3}&\frac{-7}{3} \end{pmatrix}$$
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 ### Geometry2.26 Circumcenter and Incenter Circumcenter: 1. The perpendicular bisectors of the sides of a triangle are concurrent. The point of concurrence of the perpendicular bisectors of the sides of a triangle is equidistant from the vertices of the triangle. 2. The circle passing through the vertices of a triangle is called the circumcircle. 3. The centre of a circumcircle is called circumcenter. Example: Identify the circumcircle, circumradius and circumcenter of the above triangle. Solution: Here 'c' is the circumcircle, 'AH' is the circumradius and 'A' is the in center. 5. If we draw an acute triangle and we observe that, its circumcenter falls within the triangle that is in the interior of the triangle. 6. If we draw a right triangle and we observe that its circumcenter falls on the hypotenuse of the triangle and it is midpoint of the hypotenuse. 7. If we draw an obtuse triangle and we observe that its circumcenter falls outside the triangle, that is in the exterior of the triangle. 8. The internal bisectors of the angles of a triangle are concurrent. Incenter: 1. The point of concurrence of the internal bisectors of the angles of triangle is equidistant from the sides of the triangles. 2. The circle that touches the three sides of a triangle and lies inside the triangle is called the incircle of the triangle. 3. Its centre which is the point of concurrence of the internal bisectors of the angles of the triangle is called incentre of the triangle. 4. The radius of the incircle is called the inradius of the triangle. The incentre is equidistant from the sides of the triangle. Example: Identify the incircle, inradius and incentre of the above triangle. Solution: In the above figure 'c' is the incircle, 'AE' is the inradius and 'D' is the incentre. Review Acute angled triangleRight angled triangleObtuse angled triangle Circumcenter See how the circumcenter 'A' varies in all the three types of triangles. Incentre See the incentres 'D' of all the three types of triangles. Directions: Read the above review points carefully and answer the following questions: 1. Illustrate each of the above review points by drawing a the circumcentre and incentre of triangles. 2. Explain in your own words the circumcentre and incentre of a triangle. 3. Draw the circumcentre and incentre of the triangle with sides 10 cm, 8 cm and 6 cm. Also measure the angles. Q 1: The circle that touches the three sides of a triangle and lies inside the triangle is called the ______.coincidesincentreincircle Q 2: The ______ of a triangle is the point of concurrence of the perpendicular bisectors of the sides of the triangle.circumcirclecircumradiuscircumcentre Q 3: The centre of circumcircle is the ______.circumcentrecircumradiuscircumcircle Q 4: The circle passing through the vertices of a triangle is the ______.circumradiuscircumcirclecircumcentre Q 5: The centre of the incircle, which is the point of concurrence of the internal bisectors of the angles of the triangle is the ______.incircleincentrecoincides Q 6: The radius of circumcircle is the ______.iircumradiuscircumcentrecircumcircle Q 7: In an acute angled triangle the circumcentre falls in the ______ of the triangle.exteriorinteriormidpoint Q 8: The perpendicular bisectors of the sides of a triangle are ______.circumcirclescircumcentresconcurrent Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! #### Subscription to kwizNET Learning System offers the following benefits: • Instant scoring of online quizzes • Progress tracking and award certificates to keep your student motivated • Unlimited practice with auto-generated 'WIZ MATH' quizzes • Choice of Math, English, Science, & Social Studies Curriculums • Excellent value for K-12 and ACT, SAT, & TOEFL Test Preparation • Get discount offers by sending an email to discounts@kwiznet.com Quiz Timer
# Worksheet on Math Relation \| Relations and Functions Worksheets with Answers Students who are searching to get Math Relation problems can check the Worksheet on Math Relation. Our Math Relation Worksheets available improves your preparation level and are very helpful in your practice. It included various models of questions on Math Relations. Therefore, practice all the given examples and check out the answers to cross-check your method of solving. Practice different questions related to Ordered Pair, Cartesian Product of Two Sets, Relation, Domain, and Range of a Relation in Math Relation Worksheet. See More: Sets ## Relations and Functions Questions and Answers 1. Find the values of x and y, if (x + 4, y – 8) = (8, 1). Solution: Given that (x + 4, y – 8) = (8, 1) Compare the elements of the given ordered pairs. Firstly, compare the first components of the given ordered pairs. x + 4 = 8 x = 8 – 4 = 4 So, x = 4. Now, compare the second components of the given ordered pairs. y – 8 = 1 y = 8 + 1 = 9 So, y = 9. Therefore, the value of x = 4 and y = 9. 2. If (x/5 + 3, y – 5/7) = (4, 5/14), find the values of x and y. Solution: Given that (x/5 + 3, y – 5/7) = (4, 5/14) Compare the elements of the given ordered pairs. Firstly, compare the first components of the given ordered pairs. x/5 + 3 = 4 x/5 = 4-3 Therefore, x/5 = 1 x = 1 * 5 = 5 So, x = 5. Now, compare the second components of the given ordered pairs. y – 5/7 = 5/14 y = 5/14 + 5/7 = 15/14 So, y = 15/14. Therefore, the value of x = 5 and y = 15/14. 3. If X = {m, n, o} and Y = {u, v}, find X × Y and Y × X. Are the two products equal? Solution: Given that X = {m, n, o} and Y = {u, v}, Let’s find the X × Y X × Y = {(m, u); (m, v); (n, u); (n, v); (o, u); (o, v)} Now, find Y × X. Y × X = {(u, m); (u, n); (u, o); (v, m); (v, n); (v, o)} Compare the elements of the given ordered pairs X and Y. X × Y not equal to Y × X Therefore, it is clearly stated that the two products are not equal. 4. If P × Q = {(x, 7); (x, 9); (y, 7); (y, 9); (z, 7); (z, 9)}, find P and Q. Solution: Given that P × Q = {(x, 7); (x, 9); (y, 7); (y, 9); (z, 7); (z, 9)}, We know that P is a set of all first entries in ordered pairs in P × Q. Q is a set of all second entries in ordered pairs in P × Q. Therefore, P = {x, y, z} Q = {7, 9} Therefore, the final answer is P = {x, y, z} and Q = {7, 9} 5. If M and N are two sets, and M × N consists of 6 elements: If three elements of M × N are (8, 4) (7, 3) (6, 3). Find M × N. Solution: Given that M and N are two sets, and M × N consists of 6 elements: If three elements of M × N are (8, 4) (7, 3) (6, 3). We know that M is a set of all first entries in ordered pairs in M × N. N is a set of all second entries in ordered pairs in M × N. Therefore, M = {8, 7, 6}, and N = {4, 3} Now, M × N = {(8, 4); (8, 3); (7, 4); (7, 3); (6, 4); (6, 3)} Thus, M × N contains six ordered pairs. 6. If A × B = {(m, 3); (m, 7); (m, 4); (n, 3); (n, 7); (n, 4)}, find B × A. Solution: Given that A × B = {(m, 3); (m, 7); (m, 4); (n, 3); (n, 7); (n, 4)}, We know that A is a set of all first entries in ordered pairs in A × B. B is a set of all second entries in ordered pairs in A × B. Therefore, A = {m, n}, and B = {3, 7, 4} Now, B × A = {(3, m); (3, n); (7, m); (7, n); (4, m); (4, n)} Therefore, the final answer is B × A = {(3, m); (3, n); (7, m); (7, n); (4, m); (4, n)} 7. If P = { 2, 1, 9} and Q = {4, 5}, then Find: (i) P × Q (ii) Q × P (iii) P × P (iv) (Q × Q) Solution: Given that P = { 2, 1, 9} and Q = {4, 5} (i) P × Q = {(2, 4); (2, 5); (1, 4); (1, 5); (9, 4); (9, 5)} (ii) Q × P = {(4, 2); (4, 1); (4, 9); (5, 2); (5, 1); (5, 9)} (iii) P × P = {(2, 2); (2, 1); (2, 9); (1, 2); (1, 1); (1, 9); (9, 2); (9, 1); (9, 9)} (iv) (Q × Q) = {(4, 4); (4, 5); (5, 4); (5, 5)} 8. If P = {3, 5, 7} and Q = {2, 3, 6}, state which of the following is a relation from P to Q. (a) R₁ = {(3, 5); (6, 7); (7, 2)} (b) R₂ = {(3, 3); (7, 6)} (c) R₃ = {(3, 2); (5, 6); (6, 7)} (d) R₄ = {(7, 2); (7, 6); (5, 3); (3, 3), (5, 2), (5, 7)} Solution: Given that P = {3, 5, 7} and Q = {2, 3, 6} Note: Every element of set P is associated with a unique element of set Q. No element of P must have more than one image. (a) f(1) = 3 and f(1) = 5 are not possible. so, this relation is not mapping from P to Q. (b) R₂ = {(3, 3); (7, 6)}. Every element of set P is associated with a unique element of set Q. hence, it is relation from P to Q. (c) R₃ = {(3, 2); (5, 6); (6, 7)} it’s not relation from P to Q. (d) R₄ = {(7, 2); (7, 6); (5, 3); (3, 3), (5, 2), (5, 7)} Every element of set P is associated with a unique element of set Q. hence, it is relation from P to Q. Therefore, the final answer is (b) R₂ = {(3, 3); (7, 6)} and (d) R₄ = {(7, 2); (7, 6); (5, 3); (3, 3), (5, 2), (5, 7)} 9. Write the domain and range of the following relations. (a) R₁ = {(5, 4); (7, 9); (5, 9); (1, 8); (8, 6); (1, 9)} (b) R₂ = {(p, 3); (q, 4); (r, 3); (p, 4); (s, 5); (q, 5)} Solution: Given that (a) R₁ = {(5, 4); (7, 9); (5, 9); (1, 8); (8, 6); (1, 9)} (b) R₂ = {(p, 3); (q, 4); (r, 3); (p, 4); (s, 5); (q, 5)} (a) R₁ = {(5, 4); (7, 9); (5, 9); (1, 8); (8, 6); (1, 9)} From the given information, the Domain = {1, 5, 7, 8} and Range = {4, 6, 8, 9} (b) R₂ = {(p, 3); (q, 4); (r, 3); (p, 4); (s, 5); (q, 5)} From the given information, the Domain = {p, q, r, s} and Range = {3, 4, 5} 10. Let P = {3, 4, 5, 6, 7, 8}. Define a relation R from A to A by R = {(x, y) : y = x + 1}. • Depict this relation using an arrow diagram. • Write down the domain and range of R. Solution: Given that P = {3, 4, 5, 6, 7, 8}. Define a relation R from A to A by R = {(x, y) : y = x + 1}. If x = 3, y = x + 1 = 3 + 1 = 4. x = 4, y = x + 1 = 4 + 1 = 5. x = 5, y = x + 1 = 5 + 1 = 6. x = 6, y = x + 1 = 6 + 1 = 7. x = 7, y = x + 1 = 7 + 1 = 8. x = 8, y = x + 1 = 8 + 1 = 9. R = {(3, 4); (4, 5); (5, 6); (6, 7); (7, 8)} where P = {3, 4, 5, 6, 7, 8}. Domain = Set of all first elements in a relation = {3, 4, 5, 6, 7} Range = Set of all second elements in a relation = {4, 5, 6, 7, 8} 11. Adjoining figure shows a relationship between the set P and Q. Write this relation in the roster form. What are its domain and range? Solution: The relation mentioned in the figure shows, P a domain and Q as a range. Let the relation be R. In roster form R = {(3, 6); (6, 12); (9, 18)} Domain = Set of all first elements in a relation = {3, 6, 9} Range = Set of all second elements in a relation = {6, 12, 18} 12. In the given ordered pairs (2, 4); (4, 16); (5, 7); (1, 3); (6, 36); (2, 9); (1, 1), find the following relationship: (a) Is a factor of …. (b) Is a square root of ….. Also, find the domain and range in each case. Solution: Given that the ordered pairs (2, 4); (4, 16); (5, 7); (1, 3); (6, 36); (2, 9); (1, 1). (a) Is a factor of …. Let’s find out the factor of …. from the given order pars. (2, 4); (4, 16); (1, 3); (6, 36); (1, 1) Domain = Set of all first elements in a relation = {1, 2, 4, 6} Range = Set of all second elements in a relation = {1, 3, 4, 16, 36} (b) Is a square root of ….. Let’s find out the square root of …. from the given order pars. (2, 4); (4, 16); (6, 36). Domain = Set of all first elements in a relation = {2, 4, 6} Range = Set of all second elements in a relation = {4, 16, 36} 13. Draw the arrow diagrams to represent the following relations. (a) R₁ = {(2, 2); (2, 7); (2, 8); (6, 9); (7, 4)} (b) R₂ = {(5, 11); (5, 14); (5, 17); (6, 14); (7, 17)} (c) R₃ = {(3, 4); (4, 6); (5, 8); (6, 10); (7, 12)} (d) R₄ = {(a, x); (a, y); (b, p); (b, z); (c, y)} Solution: (a) Given that R₁ = {(2, 2); (2, 7); (2, 8); (6, 9); (7, 4)} Let the two sets are P and Q. The required diagram is (b) Given that R₂ = {(5, 11); (5, 14); (5, 17); (6, 14); (7, 17)} Let the two sets are P and Q. The required diagram is (c) Given that R₃ = {(3, 4); (4, 6); (5, 8); (6, 10); (7, 12)} Let the two sets are P and Q. The required diagram is (d) Given that R₄ = {(a, x); (a, y); (b, p); (b, z); (c, y)} Let the two sets are P and Q. The required diagram is 14. Represent the following relation in the roster form. (a) (b) (c) (d) Solution: (a) R = {(a, x) (a, z) (b, y) (c, x) (c, q) (d, z)} (b) R = {(3, 7) (3, 9) (4, 7) (4, 10) (5, 9) (3, 11)} (c) R = {(2, 2) (5, 3) (10, 4) (17, 5)} (d) R = {(11, 3) (11, 6) (13, 3) (13, 4) (13, 5) (16, 4) (16, 6) (26, 6)}
Learn about equivalent ratios to obtain theclear principle on ratio. We know that we usage ratios to compare numbers. You are watching: Which ratio is equal to 15 20 How do we find two identical ratios by making use of multiplication and division? To acquire a ratio tantamount to a provided ratio we multiply or division both the regards to the offered ratio through the exact same non-zero number. We will learn how to discover the tantamount ratios that a offered ratio by composing the proportion as a fraction and then to compare by making use of multiplication and also division. Solved examples to find two equivalent ratios: 1. Give two identical ratios the 8 : 18. Solution: We will find the very first equivalent ratio of 8 : 18 by utilizing multiplication. So, very first we need to write the givenratio together fraction, = 8/18 = (8 × 2)/(18 × 2) = 16/36 = 16 : 36 (one indistinguishable ratio), So, 16 : 36 is an indistinguishable ratio that 8 :18. Now we will find one more equivalent ratioof 8 : 18 by using division. Similarly, very first we have to write thegiven ratio as fraction, = 8/18 = (8 ÷ 2)/(18 ÷ 2) = 4/9 = 4 : 9 (another identical ratio) So, 4 : 9 is an indistinguishable ratio that 8 :18. Therefore, the two tantamount ratios of 8: 18 room 16 : 36 and 4 : 9. 2.Frame two equivalent ratios that 4 : 5. Solution: To find two identical ratios the 4 : 5 we require to apply multiplicationmethod only to gain the answer in integer form. First we should write the given ratio asfraction, = 4/5 = (4 × 2)/(5 × 2) = 8/10 = 8 : 10 is one indistinguishable ratio, Similarly again, we have to write thegiven ratio 4 : 5 as fraction to get another equivalent ratio; = 4/5 = (4 × 3)/(5 × 3) = 12/15 is one more equivalent ratio Therefore, the two indistinguishable ratios of 4: 5 space 8 : 10 and 12 : 15. Note: In this inquiry we can’t apply divisionmethod to acquire the price in integer form because the G.C.F. The 4 and also 5 is 1.That means, 4 and 5 cannot be divisible by any type of other number except 1. 3. For the following ratio find the two tantamount ratios that 11 : 13. Solution: To find two equivalent ratios that 11 : 13 first we need to write the provided ratio together fraction, 11/13 = (11 × 2)/(13 × 2) = 22/26 = 22 : 26 is one tantamount ratio Similarly again, to get one more equivalent proportion of 11 : 13; 11/13 = (11 × 4)/(13 × 4) = 44/52 = 44 : 52 is another equivalent ratio Therefore, the two tantamount ratios that 11 : 13 room 22 : 26 and also 44 : 52. Note: If a : b and also x : y space two tantamount ratios, we create a/b = x/y. Solved example to find three tantamount ratios: 4. Find three equivalent ratios that 3 : 8. Solution: 3 : 8 = 3/8 = (3 × 2)/(8 × 2), = 6/16 = 6 : 16 is the first equivalent ratio. 3 : 8 = 3/8 = (3 × 4)/(8 × 4), = 12/32 = 12 : 32 is the 2nd equivalent ratio. 3 : 8 = 3/8 = (3 × 6)/(8 × 6), = 18/48 = 18 : 48 is the third equivalent ratio. Therefore, the three indistinguishable ratios that 3 : 8 room 6 : 16, 12 : 32 and also 18 : 48. See more: How Much Is 112 Grams In Pounds ? Convert 112 G To Lb 112G In Lbs And Oz 6th grade PageFrom tantamount Ratios to residence PAGE
# How far does the runner whose velocity-time graph is shown in Fig. 2-34 travel in 16 s? 7 years ago Given below is the velocity-time graph for a runner who accelerates initially for 2 s, then moves with uniform speed for 8 s , then decelerating for another 2 s and finally moving with uniform speed for 4 s. It is important to note that the length of each box in the figure above is 2 seconds along the horizontal direction and 2 m/s along the vertical. The displacement of the runner is given by the area under the velocity-time curve but the area under the magnitude of velocity and time graph gives the distance. This can be seen from the definition of distance in integral form given as: To make it easy, divide the area into squares, rectangles and triangles (as shown in the figure above). From the figure above, we have that the area under the curve is the sum of area of triangle ABC, rectangle BDEC, triangle DOF, rectangle OFGE and square FHIG respectively. Now area of triangle ABC (say A1) is: Area of rectangle BDEC (say A2) is: Area of triangle DOF (say A3)is: Area of rectangle OFGE (say A4)is: Area of square FHIG (say A5)is: The total area (say s) under the curve is: Therefore the runner has gone 100 m in 16 seconds. one year ago Dear student, The total distance covered can be obtained by calculating area under the v – t graph between t = 0 to t = 16 s Now, The whole figure can be divided into 3 figures A1 → a triangle from origin to t = 1s A2 → a trapezium from t =1 s to t = 12 s above v = 4 m/s A3 → a rectangle from t = 1 s to t = 16 s below v = 4 m/s A1 = ½ x 1 x 4 = 2 m A2 = ½ x ([12 – 1] + [10 – 2]) x (8 – 4) = ½ x 19 x 4 = 38 m A3 = (16 – 1) x 4 = 60 m Hence, The total distance covered = A1 + A2 + A3 = 2 + 38 + 60 = 100 m Hope it helps. Thanks and regards, Kushagra Kavita 13 Points one year ago Triangle from t(0) to t(2): 8 sq.m. Square from t(2) to t(10): 64 sq.m. Trapezoid from t(10) to t(12): 12 sq.m. Square from t(12) to t(16): 16 sq.m. Total= 100 sq.m. Total = 100 SQ.m. Gajendra Chilwal 13 Points 2 months ago Distance traveled by the runner = 100 sq. metre Man I soo hungry . He is aa fucking guy. Runner masterbates daily . And now he has become weak due to his bad habits. He has ha gay in his house.
