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

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hi camicat

You have been given a formula

The units match ok, seconds and kilograms, and T is already the subject of the equation. So it's just a case of putting in w = 2 and working out T. Order of steps to take:

divide w by 200

square root that number.

multiply by pi

multiply by 2

There's the required T .

round off.

Bob

hi 666 bro

I did a google search and a Guardian newspaper article had a lot on this. Best answer was this one

They are of enormous use in applied maths and physics. Complex numbers (the sum of real and imaginary numbers) occur quite naturally in the study of quantum physics. They're useful for modelling periodic motions (such as water or light waves) as well as alternating currents. Understanding complex analysis, the study of functions of complex variables, has enabled mathematicians to solve fluid dynamic problems particularly for largely 2 dimensional problems where viscous effects are small. You can also understand their instability and progress to turbulence. All of the above are relevant in the real world, as they give insight into how to pump oil in oilrigs, how earthquakes shake buildings and how electronic devices (such as transistors and microchips) work on a quantum level (increasingly important as the devices shrink.)

Gareth Owen, Crewe UK

The word imaginary wasn't chosen because these numbers are in some way 'made up' or figments of a mad, math professors brain. It comes from the Latin word 'imago' meaning a copy. The imaginary axis is a copy of the real number axis. With the benefit of hindsight maybe another word could have been chosen.

Isaac Azimov tells a great anecdote about being challenged over imaginary numbers. This quote is from his book 'Azimov on Numbers':

member of an audience: Hand me the square root of minus one pieces of chalk!” I reddened, “Well, now, wait — ” “That’s all,” he said, waving his hand. Mission, he imagined, accomplished, both neatly and sweetly. But I raised my voice. ‘Til do it. I’ll do it. I’ll hand you the square root of minus one pieces of chalk, if you hand me a one-half piece of chalk.” The professor smiled again, and said, “Very well,” broke a fresh piece of chalk in half, and handed me one of the halves, “Now for your end of the bargain.” “Ah, but wait,” I said, “you haven’t fulfilled your end. This is one piece of chalk you’re handed me, not a one- half piece.” I held it up for the others to see. “Wouldn’t you all say this was one piece of chalk?

His point is that all numbers are in a sense 'imaginary' that is you cannot actually hold any of them, and we make up rules for them to suit real world purposes. That doesn't stop them being useful in context.

Bob

hi Mitch and Phro,

After a couple of mammoth sessions I think I've cracked it and got a spreadsheet that contains:

roster number

possible rosters

lineup1

bench for lineup1

and same for lineups 2, 3, 4 and 5.

Some cases are impossible because the available players in a roster don't fit with the lineup requirements. These are shown as 'impossible'.

So there are potentially 460 cells of data. I have checked a few at random and all worked out ok … that is to say I could pick a lineup from the roster and have 4 bench players except when there wasn't enough players to fulfil the lineup and got an 'impossible' message.

I've not tried this before so I don't know if I've done it properly, but I have uploaded the sheet to my google drive and shared it with each of you. Please let me know if that has worked and then see if you can find any errors. (Fingers crossed!)

The key to calculating the number of possibilities lies with the bench player options. eg. If the lineup has a QR and there's a second QR on the bench then you have x2 possibilities for who to play. Or if the lineup has 3 WRs and they're all in the team then there are x1 possibilities as you've got no extra WRs you can swap in. I'll start working on a quick way to calculate them all.

In the UK all sporting fixtures have been cancelled so I'm guessing your league won't have any matches at the present. What a disaster!!! Let's hope things can get back to normal soon.

Best wishes, stay healthy,

Bob

I want to go back to the basic rules to make sure I'm not getting anything wrong. Odd? I cannot find the post where you gave the rules link. I can get to the 'invitational' but no rules there. I'm sure it's lurking there somewhere. It's the line up possibilities I need. Please re-post the link for me. Thanks,

Bob

No. In my head is a part program, part spreadsheet, solution to the whole problem. But lots has been happening so it has gone on the back burner. Give me a few days. It's taken me a while to fully get to grips with this. I couldn't understand what was wanted. This is what I think I'm aiming for:

(1) Find all possible rosters using the allowable position rules. (Done this … it's 46)

(2) For each identify the (up to 5) possible line ups and bench player alternatives.

