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hi gucci_venus

If a polygon is regular that means all its sides and angles are the same. The formula you give is for a 9 sided regular polygon.

Sketch your polygon inside a circle. If you draw lines radiating out from the centre to each vertex, you are dividing the shape into 9 identical isosceles triangles. You can calculate the area of one triangle by finding its height and then area = half base x height. Then you can get the area of the polygon by multiplying by 9.

Q1. For an equilateral triangle the central angle is 360/3. Use that to calculate one of the base angles of one of the smaller isosceles triangles. If you then make a right angle by splitting the isosceles triangle in half, you can use the tangent formula to get the height of the triangle. Then 0.5 x base (the side of the original polygon) x height and you have one of the three identical isosceles triangles.

Q3. Same again starting with 360/5.

and so on.

The general steps are as follows:

(1) Check it is a regular polygon with N sides, each length s.

(2) Calculate the central angle by 360/N = let's call this 'a'.

(3) Calculate one base angle by b = (180 - a)/2

(4) Height, h = s/2 times tan(b)

(5) Area of one triangle, T = 0.5 x s x h

(6) Area of polygon, P = N x T.

Hope that helps.

Others have asked similar questions so you might find more examples with worked numbers if you search.

Bob

A 2 x 2 version of a proof is straight forward enough:

Multiplying these swopped around gives

So AB - BA gives

This is skew symmetric.

The general proof for n by n matrices is going to look very much more complicated as it will involve a subscript i,j type elements. Do you want me to have a go?

Bob

hi kittysoman2013

Welcome to the forum.

Recurrence relations lead to a sequence of terms. The nth term depends on at least one earlier term, often the (n-1)th term.

Such a recurrence is called a first order relation. It looks like your example depends on the (n-1)th and (n-2)th terms. That would make it order 2.

an= an-1 + an^2-2

I'm assuming the second element is (an-2) ^2

Bob

hi 666 bro

I've not met this before so I had to find out what symmetric and skew symmetric mean.

Firstly note: these terms only apply to square matrices: n by n or rows = columns

A matrix is said to be symmetric if it equals its transpose.

and skew symmetric if negating it gives the transpose.

For a symmetric matrix the elements are 'reflected' in the leading diagonal. For a skew, the leading diagonal must have only zeros and the other elements reverse their sign when reflected in the leading diagonal.

example:

and

Suppose B is any square matrix, and consider

The leading diagonal of both B and its transpose are the same, so when subtracted the result has only zeros on that diagonal. The other elements have the property required for a skew matrix, so this new matrix is skew. This is best understood by seeing an example.

Hopefully, you can see how the elements occur again in the reflected positions with their sign changed.

Now consider the matrix 1/2(B + Btranspose). This time the result is a symmetric matrix.

Thus we have a way to make a sum of two matrices,one skew and one symmetric, to make any square matrix.

which is what you are looking for.

Bob

hi Ava

Welcome to the forum.

Is that really a question you have been set? Just multi choice with those alternatives.

Instead I'll try to teach you something useful.

Let's say a convex polygon has N sides. By convex I mean that the diagonals are always inside the shape.

From one vertex you can make N-3 diagonals. N, less 1 for the point itself, less 2 for the neighbouring vertices as joining the point to those doesn't make a diagonal.

As there are N vertices you can make N times (N-3) diagonals like this … but … every diagonal is counted twice, once A to B, and then again, B to A. So divide by two to eliminate repeats.

So the formula for the number of diagonals in any convex polygon is

where N is the number of vertices.

Bob

hi 666 bro

You turn the rows into columns:

Bob

hi camicat

It would probably help for a start if you had a look at this MIF page:

https://www.mathsisfun.com/algebra/quad … phing.html

It shows what a quadratic graph looks like (functions with x^2 in them) so that should help with 5 and 7.

