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#1 Re: Help Me ! » I need help in doing this sum » Yesterday 21:34:32

Hi Goutam,

Welcome to the forum.

The wage bill will go down by a multiplier of times 7/11 and up by times 13/10


#2 Re: Help Me ! » who have the prove of tanh(x+y)? » Yesterday 21:29:40


Do you have formulas for sinh(x + y) and cosh(x+y) ?

If yes, use tanh = sinh / cosh

Or go back to definitions in terms of "e"


#3 Re: Help Me ! » [ASK] About Polynomial » 2018-02-15 07:13:11

hi Monox D. I-Fly

I calculated a = -9 as well so I think that part is correct.

It'll be difficult to show a division on the forum so I'll try an alternative approach.  It amounts to the same thing but will allow me to proceed one step at a time.

2x^3-5x^2 -9x+18 is to be divided by (x-1) so the result will be a quadratic.  By inspection the x^2 term must be 2x^2 so that multiplying by x gives 2x^3

So progress so far : 2x^3-5x^2 -9x+18 = (x-1)(2x^2 +  ...) + .....

When -1 multiplies this quadratic term we get -1 times 2x^2 = -2x^2.  But the x^2 term is -5x^2 so we are 'short' by -3x^2.  That can only come from a linear term  multiplied by x ... implies the linear term must be -3x

So progress so far : 2x^3-5x^2 -9x+18 = (x-1)(2x^2 -3x + ...) + .....

-1 multiplied by -3x gives +3x.  We want -9x so -12x must come from somewhere.  It must come from a number multiplied by x.  So the number is -12

So progress so far : 2x^3-5x^2 -9x+18 = (x-1)(2x^2 -3x  -12) + .....

-1 multiplied by -12 = + 12.  We want + 18 so the remainder is +6

So:                        2x^3-5x^2 -9x+18 = (x-1)(2x^2 -3x  -12) + 6

Although I have not set this out as a division you should be able to check where you went wrong.

WHOOPS! That sin't any of the possible answers.  Just checking what I have done ..... Back soon.

LATER EDIT I've multiplied out my expression and get 2x^3 - 5x^2 - 9x + 18.  That's odd.  I'm going to use some heavier algebraic hammers to crack this puzzle.  Back soon.

EVEN LATER EDIT:  I've checked all my working by another method and find the same result.  Could these be the answers to another question entirely ?  Or could the original cubic be different ?


Before I started I used the remainder theorem to find the remainder.  I was expecting + 6 so that looks ok.

#4 Re: Help Me ! » help » 2018-02-15 02:11:22

hi jose jose

You can achieve mathematical layout by using LaTex.  There is a tutorial on this here:

And there is a site called WolframAlpha that will compute some integrals for you.  It's here:

But that won't help if you're trying to learn how to do them for yourself.  If you post a problem on the forum someone will try to help, but don't expect we're just going to do your homework for you smile

If you cannot create the right LaTex try saying the question in words.  We can translate those for you and post up what we think you're asking.


#5 Re: Help Me ! » question about interval and the graph of function (inc and dec).. » 2018-02-15 01:58:51

hi Hannibal lecter

Thanks for posting that.  I wanted to check the wording ... in particular "measured in years before 2008".  t might have been 2 yearly rather than one yearly.

So (b) First measure is 7.019.  t value from table is 25 so 2008 - 25 = 1983.  Your answer is correct.

Second measure is 6.957.  t value is 125 so 2008 - 125 = 1883.   Book answer is correct.  If the question said 6.992 your answer would be correct, but both posts show 6.957.

Text books do sometimes have wrong answers.  Mostly a typo, but in this case it looks like the question was changed (maybe to make it easier) but the writer forgot to alter the answer.  Then you have to look further along for the next answer ... stopping you from just going along to the next column.

