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hi Seetha Rama Raju Sanapala

Welcome to the forum.

I am replacing an earlier post with this version. My apologies to anyone who spent time on my previous version. Hopefully, I've got it right now.

Note: both intersection and union are associative.

I'll consider intersection first.

If intersection is the binary operation for a group, there must be an identity set; let's call it I.

Also we want another member; let's call it A.

For a group we require inverses; let's call the inverse of A, B. So

This means that I is contained in A {and also in B).

But I is the identity so

So A is contained in I.

So A is contained in I and I is contained in A. This means that A = I.

So every element of the group is I. Conclusion: no such groups with 2 or more members.

The case for union follows similar lines.

So A is contained in I

But I is an identity so

So I is contained in A. etc etc.

Bob

ps. Sorry but your full name is too long to fit the blue space above.

hi Monox D. I-Fly

I agree the totals are 180, and 170. To end up equal, they must each be 175. So one weight must go down by 5 and the other up by 5.

Bob

hi Zeeshan 01

Think of a function as a box that takes in a number and outputs another number.

Here you can see what happens when the number 30 is the input. Four different functions give four different outputs.

sin(30) is a single number. sin(90) is also a number [ sin(90) = 1 ]

If you have one function divided by another you **must** work out the value of each function first before you do the division:

You **cannot** separate the function from its number by cancelling or you get the wrong result.

Hope that helps,

Bob

hi samuel.bradley.99

General solution:

Draw parallelogram ABCD where AB = 1.1n and BC = n

With centre D and radius 1.1n, draw an arc to cut AB at E.

The cut required is along the line DE.

Remove the triangle coloured yellow, and move it to the position shown in orange.

AB = DC = EF = FC = 1.1n so EFCD is the required rhombus.

Bob

hi kk63353

Welcome to the forum.

I think you can say that, so well done. I do not claim that my methods are the quickest, neatest, or even correct.

That's why the forum has other members!!!

Bob

hi dazzle1230

Q1. That's a quadratic. Sketch the graph and all except the lowest point have two x values for each y.

So to make it invertible you can choose any domain that stops that happening: eg from the lowest point rightwards.

x=0 isn't the lowest point so you should be able to find the required interval to include it and also be the largest.

Q2. Sketch a random wiggly line for f(x) between x = -1 and +1.

Now sketch h(x) by just moving the wiggly line up by 1. Looks to me like the domain is the same.

Bob

hi samuel.bradley.99

This works only if A lies on that circle. Still working on a general case.

AB = DC = 1.1n

AD = BC = n

Note AC = n, so the dotted diagonal will make the cut. Translate ADC (ie. Move it maintaining the parallel directions) so that DC fits over AB as the new diagonal.

The new parallelogram has four equal sides of size n, so it is a rhombus.

Bob

If you had x**i** and y**j**, this would be a vector that starts at (0,0) and goes 'x' across and 'y' up. So it would be the vector way to show a general point which would be the same as (x,y) in coordinate form.

Bob

hi mathattack

Q1. If you are doing vector geometry, then the usual approach is to fix an origin and two base vectors. Let's say A, **AB** = **i** and **AC**= **j**

To save time I'm not going to continue using bold for the vectors.

So BC = = -i + j, and BS = 1/3 . (-i + j)

So the line AS is r = i + lambda . 1/3 . (-i + j)

Do a similar thing for the equation for BT and find where they cross, U.

This will enable you to determine the ratio.

Bob

The Earth is not exactly spherical; it is an oblate spheroid (bulges at the equator). So the diameter varies depending on where you measure it. So I wouldn't worry about small discrepancies in answer here.

Bob

Maybe the topic covers the use of logical proofs ?

Bob

Nothing to do with space, the final frontier.

A vector space could just be all the points in plane. We say that is a two dimensional space.

It could be three dimensional, it which case you would need vectors with three components in the x, y and z directions.

It could also have four, five, .... n dimensions, but you probably won't need to worry about those.

I kept it simple in my example by sticking to two dimensions.

Hope that explains that part. Next question?

Bob

I thought that was what I did. Please say **exactly** where you are confused.

Bob

-2 is not yet a factor as the 2 in the second term is + not -

But two minuses make a plus ie. -- is the same as + so I can re-write the expression as

Now -2 and y are common factors so we can move them to a single expression outside a bracket for the remainder.

Notice I have put in an extra x 1 so that there is something in the second part of the bracket and not just an empty space or worse still a zero.

You can test whether you have factorised correctly by multiplying out:

Hope that helps.

