Ekene Franklin's messageboard.
Feel free to discuss anything here with me friends! We're here to help each other!
Note by
Ekene Franklin
2 years, 9 months ago
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Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
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[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
My country's national olympiads 1st round is to be conducted tomorrow. I will most likely post all of them here for our consumption! I expect them to be pretty easy.
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Ok. Thanks and also are you writing that exam.
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It is actually meant for students two years older than I am, but I will give it a shot.
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Oh !!!
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It there a minimum age requirement to be eligible to write your country's olympiads?
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Yes, I think it is 13 or 14. People from class 8 or 9 onwards can write the exam.
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Are you also writing the olympiads soon?
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No. At present I am not interested. The main goal of student student studying in class 11 and 12 in India is to get into IIT (Indian Institute Of Technology)
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Oh, so that is what JEE and IIT are all about. I hope you get into it seamlessly. But say, do you hope to write olympiads in the future?
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Yes. I may.
This is a marvelous secret and I benefited a lot from this. If you want to write any question for the problems of the week the better option is to write a physics problem rather than maths. Out of 5 problems 3 are allotted for maths and 2 for science. In the maths category there will be a much competition from many users. But in the science category there are only 4 competitors namely Rohit Gupta, Pranshu Gaba, and two staff members Blake Farrow, Danielle Scarano. So, I posted some good physics problems and 3 of them are featured now. But I won't force you to post only physics problems then there will be much competition for science. If you have good problems in maths you can post them (as 3 of my maths problems are also featured although I say that I am basically a physics man).
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Where is the show full site option??(from ur mt. Everest Q discussion)
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It will be in the main option bar. It is the place where you can find links like "profile", "stats", "search", etc. The "view" full site option will be present there.
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Main option bar : Those 3 lines on left corner?? Or in the stats section of taskbar?
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The three line aside the option "Premium"
Oh so this is on website. Can this be done in app?
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Ohh !! You are doing in the app. Sorry, I don't know whether this is possible in app or not. But can't you find any option like this. in the screenshot.
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No :-( how can I send screenshot on this chat?
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You have to copy the link of the picture and paste it here.
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K thnx a lot for helping, and sorry for disturbing..
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It's OK. No problem. I went to brilliant only once in the mobile app so I am not much aware of the options present in it.
Mauj Rajpal, from which part of India are you from.
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Delhi ;-) and u??
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Visakhapatnam, Andra Pradesh.
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Oh nice. From where do you put your problems in sets(source) and how do you have sets of other people in yours??
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In every set there will be a option at the bottom called "save". If you click that, that particular set will be saved into your sets. Some of my problems are original whereas some other are completely taken from books or the problems and the remaining are the problems which I have taken from books and modified them.
Can you suggest me some of your book(especially maths)
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It's our college material. It is a collection of all good problems from various books.
Here is the screenshot of where is it present. Can you find it now ?
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Is this screenshot from app?
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No. It is from from the main website.
How do you send questions to them to select for weekly problem??Do they themselves select from your sets?
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No. Only staff will select the problems of the week. In general, the notification will be as if I have written the problem of the week.
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They themselves select it by seeing your sets or you send them via some forum and then they choose one of the all sent questions??
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The community members write problems and post them to community and the staff selects the best 15 questions among all the problems and they will feature them as problems of the week. There is no forum involved in this process.
I personally think Brilliant should include a biology and biochemistry section. What do you think?
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Yes.
Before brilliant used to have biology. But recently when the staff updated brilliant they removed biology topic due to lack of problems in that category.
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Actually I asked them to remove it, because the biology problems could not be seen in the community page. You can see Calvin Lin's messageboard.
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Calvin Lin's message board is currently unavailable. I searched for the website for a long time but in vain. Maybe he have locked it.
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Really? Here is the link. It's still available.
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But last time when I went it was locked
yes
Here is a link of a biology problem.
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Yes, but they cannot be found in the community page.
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I know. And there is no biology when you're making a problem.
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Previously there was, but they removed it.
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OK
Those are from a set biology is so easy-2, biology is so easy-3, biology is so easy-4, biology is so easy-5biology is so easy-6, biology is so easy-7, biology is so easy-8, biology is so easy-9, biology is so easy-10, biology is so easy-11, biology is so easy-12, biology is so easy-13, biology is so easy-14 by Mohammad Khaza.