# How to Reject Outliers in Data Outliers are data points that are outside the normal range of data. They are much higher or much lower numbers than the rest of your data. In order to draw meaningful conclusions from experimental data, you must examine your data for outliers and decide whether or not to eliminate them. ### Part 1 Part 1 of 2:Calculating Outliers 1. 1 Observe your data. Look for numbers that are much higher or much lower than the majority of your data points. • Letâ€™s imagine that you have planted a dozen sunflowers and are keeping track of how tall they are each week. • All of your flowers started out 24 inches tall. Most of your flowers grew about 8-12 inches, so theyâ€™re now about 32-36 inches tall. • But a neighboring child accidentally threw his ball into your yard, and when he ran in to get it, he crushed one of your sunflowers! • When you measure your flowers at the end of the week, the crushed one is only about 3 inches off the ground. Since the others are so much taller, you might consider this crushed flower an outlier. 2. 2 • In order, your sunflower heights in inches are 3, 32, 32, 33, 33, 33, 34, 34, 35, 35, 36, 36. 3. 3 Find the halfway point of your data. For the sunflower example, the halfway point is between 33 and 34. 4. 4 Find the first quartile, or Q1. To find Q1, determine the median number in the first half your data. The median is the number that falls in the middle of the data. • In our sunflower example, the first half of the data is 3, 32, 32, 33, 33, 33. • The middle is between 32 and 33, so the median is 32.5. • Call this Q1. • Q1=32.5 5. 5 Find the third quartile, or Q3. To find Q3, determine the median number in the second half of your data. • In our sunflower example, the second half of the data is 34, 34, 35, 35, 36, 36. • The middle is between 35 and 35, so the median is 35. • Call this Q3. • Q3=35 6. 6 Subtract Q1 from Q3. This number is the interquartile range (IQR). • Q3-Q1=IQR • 35-32.5=2.5 • IQR=2.5 7. 7 Determine whether you have an outlier beyond your upper limit. Outliers are any number that is larger than Q3+1.5(IQR) or smaller than Q1-1.5(IQR). Start with your upper limit. • Q3+1.5(IQR) • 35+1.5(2.5) • 35+3.75=38.75 • 38.75 is your upper limit. Any number higher than 38.75 is an outlier. • In the sunflower data set, no number is higher than the upper limit. 8. 8 Determine whether you have an outlier beyond your lower limit. The process is similar to finding outliers beyond the upper limit, but the formula is a little different. • Q1-1.5(IQR) • 32.5-1.5(2.5) • 32.5-3.75=28.75 • 28.75 is your lower limit. Any number lower than 28.75 is an outlier. • In the sunflower data set, 3 is less than 28.75, so it is an outlier. You can justify your decision to eliminate it from your data.[1] ### Part 2 Part 2 of 2:Deciding to Reject Outliers 1. 1 Do some quick calculations. This will help you determine whether the outliers are causing problems with your data. • Perhaps the heights of your 10 sunflowers, in inches are: 34, 32, 33, 33, 34, 3, 35, 35, 36, 36, 33, and 32. • If you include 3, the average height of your sunflowers is 31.3 inches. • If you disregard 3, the average height of your sunflowers is 33.9 inches. • If you wanted to make generalizations about your flowers sunflowers, (such as calculating the average amount that they grew over a weekâ€™s time) you may want to reject the outliers. 2. 2 Determine the cause of your outliers. If human error caused a very high or very low number (as it did in the sunflower example), this data point isnâ€™t very useful to you. Ask yourself whether this number is really a part of the data set that you intended to study. • Since someone stepped on your sunflower, the outlying data point doesnâ€™t actually tell you anything about how your sunflowers grew. [2] 3. 3 Decide whether or not to eliminate your outliers. Base your decision on whether including the number in your data set gives you helpful information or not. • In the case of the crushed sunflower, you would probably reject the 3 inch sunflower. • You might also reject outliers if you think you measured wrong or wrote down the wrong number. • On the other hand, if your sunflower was much shorter than the others because it was planted in a place where it did not receive direct sunlight, you may decide that this is useful information and include this number in your data set. 4. 4 Reject the outlier. Eliminate this number from your data. From this point forward, do your calculations without this number. 5. 5 Defend your decision. Rejecting outliers makes your data "impure." You should only reject data points if you have a very good reason. If you need to write up a report of your data, be prepared to explain why you rejected the outliers using the formulas Q3+1.5(IQR) and Q1-1.5(IQR). [3] ## Community Q&A 200 characters left ## Tips Submit a Tip All tip submissions are carefully reviewed before being published Thanks for submitting a tip for review! ## Warnings • It is not considered good statistical practice to discard outliers without strong cause. Discarding outliers without cause typically results in underestimating the actual variability of the process that generates the data. Thanks!
• date post 11-Mar-2022 • Category Documents • view 0 0 Embed Size (px) Transcript of Intercepts of Graphs - MR. JONES Today’s Vocabulary x-intercept y-intercept positive negative root zero Learn Intercepts of Graphs of Functions The intercepts of graphs are points where the graph intersects an axis. The x-intercept is the x-coordinate of a point where a graph crosses the x-axis. The y-intercept is the y-coordinate of a point where a graph crosses the y-axis. A function is positive when its graph lies above the x-axis. A function is negative when its graph lies below the x-axis. Example 1 Intercepts of the Graph of a Linear Function Use the graph to estimate the x- and y-intercepts of the function and describe where the function is positive and negative. The x-intercept is the point where the graph crosses the x-axis, ( , ). The y-intercept is the point where the graph crosses the y-axis, ( , ). A function is positive when its graph lies above the x-axis, or when . A function is negative when its graph lies below the x-axis, or when . Check Use the graph to estimate the x- and y-intercepts of the function and describe where the function is positive and negative. A. x-intercept: (-2, 0); y-intercept: (0, -6); positive: x > -2; negative: x < -2 B. x-intercept: (0, −6); y-intercept: (-2, 0); positive: x < -2; negative: x > -2 C. x-intercept: (-2, 0); y-intercept: (0, -6); positive: x < -2; negative: x > -2 D. x-intercept: (0, -6); y-intercept: (-2, 0); positive: x > -2; negative: x < -2 Study Tip: Notice that intercept can be used to refer to either the point where the graph intersects the axis or the nonzero coordinate of the point where the graph intersects the axis. y Study Tip To help remember the difference between the x- and y-intercepts, remember that the x-intercept is where the graph intersects the x-axis, and the y-intercept is where the graph intersects the y-axis. Today’s Goals Identify the intercepts of functions and intervals where functions are positive and negative. Solve equations by graphing. Lesson 3-4 • Intercepts of Graphs 167 Sample answer: The function is linear because its graph forms a straight line. 3 0 0 6 x < 3 x > 3 C THIS MATERIAL IS PROVIDED FOR INDIVIDUAL EDUCATIONAL PURPOSES ONLY AND MAY NOT BE DOWNLOADED, REPRODUCED, OR FURTHER DISTRIBUTED. M cG raw -H ill Education Explain why this function is nonlinear. Example 2 Intercepts of the Graph of a Nonlinear Function Use the graph to estimate the x- and y-intercepts of the function and describe where the function is positive and negative. y negative: x is between and . Check Use the graph of the function to determine key features. y x O Part A Determine whether each ordered pair represents an x-intercept, a y-intercept, or neither. (1, 0) (0, 1) Part B Describe where the function is positive and negative. A. positive: x < 1 and x > 4; negative: x is between 1 and 4 B. positive: x > 1 and x < 4; negative: x is between 1 and 4 C. positive: x is between 1 and 4; negative: x > 1 and x < 4 D. positive: x is between 1 and 4; negative: x < 1 and x > 4 Watch Out! The graphs of nonlinear functions can have more than one x-intercept. Go Online You can complete an Extra Example online. -4 -12 x-intercept neither D 3 Sample answer: The function is nonlinear because its graph does not form a straight line. THIS MATERIAL IS PROVIDED FOR INDIVIDUAL EDUCATIONAL PURPOSES ONLY AND MAY NOT BE DOWNLOADED, REPRODUCED, OR FURTHER DISTRIBUTED. C op yr ig The function is only graphed from 0 to 9 seconds. What can you assume about the function when x > 9? Interpret this meaning. Does it make sense in the context of the situation? Example 3 Find Intercepts from a Graph SPORTS The graph shows the height of a ball for each second x that it is airborne. Use the graph to estimate the x- and y-intercepts of the function, where the function is positive and negative, and interpret the meanings in the context of the situation. The x-intercept is . That means that the ball will hit the ground after seconds. The y-intercept is . This means that at time , the ball was at a height of feet. The function is positive when x is between and , which means that the ball is in the air for seconds. No portion of the graph shows that the function is . Check FITNESS The graph shows the number of people y at a gym x hours after the gym opens. 40 0 80 120 160 200 x y N um Gym Occupancy Part A Use the graph to estimate the x- and y-intercepts. x-intercept: ( , ) y-intercept: ( , ) Part B Which statements describe the meaning of the x- and y-intercepts in the context of the situation? Select all that apply. A. There were 20 people at the gym when it opened. B. The gym closed after 20 hours. C. The gym closed after 12 hours. D. There were 12 people at the gym when it opened. x y 10 H ei gh t ( 9 9 A, C Sample answer: I assume that the graph continues to follow the same path, so the function is negative when x > 9. This means that the ball would be at a negative height when x > 9. This does not make sense in the context of the situation because it is impossible for a ball to be at a negative height. THIS MATERIAL IS PROVIDED FOR INDIVIDUAL EDUCATIONAL PURPOSES ONLY AND MAY NOT BE DOWNLOADED, REPRODUCED, OR FURTHER DISTRIBUTED. M cG raw -H ill Education Go Online You can complete an Extra Example online. Example 4 Find Intercepts from a Table LUNCH Violet starts the semester with \$150 in her student lunch account. Each day she spends \$3.75 on lunch. The table shows the function relating the amount of money remaining in her lunch account to the number of days Violet has purchased lunch. Part A Find the intercepts. The x-intercept is where y = , so the x-intercept is . The y-intercept is where x = , so the y-intercept is . Part B Describe what the intercepts mean in the context of the situation. The x-intercept means that after buying lunch for days, Violet will have \$ left in her lunch account, or it will take Violet days to use all of the money in her lunch account. The y-intercept means that Violet’s lunch account has \$ after buying lunch for days, or the beginning balance of her lunch account is \$ . Check MOVIES Ashley received a gift card to the movie theater for her birthday. The table shows the amount of money remaining on her gift card y after x trips to the movie theater. Number of Trips Balance (\$) x y 0 90 1 81 2 72 3 63 5 45 7 27 10 0 Part A Find the y-intercept. ( , ) Part B Find the x-intercept and describe what it means in the context of the situation. A. (10, 0); The initial balance on the gift card was \$10. B. (90, 0); The initial balance on the gift card was \$90. C. (10, 0); After 10 trips to the movies, there will be no money left on the gift card. D. (90, 0); After 90 trips to the movies, there will be no money left on the gift card. Time (Days) Balance (\$) x y 0 150 2 142.50 5 131.25 10 112.50 15 93.75 30 37.50 40 0 170 Module 3 • Relations and Functions 0 40 0 150 40 0 40 150 0 150 0 90 C THIS MATERIAL IS PROVIDED FOR INDIVIDUAL EDUCATIONAL PURPOSES ONLY AND MAY NOT BE DOWNLOADED, REPRODUCED, OR FURTHER DISTRIBUTED. C op yr ig on Learn Solving Equations by Graphing The solution, or root, of an equation is any value that makes the equation true. A zero is an x-intercept of the graph of the function. For example, the root of 3x = 6 is 2. A linear equation, like 3x = 6, has at most one root, while a nonlinear equation, like x2 + 4x − 5 = 0, may have more than one. Equation Related Function 3x = 6 f (x) = 3x − 6 or y = 3x − 6 x2 + 4x − 5 = 0 f (x) = x2 + 4x − 5 or y = x2 + 4x − 5 The graph of the related function can be used to find the solutions of an equation. The related function is formed by solving the equation for 0 and then replacing 0 with f (x) or y. Values of x for which f(x) = 0 are located at the x-intercepts of the graph of a function and are called the zeros of the function f. The roots of an equation are the same as the zeros of its related function. The solutions and roots of an equation are the same value as the zeros and x-intercepts of its related function. For the equation 3x = 6: • 2 is the solution of 3x = 6. • 2 is the root of 3x = 6. • 2 is the zero of f (x) = 3x − 6. • 2 is the x-intercept of f (x) = 3x − 6. Example 5 Solve a Linear Equation by Graphing Solve −2x + 7 = 1 by graphing. Check your solution. Find the related function. −2x + 7 = 1 Subtract 1 from each side. −2x + = Simplify. Graph the left side of the equation. The related function is f(x) = , which can be graphed. y xO The graph intersects the x-axis at . This is the x-intercept, or zero, which is also the root of the equation. So, the solution of the equation is . Check your solution by solving the equation algebraically. Go Online You can complete an Extra Example online. What is the difference between a root and a zero? Suppose you first solved the equation algebraically. How could you use your solution to graph the zero of the related function? Go Online An alternate method is available for this example. Lesson 3-4 • Intercepts of Graphs 171 - 1 - 1 -2x + 6 6 0 3 3 Sample answer: A root is the solution of an equation, while a zero is the x-intercept of the related function. Sample answer: The solution of the equation is the same as the x-intercept, or zero, of the function. So, I can graph the zero by plotting a point at 3 on the x-axis. THIS MATERIAL IS PROVIDED FOR INDIVIDUAL EDUCATIONAL PURPOSES ONLY AND MAY NOT BE DOWNLOADED, REPRODUCED, OR FURTHER DISTRIBUTED. M cG raw -H ill Education Example 7 Solve an Equation of a Horizontal Line by Graphing Solve 4x + 3 = 4x − 5 by graphing. Check your solution. Find the related function. 4x + 3 = 4x − 5 Add 5 to each side. 4x = 4x Simplify. 8 = Simplify. Graph the left side of the equation. The related function is f (x) = , which can be graphed. y xO The graph does not intersect the x-axis. This means that there is x-intercept and, therefore, there is solution. Does solving the equation algebraically give a different solution? Explain your reasoning. Go Online You can complete an Extra Example online. Example 6 Solve a Nonlinear Equation by Graphing Solve x2 − 4x = −3 by graphing. Check your solution. Find the related function. x2 − 4x = −3 Add 3 to each side. x2 − 4x + 3 = Simplify. Graph the left side of the equation. The related function is f(x) = , which can be graphed. y xO . Go Online You can watch a video to see how to use a graphing calculator with this example. 172 Module 3 • Relations and Functions + 5 + 5 8 Sample answer: No; solving algebraically results in the false statement 8 = 0, which means there is no solution. + 3 + 3 0 x2 - 4x = -3 1 3 1 3 THIS MATERIAL IS PROVIDED FOR INDIVIDUAL EDUCATIONAL PURPOSES ONLY AND MAY NOT BE DOWNLOADED, REPRODUCED, OR FURTHER DISTRIBUTED. C op yr ig on Check Equations and the graphs of their related functions are shown. Write the related function and its zero(s) under the appropriate graph. y −8 −4−6 −2 8642 −8 −6 −4 x2 - 6 = x related function: zeros: Apply Example 8 Estimate Solutions by Graphing PARTY Haley is ordering invitations for her graduation party. She has \$40 to spend and each invitation costs \$0.96. The function m = 40 − 0.96p represents the amount of money m Haley has left after ordering p party invitations Find the zero of the function. Describe what this value means in the context of this situation. 1 What is the task? Describe the task in your own words. Then list any questions that you may have. How can you find answers to your questions? 2 How will you approach the task? What have you learned that you can use to help you complete the task? Go Online You can complete an Extra Example online. Lesson 3-4 • Intercepts of Graphs 173 4 f (x) = 5 f (x) = - 1 __ 2 x + 2 Sample answer: I need to find the zero of the function and describe what it means. How can I determine the meaning of the zero from a graph of the function? I can review graphing linear functions and labeling axes. Sample answer: I will graph the function by making a table of values. I will estimate the x-intercept of the graph to find the zero. I will then check my solution by solving the equation algebraically. I will use the axes labels to help me interpret my solution. (continued on the next page) THIS MATERIAL IS PROVIDED FOR INDIVIDUAL EDUCATIONAL PURPOSES ONLY AND MAY NOT BE DOWNLOADED, REPRODUCED, OR FURTHER DISTRIBUTED. M cG raw -H ill Education A m ou nt 3 What is your solution? Use your strategy to solve the problem. Graph the function. Estimate the solution. 10 20 30 40 500 10 20 30 40 50 _______ invitations. What does your solution mean in the context of the situation? 4 How can you know that your solution is reasonable? Write About It! Write an argument that can be used to defend your solution. Check DATA Blair’s cell phone plan allows her to use 3 GB of data, and she uses approximately 0.14 GB of data each day. The function g = 3 − 0.14d represents the amount of data g in GB she has left after d days. Part A Examine the graph of the function to estimate its zero to the nearest day. The graph appears to intersect the x-axis at . Part B Solve algebraically to check your answer. Round to the nearest tenth. x = Part C Describe what your answer to Part B means in this context. After days, Blair has GB left. 174 Module 3 • Relations and Functions 22 21.4 21.4 0 Sample answer: Haley can order 41 invitations with the amount of money she has to spend. Sample answer: This amount is close to the estimated zero of 42 invitations from the graph. 42 ≈ 41.67