(3) Now assume you've picked your players (14 for each of the above) label them WR1 WR2 etc.

(4) Work out how many alternative line ups that allows.

Bob

Thanks,

Bob

hi

I'm getting

Thanks for the puzzles.

Bob

hi tony123

Bob

hi renevanderlende

I have an idea. When I'm struggling with an area of maths, especially algebra, I try it with numbers to see what's happening. And with a computer program a 'dry run' checking the procedure with numbers.

So, I'd like to follow through the sums to help me get a 'handle' on this and see where it leads.

You've got this line of text:

Lorem ipsum dolor sit amet, exerci dolorem est ad. Sumo rebum prompta vim ad. Legendos expetendis id sed. Ex ius quem accusamus, pri et deleniti copiosae.

That's 154 characters. Is that the 'w' value?

The square place is 2000 by 2000 pixels.

So I calculate the font size.

In MS Paint I can create a square of that size, and choose my font size. Will that text then fit?

And can you also suggest some other values; some of which 'work' and some that don't.

Bob

hi

Still not getting it.

if a string of 1px high is 150px long (1*1*150)

then a string of 2px is 300px long (2*2*150)

then a string of 3px is 450px long (3*3*150, etc)

What does 1*1*150 represent? It's 150, but 2*2*150 = 600, so that's not 300px. And 3*3*150 = 1350, not 450px

Here's a W, expanded using MS Paint.

I've counted and it's 28 across and 20 down. So [fontHeight] = 20. [fontWidth] = 28.

If I want to put 50 of these on screen the [totalWidth] is 28 * 50 = 1400.

So I want to fit 1400 pixels across by 20 down.

Let's say my space is 1500 by 1500

Area available = 1500 * 1500 = 2 250 000

area needed = 1400 * 20 = 28,000

I can do this easily. Note. I haven't square rooted anything. In mathematical 'dimensions' notation:

[length] * [length] = [length] * [length]

If you root one side the 'dimension' of the equation becomes

[length] * [length] = [length] which isn't allowed.

What's the biggest W I could fit? I'll assume the W is 20k by 28k

By comparing areas

20k * 28k * 50 = 1500 * 1500

Solving for k I get 8.9. To be on the safe side, let's choose k = 8

So I can have a character that is 20*8 by 28*8 = 160 by 224

1500/160 = 9 across

1500/224 = 6 down

9 * 6 = 54. I wanted 50 so the text fits.

Bob

hi 666 bro

Thanks for that link to U Tube. Now I understand your difficulty.

Teacher's advice to Khan academy:

(1) Make sure your mic volume is set correctly. Even on 100% volume I could barely hear the speaker.

(2) Plan your 'board' layout in advance so you don't have wobbly lines and have to rub out bits because they won't fit.

(3) Use different colours but only those that have a decent contrast with the black background. Yellow is excellent; purple is very poor and barely readable. Have printed text not handwritten notes so that we don't have to struggle reading your writing.

Sorry 666 bro but I felt I needed to get that off my chest. No wonder you are struggling with this. It's great that the Academy do this for free but they could learn a lot from MIF. I suggest you look instead at this page:

https://www.mathsisfun.com/calculus/limits-formal.html

Compare the two and you'll see why I think MIF is such a brilliant resource.

Hope that helps,

Bob

Sorry. Still insufficient detail. What is f(x)? where did L and c come from?

hi tony123

This post and the area one.

You've posted these in 'exercises'. Did you mean to give us these to puzzle over? Or did you want help doing them yourself. It makes a difference as to how I might answer them (assuming I can do them of course )

Bob

Sorry, don't understand what is wanted here. Please give the context for this. Maybe the lines of working before and after the point where you're not understanding.

Bob

hi renevanderlende

I don't mind spending time on this at all. I'm 'confined to property' here in the UK, probably for 3-6 months, so you are performing a valuable service by helping to stop me going crazy.