6 is a cubic. I cannot find a MIF page specifically on this but there is a function grapher here:

https://www.mathsisfun.com/data/function-grapher.php

The question says don't use a program to do the graphs for you, but there is no reason why you shouldn't use one to find out what similar graphs look like. Once you've tried a few you'll get the hang of how the function relates to the shape of the graph, and then you should be able to work out what these graphs look like for yourself.

One other thing you can always do is to make a table of values. The table needs two rows; one for your chosen x values; the second for the calculated y values using the equation for y in terms of x. Then you can plot your points. If you haven't got enough points yet to see what is going on, then choose some more x values and so get more points.

Slope/intercept form is usually only used for straight line graphs and these questions are all curves. Hope that's enough to get you started. If you want your answers to be checked post them back here.

Bob

hi Knewlogik

It was your method. I'm just commenting on it. As far as I know, there is no simple way to generate all primes from an algorithm other than the following:

(1) Start the primes list with {2}

(2) Consider the next candidate. This can be 3 initially and then every odd number thereafter.

(3) Test whether any of the numbers in the primes list divide into the number under test.

(4) If they do continue to the next candidate number and repeat step 3.

(5) If not, then the number is also a prime and should be added to the primes list.

You can short cut slightly by eliminating any candidate number that ends in 5.

You can also cutdown on the number of divisions by using the fact that if f is a factor of N below square root(N) then N/f is another factor that is above the square root. Thus if you haven't found a factor by the time you reach square root(N) you can safely stop looking.

The Sieve of Eratosthenes involves eliminating non primes by going through a table of numbers crossing out every number divisible by each prime except that number and looking at what remains. When I taught this I used a grid with 1 - 10 on the top row, then 11 - 20, then 21 - 30 and so on. This makes it easier to cross out certain candidates such as the whole of the 15, 25, 35 … column. Factors of three make a satisfying diagonal pattern etc.

But it doesn't provide a way to get all the primes, as you have to keep extending the grid and testing for division by larger and larger primes.

I think your idea amounts to the same thing; just a different grid shape; which helps to cut out more non primes more quickly. Unfortunately the issue of 'when can I stop crossing out non primes' still arises, as with Eratosthenes. The answer is never!

Bob

hi Mandy,

As well as finding the maths difficult, you seem to have a problem staying motivated and completing homework. So ask yourself this: Which way of working is more likely to serve the purpose of keeping you going?

A week has gone by since you first announced that you are back to try and get a GCSE in maths. I suggested in a reply to your email that you start reworking the homework set for you in this thread. How much actual maths have you done in that time?

Bob

**bob bundy**- Replies: 0

hi 666bro,

You've asked about multiplying matrices.

I'll use an example to show how it works.

John, Mary and Paul are buying sweets.

John buys 2 choc bars, 1 fruity sweet, no chews and 3 lollipops.

Mary buys 4 choc bars, no fruity sweets, 1 chew and 1 lollipop.

Paul buys no choc bars, 1 fruity sweet, 2 chews and 1 lollipop.

We can summarise that using a 3 by 4 matrix.

Here are the prices:

choc bars 15p; fruity sweets 7p; chews 2p; lollipops 5p.

We can put that in a 4 by 1 matrix.

So now let's find out how much each child has to pay. We can multiply the two matrices. The number of columns in the first must match the number of rows in the second. The answer matrix has the same rows as the first and the same columns as the second. ie 3 by 4 times 4 by 1 gives a 3 by 1 result.

Each row of the first is combined with each column of the second by multiplying the first elements, then adding the second elements multiplied, then adding the third elements multiplied and so on.

So if the Nth row of the first consists of {a b c d e} and the Mth column of the second consists of {p q r s t}, then the Nth row Mth column value of the product is ap + bq + cr + ds + et

Hope that helps,

Bob

hi Srikantan

One way is 20 + 20 + 20 + … = 200

If you want to substitute some 10s and 5s into this, it's got to be using 5 + 5 + 5 + 5 = 20 or 10 + 10, or 10 + 5 + 5

So you ought to be able to examine all the ways of doing that.

Bob

hi Stefy,

8-3 = 5. But the diagram shows 8+3.