# top answer: Section 5.2, page 202, use pencil and paper calculations and your table of Laplace transforms to pe 1. Section 5.2, page 202, use pencil and paper calculations and your table of Laplace transforms to perform the tasks required in exercises #28 and #31. Show all the steps to convince your reader you understand the process. 2. Section 5.2, page 202, use pencil and paper calculations and your table of Laplace transforms to perform the tasks required in exercises #34 and #40. Simplify your answer as a single fraction, leaving the denominator in factored form, but writing the numerator as a single polynomial. Show all the steps to convince your reader you understand the process. Section 5.2 Lecture Notes (Part 2) The goal of this activity is to add a couple more important properties of the Laplace transform. Theorem Suppose f is piecewise continuous and of exponential order. If c is a constant, then Don't use plagiarized sources. Get Your Custom Essay on top answer: Section 5.2, page 202, use pencil and paper calculations and your table of Laplace transforms to pe Just from \$10/Page Proof: We’ll use our definition of the Laplace transform to write where F is the Laplace transform of f. Now, many find the last step of the proof difficult to understand, so let’s note that the Laplace transform of a function of t returns a function of s. That is, where F is the Laplace transform of f. Now, let’s note what this means. Suppose we substitute 5 for s. Then we would get Now, note that if we substituted for s, we would get: Let’s add this to our table. 1 Let’s try an example. Example #1: Find the Laplace transform of . Solution: By property #4 in our table, we know that: That is, Therefore, by our new theorem (item #6 in our table), we can write: Let’s check our answer with Matlab. syms s t 2 y=exp(3t)cos(5t); Y=laplace(y) Y = Y=simplifyFraction(Y) Y = Same answer! Now, here is some visual evidence that it is true. Recall from algebra, that the graph of is a shift of the graph of , c units to the right (provided ). Let’s use this fact to better understand what is going on with our new property. syms s t g=cos(5t); G=laplace(g); f=exp(3t)cos(5t); F=laplace(f); fplot(G,[0,10]) hold on fplot(F,[3,10]) grid on xlabel(‘s-axis’) title(‘Laplace transforms of g(t)=cos 5t and f(t)=exp(3t)cos 5t’) legend(‘F(s)’,’F(s-3)’,’Location’,’southeast’) hold off 3 Nice! Clearly shows that if , then , because the graph of the second is translated 3 units to the right. Theorem Suppose f is a piecewise continuous function of exponential order. Let be the Laplace transform of f. Then, Now, let’s differentiate both sides with respect to s. Note that we can bring the derivative with respect to s inside because we have an integral with respect to t. Now, we can continue with: 4 Therefore, we’ve shown that: Let’s try an example and see if this works. Example #2: Find the Laplace transform of . Solution: Using item in our table, we know that if , then: Therefore, by our new theorem, so we’ll need to differentiate to find the answer. Therefore, Now, let’s check our answer with Matlab. syms s t y=texp(5t); 5 Y=laplace(y) Y = Higher Powers of t We’ve shown that where F is the Laplace transform of f. Let’s find what is equal to. Therefore, we’ve shown that If we were to differentiate again, we would discover that: If we were to differentiate again, we would discover that where the notation means the “fourth derivative of F.” If we were to differentiate again, we would discover that If we were to differentiate again, we would discover that 6 There is a definite pattern developing here which can be finally expressed as where means the “n th derivative of F.” Let’s add this to our table. Let’s try an example. Example #3: Find the Laplace transform of . Solution: Using item #5 in our table, we know that if , then Therefore, by property #7 in our table, Therefore, we’ll have to find the second derivative of . Let’s start with the first derivative. 7 We can check this answer with Matlab. syms s t F=4/(s^2+16); diff(F,s) ans = Same answer, so we can find the second derivative. A lot of work. Let’s check this derivative with Matlab. diff(F,s,2) ans = simplifyFraction(diff(F,s,2)) ans = 8 Now, we can quickly check our answer with Matlab’s laplace command. y=t^2sin(4t); Y=laplace(y) Y = Y=simplifyFraction(Y) Y = Let’s do another example. In our previous examples, we showed a number of detailed steps to help our readers, but now we will start to take some shortcuts. Example #4: Write the following initial value problem in algebraic form, then solve for . Solution: The Laplace transform of the left side of our equation is: Note how we used the initial conditions and in our last step. Next, the right side. By property #5 in our table, Therefore, by property #6 in our table, Therefore, the Laplace transform of our equation is: 9 Now we can solve for . A lot of work, so errors are likely. Let’s check our answer with Matlab. syms s t Y f=exp(2t)sin(3t); F=laplace(f); y0=1; yp0=-1; Y1=sY-y0; Y2=sY1-yp0; sol=solve(Y2-2Y==F,Y); sol=simplifyFraction(sol) sol = Perfect! 🙂 Example #5: Write the following initial value problem in algebraic form, then solve for . Solution: We’ll first find the Lapalce transform of the left side, using our initial conditions and . Now we take the Laplace transform of the right side. First, by property #4 in our table, we have: 10 Therefore, by property #7 in our table, so we need to differentiate . We can check our differentiation with Matlab. syms s F=s/(s^2+9); Fp=diff(F,s); Fp=simplifyFraction(Fp) Fp = Same answer! Therefore, the Laplace transform of our right side is: Now, we can set up an algebraic equation for our initial value problem with these results and solve for . 11 Wow! A lot of steps, so a mistake is likely. Let’s check our answer with Matlab. Recall that the initial value problem is: syms s t Y f=tcos(3t); F=laplace(f); y0=1; yp0=0; Y1=sY-y0; Y2=sY1-yp0; sol=solve(Y2-2Y1+Y==F,Y); sol=simplifyFraction(sol) sol = 12 T 8. y(I): r: ” ro. y(il: e:, tr. )(/): cos3r 9. y(t) : 73 ll. y(tl: e-3′ /@ ,,,:,: sin5r 202 CHAPTER 5 ‘The Laplace Transform Exercises 8-13 are designed to test the validity of Proposi- tion 2. L In each exercise. (i) cornpute 4(,y’)(s) for the given function, and (ii) compute s 4(.v)(s) – )(0) for the given function. Com- pare this result to that found in part (i) to verify that t 4(,r,’)(.r): s4(.y)(s) )(0). 39. y” I y’ + 2y : e-‘ cos2t, y(0) : l, y'(0) : -1 40. y” l2y’ + 5y : t2e-‘, y(0) : l, y'(0) : -2 41. y” 5y:3e-tcos4t,y(0) = -1,y'(0):) 42. Here is an interesting way to compute the Laplace trans- form of cos ror. (a) Using only Definition 1.1, show that f,{sint}(s) : l/(s2 + l). (b) Suppose that /(/) has Laplace transform F(s). Show that Llf (arlls) = j. (;) Use this property and the result found in part (a) to show that ^C{sina,lt}(s) : -3-.s”[email protected] (c) If /(r) : sin cr.r/, then /'(r) : @ cos @t . Use Proposi- tion 2.1 to show that f{rocos<ot}(s): “: “s’+ a’ thus ensuring J’ 4,{cos ror}ts) : —,-;.s-+o’ -FIn a manner similar to that proposed in Exercises 8-13, verify -p the result of Proposition 2.4 for the functions defined in Exer- – cises l4-17.E.-F r+. 1.(r) :11 N’ ro. y(r) : sin2r 15. .v(r) : s-2l /fO )'(r) = rr + 3I + s In Exercises 18-25, use Propositions 2.1,2.4, and2.7 to trans- form the given initial value problem into an algebraic equation involving,C(y). Solve the resulting equation for the Laplace transform of v.,,A ” ).v’+ 3.v: 12. v(0) : -l19.y’-5y:””,.y(0):I F ro. )’ +5y : t2 * 2r + 3.y(0) : 0 Zt. y’ – 4y : cos2r. y{0} : -2 22. y” * 1l : sin4/, y(0) :0, y'(0) : l /@ y” 2y’ 2y : co, 2t, y(0): l, y'(0) :0 ?* y” I y’ +2y : cos2t * sin3r, )(0) : -1, y'(0) : I 49 y” + 3y’r5y : t + e-t,y(o) : -1, y'(o) : o In Exercises 26-29, we Proposition 2.12 to find the Laplace transform of the given function. 2-6. y(t) : e-t sin3t 27. y(t) : e2′ cosZt 1Q,,r, : e-2t (2t + 3) 29. y(t) = e-‘(t2 + 3t + 4) In Exercises 30-33, use Proposition 2.14 to find the Laplace transform of the given function. 30. y(r): /sin3/ r,Qr(r): re-‘ 32. y(t) : t2 cos2t 33. y(r) : 12r2t ‘ – In Exercise s 3441, use the propositions in Section 2 to trans- -‘rt1 fo.rn the given initial value problem into an algebraic equation ( N involving LO). Solve the resulting equation for the Laplace j L transform of y. iIr. y’+2y: rsinI. y(o): t t/ S- 35. y’ – y : t2e-z’ ,y(0) : 0 – fO. y’ + ]l, : e-‘ sin 3r, l(O) : O Lr l’ Sz. y’ – 2y – e2′ cos t, y(0) – -Z =: @tn”ro. mafunctionis defined by l-1a.1 : [- ,.’r’ ‘ ,lt. 1a > 0). Jo (a) Prove that f(1) : 1 (b) Prove that l(cvt 1) : af (cv). [n fact, if r is a positive integer, show that f (r * 1) : n!. (c) Show that Llt’l(s’ : f’11 “‘{u1 r Indeed, ifr is a positive integer, use this result to shou that,C{r”}(s) : nl/ s”+t. 44. Suppose that I is a continuous function for I > 0 and i. of exponential order. (a) If /(/) + F(s) is a transform pair, prove that c[ [‘ rr,rar]t,r – F”’.lJu’ \| .r Hint: Let g(r) : Ll .f trlrlr. Then note that 8′(t) = f (t) and use Proposition 2.1 to compute 4{g'(t)}(s t (b) Use the technique suggested in part (a) to find 38. y” * 4y : t2 sin4t, y(0) :0, y'(0) : -1 “,1 I I’ lr,,rt+l)l’ _-l ## Calculate the price of your order 550 words We'll send you the first draft for approval by September 11, 2018 at 10:52 AM Total price: \$26 The price is based on these factors: Number of pages Urgency Basic features • Free title page and bibliography • Unlimited revisions • Plagiarism-free guarantee • Money-back guarantee On-demand options • Writer’s samples • Part-by-part delivery • Overnight delivery • Copies of used sources Paper format • 275 words per page • 12 pt Arial/Times New Roman • Double line spacing • Any citation style (APA, MLA, Chicago/Turabian, Harvard) # Our guarantees Delivering a high-quality product at a reasonable price is not enough anymore. 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# Principle of Mathematical Induction/Warning/Example 2 ## Example of Incorrect Use of Principle of Mathematical Induction We are to prove that: $\dfrac 1 {1 \times 2} + \dfrac 1 {2 \times 3} + \dotsb + \dfrac 1 {\paren {n - 1} \times n} = \dfrac 3 2 - \dfrac 1 n$ For $n = 1$ we have: $\dfrac 3 2 - \dfrac 1 n = \dfrac 1 2 = \dfrac 1 {1 \times 2}$ Assuming true for $k$, we have: $\displaystyle \dfrac 1 {1 \times 2} + \dfrac 1 {2 \times 3} + \dotsb + \dfrac 1 {\paren {n - 1} \times n} + \dfrac 1 {n \times \paren {n + 1} }$ $=$ $\displaystyle \dfrac 3 2 - \frac 1 n + \dfrac 1 {n \paren {n + 1} }$ by the induction hypothesis $\displaystyle$ $=$ $\displaystyle \dfrac 3 2 - \frac 1 n + \paren {\dfrac 1 n - \dfrac 1 {n + 1} }$ $\displaystyle$ $=$ $\displaystyle \dfrac 3 2 - \frac 1 {n + 1}$ But clearly this is wrong, because for $n = 6$: $\dfrac 1 2 + \dfrac 1 6 + \dfrac 1 {12} + \dfrac 1 {30} = \dfrac 5 6$ on the left hand side, but: $\dfrac 3 2 - \dfrac 1 6 = \dfrac 4 3$ on the right hand side. ## Refutation The supposed sequence of terms on the left hand side starts at $n = 2$. It can be seen that it is meaningless (or has no terms for $n = 1$. Hence in the first statement: $\dfrac 3 2 - \dfrac 1 n = \dfrac 1 2 = \dfrac 1 {1 \times 2}$ the term on the right hand side is the term for $n = 2$, making this equation invalid. The correct result is Sum of Sequence of Products of Consecutive Reciprocals: $\displaystyle \sum_{j \mathop = 1}^n \frac 1 {j \paren {j +1} } = \frac n {n + 1}$ $\blacksquare$
Question Video: Determine the Magnitude of the Resultant of Two Parallel Forces Acting in Different Directions \| Nagwa Question Video: Determine the Magnitude of the Resultant of Two Parallel Forces Acting in Different Directions \| Nagwa # Question Video: Determine the Magnitude of the Resultant of Two Parallel Forces Acting in Different Directions Mathematics • Third Year of Secondary School ## Join Nagwa Classes In the figure below, 𝐅₁ and 𝐅₂ are two parallel forces measured in newtons, where 𝑅 is their resultant. If 𝑅 = 30 N, 𝐴𝐵 = 36 cm, and 𝐵𝐶 = 24 cm, determine the magnitude of 𝐅₁ and 𝐅₂. 05:24 ### Video Transcript In the figure below, 𝐅 sub one and 𝐅 sub two are two parallel forces measured in newtons, where 𝑅 is their resultant. If 𝑅 is equal to 30 newtons, 𝐴𝐵 equals 36 centimeters, and 𝐵𝐶 equals 24 centimeters, determine the magnitude of 𝐅 sub one and 𝐅 sub two. In this question, we have two parallel coplanar forces, 𝐅 sub one and 𝐅 sub two, acting in opposite directions. We are also given the resultant force 𝑅, which is equal to 30 newtons. The distance from point 𝐴 to 𝐵 is 36 centimeters, and the distance from 𝐵 to 𝐶 is 24 centimeters. The lines of action of our forces 𝐅 sub one, 𝐅 sub two, and 𝑅 are not perpendicular to the line segment 𝐴𝐶. However, as our three forces are parallel, we can add the angle 𝜃 to our diagram as shown. We can calculate the perpendicular components of these forces using our knowledge of right angle trigonometry. These are equal to 𝐅 sub one sin 𝜃, 𝐅 sub two sin 𝜃, and 𝑅 sin 𝜃. These will be useful when we come to take moments, as the moment of a force is equal to the magnitude of force multiplied by the perpendicular distance to the point at which we are taking moments. Whilst we can take moments about any point on our line, in this question, we will do so about point 𝐴. We will consider moments acting in the counterclockwise direction to be positive and those acting in the clockwise direction to be negative. This means that the moment 𝑀 sub one of the force 𝐅 sub one acts in the positive direction and is equal to 𝐅 sub one sin 𝜃 multiplied by 36. We can repeat this process for 𝑀 sub two, which is the moment of the force 𝐅 sub two. As this acts in the negative direction, this is equal to negative 𝐅 sub two sin 𝜃 multiplied by 60. Our expressions for 𝑀 sub one and 𝑀 sub two can be simplified as shown. We know that the distance 𝑥 from the line of action of the resultant force to the point at which we are taking moments is equal to the sum of the moments divided by 𝑅. In this case, we are taking moments about the point where the resultant acts. Therefore, 𝑥 is equal to zero. This means that the sum of our two moments must equal zero. 36 multiplied by 𝐅 sub one sin 𝜃 plus negative 60 multiplied by 𝐅 sub two sin 𝜃 equals zero. Since sin 𝜃 cannot be equal to zero, we can divide through by this. We can also divide through by 12 such that three 𝐅 sub one minus five 𝐅 sub two equals zero. As there are two unknowns here, we will call this equation one. We will now consider the resultant force and the fact that this is equal to the sum of the other forces. Going back to our initial diagram, if we let the positive direction be vertically upwards, we have 𝑅 is equal to 𝐅 sub one plus negative 𝐅 sub two. Since 𝑅 is equal to 30 newtons, we have 30 is equal to 𝐅 sub one minus 𝐅 sub two. Adding 𝐅 sub two to both sides of this equation, we have 𝐅 sub one is equal to 30 plus 𝐅 sub two. We will call this equation two. And we now have a pair of simultaneous equations that we can solve by substitution. One way of doing this is to substitute the expression for 𝐅 sub one in equation two into equation one. This gives us three multiplied by 30 plus 𝐅 sub two minus five 𝐅 sub two is equal to zero. Distributing the parentheses gives us 90 plus three 𝐅 sub two. The left-hand side then simplifies to 90 minus two 𝐅 sub two. By adding two 𝐅 sub two to both sides and then dividing through by two, we have 𝐅 sub two is equal to 45. Substituting this value back into equation two gives us 𝐅 sub one is equal to 30 plus 45, which is equal to 75. The magnitude of the two forces 𝐅 sub one and 𝐅 sub two are 75 newtons and 45 newtons, respectively. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
# Multiplication and Division, Part 1 ## Objective Build fluency with division facts using units of 2, 5, and 10. ## Common Core Standards ### Core Standards ? • 3.OA.A.2 — Interpret whole-number quotients of whole numbers, e.g., interpret 56 ÷ 8 as the number of objects in each share when 56 objects are partitioned equally into 8 shares, or as a number of shares when 56 objects are partitioned into equal shares of 8 objects each. For example, describe a context in which a number of shares or a number of groups can be expressed as 56 ÷ 8. • 3.OA.B.6 — Understand division as an unknown-factor problem. For example, find 32 ÷ 8 by finding the number that makes 32 when multiplied by 8. • 3.OA.C.7 — Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers. ## Criteria for Success ? 1. Solve division and unknown factor problems involving twos, fives, and tens by skip-counting, keeping track on their papers or on their fingers of how many twos, fives, or tens have been counted, stopping the skip-counting sequence when they reach the dividend, knowing that the number of twos, fives, or tens that they counted is the solution. ## Tips for Teachers ? • “For $8\times3$, you know the number of 3s and count by 3 until you reach 8 of them. For $24\div 3$, you count by 3 until you hear 24, then look at your tracking method to see how many 3s you have. Because listening for 24 is easier than monitoring the tracking method for 8 3s to stop at 8, dividing can be easier than multiplying” when using a skip-counting strategy to solve (OA Progression, p. 25). Thus, this lesson should be more accessible to students than Lesson 7. • As a supplement to the Problem Set, students can play the first version of "What's your Number?" from the Preferred Tasks in 3.OA.4 - About the Math, Learning Targets, and Rigor by the Howard County Public School System. #### Fishtank Plus • Problem Set • Student Handout Editor • Vocabulary Package ? ### Problem 1 Maureen says the skip-counting sequence “10, 20, 30, 40, 50, 60” to help her solve a problem. a. What multiplication problem might Maureen be trying to solve? How do you know? b. What if Maureen was solving a division problem? What problem might that have been? ### Problem 2 Solve. a. $12\div2 =$ _____ b. _____ $=35\div 5$ c. $90\div 10=$ _____ d. _____ $\times 5 = 45$ e. $2 \times$ _____ $=16$ ## Discussion of Problem Set ? • I think the answer in #3f is 8. Do you agree or disagree? What mistake did I make? • What do you notice about #11a and #11d? What do you wonder? • Is division commutative? How do you know? ? Solve. 1. $90 \div 10 =$ _____ 2. ____ $\times 2 = 8$ 3. $40 = 5\times$ _____ ?