I knew you wanted font height, not width; when deciding which variable to call 'x' I thought it would be easier to have [height] = [width] times [1.6], rather than [width] = [height] divided by [1.6]. That's the only reason. If you've got x, its easy enough to times by 1.6 to get the height.

I don't understand your latest formula.

fontHeight = sqrt((finalBoxWidth * finalBoxHeight) / (fullTextWidth + avgCharacterWidth))

I'm assuming the 'finals' are the total space measurements and that [fullTextWidth] is the width of all the characters strung together in one long line. I'll assume that [avgCharacterWidth] is just a small amount to allow for some spacing.

It seems to me that your math logic here is:

[area of all the characters] = [area of available space] so using your variables :

[font height] * [fullTextWidth] = [finalBoxWidth] * [finalBoxHeight]

Then dividing by [fullTextWidth] gives

[fontHeight] = [finalBoxWidth] * [finalBoxHeight] / [fullTextWidth]

So why the square root?

Bob

Do you mean this?

I'll try to put it more simply. I am assuming that the total area of all the characters will exactly fill the available area. So

[total area] = [number of characters] x [area of one character]

If the area available is a square with dimensions S and S, then [total area] = S times S = S²

If a character is x units wide and 1.6x units high then [area of one character] = x times 1.6x = 1.6x²

[number of characters] is N

So the formula not in words becomes S² = N * 1.6x² where * means multiply. (It would be even more confusing if I used a times sign when there are xs around as well)

There's a neat device you can used to re-arrange a formula of this type [number1] = [number2] times [number3]

You make a triangle, divide it with a horizontal line and divide the bottom part with a vertical line. Put [number1] in the top space and the other two bits in the bottom spaces, like this:

If you want to calculate S², multiply the two bottom elements like I have already done.

If you want N, cover it up and you have S² over 1.6x² so the re-arranged formula becomes:

If you want 1.6x² cover it up and you have S² over N so the re-arranged formula becomes:

In words: [area of one character] = [total area] divided by [number of characters]

So if you know both of the right hand side elements you can calculate 1.6x² . Then divide by 1.6 to get x² and finally square root that to get x.

If you actually did all that then the characters would fit to the very edge of the space with no margin. Also, it still might not work perfectly because either x or 1.6x might not fit the width and height respectively. But it would give what we mathematicians call 'an upper bound'. ie the actual width that is ideal cannot be greater than the calculated x. You could solve the margin problem by reducing S a little and you could test if S/1.6x and S/x both come out to be whole numbers. If not, reduce x a little until both do fit.

Bob

hi Scar

Welcome to the forum.

Sometimes it's easier to calculate the probability that it won't happen and then subtract from 1. You won't get berries if you buy all male or all female plants so you could do this:

1 - P(all male) - P(all female)

Bob

hi renevanderlende

Sorry I've been quiet for a while. The sun is shining and the garden beckons. The calculations look ok. And version 2.1 looks ok too except the box with 123456 is consistently only showing half a six. I've tried changing the page magnification (if I've understood correctly your code should allow for this) and it did. But still half a six at all sizes.

Out of curiosity I put your text into google translate. None the wiser. I was hoping for a profound peace of philosophy there.

Bob

I can think of two ways of interpreting this question. My first thought was

1, -1, 1, -1, 1, -1, ……..

The sum of terms oscillates between 1 and zero depending on whether you've reached an odd term or an even.

But then I thought: can you really refer to this as a limit if we're not getting closer and closer to a value?

Here's an example where the terms do tend to a limit but oscillating either side of that value:

1/1, 2/1, 3/2, 5/3, 8/5, 13/8, 21/13, …….

The numbers are successive numbers in the Fibonacci sequence (term r+2 is calculated by adding term r and term r+1) and tends to the golden ratio, 1.618....

Bob

hi Fruityloop

There's a useful tool that will help here. It's called a tree diagram.

So don't throw the book yet. But there's plenty of ways of setting up this problem without exhuming graves. I think the author needs psychiatric help.