Hope that helps,


#6 Re: Maths Is Fun - Suggestions and Comments » Area of a Circle by Lines » 2018-02-14 21:00:48

hi MathsIsFun

That's a wonderful moving diagram.  I love it!  Never seen that before.


#7 Re: Help Me ! » question about interval and the graph of function (inc and dec).. » 2018-02-14 20:57:58

hi Hannibal lecter

In the textbook, just before the table, there was an introduction to the question, explaining what the table shows, and what t and S stand for.  This introduction is essential for anyone trying to do the question.  Please copy that wording exactly as it is in the book.



#8 Re: Exercises » integrals? » 2018-02-09 01:12:48

hi elgordovergatario

Welcome to the forum.

Try here:


#9 Re: Maths Teaching Resources » Are these any good? » 2018-02-08 20:42:49

hi ericnumberking

I don't have any direct experience of these so I've had a look at each.  Study Pug has some free lessons so you could trial it yourself. 

How did you arrive at this forum?  Maybe you bypassed our very own teaching pages.  The owner of the site, MathsIsFun, himself,  has  written a large number of maths help pages.  They cover most topics up to calculus and each page is cross linked to related pages.  There are many interactive lessons and all are written in an easy to follow style with clear diagrams and explanations.  At the bottom of each page you will find a set of multi-choice questions to test your understanding.  As an retired teacher I think these pages are excellent!  And they're completely FREE.  So before you commit money to any on-line course, why not try out MIF's own material.

Start here:

And you can also ask for help from members of the forum. smile


#11 Re: Help Me ! » [ASK] Prove Using Induction » 2018-02-01 22:22:19

I think it is valid.  Compare with a standard induction proof for the sum of n integers.

I think the formula is sum= n(n+1)/2

step 1:  when n = 1 , sum = 1x2/2 = 1

So the formula works for n = 1

step 2: assume it is correct for n = k

sum of k integers = k(k+1)/2

The next integer is k+1.

So the sum of k+1 integers is sum of k integers + k+1 = k(k+1)/2 + k+1 = (k+1)(k/2 + 1) = (k+1)(k+2)/2

This is the same formula with k replaced by k+1, so if the formula works for n=k then it works for n = k+1

The bit in bold characters is just algebra.  So it is normal to include some algebraic steps in the proof.


#12 Re: Help Me ! » [ASK] Prove Using Induction » 2018-02-01 07:00:47

Nothing hard.  I must have explained it badly.

step (2): consider the expression k(k+1)(k+2) + 3(k+1)(k+2) = term 1 + term 2

The first term is divisible by 6 by (inductive) assumption.

The second term has 3 as a factor and 2 also as either k+1 or k+2 must be even.

So it also is divisible by 6.

Adding term 1 to term 2 must therefore be divisible by 6.

But the expression simplifies to (k+1)(k+2)(k+3) completing the induction step.

As you can see there is nothing original in this; it is just your proof written backwards.

Alg Num Theory: Hi.  I agree but the title of the post seems to require a proof by induction.  Perhaps it's in that chapter as an exercise in induction.

I once did some A level marking and one question told candidates to use a certain method.  The chief examiner ordered us to give no marks if any other method was used!!!


#13 Re: Help Me ! » [ASK] Prove Using Induction » 2018-01-31 21:55:11

hi Monox D. I-Fly

That looks good to me.  You've done the two stages: (1) showed it works for a starting value; (2) showed that true for k leads to true for k+1.   There will always be a bit of algebra at (2) and what you have done is correct and does prove stage (2).

As an alternative, you could start by saying let's consider the expression  k(k+1)(k+2) + 3(k+1)(k+2).  Show this is divisible exactly as you have done, and hence complete the stage (2) step.  Essentially it's the same proof but backwards.  I'm sure either way round is acceptable.  Well done!


#14 Re: Help Me ! » Pattern Problem » 2018-01-30 20:55:02

Today I visited your link page again and found someone calling themselves "Mohammad Matin" has copied my earlier post and submitted it as their entry.  This seems to be doubly dishonest: (1) because it was not that person's work; and (2) it claimed to be a solution but definitely was not!.