# Lesson 3 Half an Equilateral Triangle • Let’s investigate the properties of altitudes of equilateral triangles. ### 3.1: Notice and Wonder: Triangle Slices Sketch an equilateral triangle and an altitude from any vertex in the equilateral triangle. What do you notice? What do you wonder? ### 3.2: Decomposing Equilateral Triangles 1. Here is an equilateral triangle with side length 2 units and an altitude drawn. Find the values of $$x$$ and $$y$$. 2. Measure several more of these “half equilateral triangles” by drawing equilateral triangles and altitudes. Compute the ratios of the side lengths of these new triangles. 3. Make a conjecture about side lengths in “half equilateral triangles.” ### 3.3: Generalize Half Equilateral Triangles Calculate the lengths of the 5 unlabeled sides. Here is a collection of triangles which all have angles measuring 30, 60 and 90 degrees. What is the total area enclosed by the 5 triangles? ### Summary Drawing the altitude of an equilateral triangle decomposes the equilateral triangle into 2 congruent triangles. They are right triangles with acute angles of 30 and 60 degrees. These congruent angles make all triangles with angles 30, 60, and 90 degrees similar by the Angle-Angle Triangle Similarity Theorem. If we consider a right triangle with angle measures of 30, 60, and 90 degrees, and with the shortest side 1 unit long, then the hypotenuse must be 2 units long since the triangle can be thought of as half of an equilateral triangle. Call the length of the altitude $$a$$. By the Pythagorean Theorem, we can say $$a^2+1^2=2^2$$ so $$a=\sqrt3$$. Now, consider another right triangle with angle measures of 30, 60, and 90 degrees, and with the shortest side $$y$$ units long. By the Angle-Angle Triangle Similarity Theorem, it must be similar to the right triangle with angles 30, 60, and 90 degrees and with sides 1, $$\sqrt3$$, and 2 units long. The scale factor is $$y$$, so a triangle with angles 30, 60, and 90 degrees has side lengths $$y, y\sqrt3,$$ and $$2y$$ units long. In triangle $$ABC, 2y=5$$ so $$y=\frac 52$$. That means $$DB$$ is $$\frac 52$$ units and $$DC$$ is $$\frac 52 \sqrt 3$$ units. In triangle $$EGH, y\sqrt 3=4$$ so $$y=\frac{4}{\sqrt 3}$$. That means $$FG$$ is $$\frac{4}{\sqrt 3}$$ units and $$EG$$ is $$2 \frac{4}{\sqrt 3}$$ or $$\frac{8}{\sqrt 3}$$ units.
# SOLVING WORD PROBLEMS WITH CROSS MULTIPLICATION METHOD Problem 1 : Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the car travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars? Solution : Let “x km/hr” be the speed of 1st car Let “y km/hr” be the speed of the 2nd car Time = Distance/Speed Speed of both cars while they are traveling in the same direction = (x – y) Speed of both cars while they are traveling in the opposite direction = (x + y) 5 = 100/(x -y) x – y = 100/5 x - y = 20 x - y - 20 = 0 ---(1) 1 = 100/(x + y) x + y = 100 x + y - 100 = 0------(2) x/(100 + 20) = y/(-20 + 100) = 1/(1 + 1) x/120 = y/80 = 1/2 x/120 = 1/2 y/80 = 1/2 x = 120/2 y = 80/2 x = 60 y = 40 So, the speed of first car = 60 km/hr Speed of second car = 40 km/hr Problem 2 : The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle. Solution : Area of any rectangle = Length x breadth Let “x” be the length of rectangle Let “y” be the breadth of rectangle (x – 5) (y + 3) = xy – 9 x y + 3x - 5y – 15 = xy – 9 xy – xy + 3x – 5y – 15 + 9 = 0 3 x – 5 y – 6 = 0 -------(1) (x + 3) (y + 2) = xy + 67 xy + 2x + 3y + 6 – xy – 67 = 0 2x + 3 y – 61 = 0 -------(2) x/(305+18) = y/(-12 + 183) = 1/(9 + 10) x/323 = y/171 = 1/19 x/323 = 1/19 y/171 = 1/19 x = 323/19 y = 171/19 x = 17 y = 9 So, the length of rectangle = 17 units Breadth of rectangle = 9 units Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. You can also visit the following web pages on different stuff in math. WORD PROBLEMS Word problems on simple equations Word problems on linear equations Algebra word problems Word problems on trains Area and perimeter word problems Word problems on direct variation and inverse variation Word problems on unit price Word problems on unit rate Word problems on comparing rates Converting customary units word problems Converting metric units word problems Word problems on simple interest Word problems on compound interest Word problems on types of angles Complementary and supplementary angles word problems Double facts word problems Trigonometry word problems Percentage word problems Profit and loss word problems Markup and markdown word problems Decimal word problems Word problems on fractions Word problems on mixed fractrions One step equation word problems Linear inequalities word problems Ratio and proportion word problems Time and work word problems Word problems on sets and venn diagrams Word problems on ages Pythagorean theorem word problems Percent of a number word problems Word problems on constant speed Word problems on average speed Word problems on sum of the angles of a triangle is 180 degree OTHER TOPICS Profit and loss shortcuts Percentage shortcuts Times table shortcuts Time, speed and distance shortcuts Ratio and proportion shortcuts Domain and range of rational functions Domain and range of rational functions with holes Graphing rational functions Graphing rational functions with holes Converting repeating decimals in to fractions Decimal representation of rational numbers Finding square root using long division L.C.M method to solve time and work problems Translating the word problems in to algebraic expressions Remainder when 2 power 256 is divided by 17 Remainder when 17 power 23 is divided by 16 Sum of all three digit numbers divisible by 6 Sum of all three digit numbers divisible by 7 Sum of all three digit numbers divisible by 8 Sum of all three digit numbers formed using 1, 3, 4 Sum of all three four digit numbers formed with non zero digits Sum of all three four digit numbers formed using 0, 1, 2, 3 Sum of all three four digit numbers formed using 1, 2, 5, 6
# How do you write the equation of a line in slope intercept form that is perpendicular to the line y = –4x and passes through the point (2, 6)? Mar 7, 2018 $y = \frac{1}{4} x + \frac{11}{2}$ #### Explanation: $\text{Given a line with slope m then the slope of a line}$ $\text{perpendicular to it is }$ •color(white)(x)m_(color(red)"perpendicular")=-1/m $\text{the equation of a line in "color(blue)"slope-intercept form}$ is •color(white)(x)y=mx+b $\text{where m is the slope and b the y-intercept}$ $y = - 4 x \text{ is in this form}$ $\text{with slope } m = - 4$ $\Rightarrow {m}_{\textcolor{red}{\text{perpendicular}}} = - \frac{1}{- 4} = \frac{1}{4}$ $\Rightarrow y = \frac{1}{4} x + b \leftarrow \textcolor{b l u e}{\text{is the partial equation}}$ $\text{to find b substitute "(2,6)" into the partial equation}$ $6 = \frac{1}{2} + b \Rightarrow b = \frac{12}{2} - \frac{1}{2} = \frac{11}{2}$ $\Rightarrow y = \frac{1}{4} x + \frac{11}{2} \leftarrow \textcolor{red}{\text{in slope-intercept form}}$
### Slope and Y-intercept in Real World Examples ```A lesson on interpreting slope and y-intercept in real world examples Standard: MAFS.912.S-ID.3.7: Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. Problem of the Day: Solve for the slope between (-1,-5) and (6,9). m= y2-y1 x2-x1 m= 9-(-1) 6-(-1) m= 14 7 m=2 Slope intercept form- y=mx+b, where m is slope and b is the yintercept Slope- Change in y over change in x (rate of change) Y-intercept- the value of y when x is zero Example of Slope in a Real World Scenario The graph to the right shows the growth of a tree at a constant rate, over a period of four years. Interpret the slope of the line. m= Change in height Change in time Example of Slope in a Real World Scenario m= change in distance change in time . Example of Y-Intercept in a Real World Scenario For example: The yintercept in this graph is 1080, meaning it is the amount the person owes before he/she began making payments. (zero payments have been The graph then shows that over the next 24 months this debt will be paid off. Example of Y-Intercept in a Real World Scenario You have 300 items of clothing and decide to start donating to amount of clothing you have before you start donating to Goodwill every month. Solving a Real World Example  A student is eating an ice cream cone at the park that is 12.7cm tall. It is extremely hot outside and the ice cream starts to melt at a constant rate of 2cm/minute. If the student didn’t eat any of the ice cream and it started to melt, how much would be left after 3 minutes?  1st: Identify the slope and y- intercept  2nd: Plug into slope intercept form  Y=-2x+12.7 (slope is negative because it is decreasing in size)  3rd: Plug in 3 for x since we     want to know how tall it will be after 3 minutes 4th: Solve y=-2(3)+12.7 y=-6+12.7 y=6.7 Understand that after 3 minutes of melting the ice cream cone will now measure 6.7cm. Leaky Lines Project  Items you should have:  400ml of water cylinder  Empty water bottle  Stopwatch Leaky Lines Project  Get into groups of two  One person will hold the water bottle and be in charge of the stopwatch  Measure 400ml into bottle  Turn water bottle over and start timer  Every 10 seconds record how much water has accumulated in the cylinder Leaky Lines Project  Create a graph based on the data gathered graphing the time intervals on the x-axis and the amount of milliliters on the y-axis.  Solve for the rate of change between two coordinates.  Write the equation of the line.  Discussion:  Is the slope positive or negative?  What is the y-intercept? Independent Practice measuring the rate at which the water left the bottle, would the slope have been positive or negative?  What would the yintercept have been?  Write an equation expressing this linear relationship using m for slope. Review for Quiz Your family is taking a trip to Disney and is driving at a constant rate. After one hour, you have traveled 60 miles, and after 2 hours you have traveled 120 miles. How fast is the car going? 2. You are selling candy bars for a fundraiser. You have raised \$50 so far and sell each candy bar for 75 cents. How much money will you have made after selling 30 candy bars? 1. Review for Quiz Continued  The graph shows the amount of money you have at the beginning of the month. a. How much money did you begin with? b. How much money do you earn each week? c. How much money will you have after 3 weeks? ```
Future Study Point # Solutions of 10 class maths questions frequently asked in CBSE board exams ## Solutions of 10 class maths questions frequently asked in CBSE board exams Here the solutions are given of the questions of maths that are asked frequently in CBSE X board exams .These questions have been examined by a team of experts by the study of last year’s maths question papers of CBSE board exam, the expert team of future study point found that these questions are asked frequently in the last year’s 10 class CBSE board exams in maths question papers. The solutions to these questions will give an idea to the students about the type of questions asked in the exam and the way of solving them in the exam. The frequently asked question in the exam will help the students in gaining the idea that how often the maths questions are repeated and why is it important to study last year’s question papers . The study of this set of questions and answers will provide you a confidence that is necessary for achieving excellent marks in the exam. ## Solutions of 10 class maths questions frequently asked in CBSE board exams Click for online shopping Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc ## Section-A (1 marks) Q1.If the quadratic equation has two equal roots, then find the value of p. Solution. When two roots are equal in a quadratic equation we have The given equation is b = –2√5, c = 15, a = P 20 – 60p = 0 NCERT Solutions Class 10 Science from chapter 1 to 16 Q2. In fig, a tower AB is 20 m high and BC, its shadow on the ground, is 20√3 m long . Find the Sun’s altitude. Solution. Let altitude of the Sun is at θ From the fig. we have tanθ = tan30° θ = 30° Therefore the altitude of the sun is at 30°. Q3. If the area of the circle is equal to the sum of areas of the two circles of diameter 10 cm and 24 cm, then the diameter of the larger circle (in cm) is : (A) 34 (B) 26 (C) 17 (D) 14 Solution. Area of the circle is = πr² radii of two circle = 10/2 = 5 cm and 24/2 = 12 cm Let the radius of the larger circle is = R The sum of the areas of both circle = Area of the larger circle π × 5² + π × 12 ² = πR² R² = 25 + 144 R² = 169 R = 13 Hence the diameter of the larger circle = 2R = 2× 13 = 26 cm Q4. Two dice are thrown together . The probability of getting the same number on both dice is : Solution. Total outcome when die is thrown alone = 6 When two dice are thrown together then total possible outcome = 6² = 36 The favorable outcome ( the number is same on both dice) → (1,1), (2,2), (3,3), (4,4),(5,5), (6,6) which are 6 in number Q5. In fig. the sides AB, BC, and CA of a triangle ABC touches a circle at P, Q and R respectively. If PA = 4 cm ,BP = 3 cm and AC =11 cm,then the length of BC(in cm) is: Solution. AP = AR ( tangents drown from the same external points) Q6. If co-ordinates of the one end of the diameter of a circle are (2,3) and the coordinates of its centers are (–2,5) , then co-ordinates of other end of the diameter are : (A) (–6,7) (B) (6,–7) (C) (6,7) (D) (–6,–7) Solution. Let co-ordinates of another end of the diameter are = (x,y) As the center of the circle is the midpoint of the diameter so applying the section formula The coordinates of the midpoint of the diameter are = (–2,5) The coordinates of the endpoint of the diameter are (2,3) and (x,y) x = –6, y = 7 Therefore co-ordinates of another end are = A(–6,7) Q7. The sum of the first 20 odd natural numbers is: (A) 100 (B) 210 (C) 400 (D) 420 Solution. The series formed by first odd natural number is 1,3, 5,7,9………upto 20 th term d = 2 , n = 20 Applying the formula used for sum of the AP = 10 [2 +38] = 400 Hence the sum of the first 20 odd natural number is = C(400) Q8. If 1 is a root of the equations ay² + ay + 3 = 0 and y² + y + b = 0 then ab equals : (A) 3 (B) (C) 6 (D) –3 Solution. 1 is the root of the equation ay² + ay + 3 = 0 Therefore placing the value y = 1 a +a + 3 = 0 2a = –3 ⇒ a = –3/2 1 is also the root of the equation y²+ y + b = 0, putting y = 1 1² + 1 +b = 0 ⇒ b = –2 ab = –3/2×–2 = –3 The rquired answer is (D) –3 ## Section B(2 marks) Q9. If a point A (0,2) is equidistant from the points B(3,p) and C(p,5) then find the value of p. Solution. The distance between A(0,2) and B(3,p) The distance between A(0,2) and C(p,5) AB = AC 9 + 4 +P² –4P = P² + (–3)² –4P = 9 –13 –4P = –4 P = 1 Therefore value of p = 1 Q10. The volume of a hemisphere is cm³ . Find its curved surface area. Solution. The volume of hemisphere = Volume given cm³ The curved surface area of hemisphere = 2πr² Hence the curved surface area of the hemisphere is 693 cm² Q11. The tangents PA and PB are drawn from an external point P to two concentric circles with center O and radii 8 cm and 5cm respectively, as shown in fig.if AP = 15 cm , then find the length of BP. Solution. Drawing the line by joining O to P In ΔAOP, AP = 15 cm, OA = 8 cm, OB = 5 cm ∠OAP = 90° Applying the Pythagoras theorem OP² = OA² + AP² OP² = 15² + 8² OP² = 225 + 64 OP = √(289) OP = 17 In ΔOPB, ∠OPB = 90° Applying the pythogorus theorem BP² = OP² – OB² = 17² – 5² BP² = 289 – 25 = 264 BP = 2√66 Therefore the length of the tangent BP is 2√66 cm Q12.How many four digits numbers are divisible by 7. Solutions. The smallest four-digit number is = 1000 Dividing it by 7 we get reminder 6 The first 4 digit number divisible by 7 = 1000 + (7–6) = 1001 The largest four-digit number = 9999 Dividing it by 7 we get reminder 3 Subtracting it from 9999 we get 9996 the largest four-digit number divisible by 7 Therefore series of the 4 digit number divisible by 7 1001, 1008,1015, 1022……….9996 Applying the formula for n th term of an AP 9996 = 1001 + (n – 1)×7 7n – 7 = 9996 –1001 7n –7= 8995 7n = 9002 Therefore the number of 4 digit numbers divisible by 7 are 1286. Q13. Solve the following quadratic equation for x: 4√3 x² + 5x – 2√3 = 0. Solution. 4√3 x² + 5x – 2√3 = 0 The product of 4√3 × 2√3 = 8 × 3 = 24 Splitting up 5 into 8 – 3 4√3 x² + (8–3) x – 2√3 = 0 4√3 x² + 8x – 3x – 2√3 4x (√3x + 2) – √3( √3x + 2) (√3x + 2)(4x –√3) Q15. A card is drawn at random from a well-suffled pack of 52 playing cards. Find the probability that the drawn card is neither a king nor a queen. Solution. Total number of possible outcome = 52 Number of queens and kings = 4 + 4 = 8 Number of the cards in which neither a king nor a queen = 52 – 8 = 44 So, the required probabilty is . Section C(3 marks) Q16. A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder of the same diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. Find the total surface area of the vessel. Solution. The radius of cylinder (r) = 14/2 = 7 cm The height of the cylinder is (h) = 13 – 7 = 6 cm The total surface area of the vessel = inner surface area + outer surface area ∵ inner surface area = outer surface area ( here thickness of the vessel is not given so inner and outer radii are supposed the same) The total surface area of the vessel = 2(curved surface area of cylinder + curved surface area of the hemisphere) (∵ the vessel is hollow) The surface area of the vessel = 2(2πrh + 2πr²) = 4πr(h + r) The surface area of the vessel The surface area of vessel = 88 × 13 = 1154 The surface area of the vessel is = 1154 sq.cm Q17.Find the ratio in which the y-axis divides the line segments joining the points (–4,–6) and (10,12) also, find the coordinates of the point of division. Solution. The abscissa of the co-ordinates of the point of intersection of the y-axis and the line segment is 0, so let the co-ordinates in which the y-axis divides the line segment are (0,y) Applying the section formula Applying the formula for the value of x because abscissa of intersecting coordinates is known i.e o and substituting the following values. Substituting the in the formula for the value of y The required co-ordinates are and the ratio in which these coordinates divides the line segments is Q19.For what values of k, the roots of the quadratic equation (k+4)x² + (k+1)x +1 = 0 are equal. Solution. (k+4)x² + (k+1)x +1 = 0 When roots of the quadratic equation ax² + bx + c = 0 are equal then we have b² – 4ac = 0 Substituting the value of a,b and c from the given quadratic equation (k+4)x² + (k+1)x +1 = 0 (k+1)² – 4(k+4) = 0 k² + 1 + 2k – 4k –16 = 0 k² – 2k – 15 = 0 k² –5k + 3k – 15 = 0 k(k –5) + 3(k –5) = 0 (k –5)(k +3) = 0 k = 5, = –3 Q18. In fig. a circle is inscribed in triangle ABC touches its sides AB,BC and AC at points D,E and F respectively . If AB = 12 cm, BC = 8 cm and AC =10 cm,then find the length of AD, BE and CF. Solution.AB = 12 cm, BC = 8 cm, AC = 10 cm Applying the theorem that the tangents drawn from an external point to the circle are equal BD = 12 – x, BD = BE = 12 – x , CE = BC – BE = 8 – (12 – x) = 8 – 12 + x = x – 4 CE = CF = x – 4, AF = AC – CF = 10 – (x – 4) = 10 – x + 4 = 14 – x 14 – x = x 2x = 14⇒ x = 7 AD =x = 7 cm, BE = 12 – x = 12 – 7 = 5 cm. CF = x – 4 = 7 – 4 = 3 cm Therefore length of AD, BE and CF are 7 cm, 5 cm and 3 cm respectively. ## Section C (3 marks) Q19. In a circle of radius 21 cm, an arc subtends an angle of 60° at the center. Find (i) the length of the arc (ii) area of the sector formed by the arc. Solution. Length of an arc (l) is given by the following formula r = 21 cm, θ = 60° l= 22 The length of arc = 22 cm Area of the sector formed by the arc = = 231 Therefore the length of arc is 22 cm and the area of the sector is 231 square cm. Q20. The first and last terms of an AP are 5 and 45 respectively . If the sum of all its terms is 400 , find its common difference. Solution. Applying the sum of n term formula of an AP where l (last term) = a + (n – 1)d In the question, we are given a = 5, =45 and 25n = 400 n = 16 For calculating the value of common difference(d), applying the n th term formula of an AP. 45 = 5 + (16 – 1)d 15d = 40 Thus the common difference of given AP is . Q21. Prove that the line segment joining the points of contact of two parallel tangents of a circle passes through its center. Solution. Given- A circle with center O in which two parallel tangents exist XY∥PQ To prove – AB passes through O Proof – ∵XA ⊥ OA ( the angle between tangent and radius =90°) ∴ ∠XAO = 90° ∠AOC = 180°– 90° = 90°(OC∥ XA)……….