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#1 Re: Help Me ! » [ASK] Two Cones » 2018-07-06 00:26:26

hi Monox D. I-Fly

Ok.  Still puzzled.  I thought I'd try some numbers.  I fixed L1 at 10 and L2 at 5.

I chose any number for r1 and tried r2 = 5,4,3,2,and 1

In Excel there is a feature called 'goal seek'.  You specify a cell and a target value for that cell, and choose a cell that may be varied to achieve the target.

So I computed the two areas and also area1 minus 2 times area2.  The target is to make this value zero.  When it is, area1 is twice area2.  I chose to vary the r1 value to achieve this target.  Here are the results (sorry the table is a bit lopsided.  It was Ok when I copied it.)

L1    L2     r1                           r2    area1            area2               A1-2*A2                  ratio
10    5    6.180339868            5    314.159264    157.0796327    -1.34787E-06        1.236067974
10    5    4.848861716            4    226.1949133    113.0973355    0.000242224        1.212215429
10    5    3.544016352          3    150.7971241    75.39822369    0.000676761        1.181338784
10    5    2.280109903         2    87.96459494    43.98229715    6.38796E-07        1.140054952
10    5    1.082762919         1    37.69912669    18.84955592    1.48466E-05        1.082762919

The target, A1-2*A2 is not quite zero because the goal seek feature is only approximately accurate, but I think good enough to demonstrate my point.  With different r1 and r2 values the area constraint is satisfied without the ratio being the same each time.  So there is no unique answer.


#2 Re: Help Me ! » [ASK] Two Cones » 2018-07-05 20:07:26

hi Monox D. I-Fly

But we must take into account the area of the base.

Are you sure because I am also stuck with the same equations as you.  There are 4 unknowns, L1, L2, r1, and r2.  And only two constraints.  That's not enough to determine the ratio of r1 to r2.

What is meant by 'cartons' in this question.  If I was trying to make a cone I'd use a sheet of card. Maybe there's a clue there to a third constraint.


#3 Re: Help Me ! » [ASK] Water Pipe » 2018-07-05 19:55:21

hi Monox D. I-Fly

There is a circle theorem which will help for the surface area.

If chords AB and CD, for any circle,  intersect at E then AE.EB = CE.ED  You can prove this using similar triangles.

So if the width of the surface area is 2x (CD) then x^2 = 16 times 68

You can also use Pythag on triangle CEO. (x^2 = 42^2 - 26^2)

To get the volume you'd need the area of cross section .  Let O be the centre of the circle.

You can work out the area of triangle CDO.  Then you'd need the area of the sector CBD which requires the angle COD (obtuse).  I cannot see a non trig way of getting that with 68 as a measurement.  With another length it might be possible.


#4 Re: Exercises » Question » 2018-07-01 18:38:37

If the discount is 10% then how it costs 0.9x?

If you take 10% off the price, you'll be left with 90% to pay.

3x/10 is correct.  Then rearrange the formula I gave:

This answer will be in hours so you'll need to convert to minutes.


#5 Re: Exercises » Question » 2018-07-01 06:00:34

Here again you can use x.

Let the amount of work to be done be x.

Then Ali works at x/5 work units per hour and Hashim at x/10.  Add these to get their combined work rate and use the formula:


If the price is  x then a discount of 10% means it will now cost 0.9x and then multiply by a further 0.85 to apply the second discount.

If this is kx then the discount is (x-kx)/x times 100%


#6 Re: This is Cool » Proving Primality Using Minus 1 Divided By 2 » 2018-06-30 04:58:06

hi Primenumbers

That's very interesting.  I confess I'm still struggling to follow your proof.  I thought I'd collect some evidence to help with this so I wrote a program to do all the sums.

Every number the program said was a prime was indeed a prime.  But some known primes weren't detected.  Now admittedly you don't claim the algorithm will find all primes but you did say "Also if you don't know if z is prime , you can repeat the process!"

So let's say I try 23 like you but I don't know if z = 11 is a prime.

z=1   (11-1)/2 = 5
z=2   (5-1)/2 = 2
z=3 (2+11-1)/2 = 6
z=4 (6+11-1)/2 = 8
z=5 (8+11-1)/2 = 9
z=6 (9-1)/2 = 4
z=7 (4+11-1)/2 = 7
z=8 (7-1)/2 = 3
z=9 (3-1)/2 = 1
z=10 (1-1)/2 = 0

z is not prime so we cannot determine y = 11

Please would you expand a little on your proof.


