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#1 Re: Introductions » Short intro » 2018-04-14 21:10:52

hi zahlenspieler

Welcome to the forum.

Bob

#2 Re: Help Me ! » a question of a dice in probabilty » 2018-04-14 21:09:49

hi Hannibal lecter

I'm sorry you didn't get what you wanted in time for your exam.  What you have to realise is that MIF is a world wide, and free to use, forum.  While you are studying, the person helping you may be sleeping, or at work, or just digging the garden.

When you know you have an exam approaching you need to allow plenty of time for other members to respond.  I'm surprised that you knew in advance what question would be asked.  This doesn't happen in the UK;  candidates can practise similar questions but the examiners set new questions for the exam itself.  Had I known this I would have been even more reluctant to provide the actual answer since it goes against what I believe is the purpose of the exam in the first place.  If you give an answer having no idea why it is correct but just copying someone else then you give the examiners a false impression of your understanding.

This is why I try to provide general help on the topic rather than just a model answer.  It says in the rules: "This is a Forum, it is not instant chat. Leave your message, and come back later (hours or days) to see what responses you got."  And in the advice when asking for help: "We are happy to help! But we don't do your homework for you."  This could equally have said: "We won't do your exam for you."

Bob

#4 Re: Help Me ! » [ASK] P to AC » 2018-04-09 11:19:25

This is to 'bump' my previous post as I have edited it.

Bob

#5 Re: Help Me ! » [ASK] P to AC » 2018-04-09 04:09:07

Thanks.  This is an edited version of the post taking into account my new thinking on this problem. 

I've set up 3D coordinates as follows:

Origin A.
x axis AD
y axis AB
z axis AE.

As P is on AH with AP = 3.PH this means that P is 3/4 of the way along AH.

Using this, P has coordinates (6,0,6)

Let Q be on AC so that PQ is perpendicular to AQ (ie. PQ is the shortest distance between P and AC)

Let Q be (j,j,0) for some number j.

Using row vectors:

PQ = ( j-6, j-0, 0-6 ) and AQ = ( j, j, 0 )

If these are perpendicular to each other then their dot product is zero:

So Q is the point ( 3, 3, 0 )  So

Bob

#6 Re: Help Me ! » [ASK] Parallelogram in Cuboid » 2018-04-08 19:43:44

Arhh.  That's better.

FP = PH = 2.5 by Pythag.  PO = 5 and angle OPH is 90 degrees so OH is given by

Angle PHO = PBO = alpha

Bob

#7 Re: Help Me ! » [ASK] G to BH » 2018-04-08 19:25:19

With that arrangement of the letters, BH is a diagonal and AG another.  The diagonals cross at 90 degrees and bisect each other, so the distance required is half a diagonal.

Bob

#8 Re: Help Me ! » [ASK] P to AC » 2018-04-08 06:48:20

Again I'm having to assume that E is 'above' A etc.  I agree with your answer to AH but have no idea how you got PH.  Please explain.

Bob

#9 Re: Help Me ! » [ASK] Parallelogram in Cuboid » 2018-04-08 06:45:57

hi

???  FPH is a straight line and parallelograms only have four not five vertices.  I think OBFPH is a trapezium.

Bob

#10 Re: Help Me ! » [ASK] G to BH » 2018-04-08 06:36:04

hi Monox D. I-Fly

I assumed that E is 'above' A, F 'above' B and so on.  And then didn't get your answer.

So I tried E 'above' D, F 'above' C and so on.  And got a different answer to us both.

So it would seem that the exact diagram is important here.

Bob

#11 Re: Help Me ! » Geometry » 2018-04-06 05:56:54

hi Kayla,

Thanks for the complete question.  Using my lettering CDA is isosceles so CD = AD = 45, so I'm glad you said answer B for Q9.

But how are you supposed to do Q8 when the diagram has no letters at all ?  Email your  teacher and ask which points are A and B ?  As it stands the question cannot be answered as we don't know this.

Bob

ps.  Sorry.  Those instructions for Imgur were incomplete.  Log in and upload your image.  Then click your username, and then images,  to get a view of all your uploads.  Click the one you want and on the right hand side you'll see a set of codes for different purposes.  The one you need is called BBcode.  It has the  tags already in the code so just copy it in full to your post and the image should appear in the post when you submit.

#12 Re: Help Me ! » help on proving this » 2018-04-06 05:51:24

hi G123

Welcome to the forum.