Draw a line through the centre of the circle, parallel to BE. Reflect B, and E in that line.

Bob

hi Ayomide

I've would do this puzzle in a similar fashion too. The 'solution' uses a sort of averaging effect and I'm a bit suspicious about that in general, although it does seems to work here.

The question actually asks for the time to do a mile so you are one step short of that at the moment.

Bob

hi Ayomide

Welcome to the forum.

Bob

Wow! Where did that come from?

I have never met this term. I did a bit of internet searching and found two possibilities:

Compact matrix pseudogroups: https://www.researchgate.net/publicatio … eudogroups

and

Compact quantum group: https://en.wikipedia.org/wiki/Compact_quantum_group

It sounds like you'll need to study group theory first.

Bob

algebra

limits

differentiation

integration

graphs of 'families' of similar functions

rules of logarithms

B

Post 2 has the first part of a series of lessons that will lead to calculating dy/dx for any a^x. The steps to achieve this are (in outline only)

(1) Get a limit expression for the gradient at (0,1) but we cannot evaluate the limit at this stage. Call it k anyway even though we don't know what it is yet.

(2) Get a limit expression for the gradient at any point. Once again cannot evaluate it at this stage.

(3) Note that for y = a^x the gradient function dy/dx is ky.

So if we can find k, we can differentiate any a^x.

(4) Investigate the particular value of a for which k=1. Call this e.

(5) So when y = e^x, dy/dx = e^x

(6) Find a series expansion for e^x. [By putting x = 1 we can evaluate e.]

(7) Show that all functions that differentiate to give themselves differ only by a multiplier.

That's as far as post 2 goes. Still to come if you want it:

(8) Find how to differentiate y = ln(x) where ln is the natural logarithm, ie base e.

(9) Hence integrate 1/x.

(10) Use change of log base rules to differentiate y = log(x) when the log is in any base.

(11) Hence determine k for any a.

Post back if you would like to see these remaining steps.

Bob

hi Knewlogik

169 won't be removed by dividing by 5, 7 and 11. That led me to a general result:

Consider the number p^2 where p is any prime > 3.

This number is not a prime as it has more than 2 factors {1, p, p^2}.

It's odd so it won't be in columns 2, 4, or 6.

It isn't divisible by 3 so it's not in column 3.

Therefore it is in column 1 or 5.

Even if you have removed all non primes with prime factors, each of which is less than p, the number p^2 won't have need detected yet.

So, there is no highest prime that can be used to check division and remove all non primes.

Bob

hi Knewlogik

I've used a spreadsheet to create a large set of data.

If I label the columns by the first number, ie. 1 2 3 4 5 and 6, then the 2, 4, and 6 columns are always even and so will never produce primes. Similarly the 3 column always gives numbers that are divisible by 3, so will not be primes.

So the 1 and 5 columns are the ones to look at and that's the ones you are talking about … good so far.

Looking at the 1 column, we have

1, 7, 13, 19, 25, 31, 37, 43, …….

From 7 onward we do appear to be getting a lot of primes, but also some that are not, such as 25. You want to know if these can be easily removed from the list.

Well numbers that are divisible by 5 do keep appearing: 25, 55, 115, 145, and so on. We could remove these by testing for divisibility by 5. Notice that they occur regularly down the column.

But there are also some that are divisible by 7: 49, 91, 133, 175, and so on. Once again these occur at regular intervals.

Once again we could eliminate these by checking for divisibility by 7.

But there are also some that are divisible by 11: 55, 121, 187 and so on. Again at regular intervals.

Will this ever stop happening? Sorry but no. There are numbers in the column that are divisible by 13, 17, 19, 23 and so on. In fact every prime will eventually be a divisor for some number in the column,

I have a way of proving this but it involves some more advanced mathematics. I'll try to explain it if you would like.

And the same is true for the 5 column.

Thus there is no easy way to remove the unwanted non primes. Sorry.

Look up the Sieve of Eratosthenes and you'll see that what you have devised is very similar. It works, but you have to keep eliminating numbers by removing those that are divisible by a known prime.

