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#1 Re: Guestbook » 0.999... = 1 :: alternative version » 2018-09-07 19:42:21

hi Xepemu

Welcome to the forum.

That's a neat method which seems to overcome the objections some folk have about infinite digits.

How about joining the membership?

Bob

#2 Re: Exercises » Percentage » 2018-09-07 19:39:38

hi Sresta702

Welcome to the forum.

Which ans?  I like to encourage posters to work things out for themselves.

Bob

#3 Re: Introductions » Hi there! » 2018-08-25 19:45:40

hi Pablo,

Welcome to the forum.

Bob

#4 Re: Help Me ! » Does 1 = .999...? » 2018-08-25 19:42:08

The difficulty arises because it is a property of the real number system that between any two real numbers there is another real number.  For all other reals this works ok but it fails whenever the decimal representation 'ends' with an infinite string of 9s.  I have encountered two 'solutions' to the problem and both allow a consistent set of axioms.

(1) Allow that for example 0.99999999.... is the same as 1   I can understand why some people dislike this but, if you remember that the reals exist irrespective of the decimal representations, then why not?  Lots of mathematical topics work perfectly using this solution.  For example, you can convert an infinite, and recurring, decimal representation into a fraction.

(2) You can forbid the existence of such numbers from the real number system.  That also works and is the basis of the approach used by Georgi E Shilov in his book Elementary Real and Complex Analysis.

It might help if you stop thinking that numbers have a concrete existence and accept that they are just convenient abstract ideas that help us to do certain types of mathematics. (You can hold 3 apples or even a wooden shaped '3' object but you cannot hold a 3.)  So we can make numbers do what we wish according to the model we are making.  Not everything obeys the rules for real numbers.  For example add one pile of sand to another pile of sand and you have one pile of sand. You cannot have half a stick of chalk.

In summary, decide what mathematical model you are building; invent a consistent set of axioms for your model; and then use it to discover new things.

Bob

#5 Re: Puzzles and Games » Starter puzzle on this site "12 Days Of Christmas" » 2018-08-08 21:27:33

hi chinnu03

Welcome to the forum.

Good result.  But I'm afraid it has been done before.  This was, for a while, a GCSE maths project in the UK.  You'd get a grade 'C' for what you have posted but would need to show the whole proof and try possible extensions to get an 'A'. smile

Bob

#6 Re: This is Cool » Procrastinating with some graphing software » 2018-08-08 21:22:43

hi JoshNutter

Welcome to the forum.

Yes, you should be doing your chem engg project.  How long did this take?

forum rules wrote:

No Swearing or Offensive Topics. Young people use these forums, and should not be exposed to crudeness.

My first reaction was to delete it completely but then we lose all your posts. sad   I think you should do the right thing and edit it yourself to something more wholesome.

Thanks,

Bob

#7 Re: Exercises » Geometry » 2018-08-08 21:06:42

Not sure what you mean.  My answer had 'x' in it.  Isn't that algebra?  Algebra doesn't exist independently from the maths topic it occurs in.  This question requires trigonometry (in my opinion) and there's no getting away from that.  You only need to pick one of the multi choice answers.

30:30 is impossible as the angles don't add up to 90.

45:45 won't work as the opposite and adjacent would be equal in this case.

30:60 won't work as the opposite and adjacent are in the ratio root 3 : 1

Bob

#8 Re: Help Me ! » Area of a pool » 2018-08-07 19:54:08

hi Sam_thing

I'm struggling a bit with your working.  The octagon consists of 8 isosceles triangles.  If 'a' is the height of one of these then I agree with your answer.  So next I worked out the area of 8 of these triangles 8 times 2.5 times a.  Where did 40 come from ?

Please re-post your working with some notes at each line to say what you are calculating.

Bob

#9 Re: Exercises » Geometry » 2018-08-07 19:46:49

In a right-angle triangle, perpendicular is 3 times the base.

So I made a triangle with B = 90, AC = 3 times BC.

I assumed that

What is ratio of their opposite angles?

means find the ratio of B:A

Bob

#10 Re: Help Me ! » [ASK] Polynomial » 2018-08-07 19:41:16

hi Monox D. I-Fly

I also get the x^3 coefficient as -1/3  Looks like the question has a typo as none of those answers works.

I get

and this checks out correctly with the given information.

