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Hi bobbym
Oh!
Now I don't feel superior any more.
If you've got a value, substitute it in and show that LHS = RHS.
Since it's a linear equation, one solution = the solution.
Bob
Actually, cancel that. I just took an intelligent guess and got the right answer.
But I'm keeping it to myself 'cos it makes me feel superior.
:)
Bob
Hi,
I'm planning to wait until I know it's the right question.
![]()
Bob
hi bobbym
How are you today? Looks like I've lost this poster. I'm not going to attempt this one until I know the problem is as posted because it looks horrid, don't you think.
Bob
hi
Again, let's just check I've got the question
Bob
You may find it useful to learn some Latex in order to make your posts look mathematical.
You start with square bracket math the the code then finish with square bracket /math
Without the square bracket bit here's what I did for the above.
\frac{(19^{98}+ 342\times 19^{97} )}{19^{99}}=?
The top post in the help me section will teach you how to do it.
Bob
so
Extract this factor
cancel the powers of 19
I think you can finish it from there.
Bob
hi Agnishom
Did you mean
Looks like you can factorise 19^97 and cancel.
Bob ![]()
hi aqa,
I agree with your values for 'a', 'b' and 'r'.
V1 and V2 are not parallel so there will be a plane surface going through the origin* that contains them.
Any linear combination of V1 and V2 will be in that plane.
Have a look at
http://www.mathisfunforum.com/viewtopic.php?id=14934
and
http://www.mathisfunforum.com/viewtopic.php?id=14856
(especially diagram on post #20)
*(When you make a linear combination the origin is one possible combination; a = b = 0. The diagram mentioned above is for the more general case where the origin is not part of the plane.)
Bob
Hi Dave,
Good to hear from you. I'm fine. In the throws of decorating the lounge ... and relatives, 2 + 3 kids, want to come and stay.
Where to put everything? Now if I could scale things a bit (selectively of course)
Hhhhmmmm. ![]()
Bob
Hi Dave
I think that's wrong. Just because the model is built to a scale why should it be ok to scale the speed similarly.
The recorded speed is the speed. No scaling necessary or appropriate.
Otherwise when I put my daughter's Sindy doll in my son's Millenium Falcon and throw them together across the room, they are travelling at different speeds ????
And if I have a Porsche made that is twice the normal size, I don't think the police will accept that argument when I'm caught speeding.
You were right to be suspicious.
Bob
ps. Apologies for the non-politically-correct choice of toys. But don't blame me; it's what they wanted.
hi pi_muncher
You want the answer so soon? I expected at least a few days of thinking ...... ![]()
Ok. There was a hint in an earlier post so here it is again.
http://en.wikipedia.org/wiki/Euler_characteristic
Oh yes and here's a hint that is useful for many problems, mathematical and real life:
If you cannot find a solution to a problem; question your assumptions.
Bob
You are not allowed to dig tunnels.
In the real world, of course, that's exactly what you would do; but this puzzle isn't from the real world.
Bob
Hi
Further thoughts on this homework sheet.
I believe it is a requirement of the curriculum that children learn the properties of shapes and the correct names for shapes.
It sounds like this sheet is designed to get children looking at the properties of the shapes on the sheet; identifying those that a set of shapes have in common; and then finding a good 'name', 'description', or 'title' for those shapes.
There is no one correct answer.
Look at my example below. Any of the following is an answer.
Polygons.
Quadrilaterals.
Straight sided, closed shapes.
Four sided shapes with at least one pair of parallels.
So if the child puts a suitable title, she has satisfied the homework.
But, to show off how well she has met the curriculum objectives, some answers are better than others.
I wouldn't give high marks for "I called them Henry because that was what they looked like to me." although I'd probably laugh and make a mental note that here is a very clever child.
I would try to avoid using the same Title twice and, if I could see several possibilities I'd try to put them all:
eg. 'Polygons which are quadrilaterals with a pair of parallel sides.'
In fact, the best outcome from the homework has already been satisfied. Parents are taking an interest in their child's school work and showing they think it's important.
It could also be argued that showing that even parents don't always know the answers is beneficial.
Plus, they have discovered Maths Is Fun and that's got to be a good thing! ![]()
See http://www.mathsisfun.com/geometry/polygons.html and similar pages for hints.