Hello Everyone !!! As this note has got some immense response from young students I would like to continue this further. But the problem is that this note is getting more and more comments day by day due to which the process of loading and opening any notifications is taking more time. So, on the request of @EKENE FRANKLIN any further discussion can be continued in the new note Brilliant young students messageboard
Doubt : 1
If the mole fraction of a solute is changed from 41 to 21 by adding some solute in the 800 g of H2O solvent, then what will be the ratio of molarity of the two solutions ?
Doubt : 2
In the following reaction : aNO3−+bCl−+cH+→dNO+eCl2+fH2O find the value of d×fa2(b+c)
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Simply balance the redox equation and substitute the stoichiometric coefficients of the balanced reaction into your expression.( I would have posted a solution if only my device could type LaTeX.)
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Ok. Try now after keeping it in \ (....) without spaces which will give you ...
Did you tried it. What is the answer you got.
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Yeah, I solved it. My answer is 7.
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That's correct answer. Can you atleast keep the values of a,b,cd,e,f.
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a=2, b=6, c=8, d=2, e=3 and f=4
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Ok. Thanks i will check with mine.
Doubt : 3
Uncertainty in position and momentum are equal. Then what will be the Uncertainty in velocity ?
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I don't think uncertainty in position and momentum are relativistically equal. However, if a non-relativistic assumption is to be made, then from Heisenberg's, and taking p=mv, then uncertainty in velocity>=(h/2pi)X(1/(uncertainty in position X mass)). Sorry for the math ambiguity. My device doesn't type LaTeX.
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Yes you are correct. You got the correct answer. Regarding the latex you can keep it in \ (...\ ).
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The LaTeX is not still working. It prevents me from posting some reeeeeeallly cool problems.;(
Doubt : 4
In (Si4O12)8−, how many Si−O−Si linkages are there ?
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I have an answer, but I am not totally sure it is correct. Coordination structure chemistry of silicates and silicones is my weakest point in chemistry. Perhaps I need to practise more!:)
What is the answer bro?:)
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It is 4.
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I was actually sure of 4 linkages, but I was not sure of one more. Thanks anyways.
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It's OK.
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You are a great guy, Ram.
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That's my quality quantitatively speaking.
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My goal in many interactions is to make them frictionless. Trivially speaking, I resist every enmity and conduct friendliness. I have even successfully transmuted people from an angry ground state, to a happy excited state!
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Oh !!! That's good.
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Thanks. Ask if you want more!
Thanks guys. Keep sending problems and lets discuss!
Hi guys! For sure we all have noticed that the derivative of kinetic energy is momentum. Is there a physical explanation for this?
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Quantitatively speaking, momentum is the product of mass of a body and its velocity. It is denoted by P. P=mv But the above definition is valid only when velocity of the body is constant. If the velocity of the body is variable we cannot use this formula and we should go by calculus. Physically speaking, momentum is the rate of change of kinetic energy with respect to its velocity. So, you ca say it as kinetic energy at that instance. Here is how we get P=mv from kinetic energy. K.E=21m(v)2 Now, differentiate both side with respect to velocity. We get : dxdK.E=dxd(21m(v)2) As, we know dxdK.E=P, we get : P=21⋅m⋅2v So, we get : P=mv
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Perfect bro!
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Thanks. But what I should say you, brother or sister
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I am a young boy:)
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Ok
Explain the Bohr theory of atoms, combining both mathematics and physics in your explanations.
Does anyone know how to post a solution of a problem that doesn't have a posted solution yet?
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If you got it correct you click the button "No solution" under Continue and write. And if you got it wrong you can't.
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Thanks a lot, bro!
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You are welcome!
@Mohmmad Farhan why can't you come here. I think this is a best place for a student like you. You can post your quires here.
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Sorry! I am not really active these days because of exams
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Ok. No problem !!!
How do I post numerical answer problems on Brilliant?
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I think you have posted many problems in brilliant. However, when you are writng a question there will be two options, either Number or Multiple choice. You can select whatever you wish.
Gearing up for today's 1st round olympiads!
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Good luck!
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Thanks. The results will be uploaded online in about 1.5 hours. Meanwhile I am posting them here for our consumption. Solve some, post solutions and invite others to do same!
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Maybe you can post them like problems.
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Ok.
Problem 1: A password consists of 4 distinct digits such that their sum is 19 and exactly two of its digits are prime, e.g. 0397. How many possible such passwords exist?