Bob

hi peter010

Very clear. Thanks,

Bob

hi Mathegocart and thickhead

I've searched their site but cannot find any dimensions for their trucks. Strange! Wouild you buy a vehicle without knowing how big it is? Pictures with a trucker in shot suggest they are only about 12 ft tall. If a truck really was 17 ft tall it wouldn't fit under UK motorway bridges.

Bob

ps. LATER EDIT: This is how to cut off the main route from the rest of England to the Dover continental ferry port and the Channel Tunnel:

hi Zeeshan 01

I want to help but I need you to say more than ??? What don't you understand? Or give an example of what you do understand.

Bob

hi Zeeshan 01,

I am not sure what you are asking. So I will go back to basics and explain why vectors and coordinates are part of the same topic.

Maths theorists usually start with vector spaces. To keep it simple, let's make a 2 dimensional vector space.

The vectors **a** and **b** here are not parallel so you may use them as a **basis** for the space. That means that any other vector can be written as a linear combination of **a** and **b**.

To demonstrate this, I've drawn another vector **c**. It could be any vector so I just choose one at random.

I've then made two dotted lines. The first goes through O and A and extends as far as necessary in both directions. The second is parallel to OB but goes through C.

Because OA and OB are not parallel, the dotted lines must cross somewhere. I have called that point P. If you count squares you will see that

You can use this method to make any vector a linear combination of **a** and **b**.

**a** and **b** are said to form a basis for the space. Nothing special about **a** and **b** either. I just drew any two non parallel lines through (0,0).

One special basis is the one made by the unit vectors

and

If they are used as the basis then

And then we define the coordinates of C to be those two numbers and write C = (-6,6)

Bob

hi peter010

You have done a good job. A simple example helps any student to see easily what is happening before they meet much more complicated examples. I like the way you have highlighted the text as you speak. This must be a great help to any for whom English is not the first language.

Defining integration as a summation process is the way I would do it too. It is possible to prove that the area under a curve gives the same result so I would see this as a consequence of integration rather than another way of defining it. When I was at school my maths teacher said the proof was too hard and 'glossed' over it. But I looked it up and I don't think it is that tough.

Bob

Yes. He used to be the only administrator but he appointed bobbym to help out. bobbym has most administrator privileges but there are a few things that only MIF can do.

MIF will see your post. But he has lots of other things to do as well. He has to maintain the forum website and also the main mathsisfun website. He frequently adds new pages there. He has also started a science subsection and made lots of pages for that.

zetafunc has made some videos http://www.mathisfunforum.com/viewtopic.php?id=22490 so it would be worthwhile for you to look at those., If you cannot find what you are looking for, he might be willing to create a video for you.

Bob

Vectors and coordinates are part of the same topic. O = (0,0) is the origin and P = (2,3) is just another way of saying vector OP = 2i and 3j.

i is the unit vector in the x direction and j is the unit vector in the y direction. So when you describe a point as (2,3) what you are really saying is that, starting from the origin, you need to go 2 in the x direction (or 2i) and 3 in the y direction (or 3j) to reach P.

Bob

ps. My computer keeps changing i to I. This is because I have got certain auto-adjust spelling corrections set up. If you go back to the I and change it back to i the auto-adjust usually gives up.

Bob

hi Annar

Welcome to the forum.

Let's say there are n stops. So the summation becomes 1 + 2 + 3 + ... + n or n + (n-1) + ... + 3 + 2 + 1

If you add these two sums together, first terms together, then second terms together, and so on, you'll get (1+n) + (2 + n-1) + ... + (n+1). Every bracket comes to the same thing (n+1) and there are n of them. So double the total comes to n(n+1). So we have this formula

To solve this, double to get n(n+1) = 78 x 2 = 156. So you want two consecutive numbers that multiply to 156. Only one positive answer to this: n=12, as 12x13=156.

Bob

hi Zeeshan 01

MathsIsFun created the forum and is the first member.

http://www.mathisfunforum.com/profile.php?id=2

Bob

hello Bart

Welcome to the forum.

I'm a UK teacher (retired).

GIVEN - ASKED - SOLUTION - ANSWER - CHECK.

This looks good to me.

"Oplv.V.="

I don't recognise that. If I had x^2 = 9 then I might write solution set = {3,-3) but x = 3 or -3 is more common. I haven't met an abbreviation for 'solution set' in English.

One third and two thirds are correct. In the UK we say one quarter although I think most people would understand one fourth. Also a student could say 'one over three' etc.

Bob

ps. Brexit? Don't blame me; I voted 'REMAIN'.

hi Zeeshan 01

You have OP = 2i + 3j and OQ = -4i + 7j .

PQ = PO + OQ = -2i - 3j -4i + 7j = -6i + 4j

If you fraw a coordinate diagram and plot the points you'll see what 'i' and 'j' is needed to travel from P to Q.

Bob