Bob

hi Mathegocart

Not me, sorry. You best hope is zetafunc.

Bob

hi Knewlogik

Welcome to the forum. you've made two similar posts so I'll try to deal with both here.

If you start with 7 and keep adding 6 you get:

7, 13, 19, 25, 31, 37, …..

Starting with 11 gives:

11, 17, 23, 29, 35, 41, …..

Will you ever get the same number in both lists? No; it'll never happen.

Because the two lists start 4 apart and both go up at the same rate, they never share a value.

Do you know about straight line graphs? If not look here:

https://www.mathsisfun.com/equation_of_line.html

The first sequence has equation y = 7 +6x. The second is y = 11 + 6x

These two lines have the same gradient (6) so they are parallel. So they never cross.

For many, many years mathematicians have tried to find a formula for generating primes. There are a few that give some primes, and many that look like they are working but then fail, but nobody has yet come up with a formula for all primes. I suspect it doesn't exist but I don't think there is a proof that no such formula exists.

Best wishes, keep safe,

Bob

It seems that you haven't done one of the things I suggested.

Go to this page:

https://www.mathsisfun.com/data/function-grapher.php

On the top (blue) line enter (y=) 2^x

On the lower (orange) line enter 3^x

You will get two curves with similar shape. Both go through (0,1)

Try varying the 3, say try 4^x.

You'll see the effect that has on the curve.

Experiment with different values so you are trying y = a^x for many different values of a.

You should see that the bigger 'a' is, the faster the curve rises when x>0 and also the faster it drops when x<0.

Make the 'orange' graph into (y=) x + 1 so you have a straight line, gradient 1, going through (0,1)

Adjust the blue 'a' value to 2.1.... until the straight line looks like it is a tangent to the curve.

If you try y = e^x you'll see it's a perfect tangent.

Bob

hi Nasir,

I have noticed you are having fun trying the MIF puzzles. That's great!

If you want to comment on a puzzle, it would be helpful if you provided a link to the puzzle so it can be found easily. At the moment I'm not following which one you are referring to.

I've noticed that the site maker always gives a solution, but does not claim it's the only one. Recently I tried a puzzle and found 8 more solutions on top of the one given. So you may well have managed the same.

Bob

hi 666 bro,

Are you asking about post 2 ?

I was finding out dy/dx for the graph using basic calculus.

When differential calculus was invented the process for finding out a gradient function was developed. It goes like this:

Pick a general point (x,y) on the graph. And a second point close to the first.

Calculate algebraically the gradient of the line joining those points.

Analyse what happens to this gradient as the second point moves closer to the first. If a clear limit emerges that is the gradient function.

One way of showing that the second point is close to the first is to use 'delta' notation. Δx means a little bit in the x direction. This is explained very well here:

https://www.mathsisfun.com/calculus/introduction.html

Δ is a capital letter from the Greek alphabet. The small version of the same letter is:

When I pick the symbol from the ones at the top of the MIF page, only a capital is available. When I use Latex to get mathematical notation the command \delta gives small letter. So I had to switch between the two in post 2. Sorry about that. Perhaps someone knows how to get a small delta when using text only or a capital when using Latex. Please let us know if you do.

Bob

hi Ben,

Welcome to the forum.

When coordinates were invented (By Descartes) it was entirely arbitrary which went across and which up, and also which is the independent variable. It's just a matter of what is usually done.

If you plot a function y = f(x) and also it's inverse y = f-1(x) on the same graph you will notice that each is a reflection of the other in the line y = x. This can be a quite useful property, and you'd loose it if you kept x as a function of y. But I take your point; it does seem odd to change things around after you've re-arranged the formula.

But the notation is just a way of telling people what the function is so the choice of letters is arbitrary too.

y = 2x + 3

x = 2y + 3

t = 2v + 3

are all describing the same function.

If you're asked to re-arrange a formula in an exam, read the question carefully to see what is expected.

Best wishes, stay safe,

Bob