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2,3: (2,3,6,8)
2,5: (2,5,4,8)
2,7: (2,7,4,6),(2,7,1,9)
3,5: x
3,7: (3,7,1,8),(3,7,0,9)
5,7: (5,7,1,6)
There are 7×4!=168 possibilities
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How did you get all possibilities?
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Find the 7 possibilities one by one, then times 24
Problem 2: What is the maximum possible area of a quadrilateral of sides 1,4,7,8?
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I think we should use the concept of maxima and minima.
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Perhaps you should solve it. I used the maximality condition of Bretschneider's formula. My answer was 18.
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Yes I too got like that.
What is that?
Looks similar to:
https://brilliant.org/problems/hexagon-octagon-in-a-rectangle/?ref_id=1538063
Maybe a harder version perhaps! :D
Problem 3: 20 lines are drawn in a plane. What is the maximum number of parts into which the plane can be divided by the 20 straight lines?
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I think it is 211
There is a trick. First, take the 20th triangular number. 220(20+1)=210 Add 1. 210+1=211 These numbers are known as pizza numbers and are always one more than a triangular number.
P.S. @EKENE FRANKLIN did you use this method?
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Great! Just what I wanted, and what I used during the examination!
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Thumbs Up!
What is the remainder when 1241+2241+3241+4241 is divided by 5?
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0
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Of course that is the answer, but prove it.
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Since the divisibility rule of 5 only concerns the last digit, we have to find the last digits of the values. 1124 is obviously 1. Since the repetition in the last digit is (for 2): 2, 4, 8, 6 then 241÷4=k(Remainder1) the first number is 2 (in this set). Apply this to the other powers. Summing the last digits, we get 10 and 10÷5=k(Remainder0)
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Excellent
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Great job. I did modular arithmetic during the examination.
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Me too
Problem 5: Find the general term of the sequence described by an=4an−1−3an−2 given that a1=1 and a2=9.
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an=4×3n−1−3
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Great. This was what I got. But do you mind showing your solution so I can clarify mine?
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Solve the equation x2=4x−3 and get x=3 or 1. Let bn=3n−1n.
The first few terms of bn is 2,8,26,80... and bn also satisfies the recursion bn=4bn−1−3bn−2
Compare the two sequences: an=34bn−35=4×3n−1−3
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Not exactly what I did, but something similar. Thanks a lot. If I may ask, have you ever participated in an olympiad before, or do you hope to participate in one?
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No, I haven't. I knew few things about it, so I'm not sure about participating in one.
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Ok. Thanks. By the way, I really admire your problems. They are very cool. One of my major problems here is creating problems. My original problems are too easy(you can check them out). But solving others problems(of course not those extremely hard ones) is not a big problem for me.
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@X X I think this is your first problem of the week in easy catogery.
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You're right! (Ha, ha! I'm a little jealous of you because your problems are often the first problem in the easy category, and in your contributions page, the number of people are banging up:p)
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But I am not fully done. I expected that Number + Reciprocal will take me to 1 lakh (100 K) solvers. But it fell short of 7 K. But luckily this week too my question came in one of the problems of week (the first one) so it is confirmed that I will reach the milestone in one or two days.
But some of your problems came in Intermediate and Advanced catogery.
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I know. (Fewer people solves those, that is why I was so happy to see my problem as the second of the basic category)
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OH !!! Your problem is featured as the second one in the problems of the week. But my first problem (Thundering Question) is the third one.
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Oh, I remember that one! (I loved that week because a lot of my solutions appeared in that week. )
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Yes, Vamprie Fan Club. It is were I had my first individual century. After that, the next week I had got 150 upvotes for my question Rising Balloons and the next week is a memorable for me, as I had my first double century for the question Counting 0's.
Ram Mohith and X X, both of you are excellent problem contributors, some of the best here. I hope I can reach your level!
It's a Fact :
I have made a hatrick in June 9 - June 29 (3 weeks) as 3 of my problem continuously came in easy part of the POTW. Now again the chance has come for me. Last week and this week too two of my problems are featured. So, if luck favors next week to I can have a problem in POTW. I am thinking very hard on which question should I post the next.
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The sequel to the Mt Everest problem about pressure cookers?
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Should see about it.
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Is it released yet?
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No. Still under construction. Before that one I have two more questions, Raising Balloons - 2 and Popping balloons which are yet to be released. After these two the sequel for My. Everest vs Mt K 2 will be released.
