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hi Abbas0000

Arh, I understand your confusion. This result does work for this quadratic, but is not true in general. If you multiply out the expression you'll see that it is true.

The reason is that r1 and r2 are reciprocals; ie. r1 = 1/r2 and r2 = 1/r1

If you make that substitution you'll see that the expression becomes (1-r1)(1-r2) which is what you have.

Hopefully that's enough for you to sort this out. I'm in a rush right now, but I'll post more details later if you need.

Bob

hi Abbas0000

The book result comes straight from the quadratic formula http://www.mathsisfun.com/algebra/quadr … ation.html

But your result is the same. You just have a negative on both sides of the book version.

Bob

hi Kayla,

I will show you what to do for any regular polygon. This diagram shows a regular pentagon (5 sides).

You can see that lines radiating out from the centre divide the pentagon into 5 equal triangles.

(1) 360 ÷ 5 will tell you the angle at the top of one triangle.

(2) The triangle is isosceles so if you subtract the top angle from 180 and divide the result by 2 you'll have the angle at the bottom of the triangle.

(3) The triangle is split in two so that you have a right angled triangle. You can use basic trig. on this. (Half the side) x tan(angle at bottom) = height of triangle. This is the length of the dotted line.

(4) Calculate the area of the triangle using the formula half x base x height.

(5) Multiply this answer by 5 to get the total area of the pentagon.

If you give this method a try and post your answer at each stage I'll check them for you.

Bob

hi Vedanti

Q1. There will be different numeric answers, depending on the values of a, b and c; so you won't be able to give numeric coordinates for this. But an answer with a, b and c is possible.

G is the midpoint of FH, and there is a simple formula for its coordinates:

Add together the x coords of F and H and half it. This is the x coord of G. y similarly.

Q2. Again there's a formula (for the equation of a circle) that will help here:

where (a,b) is the centre of the circle, and r the radius.

So if you substitute the known values you'll get a pair of simultaneous equations in a and r. (not b? why no b?)

So you can work out both (looks like it may be easier to calculate 'a' first).

Bob

hi kayla1dance

Well done! That looks good to me.

Bob

hi kayla1dance

Ok, that's good so far. The area of the whole cake is now divided into an inner circle (area = 157.079) and an outer ring (also area = 157.079). don't worry about the sectors for now.

So if r is the inner radius:

Solve this for r.

Bob

hi kayla1dance,

That's great; thanks for the picture.

What you have written seems, to me, to be back to front. You have used the area to get the radius, whereas you must, in fact, have used the radius to get the area.

It is not relevant that 15/16 has been eaten. You can just determine how to divide the area of the whole cake in half. Then every slice will have the potential to be split in half.

I think you need these steps:

(1) Calculate the area of the cake.

(2) Calculate half this. This will give you a inner circle which has half the area and a ring which has the other half.

(3) Work backwards from this inner circle to determine what radius it must have. (note: The answer must be under 10)

Bob

Note about posts containing links. The way you have done this means any member is linked to all the imgur image posts. I don't do it like this because some of our members are too young to be looking at imgur. It has an age limit restriction. The link I gave to an earlier post shows how to create a bc code link. This image appears in the post and any viewer can see it without having to go to the imgur site.

Thanks Alg Num Theory. I seem to have lost an 'e'. Thanks for the correction. I have edited my post as a penance.

Bob

hi chen.aavaz

If n = 5, then n(n+1)/2 = 5x6/2 = 15

I'll use 0 to indicate a light OFF and 1 for ON.

Here's one possible sequence:

00000

10000

11100

11011

10100

01011

Is it always possible for all the lamps to be turned on at the end of the process?

At the penultimate step, the state has to be

00000

so can you find a way to be sure you could reach this state after n-1 people have had their go?

I'll start thinking about this. I may be 'gone' some time.

Bob

hi Trigtrigger

Welcome to the forum.

Can anyone assist with the following problem to find a where x = 0.85, k=1 and y = 11.2;

y=ae^-k*x

Please see next post by Alg Num Theory. I'll leave mine as a testament to the importance of 'reading the question' !!!

