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#1026 Re: Exercises » kind of the triangle » 2016-04-22 02:47:31
ans.8 Angle B is a right angle.
#1027 Re: Exercises » kind of the triangle » 2016-04-22 02:34:20
Ans9. one of the angles is a right angle.
#1028 Re: Help Me ! » RLC parallel circuit, complex numbers, done it but not sure » 2016-04-21 22:23:00
The problem seems to be with C which in nF is almost a short circuit. Check whether it is
#1029 Re: Exercises » Bitter triangles. » 2016-04-21 21:54:35
How to hide the text?
#1030 Re: This is Cool » Fun with 0 / 0 » 2016-04-21 19:40:12
If somebody starts with zero asset and go on multiplying his assets he will never become rich. for that he has to add assets. Of course once he adds then he can go on multiplying. 
#1031 Re: Help Me ! » Geometry question » 2016-04-21 17:51:14
Usually we start stating the triangle name with the vertex at the top second and third showing the base. So unless and otherwise stated the height means the perpendicular from first vertex.
#1032 Re: Maths Teaching Resources » Regular Icosahedron » 2016-04-21 04:55:05
Very good calculation.I was not sure about the orientation of various triangles and doubted the analysis but on close scrutiny no hole could be found.Calculation could be shortened by
as OP itself is radius.For curiosity I compared the volume of icosahedron with volume of sphere.The difference looked unrealistic at first sight the cut portion having taken away so much volume.The radius of the sphere which accommodated 20 triangles being less than one side of a triangle also looked odd. But looking around the circumference 2 complete sides of triangle like P'B and heights of 4 triangles i.e about only 6 elements along circumference removed the doubt.
#1033 Re: Exercises » Bitter triangles. » 2016-04-20 19:22:40
#1034 Re: Puzzles and Games » Is this an impossible question? » 2016-04-20 19:03:45
The question is not impossible but the answer is impossible.
#1035 Re: Help Me ! » Double Integrals Using Polar Coordinates » 2016-04-20 18:15:23
If I give the solution now after 4 months will it be relevant ?
#1037 Re: Help Me ! » Piece wise functions » 2016-04-20 16:11:31
It is a piecewise function.
equation can be written in the form
rent=2*int (t+0.9999) where int() is the function "integer part of ".
I wanted to make it int(t+1) but at time cross over it gives rise to problem e.g. int(t+1) gives value of 3 when t=2 exactly in which case charge will be for 3 hours but int(t+0.9999) will give the correct value of 2.
#1038 Re: Help Me ! » Proof by induction » 2016-04-20 04:52:46
I shall provide the proof only if it is required as so much time has passed since it was posted.
#1039 Re: Help Me ! » Geometry question » 2016-04-20 02:58:06
hi math9maniac
The key to many geometry questions is to get the right diagram.
I drew PS and QR parallel to it. Then SR and PT parallel to it. PXRS is a parallelogram where X lies on QR. As SR = PT one interpretation of the information is that T = X. The alternative is that T is on the other side of XP. I have shown this possibility as T'
But the usual convention when describing parallels is to preserve the directional sense which makes T where I have put it. Also later the question talks about PTRS. Again the convention is to take the points in the given order which confirms this interpretation. If T is at T' then the shape PTRS crosses over itself and finding its area would be silly.
http://i.imgur.com/Be2T1zI.gif
The dotted line shows you how to get the ratio of areas very simply. You can get the angle using the cosine rule and the height with simple trig.
Bob
#1040 Re: Help Me ! » Need help with the following grade 9 papers. » 2016-04-19 22:30:53
Most of the problems are simple. Do it yourself or else be specific ; which problem?
#1041 Re: Help Me ! » Hard math problem - ratios. » 2016-04-19 22:26:43
I opened the pdf of the paper shown elsewhere. the number of tourists increased by 220.
take 7x and x as the numbers of natives &tourists in 1986. Then
Solve it x=125 is the answer.
#1042 Re: Help Me ! » Hard math problem - ratios. » 2016-04-19 21:39:44
There is something wrong with the statement of the problem. Otherwise the problem is simple.
#1043 Re: Help Me ! » 2 trig problems » 2016-04-19 20:57:43
#1044 Re: Help Me ! » Algebra » 2016-04-19 15:39:34
The product of all roots is the constant in the expression. 6 can be split into 6x1x1x1 when you have 3 roots of 1.
3x2x1x1 gives two double roots =1 and one root of 2 and one root of 3. There is no possibility of 3 having double roots as the product comes to 9 at least.
#1045 Re: Help Me ! » I "know" that I am wrong, but am looking for the fault in my reasoning » 2016-04-19 05:18:19
If you were to equate the volumes you could do it easier by equating the volume above c.g equal to half of the total volume
However te method itself is wrong. the center of gravity is the point where the moment of total mass equals the sum of the moments of individual masses. In other words}
The concept of c.g. can better be understood if we take the axis of cone horizontally and hang it by a thread at c.g. it is obvious that to keep it in balance the moment of mass/weight of the right hand side(shorter cone) about the thread must be balanced by the moment of the weight/mass of the left hand side which is a frustrum.
