Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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hi MathWA

Cuts have got to leave congruent pieces so four cuts at right angles to the axis and equally spaced would be one way. This would make 5 equal shorter cylinders.

Or you could make the cuts through a diameter of the circular ends and along the axis. These cuts would have to be 45 degrees apart.

And there are several ways that mix these two types of cut. I'll leave you to find them.

Bob

hi malaysiawill

Welcome to the forum.

These look like CompuHigh questions that have been posted before. Try doing a search of the forum old posts.

Bob

hi Sisyphus

You have said it's a transformation, so maybe it will help to ask what transformation?

Bob

Thanks for clearing that up .

Bob

hi Relentless

Thanks for posting that. It made me check and edit my hint. Please read my corrected version.

Bob

Oh. To describe an angle you really need the three points that make the angle eg. MWS. I was assuming the question meant RWS.

If it's MWS + RWM then that won't be 180 as the points aren't cyclic.

Bob

hi leopragi

Welcome to the forum.

You must have some sort of input device and viewer to use this forum. So these three; <input> .... <forum> .... <view answer> form such a machine. It's a powerful one too because you can ask your question in a variety of ways and there'll usually be someone who will give you a mathematical answer. Other 'machines' are not so flexible

Bob

hi dazzle1230

The second hand is the radius and it sweeps out circles as time passes. So firstly work out the circumference of that circle (2 times pi times radius) then work out how many circuits it will do in that time . One per minute. Don't use 3.14 or other for pi ... just leave the answer a N.pi where N is a number.

Bob

MathWA: I don't see the picture either, but, do you really need it for this question?

hi lightcreatortwentyjuan

Welcome to the forum.

You should have a line of useful symbols at the top of each forum page which includes this one: ≥

Part (a). You can re-write this as a + a + ... + a (n of these) + 1 + 2 + ... + n-1.

The sum of n-1 natural numbers is given by n(n-1)/2

Part (b)

So you can put the answer for part (a) equal to 100 and look for possible whole numbered solutions.

Bob

hi nikita,

In the instruction page for asking for help it says "We are happy to help! But we don't do your homework for you."

I thin that applies to AoPS problems too.

So, if I help, I try to provide hints to the method rather than just posting an answer. If you would like such hints please post again and I'll do so.

Bob

hi MathWA

Welcome to the forum.

I followed this back to the AoPS page and got a 'lack of the right font' error message.

I agree with ishankhare's interpretation since it actually leads to a problem that can be done.

Some mathematicians use a little m to mean 'the measure of'. It can be used for either a length or an angle measure so I avoid it myself as I think it creates confusion.

Firstly you're told the (angle) measure of the arc RS. If C is the centre of the circle this means that angle RCS is 40 degrees.

There is a Euclidean theorem that the angle at the centre is twice the angle at the edge of the circumference (see http://www.mathisfunforum.com/viewtopic.php?id=17799 posts 6 and 7)

So angle RWS = 20 = angle RTS.

So the problem makes sense if it's asking what these two angles add up to.

Hope that helps,

Bob

hi pavya1

You need to calculate P(0), P(1) .... P(15) using the formula and add these up. That's the probability of safety, so then do (1 minus that value) to get the probability of jeopardy.

Bob

hi pavyas

Welcome to the forum.

If the random variable V(x)=10 , what is the value of V(2x-1)?

If that's all you're given, then V is constant for all x, so V(2x-1) = 10

Bob

hi MathsIsFun

All looks good to me

Bob

hi guys,

This problem can be done by using the 'triangle inequalty'. If x, y and z are the sides of any triangle then x < y + z

THis is an outline of what I did:

(a) AD < AB + BD

Write another expression for AD and add them together.

(b) You can get one half of the inequality by using the first part. Write two more expressions like this for BE and CF. Then add them together.

Then AB < BG + AG and similar expressions for BC and AC. Add these together and also use 3BG = 2BG and two similar.

Bob

hi apsara123

I would do Q1 just by listing all the possible combinations of 2,4 and 6 and eliminating those that don't make a triangle.

Q2. See http://www.mathisfunforum.com/viewtopic.php?id=22423

Bob

hi evene

No difference. There's no absolute authority on which maths symbols to use or even how to spell math. So you'll find both are used. It's only the squiggle on the page after all. It's the square root of minus 1 either way. Or negative one. And that could be ee-th er or ahy-th er

Bob

hi Sisyphus

Welcome to the forum.

No apology needed. Post away. Your queries may be the only ones I can do, so you'll be making me feel better at least.

Bob

Throughout my life I learnt no music nor how to play an instrument. When I retired I decided to build my own electric guitar. Still struggling to play but I did discover some interesting maths in those frets. If you like, I'll explain in easy to follow steps.

hi Agnishom,

I've not heard of that. No, my example isn't even about geometry.

Back when I taught classes, I had a sequence of lessons on 'The Nature of Proof'. It only touched the surface, of course, but I was trying to show why mathematicians want to prove things and give some ideas about how they go about it.

My introduction was this:

Consider the quadratic

Pick a counting number for n and evaluate the formula. Is the result a prime number?

n=1, formula = 43 This is a prime.

n=2, formula = 47 This is a prime.

n=3, formula = 53 This is a prime.

........

n= 20, formula = 461 This is a prime.

You might like to try a few values for n yourself.

By the time I'd been round the class and everyone had had a go, they were happy to agree that **this formula generates primes**.

If you've spotted a value of n that disproves this, well done!