So 11.2 = a^(-0.85)

If you take logs (any will do but let's use base 10 as it's handy on a calculator) you get

log(11.2) = log(a^-0.85) = -0.85 * log(a)

So re-arrange this to make log(a) the subject, and inverse log it (ie. raise 10 to the power of) and you'll get a.

eg. If log(a) = 2 then a = 10^2 = 100

Bob

hi

Looks good to me.

Bob

hi kayla1dance

I made a set of instructions back in 2015. They still work. Here's a link to the post:

http://www.mathisfunforum.com/viewtopic … 40#p352440

I'm really pleased that you have decided to join us.

Welcome again!

Bob

hi iamaditya

Advertising is not accepted on the forum. Two reasons: It's not fair that members have to put up with advertising like this. Mostly they are just an attempt to promote a product; unrelated to maths. Recently I have encountered some that are totally unacceptable when you consider that we have young members. Secondly, MIF does have some paying advertisers. These help to fund the site. They are carefully selected and are all maths related. If they are paying, why should others get a free go?

If it's a maths website then a lot depends on what links lead off from it and whether the site charges for its services. I investigated one recently that looked innocent to start with, but the side links led to some very unpleasant sites. When a new member joins any forum, they should spend a while getting to know how that forum operates and what is acceptable. The ones that have resulted in bans have been where a new member jumps straight in with unacceptable posts from day 1. Some time back a maths writer started to post about his new maths research and included a link to his book (for sale). He was messaged about this and was happy to make a pdf of his work available to members for free. That seemed a satisfactory compromise.

The administrators and moderators have been discussing this recently and are working towards a policy to cover this. The rules may be modified.

Bob

hi kayalexis

Welcome to the forum.

What is the radius of a half-piece when it is cut as shown in the picture?

I'm having trouble understanding what this picture is like. Please would you describe it in words so I can think about whether what you are doing is correct.

Bob

The calculation 1/-∞ doesn't really exist as the rules of arithmetic fail for ∞ .

I suppose you could ask what happens to 1/x as x tends to -∞ .

eg 1/-1000 < 1

1/-1000000000 < 1

etc.

It looks like 1/x tends to zero from below (from the left) as x tends to -∞ and we know that 0 < 1.

So I suppose the answer to your question (ignoring the rules for a moment) is YES.

Bob

There is a way. Don't want to advertise it (no pun intended )

Please can we let this thread fade away so it doesn't give anything away. Thanks.

B

hi ezrarexy

Welcome to the forum.

I have to find the distance value from HOG vector,,

But the vector size is to big and make the process a bit longer,,

can I reduce HOG vector dimension using PCA

I had to look up what HOG and PCA are. Personally I cannot help with this, but, maybe someone on the forum does know. It might help if you were to say more about your data.

Bob

hi Monox D. I-Fly

I am very grateful to you for drawing my attention to this member. Administrators and moderators have access to additional information about each member and also controls that allow us to keep the forum running properly for all members. I cannot give you details of what these are; sorry. If you think about it (and bear in mind that this thread is viewable by anyone, anywhere in the world) you'll realise why this is the case.

Best wishes,

Bob

thanks Monox D. I-Fly

I will remove the link myself, and wait to see if the OP responds to my post.

Later edit: Decided to implement a ban having checked this person's registration details.

Bob

hi WriterWriter,

That link could be taken to be an advert. As MIF is funded by paying (and carefully selected) advertisers, unsolicited advertising is a breech of our posting rules. There is no need for the link so please demonstrate your good intentions by editing the post to remove it.

As for the questions; there's a lot there. It would be nice to know how much you know already about ellipses. Question 1 could be answered in so many ways. Have you tried 'googling' for an answer?

Find the standard form .... ? Again this depends on what you have already been taught. One way to define an ellipse is to say it's a curve for which the equation is ......

So what definition have you been given?

Bob

See here:

http://www.mathsisfun.com/data/function … nc1=e^(-x)

Just add y = x as the second function and you'll see it does have that value as the solution.

Bob

hi tchrjoegachau

**Welcome to the forum.**

Bob

hi zetafunc,

Thanks for the explanation. Now the question and answer are both clear to me.

Bob

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Bob