It's a great shame that the work of members on this forum is being abused in this way.


#15 Re: Help Me ! » [ASK] Trigonometric Equation » 2018-01-29 20:45:53

hi Monox D. I-Fly

Sorry my post was too brief.  I was in a hurry as I had to go out.  Here's some more detail:


Let A = 2x-5 and B = x+5  => A-B = x - 10  and A+B = 3x

The two trig. formulas that will help are


so the equation becomes:

I'll leave you to finish this. smile


#16 Re: Help Me ! » Pattern Problem » 2018-01-29 07:43:53

I've received your message but I don't understand what you're wanting.  I've visited the contest page and seen this:

"Contest Brief

Please find the problem attached to the file below."

But I cannot find any such attachment.


#17 Re: Help Me ! » [ASK] Trigonometric Equation » 2018-01-29 04:43:53

try using:

cos(A-B) + cos(A+B) = 2 cos(A) . cos(B)

A = 2x-5 and B = x+5

and a similar formula with a minus sign.


#18 Re: Help Me ! » Pattern Problem » 2018-01-29 02:55:49

I think this can be done by calculating expected result for each route.

Assuming the plus and minus figures are equally likely:

the expected result for A is - (0.25 x 0.1 + 0.25 x 0.2 + 0.25 x 0.4 + 0.25 x 0.4)  In your post you have 0.4 twice.  If that is a typo then just substitute the correct value instead. = - 0.25 x (0.1+0.2+0.4+0.4) = 0.25 x 1.1 = - 0.275

And for B = + 0.2 x 1 + 0.2 x 1.3 + 0.2 x 1.4 + 0.2 x 0.7 + 0.2 x 0.5 = 0.2 x (1+1.3+1.4+0.7+0.5) = 0.2 x 4.9 = + 0.98

So for A = Y + 1.5 - 0.275 = Y + 1.225

and for B = Y - 0.5 + 1.225 = Y + 0.725

Neither has an expected result within the limits.  Am I misunderstanding something here?


formula for an expected result is

where the Ps are the probabilities and the Ws are the 'weights' for each possibility.

#19 Re: Help Me ! » Pattern Problem » 2018-01-28 21:23:21

hi MrGiggles

Welcome to the forum.

I'm not sure I can help as I'm a bit confused about the problem.  (no diagram seen)

Here's what I think you're asking:

There's a line connection between point X and point Y.  Then the signal can either be sent to A, or to B.  Either is acceptable if the signal stays within 0.5 of the original from X ( not higher nor lower than this limit)

There's no loss between X and Y (or it's irrelevant since we have to use this connection).

If A is chosen the signal will go up to value at Y + 1.5
If B is chosen the signal will drop to value at Y - 0.5

In addition transmission along YA and YB are subject to fluctuations as follows A route: minus 0.1,0.2,0.4,0.4    B route: plus  1,1.3,1.4,0.7,0.5

Is this correct so far?

Are these fluctuations the only ones possible?  And is each equally likely?


#20 Re: Introductions » Introduction » 2018-01-28 21:06:24

hi Leesa

Welcome to the forum.

I have removed the advert from your signature.  Please read this:


#21 Re: Help Me ! » Quadratic equations (1-x-x^2) » 2018-01-21 21:38:56

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.


#22 Re: Help Me ! » Quadratic equations (1-x-x^2) » 2018-01-21 07:10:47

hi Abbas0000

The book result comes straight from the quadratic formula … ation.html

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


#23 Re: Help Me ! » Math » 2018-01-18 06:47:29

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.


#24 Re: Help Me ! » Homework Problems - Please Help! » 2018-01-17 20:19:48

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).


#25 Re: Help Me ! » Math » 2018-01-17 08:35:40

hi kayla1dance

Well done! That looks good to me. smile


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