Bob
hi dolgopolov
You are very welcome.
That's older than I had imagined but I stay with my suggestions.
Do you know the puzzle about 3 houses that need electricity, gas and water pipes. How can you connect them across the surface without crossing pipes?
Picture below.
At the start of this topic you could set this challenge for your class.
Later, when they assure you it's impossible, you can use topology to show them a surprising answer.
I leave this as a puzzle. Post back if you cannot find a solution. He he !:lol:
Bob
hi dolgopolov
What age are your kids?
I'd start with nets for tetrahedron, cube, octahedron, dodecahedron and icosahedron and get them to make them and discover their properties.
See http://en.wikipedia.org/wiki/Euler_characteristic
Then start an investigation: (i) Draw a plane shape or network (ii) count nodes, arcs and regions (iii) summarise results in a table (iv) ask "What do you notice?"
You might look up http://en.wikipedia.org/wiki/Schlegel_diagram as well.
Bob
hi allyson5
I thought this sounded familiar. It's surfaced every year since at least 2005. There are 73 posts of discussion with lots of suggestions on
http://www.mathisfunforum.com/viewtopic.php?id=11&p=3
One day someone will post the sheet which would be a big help.
edit: I've emailed the company and asked for a copy. If I get one, I'll post it (if they'll let me; copyright and all that)
Bob
Obviously, my abstract algebra isn't up to it; but his is; it's his specialism; so it seemed appropriate. And I loved that little touch at the end .... just finish this off .... he does that to me .... and three days later .... I think I need another hint please.
He once solved one of those string and rod with holes in puzzles that I'd been struggling with by watching me and just doing it straight off once I stopped moving it about and let him have it!
Not that I'm suggesting I'm Sherlock or anything, that would be vain, but he is definitely my Mycroft.
Bob
Thanks bobbym. ![]()
Did you notice I asked David to post a reply?
Bob
Oh drat! ![]()
They are good explanations. Any chance of viewing them on someone's computer?
Bob
hi Magniack
I've calculated it and agree with all the simulations.
My calculations are in Excel so the screen shot below shows the formulas I used, and then the probabilities.
I've cropped the column headings. Whoops. Column E has the first throw results and G the second.
Bob
hi again,
The link above that one is good too for how to find a det.
B
hi Superlynx,
Determinant short cuts at
http://www.mathedup.co.uk/Key_stage_5_FP4.htm
Click the 'Manipulating determinants' link
It's a PowerPoint.
Bob
hi All_Is_Number
Any rule that you make that 'connects' values in the domain to values in the range is a relation.
These become functions when they are 'well defined'; ie. from a start point in the domain you have one path only leading to the range, so you know where you're going.
An inverse is something that takes you back from the co-domain to the domain.
Notice I didn't use the word range this time; I deliberately said co-domain. There's a difference.
If the domain is all positive integers and the co-domain is all positive integers and the function is square, then the range is only part of the co-domain because for example you cannot get to 17 in the co-domain by squaring an integer.
If I try to create an inverse function on the co-domain, I cannot make it 'well defined' because there's no √17 in the domain.
But if I change the co-domain to, say, the square numbers, then the function square root becomes 'well defined'.
(because I've made the co-domain = the range)
So whether an inverse function is possible doesn't just depend on the function; it also depends on the domain and co-domain you choose. By setting a limit as I did in my example, I can turn a relation into a function or not. That's why I took care to specify my sets in the earlier post.
Does that help to clear this up?
Bob
hi Superlynx,
Oh boy! I guess I asked for it so thanks for the flow chart.
Where did you get it from? I cannot connect it to what has gone before; nor see where it is leading.
If I call P the position vector and N the normal vector (normal to what?)
and use f(x) for the function (what function?) R for the rescale factor ( why does it need two inputs?) and W for the 'weighting factor' (what again?) then I can reduce the flow chart to this formula:
new vector = W.R.f(x).N + P
but I have absolutely no idea what it's for or why you would want to struggle with it.
And why is 8.78 called a 'result' when it's at the start and what does it do anyway? Is it the scaling factor?
Given the problems you had with determinants, I think the best idea might be to put this 'on the back burner'; ie. leave it until later.
Bob ![]()