Bob

hi renevanderlende

I haven't been able to execute this code.

Bob

hi 666 bro,

In general you couldn't do this by hand because the calculations would be too complicated.(power series for log) But some values can be determined.

eg. log 10 = 1

log 100 = 2

log 1000 = 3

log(0.1) = -1

log √10 = 0.5 because to multiply two numbers you add their logs.

When I was at school (before modern calculators!) we had to learn that log 2 ≈ 0.3010, log 3 ≈ 0.4771 and log (pi) ≈ 0.4972

So I could get log 4, log 6, log 9, log 2pi etc.

As a class exercise when I became a teacher, prior to 'doing' logs for the first time, we would make our own slide rules by making two strips and marking 1 at one end and 1000 at the other. You can mark on 10 and 100 at equal intervals between, so, in effect, the ruler is 'measuring' logs. Then estimate the position for 32 by saying 1024 ≈ 1000 so 32 is half way along. And 2 is 1/5 of the way between 1 and 32 (2^5 = 32) so then you can use that to estimate 5 (log 5 = log 10 - log 2). By marking the same log scale on both strips you can transfer measurements from one to the other and hence get 64, 50, 500, 640 and so on. Estimate the position for 49 as it's nearly 50, and hence 7. 81 is close to 80, so then 9 and then 3. All the time practising the rules of logs by discovery.

Bob

hi renevanderlende

Thanks. Those examples have helped a lot. Thinking about the long line of text, there's also a problem with breaks in a word. Newspapers sometimes have to break a word, with a - , and I wonder how you'd want to handle this. As a start I thought I'd get a best case answer like this:

Say a character has dimensions 1.6x and x. And the word space is a square S x S. To fit N words the ideal would be if all the available area is filled exactly, ie. 1.6x^2 x N = S^2 So then I investigated the graph N = s^2/(1.6x^2). Here's the result:

If you then choose a value for N, you can see what value of x will enable this. It's not a perfect answer of course because x probably won't be a value for a font and also the characters have got to actually fit across and down the space, but it's given me an upper bound for the font size.

Maybe from there you could shrink down until you get a fit that actually meets the font size and need to have a whole number of characters per line.

I'll keep working on it.

Is it possible you could build something like this into an example to see what happens and post a picture. Then I can see the snags and plan an improvement.

Bob

Welcome to the forum.

hi claudineskiles

Q1. If two lines are parallel then the numbers in front of x and y must be the same. So you can rewrite Ax + By = 7 as 2x – 7y = C. Then put x = 1 and y = 1 in the expression 2x - 7y to answer the question.

Q2. I suggest you make up a number for how many games there will be in the season. It won’t affect the percentage answer you get**. So let’s keep it easy and say there are 100 games to play. 80 have been played so work out 40% of that to get the number of wins so far. The question requires that 50 games are won and 50 lost so you can work out how many more wins are needed and write that as a percentage. ** There’s a way you can demonstrate this for yourself by re-working the problem with a different choice for the total number of games.

Q3. Add up the fractions to get how much water they have in total. Divide by 3 to share it equally. Both James and Karen have enough to give Lou some of their water … but different amounts. So work out how much Lou will get from each and that gives you a way to determine how to divide the $2 fairly.

I’ve left you with some work to do with each question. Please post back your working and I’ll check it. If you get stuck partway I’ll be able to help you complete the question.

Bob

My mistake. I missed that on the rules page. Amended result:

QB RB RB WR WR TE PK TM + WR WR = 1

QB RB RB WR WR TE PK TM + RB WR = 2

QB RB RB WR WR TE PK TM + RB TE = 4

QB RB RB WR WR TE PK TM + WR TE = 5

I got these in a different order but I've re-arranged them to fit the stated options on the rules page. Notice = 3 is missing because you haven't got 4 RBs in this roster. So far so good.

But I still don't get the spreadsheet. What does 56 measure then? I thought it was measuring the above. But obviously not. Pretend I'm really thick (easy to do in the current situation) and explain what 56 is a measure of.

Thanks,

Bob