(i)(sum of co-interior angles) ∵ OB⊥ PB ( the angle between tangent and radius =90°) ∴ ∠PBO = 90° ∠BOC = 180°– 90° = 90°(OC∥ PB)………(ii)(sum of co-interior angles) From (i) and (ii) we have ∠AOC + ∠BOC = 90° + 90° =180° ∴ AOB is a straight line which passes through O. Q22. Rahim tosses two coins simultaneously. Find the probability of getting. (i) At least 1 tail Solution. The total possible outcomes after tossing two coins =(H,H),(T,T),(H,T),(T,H) (i) The outcomes of at least 1 tail =(T,H),(H,T),(T,T) Hint: at least means 1 or more than 1 tail (ii) The outcomes of almost two heads = (H,H),(T,T),(T,H),(H,T) (iii) The outcomes of exactly two head =(H,H) Q23. Solve the following equation. Solution. (4–3x)(2x + 3) = 5x 8x + 12 –6x ²– 9x = 5x –6x² –6x + 12 = 0 6x² + 6x – 12 = 0 x² + x – 2 = 0 x² + 2x – x – 2 = 0 x(x + 2) –1(x +2) =0 (x + 2)(x –1) = 0 x = –2, 1 Therefore the solution of the given equation is –2 and 1 Q24. If the coordinates of the points A and B are (–2,–2) and (2,–4) respectively, find the co-ordinates of P such that where P lies on the line segment AB. Solution. We are given that The P point lies in the line segment AB ,so getting its posion we have to determine AP : BP ∵ AP +BP = AB⇒AB–AP = BP P divides the line segment joining the points A(–2,–2) and B(2,–4) in ratio of 3: 4, let co-ordinates of P are(x,y) Hence co-ordinates of P are Click for online shopping Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc ## Section D (4 marks) Q25. The probability of selecting a red ball at random from a jar that contains only red, blue and the orange ball is . The probability of selecting a blue ball at random from the same jar . If the jar contains 10 orange balls, find the total number of the balls in the jar. Solution. Let the number of blue balls in the jar = x The number of orange balls in the jar = 10 The number of red balls = y Total number of balls in the jar = x +y+ 10 According to question 3x = x +y +10 2x –y = 10………….(i) according to question 4y = x +y +10 3y –x = 10…………(ii) Solving equation number (ii) for y Substituting the value of y in equation number (i) 6x – 10 –x = 30 5x = 40 x = 8 Putting the value x in eq.(i),we get y= 6. Therefore the number of blue balls are 8 and the number of red balls are 6 and total balls in the jar =x+y+10=8+6+10=24 Q26.A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of each bottle, if 10% liquid is wasted in this transfer. Solution. Let the height of cylindrical bottles is h cm The volume of each bottle =πr²h r = 6 /2 = 3cm The volume of each bottle =π ×3²h =9πh The volume of 72 bottles = 72×9πh=648πh Volume of spherical bowl 10% of the volume of a spherical bowl = 388.8π cubic cm The water-filled in the bottles = Volume of bowl – Water wasted = 3888π – 388.8π =3499.2π cubic cm The volume of 72 bottles =648πh = 3499.2π h = 5.4 Hence height of bottle = 5.4 cm Q27.Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60°. Also, find the area of the corresponding major segment(use). Solution. Area of minor segment = Area of minor sector – Area of ΔAOB In ΔAOB , ∠AOB = 60° Since OA = OB (radii of the same circle) ∠BAO = ∠ABO=x (OA = OB) x + x + 60 = 180 x =60°, therefore ΔAOB is an equilateral triangle OA = 14 cm The area of major segment is Q28.If the point P(x,y) is equidistant from the points A(a+b,b–a) and B(a –b,a +b), prove that bx = ay. Solution.As the point P(x,y) is equidistant from the points A(a+b,b–a) and B(a –b,a +b) ∴PA = PB [x –(a +b)]² + [y –(b–a)]² =[x –(a –b)]² + [y –(a +b)]² x² + (a +b)² –2x(a +b) + y² + (b–a)²– 2y(b– a) =x² + (a –b)² –2x(a– b) + y² + (a +b)²– 2y(a +b) –2x(a +b) + (b–a)²–2y(b–a) = (a –b)²–2x(a– b)– 2y(a +b) –2xa –2xb + b² + a² –2ab –2yb + 2ya = a² + b²–2ab –2xa +2xb –2ya –2yb –2xb–2xb = –2ya–2ya –4xb = –4ya bx = ay, Hence proved. Q29. A motorboat whose speed is 24 km/hr in still water takes 1 hr more to go 32 km upstream than to return downstream to the same spot. Find the speed of the stream. Solution. Let the speed of the stream = x km/hr Distance covered by the boat in upstream or downstream =32 km The speed of the river in still water = 24 km/hr The speed of the boat in upstream = (x – 24) km/h The time is taken by the boat to go up to 32 km in upstream = The time is taken by the boat to go down to 32 km in downstream = According to question 64x = 576 –x² x² + 64x – 576=0 x² + 72x – 8x – 576 =0 x(x + 72) –8(x +72) =0 (x + 72)(x –8) =0 x = –72, 8 Neglecting the negative sign, therefore the speed of the stream =8 km/hr Q30.In fig.the vertices of ΔABC are A(4,6), B(1,5) and C(7,2).A line segment DE is drawn to intersect the sides AB and AC at D and E respectively such that .Calculate the area of ΔADE and compare it with the area of ΔABC. Solution. We are given that The line DE is intersecting the sides AB and AC of the Δ ABC in the same ratio of 1 : 3 ∴ DE ∦BC (opposite of BPT theorem) ∠AED = ∠ACB (corresponding angle) ∠A = ∠A (common) According to the theorem that the ratio of areas of two similar triangles is in the ratio of the square of their corresponding sides. …………….(i) The coordinates of the vertices of ΔABC are A(4,6), B(1,5) and C(7,2). ## NCERT Solutions of Science and Maths for Class 9,10,11 and 12 ### NCERT Solutions for class 10 maths CBSE Class 10-Question paper of maths 2021 with solutions CBSE Class 10-Half yearly question paper of maths 2020 with solutions CBSE Class 10 -Question paper of maths 2020 with solutions CBSE Class 10-Question paper of maths 2019 with solutions ### NCERT Solutions for class 11 maths Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability CBSE Class 11-Question paper of maths 2015 CBSE Class 11 – Second unit test of maths 2021 with solutions ### NCERT Solutions for Class 11 Physics Chapter 1- Physical World chapter 3-Motion in a Straight Line ### NCERT Solutions for Class 11 Chemistry Chapter 1-Some basic concepts of chemistry Chapter 2- Structure of Atom ### NCERT Solutions for Class 11 Biology Chapter 1 -Living World ### NCERT solutions for class 12 maths Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2 Class 12 Maths Important Questions-Application of Integrals Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22 Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22 Solutions of class 12 maths question paper 2021 preboard exam CBSE Solution Scroll to Top Optimized by Seraphinite Accelerator Turns on site high speed to be attractive for people and search engines.
## Ms. Dawn ### Target 1 ###### Lesson Type: New Number Operation : Integer Composition Understand that the digits in a multi-digit number represent the amounts in each place-value. ###### 1: Understand the decimal point is written as “and”. ###### 2: Understand that a decimal point in a number separates the whole number from the fractional part. ###### 3: Identify place-value for a decimal number. 4th ###### Vocabulary: Decimal Point, "And," Place Value, Tenths, Hundredths, Thousandths, Multi-Digit, Whole Number, Fractional Part Activities: 1. Students played a game of Bingo. Students chose different decimal numbers and placed it wherever they wanted to on their Bingo card. Teacher called out the numbers and students need to show they knew the number by covering it with a colored counter (ex. 2.34 was said as 2 and 34 hundredths). 2. Students rolled 6 dice to create a decimal number. After creating their number, they needed to read it to me using the correct vocabulary when reading decimals. ## Absent Students: ### Target 2 : ###### 1: Place decimals on a number line. ###### 2: Understand that decimals on the number line are smaller farther to the left and greater farther to the right. 4th ###### Vocabulary: Activities: 1. Students were able to label a variety of number lines. It started out where the teacher called out a variety of decimal numbers. Students plotted where it was on the number line. 2. Students were given number lines that skip counted by 0.1's, 0.2's, and 0.25's. They labeled it and then were able to place the given decimal numbers on the number line. ### Target 3 : ###### 1: Compare decimals (up to the hundredths place). ###### 2: When comparing decimals, understand to look at the greatest place-value first. 4th ###### Vocabulary: Decimal, Compare, Least, Greatest, Place Value, Tenths, Hundredths, Thousandths, Ones, Tens, Hundreds Activities: 1. Students were given three decimal cards and needed to place it in order from least to greatest or greatest to least. 2. Students played a game by rolling a die with a variety of decimals on it. They needed to determine who had the bigger decimal number. First player to get 10 tally marks was declared the winner.
## Want to keep learning? This content is taken from the National STEM Learning Centre's online course, Maths Subject Knowledge: Understanding Numbers. Join the course to learn more. 2.6 ## National STEM Learning Centre Skip to 0 minutes and 7 secondsPAULA KELLY: In the last two steps, we looked at multiples. Is it possible for a number to be a multiple of more than one number? If you look back to the times table grid, the answer is clearly yes. 8, for example, appears four times, in the 1 times table, the 8 times table, the 2, and also the 4 times table. This tells us 8 is a common multiple of 1, 2, 4, and 8. Which times tables does the number 12 appear in? The number 12 appears in the following times tables, 1, 2, 3, 4, 6, and 12. So 12 is a multiple of 1, 2, 3, 4, 6, and 12. Skip to 0 minutes and 52 secondsSo the number 12 can be made by combining these numbers, so 1 multiplied by 12, 2 multiplied by 6, and 3 multiplied by 4. We call the numbers 1, 2, 3, 4, 6, and 12 the factors of 12. So 12 is a multiple of 1, 2, 3, 4, 6, and 12. So 1, 2, 3, 4, 6, and 12 are the factors of 12. We can see in this diagram how the factors of 12 pair up. So all numbers seem to have an even number of factors, as you could always pair them up. This is a very sensible suggestion and true for most numbers. However, let's have a look at the number 16. Skip to 1 minute and 40 seconds16 appears in the following times tables, 1, 2, 4, 8, and 16. This means that 16 is a multiple of 1, 2, 4, 8, and 16, but also means that 1, 2, 4, 8, and 16 are factors of 16. Let's try to pair them up. 1 and 16 pair up, as 1 multiplied by 16 gives us 16. 2 and 8 pair up, as 2 multiplied by 8 gives us 16. This leaves 4 without a number to pair it with. This is because 4 multiplied by 4 gives us 16. We didn't include the number 4 twice, so 16 has an odd number of factors. Skip to 2 minutes and 32 secondsFind all the factors of 36, place them in ascending order, pair them up, and see which factor is the odd one. All square numbers have an odd number of factors. Only square numbers have an odd number of factors. If a number has an odd number of factors, it must be a square number. # Multiples and factors It is easy to get the definitions of ‘multiples’ and ‘factors’ mixed up. Think about the numbers 4 and 12. How would you complete the following sentences using ‘multiple’ and ‘factor’? 12 is a ______ of 4 because 12 is in the 4 times table. 4 is a ______ of 12 because 4 divides exactly into 12 without leaving a remainder. In this video Paula explores the link between factors and multiples and shows that in most cases the factors of a number can be paired up. However, something unusual happens when we attempt to pair up the factors of a square number. ## Problem worksheet Now complete question 4 from this week’s worksheet. ## Teaching resource This collection contains resources covering topics including factors and multiples. Each topic is introduced in the form of a puzzle, many of which have several different solutions.
# Factors of 345 The factors of 345 and the prime factors of 345 differ because three hundred and forty-five is a composite number. Also, despite being closely related, the prime factors of 345 and the prime factorization of 345 are not exactly the same either. In any case, by reading on you can learn the answer to the question what are the factors of 345? and everything else you want to know about the topic. ## What are the Factors of 345? They are: 345, 115, 69, 23, 15, 5, 3, 1. These are all the factors of 345, and every entry in the list can divide 345 without rest (modulo 0). That’s why the terms factors and divisors of 345 can be used interchangeably. As is the case for any natural number greater than zero, the number itself, here 345, as well as 1 are factors and divisors of 345. ## Prime Factors of 345 The prime factors of 345 are the prime numbers which divide 345 exactly, without remainder as defined by the Euclidean division. In other words, a prime factor of 345 divides the number 345 without any rest, modulo 0. For 345, the prime factors are: 3, 5, 23. By definition, 1 is not a prime number. Besides 1, what sets the factors and the prime factors of the number 345 apart is the word “prime”. The former list contains both, composite and prime numbers, whereas the latter includes only prime numbers. ## Prime Factorization of 345 The prime factorization of 345 is 3 x 5 x 23. This is a unique list of the prime factors, along with their multiplicities. Note that the prime factorization of 345 does not include the number 1, yet it does include every instance of a certain prime factor. 345 is a composite number. In contrast to prime numbers which only have one factorization, composite numbers like 345 have at least two factorizations. To illustrate what that means select the rightmost and leftmost integer in 345, 115, 69, 23, 15, 5, 3, 1 and multiply these integers to obtain 345. This is the first factorization. Next choose the second rightmost and the second leftmost entry to obtain the 2nd factorization which also produces 345. The prime factorization or integer factorization of 345 means determining the set of prime numbers which, when multiplied together, produce the original number 345. This is also known as prime decomposition of 345. Besides factors for 345, other searched terms on our website include: We did not place any calculator here as there are already a plethora of them on the web. But you can find the factors, prime factors and the factorizations of many numbers including 345 by using the search form in the sidebar. To sum up: The factors, the prime factors and the prime factorization of 345 mean different things, and in strict terms cannot be used interchangeably despite being closely related. The factors of three hundred and forty-five are: 345, 115, 69, 23, 15, 5, 3, 1. The prime factors of three hundred and forty-five are 3, 5, 23. And the prime factorization of three hundred and forty-five is 3 x 5 x 23. Remember that 1 is not a prime factor of 345. No matter if you had been searching for prime factorization for 345 or prime numbers of 345, you have come to the right page. Also, if you typed what is the prime factorization of 345 in the search engine then you are right here, of course. Taking all of the above into account, tasks including write 345 as a product of prime factors or list the factors of 345 will no longer pose a challenge to you. If you have any questions about the factors of three hundred and forty-five then fill in the form below and we will respond as soon as possible. If our content concerning all factors of 345 has been of help to you then share it by means of pressing the social buttons. And don’t forget to bookmark us. Thanks for your visit. 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# For what x an y is y / ( x + 3 )^2 > 2/(x-y-3)? Jul 3, 2016 There is no pair $\left\{x , y\right\}$ satisfying the condition. #### Explanation: The pairs $\left\{x , y\right\}$ satisfying $\frac{y}{x + 3} ^ 2 > \frac{2}{x - y - 3}$ also satisfy $y \left(x - y - 3\right) > 2 {\left(x + 3\right)}^{2}$ and also satisfy $y \left(x - y - 3\right) - 2 {\left(x + 3\right)}^{2} > 0$ Calling now $f \left(x , y\right) = y \left(x - y - 3\right) - 2 {\left(x + 3\right)}^{2}$ let us find the local minima/maxima: Solving the conditions $\nabla f \left(x , y\right) = \vec{0}$ or { (12 - 4 x + y = 0), (-3 + x - 2 y = 0) :} with solution $\left\{x = 3 , y = 0\right\}$ for $f \left(x , y\right)$ we have ${\nabla}^{2} f \left(x , y\right) = \left(\begin{matrix}- 4 & 1 \\ 1 & - 2\end{matrix}\right)$ with eigenvalues $\left\{- 3 - \sqrt{2} , - 3 + \sqrt{2}\right\}$ so $f \left(x , y\right)$ has a maximum which is a global maximum, at $\left\{x = 3 , y = 0\right\}$. Evaluating $f \left(3 , 0\right) = 0$ we see that there is no pair $\left\{x , y\right\}$ which obeys strictly the condition.