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OH!
How is my new problem, check it here. It is posed by B.D
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Cool video!
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Yes. I liked that so much that I framed the question based on that.
Don't worry mister !!! You have the skill to post and solve tough questions (level 4 and 5).
I made the cut for the 2nd round! I scored 76.70 out of a maximum possible 110. Hurray!
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Oh !!! Congrats.
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Thanks a lot. But no one apart from us is solving the problems. I want to check my solutions and my faulty areas. Please direct the problem solving masters of Brilliant here to help me out.
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If you want to direct anyone to this place just keep the symbol @ and type the name of the person. Immediately a notification will be sent to that particular user.
Hi guys !!! I want to tell you some secret. I think next week would be a week of physics. Because this week I have found six problems which have equal chances in coming to problems of the week. There are Rohit Gupta's Trick Spring Balance and Spring Balance, Pranshu Gaba's Three blocks Sliding and Relative Stuntmen Motion, Danielle Scarano's Thermographic Dog and my Mt. Everest vs Mt. Godwin Austin. I think these question will be featured definitely this week if not possible then next week.
@Calvin Lin
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I think you there is some error in spelling or you have kept a gap between the symbol and name. Once check again. If it is correct then you should get it in blue color link and not just normal text.
For example you should type it as @ _ Ekene Franklin ( _ symbol means no gap) then you will get it as @EKENE FRANKLIN
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It is not still working. Do I need to add something like a code or anything, or should I just type it?
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You should select from the list which will appear when you type the name. Also the name should be exactly matching with that particular user even capital letters should be taken into consideration.
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Which list?
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Ok. If you want I will do it on behalf of you.
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Of course I want! Thanks!
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Tell the names of whom do you want to be called to this note.
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@Calvin Lin, @Harsh Shrivastava, @Chew Seong Cheong, @Michael Mendrin, @X X, @Aniket Sanghi.
@Calvin Lin
@Calvin Lin.
@X X, @Michael Mendrin, @Chew-Seong Cheong, @Mohmmad Farhan,and the rest of all, Can you please participate in this message-board (as requested by Ekene Franklin)
What courses would you recommend for a 4th grader like me?
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You are 4 the grade ????!!!!!
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10 years old. I told you
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But, I didn't think you are 4 th grade.
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I am! REALLY
I am going to be 5th grade
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Oh !!! Good. All the best
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Actually I am 1 year away from board exam (in 2020)
My first trophy was in Grade 1(third)[english], Second in grade 2 (1st)[first scrabble competition] and third in grade 2(Best school player award)[first scrabble competiton]
BrilliantForLife
Dare you to toggle LATEXI have a proof for the existence of infinite dimensions.
On a line (1 Dimension), a number can be represented by a point. On a grid (2 Dimensions), an intersection is represented for the data in a cell. In a cube (3 Dimensions), data can still be represented on a point (where straight lines intersect). Following this pattern, there are infinite dimensions (because there are ℵ0 numbers).
Problem 6: Find all primes which are the sum of two primes and difference of two primes.
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5=7-2=3+2,this is the only possibility.
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Exactly. Just what I did during the exam, using the logic that if a prime satisfies those conditions, then the only even prime, 2 must be in the sum and difference.
Problem 7: Simplify S=(cot10∘−33)(cosec20∘+2cot20∘)
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Why are you keeping square brackets. For square root you just need to keep flower brackets.
For trigonometric functions, logarithms, limts, etc you must keep a slash before them. Like \cot will be as cot and \log will be as log.
Problem 8: A unit square is completely covered by 3 identical circles. Find the smallest possible diameter of the circles.
Problem 9: How many 6 digit numbers can be formed using only odd digits?
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Is it 5^6?
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Correct. Nice stuff. What did you do?
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There are 1,3,5,7,9 five odd digits. There are six digits, each digit has 5 possibilities, so there are 5^6 possibilities
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Bro, u are amazing
DO YOU SERIOUSLY HAVE TO GIVE ME 46 NOTIFICATIONS OVERNIGHT!
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Is that bad, Mohmmad? I am usually on Brilliant throughout the night.
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I HAVE EXAMS
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I see. But you can use Brilliant to prepare.
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English EXAMS
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Farhat, from which part of India are were from ?
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Andhra Pradesh, From Khammam
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I am also from Andhra Pradesh. From Visakhapatnam.
