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Re: Nairaland Mathematics Clinic by kidrossy(m): 10:56am On Aug 25, 2015 |
pls anyone
5(x-1)+3(2x+9)-2=4(3x-1)+2(4x+3)...@sap |
Re: Nairaland Mathematics Clinic by dejt4u(m): 11:14am On Aug 25, 2015 |
kidrossy:are u sure about this question?? Check it again pls 'cos this is JSS1 maths question |
Re: Nairaland Mathematics Clinic by kidrossy(m): 11:46am On Aug 25, 2015 |
dejt4u:I don't think the options are correct...that is why I brought it here... |
Re: Nairaland Mathematics Clinic by maverickboy: 12:33pm On Aug 25, 2015 |
2! |
Re: Nairaland Mathematics Clinic by Nobody: 12:58pm On Aug 25, 2015 |
maverickboy:2! = 2* 1 =2 |
Re: Nairaland Mathematics Clinic by Nobody: 1:00pm On Aug 25, 2015 |
kidrossy:5 (x-1)+3 (2 x+9)-2 = 4 (3 x-1)+2 (4 x+3) Solve for x: -2+5 (x-1)+3 (2 x+9) = 4 (3 x-1)+2 (4 x+3) 5 (x-1) = 5 x-5: -2+5 x-5+3 (2 x+9) = 4 (3 x-1)+2 (4 x+3) 3 (2 x+9) = 6 x+27: 6 x+27+5 x-5-2 = 4 (3 x-1)+2 (4 x+3) Grouping like terms, 6 x+5 x-5-2+27 = (5 x+6 x)+(-2-5+27): (5 x+6 x)+(-2-5+27) = 4 (3 x-1)+2 (4 x+3) 5 x+6 x = 11 x: 11 x+(-2-5+27) = 4 (3 x-1)+2 (4 x+3) -2-5+27 = 20: 11 x+20 = 4 (3 x-1)+2 (4 x+3) 4 (3 x-1) = 12 x-4: 11 x+20 = 12 x-4+2 (4 x+3) 2 (4 x+3) = 8 x+6: 11 x+20 = 8 x+6+12 x-4 Grouping like terms, 12 x+8 x-4+6 = (12 x+8 x)+(-4+6): 11 x+20 = (12 x+8 x)+(-4+6) 12 x+8 x = 20 x: 11 x+20 = 20 x+(-4+6) 6-4 = 2: 11 x+20 = 20 x+2 Subtract 20 x from both sides: (11 x-20 x)+20 = (20 x-20 x)+2 11 x-20 x = -9 x: -9 x+20 = (20 x-20 x)+2 20 x-20 x = 0: 20-9 x = 2 Subtract 20 from both sides: (20-20)-9 x = 2-20 20-20 = 0: -9 x = 2-20 2-20 = -18: -9 x = -18 Divide both sides of -9 x = -18 by -9: (-9 x)/(-9) = (-18)/(-9) (-9)/(-9) = 1: x = (-18)/(-9) The gcd of -18 and -9 is -9, so (-18)/(-9) = (-9×2)/(-9×1) = (-9)/(-9)×2 = 2: Answer: | | x = 2 |
Re: Nairaland Mathematics Clinic by dejt4u(m): 1:23pm On Aug 25, 2015 |
Karmanaut:modified... My answer to question is x = 2 |
Re: Nairaland Mathematics Clinic by maverickboy: 1:59pm On Aug 25, 2015 |
Karmanaut:just remember I shouldn't have used the sign (!). the answer is 2 and not 2! (2 factorial) 1 Like |
Re: Nairaland Mathematics Clinic by naturalwaves: 2:14pm On Aug 25, 2015 |
kidrossy:X equals 2 not 3/2 |
Re: Nairaland Mathematics Clinic by personal59: 6:08pm On Aug 25, 2015 |
Please help All workings are needed (step to step workings with full details) on how it is solve Thanks in anticipation 1) Using 4 sub intervals in the composite trapezium rule approximate intergral 5-1 I.e integral of 5 upper limit and 1 lower limit (square root Xdx) 2) The function f is known to have a second derivative with the property that /f"(x)/ < 12 For x between 2 and 3. Using the error bound giving earlier in this section determine how many sub intervals are required so that the composite trapezieum rule used to approximate integral 5-2 f(x)dx 3) Using 4 sub intervals in the composite simpson rule approximate. Integral 5-1 square root xdx 4) The function f is known to have a forth derivative with the property that f^(iv) (x)<6 For x between -1 and 5. Determine how many sub intervals are required so that the composite trapezium rule use to approximate integral 5 - (-2) f(x)dx incurs an error that is less than 0.001.