Get Dream IIT in Drop Year \| Up to 70% OFF \| Limited Seats # NCERT Solutions For Class 9 Maths Chapter 6 Exercise 6.1 Lines and Angles - PDF Download Class 9 NCERT solutions for class 9 maths chapter 6 ex 6.1 Lines and Angles provides an overview of the fundamental concepts and definitions associated with lines and angles, intersecting lines and non - intersecting lines, pairs of angles, theorems and their properties. By practicing questions of ex 6.1, students will be able to gain a comprehensive understanding of these fundamental concepts. Class 9 maths chapter 6 exercise 6.1 ncert solutions will provide the students with the necessary instruction to differentiate between line segments, ray, parallel line, perpendicular line, angles, right angle, and associated concepts and theorem. NCERT solutions class 9 maths chapter 6 ex 6.1 consists of 6 questions where you have to prove a particular condition according to the properties of the lines and angles. You can practice all the questions of ex 6.1 that are available in PDF format. You can also learn these concepts by downloading the free PDF of NCERT solutions from the eSaral website. ## Topics Covered in Exercise 6.1 class 9 Mathematics Questions Ex 6.1 class 9 maths solutions covers topics based on lines and angles, intersecting and non-intersecting lines, pair of angles etc. You will learn these concepts in detail which are provided below. 1 Basic Terms and Definitions 2 Intersecting Lines and Non - intersecting Lines 3 Pair of Angles 4 Theorem 6.1 1. Basic Terms and Definitions • Point - A point that does not contain any elements. It is denoted by a dot. • Lines - When two different points are connected, a line is formed. A line does not have any finite length and can be extended infinitely. • Line Segment - It is the part of the line which has two endpoints. • Ray - Ray is also part of the line that only has one endpoint and there's no end at the otherside. • Collinear and Non-collinear points - If two or more points are on the same line, they're called collinear points. If they're not, they're called non-collinear points. Type of Angles 1. Acute Angle - An angle measures between 0° and 90°. 2. Right Angle - An angle measures exactly equal to 90°. 3. Obtuse angle - An angle greater than 90° but less than 180°. 4. straight angle - An angle which is exactly equal to 180°. 5. Reflex angle - An angle which is greater than 180° but less than 360° 6. Complementary angles - Two angles whose sum is 90° are called complementary angles. 7. Supplementary angles - Two angles whose sum is 180° are called supplementary angles. Relation between two Angles Adjacent Angles - If two angles have a common vertex, a common arm and their non-common arms are on different sides of the common arm, it is called adjacent angle. Linear Pair of Angles - If two angles share a common vertex and a common arm, but the non-common arms form a line, then these are referred to as linear pairs of angles. Vertically Opposite Angles - If two lines intersect at a point, then the opposite angles are called vertically opposite angles. 2. Intersecting Lines and Non - intersecting Lines - Intersecting lines are the lines that intersect each other from a specific point. Non-intersecting lines are lines that don't intersect each other. They're called Parallel lines, and the common length between two lines is the distance between parallel lines. 3. Pair of Angles Axiom 1 - If a ray stands on a line, then the sum of two adjacent angles so formed is 180°. Axiom 2 - If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line. 4. Theorem 6.1 - If two lines intersect each other, then the vertically opposite angles are equal. ## Tips for Solving Exercise 6.1 class 9 chapter 6 Lines and Angles In ex 6.1 class 9 maths chapter 6 NCERT solutions, students will learn some important tips on how to solve questions related to lines and angles. 1. Practicing the questions in NCERT solutions class 9 maths chapter 6 ex 6.1 on a regular basis will help you get the basics of lines and angles. Most of the questions in this exercise revolve around the core concept of lines and angles, so it's important to clarify all the doubts and understand all the concepts before solving the questions. 2. Exploring the examples and comprehending the theorems and concepts presented in these solutions will enable students to develop a fundamental understanding of the questions included in ex 6.1. 3. NCERT solutions also provide ways to easily memorize the terms, definitions, theorems and properties associated with lines and angles. ## Importance of Solving Ex 6.1 class 9 Maths chapter 6 Lines and Angles Using NCERT solutions to solve ex.6.1 questions has a lot of benefits. 1. Using the solutions provided by the academic team of mathematics, you will be able to gain a better understanding of the different concepts of lines and angles. 2. A comprehensive analysis of the complex questions from each part of the exercise is included in these solutions. 3. Class 9 Maths exercise 6.1 provides an introduction to parallel lines and the concept of pairs of angles. 1. These solutions are error-free and thus enable students to gain a thorough understanding of ex 6.1. Question 1. What are collinear points and non-collinear points? Answer 1. If two or more points are on the same line, they're called collinear points. If they're not, they're called non-collinear points. Question 2. What are the various types of angles? Answer 2. There are various types of angles. 1. Acute angle 2. Right angle 3. Obtuse angle 4. Straight angle 5. Reflex angle Question 3. What is a linear pair of angles? Answer 3. If two angles share a common vertex and a common arm, but the non-common arms form a line, then these are referred to as linear pairs of angles.
## Measures of Relatedness Bivariate data involves a relationship between two random variables. When those variables are quantitative, each piece of data consists of an ordered pair of numbers. These values can be plotted on a graph (that is, a scatterplot), and inspected visually to see if there is a relationship. Often, data will have a linear trend, and linear regression is used to describe that trend. ### A Bivariate Example Suppose we sample five school-age boys, and determine their age in years and their height in inches. We want to determine if there is a relationship between age and height. The sample data is: (7, 46), (9, 55), (11, 53), (11, 56), (12, 60) Admittedly, this is a very small sample, but small samples allow us to more easily see the basic issues that need to be addressed. The scatterplot of the data follows. From the data, we can easily compute some basic descriptive statistics for each variable. • For the ages, we have $\bar{x} = 10$ years, $s^2 = 4$ square years, and $s = 2$ years. • For the heights, we have $\bar{y} = 54$ inches, $s^2 = 26.5$ square inches, and $s \approx 5.15$ inches. But these statistics do not tell us anything about how the $x$ (age) and $y$ (height) variables might be related. For that, we need a different type of measure. ### Covariance In some sense, we are interested in the spread of the data in two dimensions, as opposed to the one-dimensional measures we previously created. One of the measures of spread in one dimension was variance, defined (for a sample) to be $s^2 = \dfrac{ \sum\limits_{i=1}^n (x_i - \bar{x})^2}{n - 1}$. With a slight alteration, we can define a two-dimensional measure of spread called the covariance. As does the variance, it comes in two forms, one definition for a sample, and one for a population. And as with the variance, the reason is that the sample formula is an unbiased estimator of the population covariance. Population Formula Sample Formula $Cov(x,y) = \dfrac{ \sum\limits_{i=1}^N (x_i - \mu_x)(y_i - \mu_y)}{N}$ $Cov(x,y) = \dfrac{ \sum\limits_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{n - 1}$ If you compare the variance and covariance formulas, you see that quite often we are interested in the sum of the squared differences of data from the mean. These quantities are important enough to be identified with their own variable names. In general, the expression $S_{ab}$ means sum of the deviations of the product $ab$. More specifically, we have the following three quantities: • $S_{xx} = \sum\limits_{i=1}^n (x_i - \bar{x})^2$ • $S_{yy} = \sum\limits_{i=1}^n (y_i - \bar{y})^2$ • $S_{xy} = \sum\limits_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})$ With these expressions, we could rewrite the sample variance and covariance. • $Var(x) = s_x^2 = \dfrac{S_{xx}}{n-1}$ • $Var(y) = s_y^2 = \dfrac{S_{yy}}{n-1}$ • $Cov(x,y) = \dfrac{S_{xy}}{n-1}$ Here are the computations that produce the covariance of our example above. The computation for each point takes a row of the table, and the last row of the table includes the totals for a few of the columns, including the values of $S_{xx}$, $S_{yy}$, and $S_{xy}$. $x_i$ $y_i$ $x_i - \bar{x}$ $y_i - \bar{y}$ $(x_i - \bar{x})^2$ $(y_i - \bar{y})^2$ $(x_i - \bar{x})(y_i - \bar{y})$ 7 46 $-3$ $-8$ 9 64 24 9 55 $-1$ 1 1 1 $-1$ 11 53 1 $-1$ 1 1 $-1$ 11 56 1 2 1 4 2 12 60 2 6 4 36 12 $\overline{50}$ $\overline{270}$ $\overline{16}$ $\overline{106}$ $\overline{36}$ From this table, we see that $S_{xx} = 16$, $S_{yy} = 106$, and $S_{xy} = 36$. Together with the sum of the individual coordinates, we can easily determine the means, variances, standard deviations, and covariance. • The mean age is $\bar{x} = \dfrac{ \sum\limits_{i=1}^n x_i}{n} = \dfrac{50}{5} = 10$ years. • The variance in the ages is $s_x^2 = \dfrac{ S_{xx}}{n-1} = \dfrac{16}{5-1} = 4$ square years. • The standard deviation of the ages is $s_x = \sqrt{4} = 2$ years. • The mean height is $\bar{y} = \dfrac{ \sum\limits_{i=1}^n y_i}{n} = \dfrac{270}{5} = 54$ inches. • The variance in the heights is $s_y^2 = \dfrac{ S_{yy}}{n-1} = \dfrac{106}{5-1} = \dfrac{53}{2} = 26.5$ square inches. • The standard deviation of the heights is $s_y = \sqrt{26.5} = \dfrac{ \sqrt{106}}{2} \approx 5.15$ inches. • The covariance between the ages and heights is $Cov(x,y) = \dfrac{S_{xy}}{n-1} = \dfrac{36}{5-1} = 9$ inch-years. We obtained a number for covariance, and we notice that it has units that are a product of the units for age and height. So how can we interpret this value? Below we have redrawn the scatterplot, and have included green lines indicating the means of each variable, and red lines from each point to the mean of each variable. The two green lines intersect at the point $(\bar{x},\bar{y})$, the point which has both average age and average height. For each data point, the quantity $(x_i - \bar{x})(y_i - \bar{y})$ is the area of the rectangle formed by the two red lines, together with the green lines back to their intersection at $(\bar{x},\bar{y})$. So each value of $(x_i - \bar{x})(y_i - \bar{y})$ uses an "area" to measure how different a particular data point is from average. We notice that those points in the upper right and lower left regions will have a positive value for the "area" $(x_i - \bar{x})(y_i - \bar{y})$, while points in the upper left and lower right regions will have a negative value for the "area" $(x_i - \bar{x})(y_i - \bar{y})$. So if there are more points trending from lower left to upper right, the sum of the "areas" produced will be positive. And if there are more points trending from upper left to lower right, the sum of the "areas" produced will be negative. ### Correlation The strange units for covariance make it somewhat difficult to interpret, since its value will depend on the scale used. We can remove this dependence on the scale by dividing by the standard deviation of each variable. This yields a quantity called the correlation. Population Formula Sample Formula $\rho = Corr(x,y) = \dfrac{ Cov(x,y) }{\sigma_x \sigma_y}$ $r = Corr(x,y) = \dfrac{ Cov(x,y) }{ s_x s_y}$ All correlations satisfy the inequality $-1 \le r \le 1$. Correlations which are close to $-1$ or $1$ indicate a very strong linear relationship between the two variables. Correlations which are close to zero indicate essentially no linear relationship between the variables. Values falling between zero and the extremes are often stated to indicate a weak, moderate, or strong relationship. For our example, we found $r = \dfrac{9}{\sqrt{4} \sqrt{26.5}} = \dfrac{9}{\sqrt{106}} \approx 0.8741$, which would imply a rather strong positive relationship between age and height. Or in other words, as the age of a group of school boys increase, their heights also tend to increase. Two alternative formulas for correlation are $r = \dfrac{ S_{xy} }{\sqrt{S_{xx} S_{yy}}}$ and $r = \dfrac{1}{n-1} \sum \limits_{i=1}^n \left( \dfrac{x_i-\bar{x}}{s_x} \right) \left(\dfrac{y_i-\bar{y}}{s_y} \right)$, both of which can be easily verified from the relationships given above. The second formula implies that correlation is an average z-score product of the data points. ### The Line of Best Fit and the Coefficient of Determination Having determined that there is a trend, we could ask about the equation of the line that would best describe the trend. We would expect that such a line would give better estimates for our heights than simply using the mean height. In fact, in the scatterplot above, we can see that none of the data points fell on the horizontal line that represented the mean height. More explicitly, the "errors" or deviations (differences between the actual y-value and the y-value of the line) are rather large at the left and right ends of the data, as is shown in the scatterplot below. Could we draw a line which minimizes the errors? Our five points do not lie in a line, so we cannot draw a line for which all of the errors would be zero. And we need a way to deal with the five errors simultaneously. So we might consider minimizing the sum of the errors. As it turns out, it is quite possible to draw a line for which the sum of the errors is zero. Two such lines are illustrated below. For the record, the two lines in the graph have the following equations. • Blue Line: $y = 2x + 34$ • Red Line: $y = -x + 64$ We could easily verify that the sum of the errors in each case is zero. But doing so, or just by looking at the two lines and their errors, we might recognize the problem with our approach. Clearly, the red line was much less appropriate than the blue line, in spite of the fact that their sum was still zero, because its errors were each individually larger. That is because errors can be negative or positive, and the signs are what allow a sum of zero. We could use absolute values of the errors. But you should recall that we encountered this issue before, when trying to define a measure of dispersion. Absolute values will omit signs, but they do have other algebraic complications. And in fact, when we discuss the estimation of populations from samples, we will find that they do not allow us to obtain the best estimators. Therefore, we shall use the Least Squares Criterion to obtain our line. This criterion states that the sum of the squared errors (SSE) will be minimized. The line which satisfies the Least Squares Criterion will be unique, and it is called the regression line. The regression line is often called the line of best fit, which implies that the Least Squares Criterion is essentially the definition of what makes a best fit. For our example, the regression line has the equation $y=2.25 x + 31.5$, and is graphed in the next scatterplot. Visually, this line looks closer overall to the points, but the quantity that was minimized is not the sum of the errors, but rather the sum of the squared errors. Specifically, if $y_i$ are the y-values of the data points, and $\hat{y}_i$ are the values predicted by the regression line, then the sum of the squared errors (SSE) has the following formula. $SSE = \sum\limits_{i=1}^n (y_i - \hat{y}_i)^2$ Here are the numerical details for our example. Data Point Point onRegression Line Error:$y_i - \hat{y}_i$ Squared Errors (7, 46) (7, 47.25) $-1.25$ 1.5625 (9,55) (9, 51.75) $3.25$ 10.5625 (11,53) (11, 56.25) $-3.25$ 10.5625 (11,56) (11, 56.25) $-0.25$ 0.0625 (12,60) (12, 58.5) $1.5$ 2.25 The sum of the squared errors (SSE) in our example is exactly 25, and this is the least possible value for this sum. For the record, the SSE of the red line was 194, the SSE of the blue line was 26, and the SSE for the green horizontal line was 106. Now the green line represented the mean y-value, before we considered a non-horizontal trend. This "previous" SSE is usually called the total sum of the squares (abbreviated SST), and in fact is equivalent to $S_{yy}$. Using the regression line has been a substantial improvement over the mean, in fact it reduced the sum of the squared errors from 106 to 25, which is a 76.42% reduction. Or in other words, 76.42% of the variation in the heights is explained by the regression line. This quantity is called the coefficient of determination. There are actually three quantities of the form SSQ, which we can describe as the sum of the squared "quantity Q". They are: • $SSE = \sum\limits_{i=1}^n (y_i - \hat{y}_i)^2$, the Sum of the Squared Errors • $SST = \sum\limits_{i=1}^n (y_i - \bar{y})^2$, the Sum of the Squared Total (deviation) • $SSR = \sum\limits_{i=1}^n (\hat{y}_i - \bar{y})^2$, the Sum of the Squared (quantity explained by the) Regression It should be noted that there are some relationships between these six quantities. Most notably: • $SST = SSR + SSE$ • $SST = S_{yy}$ • $SSR = \dfrac{S_{xy}^2}{S_{xx}}$ Therefore, the coefficient of determination is the quantity $\dfrac{SSR}{SST} = \dfrac{S_{xy}^2}{S_{xx} S_{yy}} = r^2$. In other words, the coefficient of determination is the square of the correlation coefficient. And finally, the formulas for the coefficients of the regression line $\hat{y} = \hat{\beta}_1 x + \hat{\beta}_0$ (and the coefficient of determination $r^2$) are: \begin{align} \hat{\beta}_1 &= \dfrac{S_{xy}}{S_{xx}} \\ \hat{\beta}_0 &= \bar{y} - \hat{\beta}_1 \bar{x} \\ r^2 &= \dfrac{S_{xy}^2}{S_{xx} S_{yy}} = \dfrac{SSR}{SST} \end{align} Typically, the sample coefficients of the regression equation are denoted $\hat{\beta}_1$ and $\hat{\beta}_0$, rather than $m$ and $b$ respectively, while the population coefficients are denoted $\beta_1$ and $\beta_0$. We should also note that the formula for $\hat{\beta}_0$ implies that the point $(\bar{x}, \bar{y})$ is a point on the regression line. In other words, the regression line passes through the intersection of the two averages. For our example, the computations for these coefficients would be: • $\hat{\beta}_1 = \dfrac{S_{xy}}{S_{xx}} = \dfrac{36}{16} = 2.25$ • $\hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{x} = 54 - 2.25 (10) = 31.5$ • $r^2 = \dfrac{S_{xy}^2}{S_{xx} S_{yy}} = \dfrac{36^2}{16(106)} = \dfrac{81}{106} \approx 0.7642$ And for our regression line, we also have $SSE = 25$, $SST = 106$, and $SSR = 81$. ### Derivation of the Coefficients of the Regression Line Suppose the data points are given by $(x_i, y_i)$, the regression line has the equation $\hat{y} = \hat{\beta}_1 x + \hat{\beta}_0$, so the points on the regression line are $(x_i,\hat{y}_i)$. Then we have \begin{align} SSE &= \sum\limits_{i=1}^n (y_i - \hat{y}_i)^2 \\ &= \sum\limits_{i=1}^n (y_i - \hat{\beta}_1 x_i - \hat{\beta}_0)^2 \end{align} To minimize SSE is to find the values of $\hat{\beta}_1$ and $\hat{\beta}_0$ for which SSE is least, and we can do that through the use of derivatives and calculus. We shall first take a partial derivative with respect to $\hat{\beta}_0$ to find $\hat{\beta}_0$. \begin{align} \frac{\partial}{\partial \hat{\beta}_0} \sum\limits_{i=1}^n (y_i - \hat{\beta}_1 x_i - \hat{\beta}_0)^2 &= \sum\limits_{i=1}^n \frac{\partial}{\partial \hat{\beta}_0} (y_i - \hat{\beta}_1 x_i - \hat{\beta}_0)^2 \\ &= \sum\limits_{i=1}^n (-2)(y_i - \hat{\beta}_1 x_i - \hat{\beta}_0) \end{align} Setting this quantity equal to zero leads to the following sequence of equations. \begin{array}{c} \sum\limits_{i=1}^n (y_i - \hat{\beta}_1 x_i - \hat{\beta}_0) = 0 \\ \sum\limits_{i=1}^n y_i - \hat{\beta}_1 \sum\limits_{i=1}^n x_i - n \hat{\beta}_0 = 0 \\ \hat{\beta}_0 = \dfrac{\sum\limits_{i=1}^n y_i}{n} - \hat{\beta}_1 \dfrac{\sum\limits_{i=1}^n x_i}{n} \\ \hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{x} \end{array} Note that this argument also implies that for a regression line, the sum of the (unsquared) errors will be zero. That is, $\sum\limits_{i=1}^n (y_i - \hat{y}_i) = 0$. A similar argument occurs when taking a partial derivative with respect to $\hat{\beta}_1$. However, we will substitute the result $\hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{x}$ to simplify the problem. \begin{align} \frac{\partial}{\partial \hat{\beta}_1} \sum\limits_{i=1}^n (y_i - \hat{\beta}_1 x_i - \hat{\beta}_0)^2 &= \sum\limits_{i=1}^n \frac{\partial}{\partial \hat{\beta}_1} (y_i - \hat{\beta}_1 x_i - \bar{y} + \hat{\beta}_1 \bar{x})^2 \\ &= \sum\limits_{i=1}^n \frac{\partial}{\partial \hat{\beta}_1} ((y_i - \bar{y}) - \hat{\beta}_1 (x_i - \bar{x}))^2 \\ &= \sum\limits_{i=1}^n (-2)(x_i - \bar{x}) ((y_i - \bar{y}) - \hat{\beta}_1 (x_i - \bar{x})) \end{align} Setting this quantity equal to zero leads to the following sequence of equations. \begin{array}{c} \sum\limits_{i=1}^n (x_i - \bar{x}) ((y_i - \bar{y}) - \hat{\beta}_1 (x_i - \bar{x})) = 0 \\ \sum\limits_{i=1}^n (x_i - \bar{x}) (y_i - \bar{y}) - \hat{\beta}_1 (x_i - \bar{x})^2 = 0 \\ \sum\limits_{i=1}^n (x_i - \bar{x}) (y_i - \bar{y}) = \sum\limits_{i=1}^n \hat{\beta}_1 (x_i - \bar{x})^2 \\ \hat{\beta}_1 = \dfrac{ \sum\limits_{i=1}^n (x_i - \bar{x}) (y_i - \bar{y}) }{ \sum\limits_{i=1}^n (x_i - \bar{x})^2 } \\ \hat{\beta}_1 = \dfrac{S_{xy}}{S_{xx}} \end{array} Before proceeding, we should also note that had we taken the derivative of the expression $\sum\limits_{i=1}^n (y_i - \hat{\beta}_1 x_i - \hat{\beta}_0)^2$ directly, then our work above would have found that $\sum\limits_{i=1}^n x_i (y_i - \hat{y}_i) = 0$, a result that we use below. ### Proof of the Connection between Correlation and the Coefficient of Determination The equation $SST = SSR + SSE$ is an immediate result of the regression line having a zero sum of the errors. \begin{align} SST &= \sum\limits_{i=1}^n (y_i - \bar{y})^2 \\ &= \sum\limits_{i=1}^n (y_i - \hat{y}_i + \hat{y}_i - \bar{y})^2 \\ &= \sum\limits_{i=1}^n (y_i - \hat{y}_i)^2 + 2 \sum\limits_{i=1}^n (y_i - \hat{y}_i)(\hat{y}_i - \bar{y}) + \sum\limits_{i=1}^n (\hat{y}_i - \bar{y})^2 \\ &= SSE + 2 \sum\limits_{i=1}^n (y_i - \hat{y}_i)(\hat{\beta}_1 x_i + \hat{\beta}_0 - \bar{y}) + SSR \\ &= SSE + 2 \hat{\beta}_1 \sum\limits_{i=1}^n x_i (y_i - \hat{y}_i) + 2(\hat{\beta}_0 - \bar{y}) \sum\limits_{i=1}^n (y_i - \hat{y}_i) + SSR \\ &= SSE + 2 \hat{\beta}_1 (0) + 2(\hat{\beta}_0 - \bar{y})(0) + SSR \\ &= SSE + SSR \end{align} And the derivation of the formula for SSR in terms of $S_{xy}$ and $S_{xx}$ (and therefore the connection between the coefficient of determination and the correlation coefficient) is as follows: \begin{align} SSR &= SST - SSE \\ &= SST - \sum\limits_{i=1}^n (y_i - \hat{y}_i)^2 \\ &= SST - \sum\limits_{i=1}^n (y_i - \hat{\beta}_1 x_i - \hat{\beta}_0)^2 \\ &= SST - \sum\limits_{i=1}^n (y_i - \hat{\beta}_1 x_i - (\bar{y} - \hat{\beta}_1 \bar{x}))^2 \\ &= SST - \sum\limits_{i=1}^n (y_i - \bar{y} - \hat{\beta}_1 (x_i - \bar{x}))^2 \\ &= SST - \left( \sum\limits_{i=1}^n (y_i - \bar{y})^2 - 2 \hat{\beta}_1 \sum\limits_{i=1}^n (x_i - \bar{x})(y_i - \bar{y}) + \hat{\beta}_1 ^2 \sum\limits_{i=1}^n (x_i - \bar{x})^2 \right) \\ &= SST - (SST - 2 \hat{\beta}_1 S_{xy} + \hat{\beta}_1^2 S_{xx}) \\ &= 2 \left( \dfrac{S_{xy}}{S_{xx}} \right) S_{xy} - \left( \dfrac{S_{xy}}{S_{xx}} \right)^2 S_{xx} \\ &= \dfrac{S_{xy}^2}{S_{xx}} \end{align} Therefore $\dfrac{SSR}{SST} = \dfrac{S_{xy}^2}{S_{xx} S_{yy}} = r^2$.