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I am Telugu. You are Telugu right because you are from Andhra Pradesh right?
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Yes. I am born in Visakhapatnam.
Don't call me Mohmmad
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Sorry, but why?
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See, My brother's name is Farhan but I am Farhat. And Mohmmad is his surname and Mohammad is my surname. (Stupid Passport errors)
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Mohammad Farhat
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YES! FINALLY!
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You can change your name in the account settings
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I don't bother
Did you know that :
1+ω=ω but ω+1=ω
Freakin' cool right?
P.S. ω is the second lowest level of ∞ after ℵ0
Finally I reached it. I have reached 1 lakh (100 k) solvers in 91 problems.
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Congratulations!
Do you know the minimum people to reach?
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I didn't get your point !!! Are you asking about minimum people required to get contribution page.
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Yes he is.
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I think you must minimum have 5K people reached. But the main parameter I think is that the contribution page is mostly obtained either by up votes or by solvers to the question (more than 1K). When I got my contribution page I have around 120 upvotes, around 300 solvers, around 30 views for my notes.
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OK. Thanks!
@X X, I want your help. I am trying to write a new problem but I got struck up at one point. Here is my doubt :
Suppose lets say a man jumps from a building with initial velocity u. After x seconds another man freely falls from the same building and from the same height. Now, we can say that the displacement of both the man's are equal as they both reach the ground simultaneously after they jump from same point, so we can equate their distances. Here is what I did : ut+21gt2=21g(t−x)2 where t is the time taken by the first person to reach the ground. The main part of my doubt is that whether we should write the time taken by second person as (t−x) or (t+x).
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I'm not good at physics, though. I think the time taken by the second person should be as same as the first one, maybe?
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What I think is that as the second person jumps after x seconds his time period will be (t−x) seconds. Let's ask it to staff.
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So, are you caculating " when the first man fell to the ground, what is the distance between the two man"?
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No. Both man reach the ground simultaneously so can we write the equation as in my before comment in which we equate displacements.
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Oh! Sorry, I misread the problem.
Yes, I too think the time taken by the second person is t−x
It should be t+x as he is free falling and the previous one jumped with velocity u
DO or DIE
You are lost in a forest on a cold midnight day. The temperature is quite low and it is bitter cold. Now you have only two chances :
You should lie on the ground and should shrink your body like a tight pack.
Go and stay in the water in a pond nearby.
There not much time left. Take your decision before cold kills you. So, what wold you choose ?
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First option.
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But the second one is the correct choice you have to make.
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Why? Because of anomaly in expansion and contraction of water.?
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And also due to the high specific heat capacity of water. This question is a slight modification of air currents (due to which it is cool during the day and warm during night near coastal areas) and I am going to post this very soon.
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I thought as much.
Hello to everyone! I have not been around for some time now on Brilliant. However, I met on my return a delightful change in the method of posting options for multiple choice questions on Brilliant. Kudos to Calvin Lin and the truly brilliant Brilliant team!
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Which questions would you handpick for people? (ie. Your favourite questions)
@Calvin Lin Sir, I am very much happy. But I hope multiple correct answer format will also be soon available for the community.
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I think and hope this change is a step up to the multiple correct answer releasing.
Actually, I sort of did that. So, I made a correct answer then removed it (crossing it with the x) and then the two dummy answers above it, became the correct answers.
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But after that it says only 1 correct answer must give.
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OH!
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See my set: LATEX Help
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Good one.
Hello guys! Which of my recent problems did you like the most?
To be honest, this note has been quiet for a while
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Yup
Hi guys. Why can't we together contribute to some wiki's. At present I am working on Heat Transfer, Types of Matrices, JEE Complex Numbers, Midpoint of a Line Segment and Section Formula. I need some help to build them so why can't you join me ad show brilliant the power of young students. If you have any such wiki's to build please share them to and I will try my best to contribute them. Thank You !!!
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You keep on changing the topic of your problems and they keep on resurging and I have to say: It is quite annoying
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I didn't get your point. Are you asking anything about my set.
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No. It is about the 'new' community problems
OH. You are saying about me reposting the problems again and again. What to do ? It's my natural habit that I want all of them to see and solve my problems. Man is made up of desires and everybody have their own desires. You cannot stop them and they go on flowing continuously.
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What will you then say of those who have solved it before?
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What they can do ? They will just leave it off. If all are having problem of me reposting problems then I will stop it. Don't make this as a big issue.