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Re: Nairaland Mathematics Clinic by agentofchange1(m): 6:16pm On Aug 25, 2015 |
worthing:I doubt if an analytical solution to this exist. 1 Like |
Re: Nairaland Mathematics Clinic by Seamione(m): 6:46pm On Aug 25, 2015 |
Good evening guys, Please take a look at this question. Three digit numbers re formed from the digits 1,2,4,5,6. If repetition is not allowed and a number is picked at random, find the probability that it's a multiple of 5 or an odd number. From my calculations, I got 2/5 but the text says it's 3/5. Who is wrong? |
Re: Nairaland Mathematics Clinic by personal59: 8:46pm On Aug 25, 2015 |
Karmanaut: personal59: agentofchange1: personal59: Pls help me with my question pls The details is here and also before this post |
Re: Nairaland Mathematics Clinic by Nobody: 10:18pm On Aug 25, 2015 |
personal59:Solution is attached.
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Re: Nairaland Mathematics Clinic by Sirneij(m): 1:13am On Aug 26, 2015 |
No one has yet attempted my question? |
Re: Nairaland Mathematics Clinic by personal59: 2:05am On Aug 26, 2015 |
Karmanaut: Thanks brother but the question u solve first I can't see some of the f() very well nd u solve most with it I don't know if u can help m rewrite it clearly And also I need answer to other questions sir thanks |
Re: Nairaland Mathematics Clinic by jackpot(f): 4:56am On Aug 26, 2015 |
Show that the curve x2+xy+y2-2=0 is an ellipse. Hence, obtain it's (i) centre (ii) vertices (iii) co-vertices (iv) foci (v) axes of symmetry (vi) semi-major axis length tags: Karmanaut, Laplacian, dejt4u, jaryeh, agentofchange1, AlphaMaximus, doubleDx, etc |
Re: Nairaland Mathematics Clinic by Nobody: 5:11am On Aug 26, 2015 |
personal59:Noted. I also made an error concerning the width of subintervals I used; I didn't read the question well. Will correct it and do the rest at my leisure. 1 Like |
Re: Nairaland Mathematics Clinic by Nobody: 5:38am On Aug 26, 2015 |
jackpot: Solution: Foci: (2/√3, -2/√3) Vertices: (√2, -√2) Center: (0,0) Semi major axis length: 2 Semi minor axis length: 2/√3 |
Re: Nairaland Mathematics Clinic by Nobody: 10:06am On Aug 26, 2015 |
@Personal59, the solution for question 1.
I trued to make my writing bigger. 1 Like
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Re: Nairaland Mathematics Clinic by personal59: 10:28am On Aug 26, 2015 |
Karmanaut:ok thanks bro Its very clear and straightforward Checking regularly too am very grateful opful for the other answers too bro |
Re: Nairaland Mathematics Clinic by Nobody: 10:51am On Aug 26, 2015 |
personal59:Cheers. Doing question 3 now. I'll research 2 & 4 later. Edit: Solution to Question 3 attached. 1 Like
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Re: Nairaland Mathematics Clinic by personal59: 1:42pm On Aug 26, 2015 |
Karmanaut: Thanks bro am very very grateful |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 2:36pm On Aug 26, 2015 |
Seamione: total ways of selection = 5C3 = 10ways sample space S={(124),(125),(126),(245),(246),(145),(156),(645),(641),(265)} check where we have an odd number or multiple of 5 thus, we get 6/10=3/5 hope you get? |
Re: Nairaland Mathematics Clinic by personal59: 4:28pm On Aug 26, 2015 |
Karmanaut: Thanks bro am very very grateful 1 Like |
Re: Nairaland Mathematics Clinic by Seamione(m): 7:12pm On Aug 26, 2015 |
agentofchange1:No, I kinda disagree with your sample space, it can't just be 5C3. For instance, where Is 421, 521 and 621 in your sample space? At least, they're also of 3 digits which are odd numbers.. |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 7:31pm On Aug 26, 2015 |
Seamione: c'mmon bro, the question says no repetition 5C3 means total number of ways we could form the 3-digit number without repetition 421 (124) ,521(125) , 621( 126) already exist in the sample space. 1 Like |
Re: Nairaland Mathematics Clinic by jackpot(f): 7:39pm On Aug 26, 2015 |
Karmanaut:please, kindly show full workings. |
Re: Nairaland Mathematics Clinic by Seamione(m): 7:50pm On Aug 26, 2015 |
agentofchange1:Oga General Ben, I think the "no repetition" in the question means we can't have numbers like 111, 222, 444, 555 and 666. Moreover, 421 and 124 are not exactly the same number, they're just numbers with the same digits so we can't consider that as repetition because if that's the case, I don't think we should also have 124 and 125 because 1 and 2 have been repeated. I hope you understand what I'm trying to say 1 Like |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 8:15pm On Aug 26, 2015 |
Seamione: hmmm, i won't say much bro let others shed more light. |
Re: Nairaland Mathematics Clinic by Seamione(m): 8:18pm On Aug 26, 2015 |
agentofchange1:Exactly, waiting for others to come in on this |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 8:26pm On Aug 26, 2015 |
Seamione: u definitely know little about probability.. hmmm. its ok then.. |
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