# Fraction calculator The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression. ## Result: ### 5 + 1 - 2 = 4/1 = 4 Spelled result in words is four. ### How do you solve fractions step by step? 1. Add: 5 + 1 = 6 2. Subtract: the result of step No. 1 - 2 = 6 - 2 = 4 #### Rules for expressions with fractions: Fractions - simply use a forward slash between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part. The slash separates the numerator (number above a fraction line) and denominator (number below). Mixed numerals (mixed fractions or mixed numbers) write as integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2. Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3. Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45. The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3. An asterisk * or × is the symbol for multiplication. Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses. The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2 #### Examples: subtracting fractions: 2/3 - 1/2 multiplying fractions: 7/8 * 3/9 dividing Fractions: 1/2 : 3/4 exponentiation of fraction: 3/5^3 fractional exponents: 16 ^ 1/2 adding fractions and mixed numbers: 8/5 + 6 2/7 dividing integer and fraction: 5 ÷ 1/2 complex fractions: 5/8 : 2 2/3 decimal to fraction: 0.625 Fraction to Decimal: 1/4 Fraction to Percent: 1/8 % comparing fractions: 1/4 2/3 multiplying a fraction by a whole number: 6 * 3/4 square root of a fraction: sqrt(1/16) reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22 expression with brackets: 1/3 * (1/2 - 3 3/8) compound fraction: 3/4 of 5/7 fractions multiple: 2/3 of 3/5 divide to find the quotient: 3/5 ÷ 2/3 The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are: PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction. GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction. Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right. ## Fractions in word problems: • Rolls Mom bought 13 rolls. Dad ate 3.5 rolls. How many rolls were left when Peter yet ate two at dinner? • In dividing In dividing fractions, get the reciprocal of the divisor and change division symbol to multiplication symbol. 2/3 : 5/6 • The recipe The recipe they are following requires 7/8 cups of milk, Tom already put 3/8 cups of milk. How much milk should Lea add to follow the recipe? • Division by reciprocal What is the corresponding illustration/model of 7÷ 1/3? • New bridge Thanks to the new bridge, the road between A and B has been cut to one third and is now 10km long. How much did the road between A and B measure before? • Third of an hour How many minutes is a third of an hour? Do you know to determine a third of the lesson hour (45min)? • Reciprocals Which among the given reciprocal is correct a. 3/15x1/3= 1 b. 3/20x20/3=1 c. 7/14x7/7=1 d. 34/3x34/34=1 • Homework In the crate are 18 plums, 27 apricot and 36 nuts. How many pieces of fruit left in the crate when Peter took 8 ninth: 1. nuts 2. apricots 3. fruit 4. drupe • Difference mixed fractions What is the difference between 4 2/3 and 3 1/6? • Ricky Ricky painted 3/5 of the side of the garage. When he repainted ½ of this part, what part of the side of the garage did he painted twice? • Fruits Amy bought a basket of fruits 1/5 of them were apples,1/4 were oranges, and the rest were 33 bananas. How many fruits did she buy in all? • Equation with x Solve the following equation: 2x- (8x + 1) - (x + 2) / 5 = 9 • The rod The rod is painted in four colors. 55% of the bar is painted in blue, green 0.2 of the rod, 1/8 is brown, and the remaining 45 cm of white. How long is the rod?
# External Secant Segment: Example, Proof An external secant segment is the part of a secant line lying outside a circle. The Exsecant function is used to find the length of the external secant segment. ## Finding Unknown Lengths with External Secant Segment The following ratio allows you to solve for unknown lengths of secant lines, external secant segments, or internal segments (the parts of secant lines that are inside the circle): This relationship is formally called Steiner’s Theorem [1] can be stated in words as: If two secant segments are drawn from an external point to a circle, then the products of the lengths of each secant line with its external segment are equal. The theorem is rarely called by its proper name, perhaps to avoid confusion with the other Steiner’s Theorem which is used in physics to determine moments of inertia. ## Proof of Steiner’s Theorem Example question: Prove that PQ * PR = PS * PT. Step 1: Draw RS and QT: From this image, you can deduce that ∠R and ∠T are congruent* because they intersect the same arc (the arc between points Q and S): ∠R ≅ ∠T Congruent objects have the same shape and dimensions. Step 2: Write that ∠P is congruent to itself; This is because of the reflexive property of congruence (which simply states that any shape is congruent to itself). This is an obvious step, but it’s needed in a formal proof. So we have: ∠P ≅ ∠P. Step 3: State that two triangles △PRS and △PQT are equivalent. This follows from Steps 1 and 2 plus the AA similarity theorem which states that two triangles are similar if two pairs of angles are congruent. In other words, we proved that two angles in Steps 1 and 2 are congruent, so the third angle must be congruent as well: △PRS ~ △PQT Step 4: Derive the cross product of side lengths. Since we have proved that the two triangles are similar in Step 3, the side lengths must be proportional: We can rewrite this with the Cross Product Property: PQ PR = PS * PT. That’s it! ## References [1] Jones, J. Steiner’s Theorem. Retrieved April 15, 2021 from: http://jwilson.coe.uga.edu/EMT669/Student.Folders/Jones.June/steiner/steinerthm.html
× # Probability Worksheets Top FAQ ## CBSE Probability Worksheets View Notes For scoring good marks in the Mathematics examination one needs a lot of practice. Here we provide a geometric probability worksheet pdf on all of the important math topics for CBSE Classes 6, 7, 8, 9, 10, 11, and 12. With the help of these geometric probability worksheets pdf, the students can practice very well and improve their preparation level for the final exams. Go through all the important topics that all the student needs to know, which includes integers, algebra, decimals, geometry, arithmetic, trigonometry, time, measurement, and much more. Students can match the solutions with the answer keys and get appropriate feedback to analyze mistakes and correct them. ### What is Probability? Probability is known as the chance of some of the events to occur. When we need to know about the probability of a particular event which has to take place, we think of the chances that we expect in accordance to every possibility that can happen. When you aren't sure about the result of a specific event, you'll specify the chances or how likely the result. Analyzing the events that are governed by probability is known as Statistics. Statistics Problems can be solved using math formulas. A simple example that defines the basics of probability is flipping a coin. We will get two possible outcomes when a coin is flipped, i.e, • Tails What will be the probability for a fair coin (which has two different faces) that it lands on the Heads? Since there are only two possible outcomes which will occur out of 1, hence the probability of the coin landing on Heads would be Hence, P(H) = 0.5 or 50% Here, we bring you the geometric probability worksheet pdf to assist you in improving within the probability concepts that also include applied mathematics, probability statistics along with applications of probability. ### Probability Formula The probability formula is defined because of the likelihood of an occasion to happen. It is equal to the ratio of the number of favorable results and the total number of outcomes. The formula for the probability of an occasion to occur is given by: P(E) = Total Number of the favorable outcomes/Total Number of the outcomes ### Probability Worksheet for Class 9 1. Two coins are tossed at the same time for 400 times and we get 2 times of heads which equals 180 times, one head = 148 times, and no head = 70 times. If two coins are tossed at random, what is the probability of getting 2 heads, 1 head and 0 heads. 2. According to the meteorological report for 300 consecutive days during a year, its weather outlook was correct 180 times. Out of those days, at some point was chosen randomly, what's the probability that the weather outlook was correct thereon day and not correct there on the day. 3. In a match, a batsman hit the boundary 5 times out of 40 balls played by him. Find the probability that the boundary isn't hit by the ball. 4. In a survey of 200 ladies, it had been found that 142 like coffee, while 58 dislike it. Find the probability that a woman chose at the present likes coffee and dislike coffee. 5. In an ongoing cricket match, a batsman hits boundary 6 times in 30 balls he plays. Find the probability that he didn't hit a ball. ### Probability Worksheet for Class 10 The 10th-grade probability worksheets are mentioned as follows: 1. A coin is tossed once, what's the probability of getting a head. 2. A die is thrown only once, so find the probability of getting a decent number and a multiple of three, 3. Two dice are thrown at an equivalent time, find the probability that the sum of two numbers appearing on the highest of the dice is quite nine. 4. A bag has 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is supposed to be three times that of the red ball, then what will be the number of blue balls within the bag. 5. One card is drawn randomly from a well-shuffled deck of 52 cards. Find the probability that the card that is drawn may be a king, a red 8, a spade, a red card, the six of the club, and a face card. Above are the mentioned points for the 10th-grade probability worksheets. Question 1) A Bag has a Red Ball, a Blue Ball, and a Yellow Ball, all the Balls are the same Size. Kritika takes a Ball Out of the Bag Without Looking Into it. What is the Probability that She Picks the: (i) Yellow Ball? (ii) Red Ball? (iii) Blue Ball? Solution) Kritika takes a ball from the bag without looking into it. So, it is very likely that she will take out any one of them from the bag. Let Y be the event and the ball which is taken out is yellowâ€™, B be the event and the ball taken out is blueâ€™, and R will be the event and the ball which is taken out is redâ€™. The number of possible outcomes equals Number of balls in the bag which is = n(S) = 3. (i) The number of the outcomes favorable to the event Y will be = n(Y) = 1. So, P(Y) equals n(Y)/n(S) = â…“ So, (ii) P(R) = â…“Â and (iii) P(B) = â…“ Question 2) One Card is Drawn from a Well-Shuffled Deck of 52 Cards. Calculate the Probability that the Card will (i) be an Ace, (ii) not be an Ace. Solution) Well-shuffling gives us the possibility for equally likely outcomes. (i) Card drawn is an ace There are four aces in a deck. Let E be the event where the card is an aceâ€™. The number of the outcomes favourable to E will be = n(E) = 4 The number of possible outcomes equals Total number of cards which is = n(S) = 52 Therefore, P(E) = n(E)/n(S) = 4/52 = 1/13 (ii) Card which is drawn is not an ace Let F be the event for the card which is drawn is not an aceâ€™. The number of the outcomes which are favourable for the event F = n(F) = 52 â€“ 4 = 48 So, P(F) = n(F)/n(S) = 48/52 = 12/13
# Worksheet: Proving Numbers Are Even or Odd Q1: State whether the answer to the sum is even or odd: . • A even • B odd Q2: Let us think about even numbers. 2, 4, 6, 8, and 10 are examples of even numbers. Notice that they are all doubles. Which of the following sums has an answer that is an even number? • A 8 + 8 • B 8 + 8 + 1 Q3: Ethan has drawn this picture to show that 16 is an even number. His teacher asks him to complete the addition sentence that shows that 16 is even. He knows that he needs to write the same number in both boxes. What number is this? Q4: If a number is odd, then we can divide it into two equal groups with one left over. This means that every odd number can be written as an addition sentence that is a doubles fact plus one. For example, we know that 11 is odd because . Which of these addition sentences shows that 9 is odd? • A • B • C • D Q5: Is the answer to the following sum odd or even: ? • A odd • B even Q6: If a number is even, then we can divide it into two equal groups. This means that every even number can be written as a doubles addition sentence. For example, we know that 10 is even because . Which of these addition sentences shows that 12 is even? • A • B • C • D Q7: An even number plus an odd number is . • Aan odd number • Ban even number Q8: An odd number plus an odd number is . • Aan even number • Ban odd number Q9: Anthony brought 12 cupcakes to a party and Madison brought 7 cupcakes. At the end of the party, there were 5 cupcakes left. Is the number of cupcakes that were eaten even or odd? • A even • B odd Q10: An even number plus is an odd number. • Aan odd number • Ban even number Q11: Madison has drawn this picture to show that 17 is an odd number. Her teacher asks her to complete the addition sentence that shows that 17 is odd. She knows that she needs to write the same number in both boxes. What number is this?
# Concept of a Line Graph ## Definition Line graph: A line graph displays data that changes continuously over periods of time. ## Notes ### Concept of a Line Graph: A line graph displays data that changes continuously over periods of time. 1. Consider the following data, given in tabular form. Time 6 a.m. 10 a.m. 2 p.m. 6 p.m. Temperature(°C) 37 40 38 35 • The horizontal line (usually called the x-axis) shows the timings at which the temperatures were recorded. • Temperature(°C) is labelled on the vertical line (usually called the y-axis). • By this line graph, we can easily understand the increase and decrease of temperature over the course of 12 hours from 6 AM to 6 PM. #### Interpretation of line graph: 1) The given graph represents the total runs scored by two batsmen A and B, during each of the ten different matches in the year 2007. Study the graph and answer the following questions. (i) What information is given on the two axes? (ii) Which line shows the runs scored by batsman A? (iii) Were the run scored by them the same in any match in 2007? If so, in which match? (iv) Among the two batsmen, who is steadier? How do you judge it? ### Solution: (i) The horizontal axis (or the x-axis) indicates the matches played during the year 2007. The vertical axis (or the y-axis) shows the total runs scored in each match. (ii) The dotted line shows the runs scored by Batsman A. (This is already indicated at the top of the graph). (iii) During the 4th match, both have scored the same number of 60 runs. (This is indicated by the point at which both graphs meet). (iv) Batsman A has one great “peak” but many deep “valleys”. He does not appear to be consistent. B, on the other hand, has never scored below a total of 40 runs, even though his highest score is only 100 in comparison to 1 15 of A. Also, A has scored a zero in two matches, and in a total of 5 matches, he has scored less than 40 runs. Since A has a lot of ups and downs, B is a more consistent and reliable batsman. ## Example The given graph describes the distances of a car from a city P at different times when it is travelling from City P to City Q, which are 350 km apart. Study the graph and answer the following: (i) What information is given on the two axes? (ii) From where and when did the car begin its journey? (iii) How far did the car go in the first hour? (iv) How far did the car go during (i) the 2nd hour? (ii) the 3rd hour? (v) Was the speed same during the first three hours? How do you know it? (vi) Did the car stop for some duration at any place? Justify your answer. (vii) When did the car reach City Q? (i) The horizontal (x) axis shows the time. The vertical (y) axis shows the distance of the car from City P. (ii) The car started from City P at 8 a.m. (iii) The car travelled 50 km during the first hour. (iv) The distance covered by the car during (a) the 2nd hour (i.e., from 9 am to 10 am) is 100 km, (150 – 50). (b) the 3rd hour (i.e., from 10 am to 11 am) is 50 km (200 – 150). (v) From the answers to questions (iii) and (iv), we find that the speed of the car was not the same all the time. (In fact, the graph illustrates how the speed varied). (vi) We find that the car was 200 km away from city P when the time was 11 a.m. and also at 12 noon. This shows that the car did not travel during the interval of 11 a.m. to 12 noon. The horizontal line segment representing “travel” during this period is illustrative of this fact. (vii) The car reached City Q at 2 p.m. If you would like to contribute notes or other learning material, please submit them using the button below.