#7 Re: Exercises » Percentage » 2018-06-26 21:34:16

hi Zeeshan 01

There's lots on percentages here: … hange.html

and there are links to other useful pages.

I have looked at all the questions.  But you have got to learn how to do them so I'm not going to give you answers to them all.

I suggested a method that would work for them all.  Call one of the amounts x.

eg. Q1

Call the third amount x.

Then the first is 1.20x and the second is 1.50x.

Now you can work out the answer.  Here, like the one I showed before, the 'x' cancels out of the final answer.


#8 Re: Exercises » Percentage » 2018-06-26 03:01:32

If Ali salary is 35% more than that of Hashim

So Ali's salary is bigger than Hashim's.

If Hashim earns $100 the Ali earns $100 + 35/100 times 100 = $135.

If Hashim earns x then Ali earns x + 35/100 times x = x + 0.35x = 1.35x


#9 Re: Exercises » Percentage » 2018-06-25 19:29:44

difference is 1.35x - x = 0.35x

So difference over reference pay = 0.35x / 1.35x = 0.35/1.35

Then times by 100 to get a percentage.  Sorry I used x again instead of 'times' for the multiply.


#10 Re: Help Me ! » Transformations » 2018-06-25 01:21:52

hi Jesus1968

You seem to be getting very confused.  Q17 gives you the dilation and asks for the new coordinates.

Just multiply the old ones by -7 to get six new coordinates (x,y) three times.


#11 Re: Exercises » Percentage » 2018-06-23 21:08:10

In these call one amount x and construct an equation from that.

E.g.. Q2 let Hashim be x

Then Ali = 1.35x

percentage less = difference / Ali x 100 = 0.35/1.35 x 100


#12 Re: Help Me ! » Transformations » 2018-06-23 20:57:03

Hi Jesus1968,

These are all centred on the origin so look at any coordinate before and after the dilation using

After = before Times dilation



9 = -36 x d. so d = -1/4


Ps. A translation is a different thing ... the general term is transformation.

#13 Re: Help Me ! » Real life applications for every math concept » 2018-06-11 19:06:47

hi Monox D. I-Fly

I have only 'scratched the surface' of this topic.  You could start by looking at this article:


#14 Re: Help Me ! » Sequences » 2018-06-10 19:18:16

hi Cliff

Welcome to the forum.

I think the answer to (a) is r=1

Here's my logic:

Certainly 1 is a possibility.

eg.  a, a + π, a + 2π + a + 3π, …. where  a is an integer and π = pi, is such a sequence.  'a' doesn't have to be the first term; you could have many terms leading up  to 'a' and then as shown.

So what about r > 1 ?

Let's say a and b are two integer terms in a sequence, m terms apart

so b - a = md where d id the common difference.  As a and b are integers, so is md.

so another integer term is c = b + md.

So if there are r such integer terms then it is always possible to find one more by adding md to the last.  So if r > 1 there will be an infinite number of such terms.

I'll post again if I find an answer to (b)


I think r = 1 or r = 2 are the only possibilities, but I'm still working on a proof.


#15 Re: Help Me ! » finding areas of polygons given only sides » 2018-06-09 19:52:36

hi Joe walsh,

Welcome to the forum.

Sounds to me like you also need a general way to do this for any regular polygon.

Here's a diagram for a heptagon (7 sides).  You can easily adapt it for n sides.


(1) Provided the polygon is regular you can calculate the central angle (AOB).  Divide 360 by 7 ( or n).

(2) Half this to get AOH and then subtract from 90 to get OAH.

(3) In the green triangle AOH, AH is adjacent to angle OAH and OH is opposite so OH = AH x tan(OAH)

(4) The yellow area is then 1/2 AB x OH

(5) Multiply by 7 (n) to get the area of the whole polygon.

Hope that is useful for you,


#16 Re: Help Me ! » Letter-arranging » 2018-06-09 19:38:49

hi Amartyanil

First you could calculate the number of arrangements where there is no restriction.