Yes, you can use the rational root theorem.  You'll find it here:

https://en.wikipedia.org/wiki/Rational_root_theorem

If that leaves your head spinning let's simplify by looking at a quartic equation.

If p/q is a rational solution, ie p and q are integers

and if we times by q^4

If we assume p/q is in its lowest terms then =>  q divides a and p divides e.

But you are told that the solutions are all integers; as all integers are rationals this means that q = 1 and p divides e.

So could p be 3 ?  Yes as 3 divides 6

Could p be 9 (ie a repeated root of 3) ? No because 9 doesn't divide 6.

Hope that helps,

Bob

#13 Re: Help Me ! » Geometry » 2018-04-05 20:19:11

Kayla,

Yes.  So maybe we aren't doing the right question.  It bothered me that your diagram had no lettering and I had to make up which points are which.  I'm guessing that there were other questions that also related to the diagram.  Otherwise what is 'x' doing there.  So have a look back and post the whole set of questions that relate to the diagram and also try to sort out what A and B should be in the diagram.  Then I'll try to make sense of this.

Bob

#15 Re: Help Me ! » Help, please! » 2018-04-03 23:36:31

hi BLANK,

Welcome to the forum.

I think this has been asked before and fairly recently but I haven't managed to find it yet.  You have typed

wz!= -1

which seems like a typo.  ! would be 'factorial' and that makes no sense in this context.  If we drop the ! then wz = -1 and that also makes no sense as it makes the denominator of the expression zero.

Please respond with a clarification.

Bob

#16 Re: Help Me ! » Geometry » 2018-04-03 01:10:11

hi Kayla,

I've had  a chance to view the diagram.  It has no letters so I've added some to the points in a way that I think fits the question:

2dMLdjl.gif

If you compare triangles ACD and BCD:

They have  a side in common, CD
They have equal angles CDA = CDB = 90
They have equal sides AD = DB = 45

So these triangles are congruent (ie. exactly equal in all measures)

So angle CAD = CBD (=45 degrees)

Once again I have to criticise the question as set.  Why have lengths that are equal numerically to angles.  The question still works if AD = DB = 87 (say).

Bob

ps.  If you had put [img].....[/img]  around your https link the diagram would have appeared in your post.

#17 Re: Help Me ! » Geometry » 2018-04-02 08:34:13

Yes, I think so.  I just wish the question was clearer.

Bob

#18 Re: Help Me ! » Geometry » 2018-04-02 08:31:56

Q12 yes.

I'll look at the diagram tomorrow when I have my laptop.

Bob

#19 Re: Help Me ! » Proofs » 2018-04-02 08:29:00

Q4'  Yes.  The explanation says sides=4, so diagonal = 1 x 4 / 2 = 2

Q12. I see why you thought supplementary at first but supplementary only applies to a pair of angles and this is about a triangle.  D is correct.

Q17'  I agree.

Bob

#20 Re: Help Me ! » [ASK] Tangent of a Circle » 2018-04-01 02:45:02

Back in post 1, when you put in x=1, you changed -8y to +8y. Apart from that the working is good.

Bob

#21 Re: Help Me ! » [ASK] Circle Equation » 2018-04-01 00:47:16

In post 1, Monox does exactly that but then goes on to ask what if one couldn't deduce an answer from the multi choice answers.  I think it's worth considering such things as it improves one's techniques.

Bob

#22 Re: Help Me ! » [ASK] Tangent of a Circle » 2018-03-31 21:03:11

Took me a while to find it. Y = +1. Or +7

Bob

#23 Re: Help Me ! » please help me with that question as you can 9:05 bob bundy » 2018-03-31 20:13:35

In circular type questions the first red may be anywhere as all spaces are equal.  So now consider how to place the rest.

The space to the left of the red may be any of 3.  We'll assume we use one of these. The space to the right of the red may be any from the remaining two.  Now to place the rest.  Two ways to do that.So that's three x two x two = 12 ways.
Now to consider the total number of possibilities.

We've placed one red.  Left may be one from four and right one from three and once again two ways to place the rest.  That's 24 ways.

So P = 12/24

Bob

Good luck with the exam.

#24 Re: Help Me ! » a question of a dice in probabilty » 2018-03-31 20:02:28

I think it's this:

36 possibilities of which 11 have a five:51, 52, 53, 54, 55, 56, 15, 25, 35, 45, 65 don't count 55 twice.

Of these 53 and 35 add to eight.

So P = 2/11

Bob

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