Finding and evaluating integrals!! Printable View • Mar 22nd 2013, 09:25 AM maximus Finding and evaluating integrals!! Hi Guys, I have attached the two questions i need help with.... Really need a step by step explanation of how to complete please :) Please help!!!! • Mar 22nd 2013, 09:43 AM Siron Re: Finding and evaluating integrals!! 1. $\displaystyle \int (2x-3)(x+4)dx$. Expand the product and integrate term by term. 2. $\displaystyle \int_{-2}^{-1} \frac{dx}{(x-2)^3}$ Use the substitution $\displaystyle t=x-2$ (be careful with the integration bounds!) • Mar 29th 2013, 06:54 PM x3bnm Re: Finding and evaluating integrals!! For first question: \displaystyle \begin{align}\int (2x-3)(x+4)\,\,dx =& \int (2x^2 + 5x - 12)\,\,dx\\ =& \frac{2x^3}{3} + \frac{5x^2}{2} - 12x + C\end{align} You can see if the result is correct here http://www.wolframalpha.com/input/?i=integration+\int+(2x-3)(x%2B4)+dx For Second question let $\displaystyle t = x - 2 \text{ and } \frac{dx}{dt} = 1$: $\displaystyle \text{for lower limit } x = -2 \text { So } t = -2 -2 = -4 \text{ for upper limit } x = -1 \text{ so } t = -1 -2 = -3$ \displaystyle \begin{align}\int_{-2}^{-1} \frac{dx}{(x-2)^3} =& \int_{-4}^{-3} \frac{dt}{t^3}\\ =& \frac{t^{-2}}{-2}\Big]_{-4}^{-3}\\ =& \frac{(-3)^{-2}}{-2} - \frac{(-4)^{-2}}{-2} \\ =& \frac{-1}{18} + \frac{1}{32} \\ =& \frac{-7}{288} \end{align} You can check it here: http://www.wolframalpha.com/input/?i=integration+1%2F((x-2)^3)+dx+from+x+%3D+-2+to+-1
# Question Video: Finding an Unknown Matrix Using the Inverse of a Matrix Mathematics Given that (π΄π΅)β»ΒΉ = (1/6)[5, β3 and β33, 21], π΄ = [β2, β1 and β3, β2] determine π΅β»ΒΉ. 02:25 ### Video Transcript Given that π΄π΅ inverse equals one-sixth multiplied by five, negative three, negative 33, 21 and π΄ equals negative two, negative one, negative three, negative two, determine π΅ inverse. Letβs start with a quick reminder of what the matrix inverse is. The inverse of a square matrix π΄, π΄ inverse, is the matrix such that π΄ multiplied by π΄ inverse gives us the identity matrix. And one property of the matrix inverse which is going to prove useful to us here is π΄π΅ inverse is equal to π΅ inverse multiplied by π΄ inverse. So because weβre told the inverse of the product π΄π΅ is one-sixth multiplied by the matrix five, negative three, negative 33, 21, then from this property of the matrix inverse, we can say that this is the same as π΅ inverse multiplied by π΄ inverse. But how is this going to help us find π΅ inverse? Well, thereβs a little bit of a trick that we can apply here. And all it requires is remembering the definition of the matrix inverse. We can find the matrix π΅ inverse, π΄ inverse, π΄ by multiplying π΅ inverse, π΄ inverse by π΄ on the right. So letβs now multiply these two matrices together to see what we get. We do this in the usual way of multiplying two two-by-two matrices together. And then we can simplify each entry. And we end up with one-sixth multiplied by the matrix negative one, one, three, negative nine. We then remember when we have a scalar multiplied by a matrix, we can just multiply each entry by that scalar. And that gives us the matrix negative one over six, one over six, one over two, negative three over two. But how has this actually helped us find the matrix π΅ inverse? Well, what we found is the matrix π΅ inverse multiplied by π΄ inverse multiplied by π΄. And from the definition of the matrix inverse, π΄ inverse multiplied by π΄ gives us the identity matrix. So what weβve actually found is the matrix π΅ inverse multiplied by the identity matrix. But multiplying any matrix by the identity matrix just gives us that matrix. So what weβve found is the matrix π΅ inverse. So by using the definition of the matrix inverse and one of the properties of the matrix inverse, we were able to find an unknown using the matrix inverse.
Collection of recommendations and tips # How do you calculate scale ratio? ## How do you calculate scale ratio? The scale factor is commonly expressed as 1:n or 1/n, where n is the factor. For example, if the scale factor is 1:8 and the real measurement is 32, divide 32 ÷ 8 = 4 to convert. To convert a measurement to a larger measurement, simply multiply the real measurement by the scale factor. ## What does a 1/4 scale mean? A 1/4″ scale means that each 1/4″ (inch) on the plan counts for 1′ (feet) of actual physical length. To scale a blueprint in imperial units to actual feet. multiply the measurement on the drawing (in inches decimal equivalent) with the denominator. What is the formula for scale factor? The basic formula to find the scale factor of a figure is: Scale factor = Dimensions of the new shape ÷ Dimensions of the original shape. This can also be used to calculate the dimensions of the new figure or the original figure by simply substituting the values in the same formula. How do you make a scale model? 1. Measure the dimensions of each aspect of the object you are modeling. 2. Scale the dimensions down to model size. 3. Draw a plan for your model. 4. Use your knife to carve out each piece from your building material. 5. Paint each piece the correct color or colors. ### What is an example of a scale model? A scale model is an enlarged or reduced representation of an object that has the exact same proportions as the actual object. Maps and floor plans are great examples of scaled models (drawings), as is the typical plant and animal cell illustrations in textbooks and online. ### What is the scale factor of 4? Suppose you have two similar figures , one larger than the other. The scale factor is the ratio of the length of a side of one figure to the length of the corresponding side of the other figure. Here, XYUV=123=4 . So, the scale factor is 4 . How do you find the scale factor with coordinates? If the center of dilation is the origin, then the coordinates are multiplied by the scale factor: (x,y) -> (kx, ky) where k is the scale factor. To solve a problem like the one you presented, determine the scale factor by dividing the coordinates of X’ by the corresponding coordinates of X. What size is a 1/4 scale? A 1/4″ scale means that each 1/4″ (inch) on the plan counts for 1′ (feet) of actual physical length. ## How big is a quarter scale ship model? Scale Inches = Actual Dimension (in inches) ÷ Scale = 3 x 12 ÷ 48 = ¾” For the above examples, the model built on a ¼” to foot scale would be 1/48 the size of the real-life vessel. It is, therefore, a quarter-scale model. The term scale should not be confused with size. ## How big is a quarter size architectural model? Common scale for architectural modelling. 1:48: 1 ⁄ 4 ″ (0.25″) 6.350 mm: For dollhouse applications, 1:48 is commonly known as quarter scale (as it is one-quarter of the 1:12 “standard” dollhouse scale). Mainly military aircraft, but in 2005 Tamiya launched a new series of How big is a 1 / 64 scale model car? A 1/64 scale replica would be about 2.8 inches long. In general, this means that for the same model car, a 1/18 scale version will be larger than a 1/24 scale version, which is larger than a 1/32, which is larger than a 1/43, which is larger than a 1/64, and so on. How big is a 1 / 18 scale replica of a car? A 1/18 scale replica of the same car would be 10 inches long (180 divided by 18). A 1/64 scale replica would be about 2.8 inches long. In general, this means that for the same model car, a 1/18 scale version will be larger than a 1/24 scale version, which is larger than a 1/32, which is larger than a 1/43, which is larger than a 1/64, and so on.
# Polynomial Curve Fitting & Interpolation This eleventh article of the mathematical journey through open source, explains curve fitting & interpolation with polynomials in octave. << Tenth Article In various fields of physics, chemistry, statistics, economics, … we very often come across something called curve fitting, and interpolation. Given a set of data points from our observations, we would like to see what mathematical equation does they follow. So, we try to fit the best curve through those data points, called the curve fitting technique. Once we have the equation, the main advantage is that then we can find the values at the points, we haven’t observed by just substituting that value in the equation. One may think of this as interpolation. But this is not exactly interpolation. As in interpolation, we are restricted to finding unobserved values only between two observed points, using a pre-defined curve fit between those two points. With that we may not be able to get values outside our observation range. But the main reason for using that is when we are not interested or it is not meaningful or we are unable to fit a known form of curve in the overall data set, but still we want to get some approximations of values, in between the observed data points. Enough of gyaan, now let’s look into each one of those. ## Curve Fitting Let’s say for any system, we have the following data points: For the inputs of 1, 3, 6, 10, 20, we get the outputs as 2.5, 7.5, 15.5, 24, 45, respectively. Then, we would like to know as how is input is related to the output. So, to get a feel of the relation, we first plot this points on a x-y plane, say inputs as x and outputs as y. Figure 6 shows the same. And the octave code for that is as follows: ```\$ octave -qf octave:1> x = [1 3 6 10 20]; octave:2> y = [2.5 7.5 15.5 24 45]; octave:3> plot(x, y, ""); octave:4> xlabel("Inputs"); octave:5> ylabel("Outputs"); octave:6>``` Figure 6: Plot of the data points By observing the plot, the simplest fitting relation looks to be a straight line. So, let’s try fitting a linear polynomial, i.e. a first order polynomial, i.e. a straight line. Here’s the code in execution for it: ```octave:1> x = [1 3 6 10 20]; octave:2> y = [2.5 7.5 15.5 24 45]; octave:3> p = polyfit(x, y, 1) p = 2.2212 1.1301 octave:4> polyout(p, "x"); 2.2212x^1 + 1.1301 octave:5> plot(x, y, "", x, polyval(p, x), "-"); octave:6> xlabel("Inputs"); octave:7> ylabel("Outputs"); octave:8> legend("Data points", "Linear Fit"); octave:9>``` polyfit() takes 3 arguments, the x values, the y values, and the order of the polynomial to fit – 1 for linear. It then returns the coefficients of the fitted polynomial p. polyout() displays the polynomial in more readable format. So, the fitted equation is y = 2.2212 x + 1.1301. polyval() takes the polynomial p and the input values x, and calculates the output values y, as per the polynomial equation. And then we plot them, along with the earlier data points, on the same plot. Figure 7 shows the same, 2.2212 being the slope / gradient of the straight line, and 1.1301 being the intercept. Now, we notice that the straight line does not fit exactly, and there is some deviation from the observed points. polyfit() fits the polynomial with the minimum deviation for the order we request, but still there is a deviation, as that is the minimum possible. If you want to get the standard deviation for the particular fit, you just need to do the following: ```octave:9> std_dev = sqrt(mean((y - polyval(p, x)).^2)) std_dev = 0.72637 octave:10>``` Figure 7: Linear fit of data points And we see that it is a small value compared to our output values. But say we are not fully satisfied with this, we may like to try higher order polynomials, say of order 2, a quadratic, and observe the results from it. Here’s how one would do this: ```octave:1> x = [1 3 6 10 20]; octave:2> y = [2.5 7.5 15.5 24 45]; octave:3> p = polyfit(x, y, 2) p = -0.018639 2.623382 -0.051663 octave:4> polyout(p, "x"); -0.018639x^2 + 2.6234x^1 - 0.051663 octave:5> plot(x, y, "", x, polyval(p, x), "-"); octave:6> xlabel("Inputs"); octave:7> ylabel("Outputs"); octave:8> legend("Data points", "Quadratic Fit"); octave:9> std_dev = sqrt(mean((y - polyval(p, x)).^2)) std_dev = 0.26873 octave:10>``` Figure 8 shows the data points and fitted quadratic curve. All the values above can be interpreted in the similar way as earlier. And we note that the standard deviation has come down further. In fact, as we increase the polynomial order further and further, we will definitely note that the standard deviation keeps on decreasing, till the order is same as number of data points – 1, after which it remains 0. But, we may not like to do this, as our observed data points are typically not perfect, and we are typically not interested in getting the standard deviation zero but for a fit more appropriate to reflect the relation between the system’s input and output. Figure 8: Quadratic fit of data points ## Interpolation Now, say we have the following wierd data points, listed as (x, y) pairs: (1, 1), (2, 2), (3, 3), (4, 2), (5, 1), (6, 3.75), (7, 7), (8, 3), (9, 0), (11, 8). Figure 9 shows, how wierd they look like. Figure 9: Wierd data points Trying to fit a polynomial, would be quite meaningless. But still they seem to have some correlation in between smaller ranges of input, say between 1 and 3, 3 and 5, 5 and 7, and so on. So, this is best suited for us to do interpolation for finding the values in between, say at 1.5, 3.8, 10, … Now interpolation is all about curve fitting between two neighbouring data points and then calculating the value at any intermediate point. So, the typical varieties of techniques used for this “piece-wise curve fitting” are: • nearest: Return the nearest data point (actually no curve fitting) • linear: Linear/Straight line fitting between the two neighbouring data points • pchip: Piece-wise cubic (order 3) hermite polynomial fitting • cubic: Cubic (order 3) polynomial fitted between four nearest neighbouring data points • spline: Cubic (order 3) spline fitting with smooth first and second derivatives throughout the curve All these may look little jargonish – so don’t worry about that. Let’s just try out some examples to drive the point home. Here’s the code to show the nearest & linear interpolation for the above data: ```octave:1> x = [1 2 3 4 5 6 7 8 9 11]; octave:2> y = [1 2 3 2 1 3.75 7 3 0 8]; octave:3> xi = 1:0.1:11; octave:4> y_n = interp1(x, y, xi, "nearest"); octave:5> y_l = interp1(x, y, xi, "linear"); octave:6> plot(x, y, "", xi, y_n, "-", xi, y_l, "-"); octave:7> xlabel("Inputs"); octave:8> ylabel("Outputs"); octave:9> legend("Data points", "Nearest", "Linear"); octave:10>``` Figure 10 shows the same. Figure 10: Nearest and linear interpolations of the data points Now, if we want to get the interpolated values say at the points 1.5, 3.8, 10, instead of giving all the intermediate points xi in the above code, we may just give these 3 – that’s all. Here goes the code in execution for the that: ```octave:1> x = [1 2 3 4 5 6 7 8 9 11]; octave:2> y = [1 2 3 2 1 3.75 7 3 0 8]; octave:3> our_x = [1.5 3.8 10]; octave:4> our_y_n = interp1(x, y, our_x, "nearest") our_y_n = 2 2 8 octave:5> our_y_l = interp1(x, y, our_x, "linear") our_y_l = 1.5000 2.2000 4.0000 octave:6>``` Now, before closing on interpolation, just to give a feel of all of the interpolation techniques, here goes an example with the points on a sine curve: ```octave:1> x = 0:2pi; octave:2> y = sin(x); octave:3> xi = 0:0.1:2pi; octave:4> y_n = interp1(x, y, xi, "nearest"); octave:5> y_l = interp1(x, y, xi, "linear"); octave:6> y_p = interp1(x, y, xi, "pchip"); octave:7> y_c = interp1(x, y, xi, "cubic"); octave:8> y_s = interp1(x, y, xi, "spline"); octave:9> plot(x, y, "*", xi, y_n, "-", xi, y_l, "-", xi, y_p, "-", xi, y_c, "-", xi, y_s, "-"); octave:10> xlabel("x ->"); octave:11> ylabel("y ->"); octave:12> legend("Data points", "Nearest", "Linear", "Pchip", "Cubic", "Spline"); octave:13> print("-dpng", "all_types_of_interpolations.png"); octave:14>``` Figure 11 shows the visualization of the same for your eyes. Figure 11: All types of interpolations for sinusoidal points ## What’s next? With today’s curve fitting & interpolation walk through, we have used some basic 2-D plotting techniques. But there are many more features and techniques to plotting, like marking our plots, visualizing 3-D stuff, etc. Next, we’ll look into that. Twelfth Article >> Send article as PDF
Goseeko blog # What is the principle of mathematical induction? Mathematical induction is a mathematical technique is used to prove a statement, a theorem or a formula is true for every natural number. An essential property of the set N = {1, 2, 3, …} of positive integers follows: Principle of Mathematical Induction I: Let P be a proposition defined on the positive integers N; that is, P(n) is either true or false for each n ∈ N. Suppose P has the following two properties: (i) P(1) is true. (ii) P(k + 1) is true whenever P(k) is true. Then P is true for every positive integer n ∈ N. We shall not prove this principle. In fact, this principle is usually given as one of the axioms when N is developed axiomatically. Principle of Mathematical Induction II: Let P be a proposition defined on the positive integers N such that: (i) P(1) is true. (ii) P(k) is true whenever P(j) is true for all 1 ≤ j < k. Then P is true for every positive integer n ∈ N. Note: Sometimes one wants to prove that a proposition P is true for the set of integers {a, a + 1, a + 2, a + 3, . . .} where a is any integer, possibly zero. This can be done by simply replacing 1 by a in either of the above Principles of Mathematical Induction. The technique has two steps to prove any statement- 1. Base step (it proves that a statement is true for the initial value) 2. Inductive step (it proves that is the statement is true for n’th iteration then it is also true for (n+1)’th iteration. Example: Sol. here for n = 1, 1 = 1² Now let us assume that the statement is true for n = k Hence we assume that is true. Now we need to prove that- So that which satisfies the second step. Hence- Example-2:Prove 1+2+…+n=n(n+1)/2 using a proof by induction Sol. Let n=1: 1 = 1(1+1)/2 = 1(2)/2 = 1 is true, Step-1: Assume n=k holds: 1+2+…+k=k(k+1)/2 Show n=k+1 holds: 1+2+…+k+(k+1)=(k+1)((k+1)+1)/2 Substitute k with k+1 in the formula to get these lines. Notice that I write out what I want to prove. 1+2+…+(k+1)=1+2+…+k+(k+1) =k(k+1)/2 + (k+1) =(k(k+1) + 2(k+1))/2 by 2/2=1 and distribution of division over addition =(k+2)(k+1)/2 by distribution of multiplication over addition =(k+1)(k+2)/2 by commutativity of multiplication. Example-3: Prove the following by using the principle of mathematical induction for all n N- Sol. Here, n = 1, which is true Step-1: Assume n = k holds- Now show n = k + 1 also holds- Consider- Which is also true for n = k + 1. Hence proved.

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