Then imagine EA locked together as a single 'letter' and calculate how many ways you can arrange these five letters.  Do a similar thing with AE etc*.  Then subtract.


*You'll need to take care to avoid repeats caused by all three vowels together.

#17 Re: Computer Math » Calculating Pi » 2018-06-09 19:34:00

hi Amartyanil

What exactly do you mean by "it's not working properly" ?  This series converges only very slowly.  I needed over 1200 terms before I got 3.14....


#18 Re: Help Me ! » Inductive and Deductive Reasoning » 2018-06-06 20:11:09

OK.  Q15 is straight forward.  An angle = "To define an angle, you need to be able to define both rays, and they need to have the same endpoint."

So let's say the ray is ry_AB and C is a point not on AB so that <BAC is the angle.

Extend the ray to make an line, extending beyond point A.  Then we have a line and a point not on the line.  This satisfies "a line and a point not lying on the line" so we have a plane.

Q14 is more problematic.  Let's call the lines l, m and n.  Since we have l and m parallel this satisfies "two lines which intersect in a single point or are parallel"

The reason I think this problem is problematic is that the third line, n, may or may not be in the same plane as the other two.  Does this matter?  Strictly, no.  We can say yes I have a plane, in fact I have three so I certainly have one.  Since we are 'in court' it seems odd not to reveal the whole truth and say we have one or three planes.  That's why I'm curious about the course you are doing.  There are two on-line courses that I know of.  Sometimes they set silly questions and I wondered if I could add this question to the list.  Ha ha!


#19 Re: Help Me ! » Conditional Statements and Venn Diagrams » 2018-06-06 19:45:09

hi riaaaa

If someone makes a statement and you can see it is untrue because you can think of an example that proves it's wrong this example is called a 'counter example'.  In English we could say this example runs counter to the statement, meaning it is against the statement.

eg.  Statement: "All swans are white"  I go to a zoo and see an Australian black swan.  It is not white.  This swan is a counter example to the statement.


#20 Re: Help Me ! » Conditional Statements and Venn Diagrams » 2018-06-02 18:26:55

hi riaaaa

For number 7, I'm thinking it could be counterpositive?

There's a mathematical term 'contrapositive'.  Is that what you mean?  What course are these from?

The contrapositive of x^2 > 10 => x >0  would be ' not x>0 => not x^2 > 10

I tried to help with this one in my last post.  Have another look at it.

As for number 13, I'm thinking E.


And for number 3, I'm thinking F.



#21 Re: Help Me ! » Conditional Statements and Venn Diagrams » 2018-06-01 20:12:59

hi riaaaa

Welcome to the forum.

statement wrote:

So I'll choose a value for x^2 that makes x^2 > 10.

Let's try x^2 = 16.

We know that positive numbers have two roots; in this case x = 4 and x = -4.

So it is not true that x must be > 0.

So what is it called when you find a value that proves a conclusion is wrong ?


#22 Re: Help Me ! » Algebra » 2018-05-28 00:45:12

hi Doggone

Welcome to the forum.

That has to be true for all values of x, y and z, so I chose a set of values that makes life easy:

If x = y = 1 and z = -2, then x + y + z = 0 and the equation becomes

From here you can get k.


#23 Re: Help Me ! » algebra » 2018-05-28 00:37:34

hi Loki,

Re-arrange the equation to make y the subject.  As x is an integer so is X+7 so replace x by X+7 and simplify that equation for y.

It should be fairly easy to work out which values of X (I can only see three and three more negative that will work) so you can then convert back to get values of x.


#24 Re: Help Me ! » Long division on these polynomials? » 2018-05-28 00:18:36

hi mrpace,

So what answer are you getting?

Note:  x^2 is a common factor of 'top' and 'bottom' and so can be cancelled.


#25 Re: Help Me ! » Initial Velocity Vector and Diagram from Parametric Equation » 2018-05-28 00:09:15

hi Oran2009

The horizontal component of velocity will be a linear expression and the vertical a quadratic so those equations look right.  But the values will depend on your initial throw (magnitude and direction) so I'll have to assume you have those right.

Your initial velocity calculations are then correct.

I do not use Desmos but I'm sure any graph plotter will allow you to enter those equations to get a graph of velocity against time. 


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