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Blan Morrison agrees with me and Ekene Franklin
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He says that this not a big issue.
I will do what I can.
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Thank you.
Hello guys. I just completed the second stage of my country's olympiads. Much more difficult.
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Good for you.
Um, Ram, I think I need some clarification on your problem 'Competition in Leaf'. Transpiration occurs on the upper part, right? So cobalt (II) chloride should come in contact with more water.
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No. Stomata is present on the lower surface of leaf and hence more transpiration will occur from lower surface. Search in internet if you want.
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But this is the case in dicots only. In monocots, it is on both surfaces
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Exactly, Manuj.
Ok. I will edit the question accordingly. Thank you.
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You are welcome.
Hello, I have been off from Brilliant for a month as my sports championships were coming close.
And I have term tests starting in a fortnight so I have to pack my self up. Similarly, my IGCSE ICT exam is coming closer.
However, I have a doubt in one of the questions of IGCSE Further Pure Mathematics... Can you help?
Q) a) Show that a3−(a−1)3≡3a2−3a+1⟹ I know how to do this. But the next one uses this, so I have given here.
b) Hence show that
r=1∑nr3−(r−1)3=3r=1∑nr2−21(3n2+n)
c) Show that
r=1∑nr3−(r−1)3=n3
d) Hence deduce that
r=1∑nr2=61n(n+1)(2n+1)
Please help me from (b) onwards.
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Isn't last one n(n+1)(2n+1)/6
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Yeah he probably forgot...
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Yes; my bad
(b)
r3−(r−1)3=3r2−3r+1
r=1∑n(r3−(r−1)3)=r=1∑n(3r2−3r+1)
r=1∑n(r3−(r−1)3)=r=1∑n3r2−3r=1∑nr+r=1∑n1
r=1∑n(r3−(r−1)3)=r=1∑n3r2−3×2n2+n+n
r=1∑n(r3−(r−1)3)=3r=1∑nr2−23n2+n
(c)
r=1∑n(r3−(r−1)3)
=r=1∑nr3−r=1∑r=n(r−1)3
=r=1∑nr3−r−1=0∑r−1=n−1(r−1)3
=r=1∑nr3−r=0∑n−1r3 (Change r−1 to r)
=(r=1∑n−1r3+n3)−(03+r=1∑n−1r3)
=n3
(d)
According to (b) and (c),
r=1∑n(r3−(r−1)3)=3r=1∑nr2−23n2+n
n3=3r=1∑nr2−23n2+n
n3+23n2+n=3r=1∑nr2
22n3+3n2+n=3r=1∑nr2
62n3+3n2+n=r=1∑nr2=6n(n+1)(2n+1)
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In this how does this happen, isn't it supposed to be 1?
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It means that you are adding 1 n times so it will be like, (1 + 1 + 1 + .... n~times \ \implies n (1) = n]
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Oh okay thx a lot
Similarly, I don't understand the blue, red, green and yellow highlighted stuff
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The first red box is just like we write r=1 down, we write r= n up (blue) Second red box, down as r =1,r-1=0 and up(blue) , r=n therefore r-1=n-1 Yellow box, वे take the upper limit as n-1 and add n^3 in first and add 0^3 to make the lower limit as 1 in second Green : still thinking
You can do it as : r^3 - (r-1)^3 = (r-r+1)(r^2+(r-1)^2+r(r-1)). (using a^3-b^3 identity)
So it will become §(3r^2 - 3r - 1) = 3§r^2 - 3§r - § 1
= 3n(n+1)(2n+1)/6- 3n(n+1)/2 - n
=n^3
Sorry i didn't have submission sign so I used §
Another question : Find x
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What is the angle of projection?
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Not mentioned.. :( Just plain theta
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Then you will get x also in theta.
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Yeah that's true... Hey @Ram Mohith Give your ans
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I have already mentioned it. It is 45(sec2θ−tanθ)
I got the answer as 45(sec2θ−tanθ)
Here is my step wise solution :
Lets first define our variables in x-direction and y-direction respectively. Also, lets assume that the body takes time t to complete the projectile.
x - direction(→+)ux=15cosθ m/sax=0Sx=45 my - direction(↓+)uy=−15sinθ m/say=gSy=x m
As, S=ut+21at2 we will apply this in x and y direction separately. Sx⟹tSyx⟹=uxt+21(0)t2=15cosθ⋅t=15cosθ45=cosθ3=uyt+21gt2=(−15cosθ)t+21×10×t2=−15sinθ×cosθ3+5×cos2θ9=−45tanθ+45sec2θx=45(sec2θ−tanθ)
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@Syed Hamza Khalid is my answer correct or not.