I did this by substituting x = -3, 1, 0 and -2 into ax^3 + bx^2 + cx + d and then solving for a, b, c, and d.

When you put x = -2 , 3x - 2 becomes 3 times -2 -2 = -8.   

Bob

#12 Re: Exercises » Geometry » 2018-08-06 19:25:20

hi Zeeshan 01

Let's call the two angles A and B.  3x and x are the opposite and adjacent, and the adjacent and opposite for these angles.

A/B = atan(3/1) / atan(1/3).  This evaluates to about 3.6

Bob

#13 Re: Help Me ! » Inductive and Deductive Reasoning » 2018-08-06 19:22:39

Thanks Alg Num Theory.  Hadn't thought of that.

But I think I can modify the plan to cover this.  Draw a plane that intersects the three lines in A, B and C.  Then choose D on that line as any point other than C.  Then as before.

Bob

#14 Re: Help Me ! » Area of a pool » 2018-08-06 19:17:51

This has been asked before.  See http://www.mathisfunforum.com/viewtopic.php?id=23304

What on-line course is this from?

Bob

#15 Re: Help Me ! » Inductive and Deductive Reasoning » 2018-07-30 22:46:31

hi Strangerrr

Draw three distinct, parallel lines and another line (a transversal) to cross them all.  Let's say it cuts them at points A, B and C.  Choose any other point on the third parallel that is not C, say D.  Then A, B and D are non colinear.

The parallel through A is a 'line' and any of B, C and D may be chosen as a point not on that line.

Bob

#16 Re: Introductions » Greetings. » 2018-07-17 06:37:18

Hi Joe,

You are indeed very welcome!

Bob

#17 Re: Help Me ! » Questions » 2018-07-17 06:35:28

Number of days in a year is 9 times 15.  Divide 700 by this, ignoring any remainder.

Bob

#18 Re: Help Me ! » Questions » 2018-07-17 06:31:44

Hi ALI1

The way I do questions like these is to call the unknown number x, and then build an equation using the information given.

So for Q1, let the number be x.  Then the true answer is 7x/8 and the wrong answer is 8x/7.  These differ by 15/14.  So make an equation out of this and solve for x.

If you post your attempt we can move on to the others.

Bob

#19 Re: This is Cool » A difficult problem from 1957 » 2018-07-16 06:19:15

Hi Thomson Lee,

Did you want me to provide a solution or what?  No square brackets (to hide) on my Kindle so you just have to close your eyes.

Also no diagram making, sorry.

This is 3D so I'll describe the diagram.

Let TB be the tower in the 'z' direction and BE be an easterly direction with the first observation at E (x direction).

Draw an oblique line NES to indicate the North/South direction (y) with ES= 42.4 ft.

As TEB = 45, BE = TB = h, the height of the tower.

And as TSB= 30, BS = root 3 h.

So in triangle BES we have BE = h, BS = root 3 h and ES = 42.4 with a right angle at E.

So it would be easy to use Pythag to get h (approx) 30 ft I think.  But here's a construction for it.  I have no paper nor instruments handy so the scale might not work.

Draw a verticle line BE' in the top right corner of your page. I'll make it 5cm to represent 10 ft. Construct a right angle at E' and extend the line right across the paper.

On a fresh sheet, construct an equilateral triangle sides 5 cm.  Construct a perpendicular bisector so the height = 5 root 3 and set this as a radius.  On the first diagram, centre B, make an arc to cut the horizontal line at S'

The triangle BE'S' is the right shape for BES but needs to be scaled up.

On E'S' produced Mark X so that E'X = 21.2 cm.

Draw XS parallel to BE' with BS' extended crossing it at S. SE parallel to S'E' will fix E and hence the correct size for triangle BES.  h = BE.

Bob

#20 Re: Exercises » Mathematical Induction » 2018-07-16 02:24:07

Numbers of the form 2n - 1 are always odd.  You are being asked to use induction to add a set of odd numbers. When the set is just 1 + 3, there are two numbers so n = 2, but the end number is 3, not 2.  The formula 2n - 1 with n=2, has the required value 2 times 2 - 1 = 3.  In the same way when  n  = 3, the last term is 2 times 3 - 1 = 5 and so on.

Use ANT's hint to write a general formula for the sum, then prove it works by induction.

Bob

#21 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.

Bob

#22 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.

Bob

#23 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.

Bob

#24 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.

Bob

#25 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%

Bob

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