#1046 Re: Help Me ! » about statistics » 2016-04-18 22:55:27
To corroborate the valid number of cards (which I showed as 44) I made a chart for invalid draw of cards.
A B C D E people at party
A B C D E (all 5 getting invalid cards)
3 people getting invalid cards (total 10 draws )
A B C E D
A B E D C
A B D C E
A E C D B
A D C B E
A C B D E
E B C D A
D B C A E
C B A D E
B A C D E
2 persons getting invalid cards (total 20 draws)
A B D E C
A B E C D
A D C E B
A E C B D
A C E D B
A E B D C
A C D B E
A D B C E
D B C E A
E B C A D
C B E D A
E B A D C
C B D A E
D B A C E
B E C D A
E A C D B
B D C A E
D A C B E
B C A D E
C A B D E
1 person getting invalid card ( chart is incomplete; it shows draw for only A to get invalid card;
Number here is 9 but altogether it is 45)
A C B E D
A C D E B
A C E B D
A D B E C
A D E B C
A D E C B
A E B C D
A E D B C
A E D C B
In all 76 card draws are invalid. However total no. of draws is 5!=120. So valid no. of draws =120-76=44 which was shown in the previous comment.
#1047 Re: Help Me ! » about statistics » 2016-04-18 05:05:53
We need to think of loops.5 people can form a loop of 5 among whom cards are exchanged.In this case no body will be left out when the names are called.Let the 5 people be A B C D &E.In order to find successful permutations start with A. He can have cards of any other 4.It makes 4 different ways.Suppose he gets card of C. for logical clarity now go to C.He should not have his card is already taken into account since A has selected it. In addition A's card is also taboo as it would prematurely close the loop and call for penalty. so he has only 3 choices. Suppose he chooses/gets card of E. e will have to avoid card of A,C and E. he has 2 choices. Say he gets B. Now only B is left over and he has to get the remaining card of A.This closely follows the rules given in the problem. A calls the name of C,C that of E,E that of B and finally B that of A. Even if somebody in the middle say C is picked up to start, it won't matter.So the number of ways of getting the cards comes out to be 4*3*2*1=4!=24 (factorial of one less than the number of people.) In order to find the number of ways which leads to penalty we have to think of smaller loops.
5 people can form 2 loops one of 3 and the other of 2. the number of ways these loops are formed is 5C3x2C2 =10;In each of these we have 2! ways for loop of 3 and 1! ways for the loop of 2.So total number of permutations with loops of 3 and 2 is 10x2!x1!=20.If the announcement of names comes from loop of 3 ,those in loop of 2 are left out and vice versa.There can not be loop of 4 because the 5th person will be having his own card.
So probability of not getting punishment is 24/(24+20)=24/44=6/11.
Extending this to any number of people n the no. of ways to avoid penalty is (n-1)!
however to find the number of people expecting penalty would be difficult as it it is not possible to scan through all loops of smaller size. A different algorithm is to be found out.
#1048 Re: Help Me ! » statisitc probability » 2016-04-17 20:52:13
(b) p1=prob. Eva & Brett meeting in 1st round=1/63
q12=prob. of both going to 2nd round=62/63*1/4 (factor 1/4 is for both to win their games)
p2=prob. Eva & Brett meeting in 2nd round= q12*1/31
q23=prob. of both going to 3rd round=q12*30/31*1/4
p3=prob. Eva & Brett meeting in 3rd round= q23*1/15
q34=prob. of both going to 4th round=q23*14/15*1/4
p4=prob. Eva & Brett meeting in 4th round= q34*1/7
q45=prob. of both going to 5th round=q34*6/7*1/4
p5=prob. Eva & Brett meeting in 5th round= q45*1/3
q56=prob. of both going to 6th round=q45*2/3*1/4
p6=prob. Eva & Brett meeting in 6th round= q56
P=total prob. of Eva & Brett meeting in the tournament=p1+p2+p3+p4+p5+p6
#1049 Re: Help Me ! » statisitc probability » 2016-04-17 20:13:15
For (a) No. of combinations in first round is 64C2 but one draw is 32 pairs.
the no. of ways in which draw is made is 64C2/32=63
In subsequent rounds it will be 31,15,7,3 and 1
#1050 Re: Help Me ! » Point locus, related to angle bisectors » 2016-04-17 17:07:43
Consider A as such point and draw perpendiculars BG and CH to segment MA.Triangles AGB & AHC are similar. therfore
AG/AH =BG/CH. So A divides HG externally in the ratio CH/BG.
LOCUS:: draw a line through M and draw BG & CH perpendiculars. Divide HG externally in the ratio CH/BG to locate point A.