@X X is your name really XX ?
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How did you get that 5592707:X X?
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I too don't know how I got it. I just types your name "X X".
Not really.
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Then what is it ? Just a convention or your nick name ?
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When I first entered Brilliant, I just typed a short name randomly, and this becomes XX.
After I started to become active on Brilliant, I think probably a short name is easier for people to memorize, so I didn't change it.
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Ok. But what is your actual name.
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Just leave it a mystery...
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Ok
Data protection, right?
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Yeah, you're right.
The comments in this note are now more. This has started to affect this note as whenever we post a comment the comment is being posted slowly. Also, when we open any notification from this note it is now opening slowly. I think it is better to delete some comments. What do you feel ?
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Yeah, you are right. I will create a new messageboard.
Hello
I accidentally dropped the milk box (from paper) and it fell on cap even when I dropped with the bottom to the ground.
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Wow, that's funny. What did you mean?:)
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Because I dropped it with the bottom to the ground I was wondering why it fell on cap.
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Are you inviting physics?
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Yes it is physics.
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Can you give a shot at explaining it?
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No. That's why I ask here.
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The air in the box initiates a couple, whose direction depends on the side that has greater aerial density. Thus the block may either fall with its top or its bottom.:)
You can play a game in brilliant called Word problem Here
Brilliant DATE OF BIRTH?
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B D : Brilliant Date :p
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X X said it
2004?
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I think it will be 10 years
Pls see my problem 'the blue one' and tell how I can make it look better.
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At the end of the question where you have kept Bonus there is a latex mistake I think. If you want it to be in bold letters you should keep in in between two * such that there is no space between * and the next letter or the previous one.
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How can I draw a figure for my question (on website in phone)?
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We cannot draw a figure. We can only keep pictures or gif's.
But if you want you can draw the figures in Microsoft, Paint or any other drawing apps and keep that picture.
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kk
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Any mobile app you know??
Hi guys !!! How is my note Square Root Of a Complex Number. It is a quite interesting and good technique to find square root of a complex number.
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Why can't I open it?
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I have now kept the correct link. I think the staff has changed the link address. You can check now it will open.
I want your help. I am trying to write a new problem but I got struck up at one point. Here is my doubt :
Suppose lets say a man jumps from a building with initial velocity u. After x seconds another man freely falls from the same building and from the same height. Now, we can say that the displacement of both the man's are equal as they both reach the ground simultaneously after they jump from same point, so we can equate their distances. Here is what I did : ut+21gt2=21g(t−x)2 where t is the time taken by the first person to reach the ground. The main part of my doubt is that whether we should write the time taken by second person as (t−x) or (t+x).
What do you think @Blake Farrow , @Calvin Lin , @Danielle Scarano , @Brilliant Physics , @Pranshu Gaba, @Rohit Gupta and others ??
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It should be 't+a' as 1st man had jumped with velocity u so he will reach earlier than the freely falling man and it isn't necessary that the extra time 2nd man takes is equal to 'x'
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Doesn't really matter, as long as you're clear on what the definition of "positive x" is. IE If you used t+x and then solved for x and got that it is negative, what is the interpretation?
In a similar manner, if we say that gravity acts downwards in the positive direction, then the "height at which the ball is thrown upwards" will (likely) be negative. Or vice versa.
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Yes. You are correct. I understood it now. Thanks for clarifying.
Anyone knows some app for drawing geometry figures??
Pls see my problem 'The Strange Angle' and also how to use latex
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Here is a note how to use LaTeX.
The multiple choice of a problem brilliant doesn't order them any more you chose on which place is the correct answer.
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We know that
How do you write about in your profile?
Did you notice that brilliant is checking your browser before ascending into some pages of brilliant?
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Yeah. Good for us.
As this note is getting more and more comments it takes a longer time to load it.
Why does nobody in B D's note for the word game can't solve this: It starts with "B" and its a type of an alphabet. I only accept answers here.
What do you think happens if you solve every problem from today.
Do you know that brilliant staff has been removed the leader board from all the sets?
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Really.
I really like this note that is why I saved it!