If you set it up on a spreadsheet, you can try out a lot of values. You will find it is 'prime heavy', at least for quite a while; I haven't tried all values of n.

And that's the point. Just because a result keeps on happening, doesn't mean it will keep on happening.

What is the first value of n which doesn't give a prime? And the next ? Say this one is 'm'. You'll find that m+1 gives another prime, and then there's another long sequence of primes before the next non prime.

Conclusion. Just because an 'experiment' generates lots of confirming results doesn't prove that result is true.

I've probably told this before (sorry) but in another lesson a pupil taught me something. For years I had done the following:

Get everyone to draw a triangle (any will do); measure the angles and add them up. If you have a class of 30, you'll get lots of answers such as 178, 179, 180, 181, 182. Up until this particular day the classes had always been happy to conclude that the angles of a triangle add up to 180 and the variation is just experimental error chiefly because of the thickness of the datum lines on the protractor. But this one time a pupil came up with a different answer. He said the angles of a triangle always add up to a number that is close to 180. On the basis of the data you cannot fault him. Worth a gold star I think.

Anyway, do tell me about the Euclid conjecture. Thanks,

Bob

I used Euclidean geometry and algebra:

Let AXB = ABX = x : XBC = y : and ACB = z

x = y + z (external angle of a triangle XBC is sum of two internal opposites) ..... (1)

x + y - z = 39 (given)

Substitute (1) into the above

y + z + y - z = 39 => 2y = 39 => y = 19.5

Bob

ps. Would you like to see an example to show why experimental maths is not always reliable ?

hi NakulG

In your final line you have y = 3 or -1. Clearly the negative cannot be a solution to the original.

Incorrect answers can arise when you use squaring as a way to solve an equation. Consider this simple example:

x = 3 => x^2 = 9 => x^2 - 9 = 0 => (x-3)(x+3) = 0 => x = 3 or x = -3

So it is important to consider if an apparent answer actually fits the problem set. -1 doesn't. There is no more significance to that than x = -3 in my example.

Here's another: A rectangle is 2 units longer than it is wide. Its area is 15. What are its measurements?

Let the width be x. Then the length is x + 2 and x(x + 2) = 15 => x^2 + 2x - 15 = 0 => (x + 5)(x - 3) = 0 => x = 3 or -5

So the measurements are 3 and 5, or -5 and -3. Clearly this second 'answer' cannot be right as a rectangle cannot have a negative measurement.

This last example shows again how spurious answers can crop up.

Solve

Squaring

squaring again

I'll leave you to work out which is a solution and why the other appears.

Bob

hi Melcalci

The simple answer is that it is defined that way. But think first about the median and you'll see the logic behind it.

The median is the middle number when n is odd and half way between the two middle numbers when n is even. You can express this algebraically by saying it's the (n+1)/2 number in the list. So, eg., when n = 13 we compute (13+1)/2 = 7 and this correctly identifies the middle number as half way along the list; 6 before it and 6 after it.

The quartiles must be half way between the start and the median and half way between the median and the end. So using (n+1)/4 and 3(n+1)/4 seems to do the trick.

Here's an example where the list does divide neatly into 4 sets:

1,3,8,9,11,13,16,17,19,22,24,26.

There are 12 numbers in the list so the positions are LQ at 13/4 = 3.25, M at 13/2 = 6.5 and UQ at 39/4 = 9.75

Those markers split the list into (1,3,8), (9,11,13), (16,17,19), and (22,24,26) That gives you equal quarters as you'd want.

If you use n/4 = 12/4 = 3 then 8 lies on the border and isn't in either the bottom quartile nor in the next.

Of course it gets harder to see the logic where n+1 doesn't neatly divide by 4 but you still get the best split available . Eg. for A the LQ = 3.75, M = 7.5 and UQ = 11.25

That puts 3 numbers in the lower quartile, 4 numbers in the next quartile, 4 numbers in the next and 3 in the upper quartile. The sets are not equal but they never were going to be with n = 14. A split of 3-4-4-3 is reasonable and the LQ is half way between 0 and 7.5

Hope that helps,

Bob

ps. I did a google search for 'formula for lower quartile'. Both BBC bytesize and mathsteacher.com.au give the formula as I stated. The Wikipedia page gives 3 methods for calculating the quartiles which do not all give the same results.

The maths national curriculum (UK) does not specify how you're meant to calculate a quartile position, nor do the exam boards AQA, Edexcel and OCR. If you are preparing for an exam you really need to check with the board concerned. If you say what exam I'll try to research this further.

hi Melcalci

Welcome to the forum.

Had to look at the 'original' imgur image to clarify that scale.

The 'formulas' for determining the fixed points are lower quartile = (n+1)/4 median = (n+1)/2 and upper = (n+1)x3/4

For A the lower quartile is at 55 and n = 14. From above the LQ is at 15/4 = 3 and 3/4 so the scores above this are 4th,5th,6th,7th,8th...14th = 11 scores in total.

For B the median is at 55 and n = 13. median is at (13+1)/2 = 7th score. So there are 6 scores above this.

So the total number of scores above or equal to 55 is 11 + 6 + 1 (the median for B) = 18.

Bob

I realise in your diagram that triangle PQT is right-angled;

Where did you get that idea from. There's no right angles in my diagram.

My diagram wasn't meant to be accurately drawn anyway but is like phrontister's.

Other problem:

Let the width be w, then L = 2x + 2w and xw = 7.5

So you can eliminate w and make L the subject.

Bob