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#1 Re: Help Me ! » Circle Problems » Today 21:05:08

Bob

hi tatekuhic

Welcome to the forum.

If you have worked out h (well done!) then you know the full equation of both circles, and hence the centre of C2.

This means you can work out the equation of AB, and so you can substitute for y in each circle equation and so work out the coordinates of A and separately, B.  This will also give you the coordinates of the second point on AB where it crosses C2, lets call it point D.

Because AB gives the line that is a diameter of each circle, the tangents must go through A and D, so you have all you need to form the equations of these new circles, centre and a point on the circumference.

I haven't tried the question yet.  If you are still having trouble with any part, post again and I'll try to help some more.

Bob

#2 Re: Exercises » Find the real solutions » 2022-01-17 21:50:08

Bob

And now for Q1:

First some preliminaries:

If

... A

and if

... B

And now the question

All logs in base 7

using A becomes

Divide by x

Replace 1 by log(7) and simplify the logs

Using A again gives

I will replace log(x) with X

Now what does the graph of the left hand side function look like?

It has two components. The first is a standard power graph, going through (0,1) and rising thereafter.  The second is is similar but, as 3/7 is under 1, it will come down from large y values to the left of the y axis, again go through (0,1) and then drop off to zero as X gets larger.

Together the second component will dominate when X is negative; then the first will dominate.  So the graph will start high, drop down to (0,2) and at some stage rise again.  Note that (0,2) gives one solution to the question.

Differentiating using B

The ln(3/7) is negative so this will give one turning point. It must be a minimum because of the overall shape already established; (0,2) could be the minimum; otherwise there will be exactly one more solution.

I can see that X = 1 gives x = 7 leading to 11/7 + 3/7 = 14/7 = 2. So here is the second solution.

So the equation has two solutions, x = 1 and x = 7.

Bob

#4 Re: Maths Is Fun - Suggestions and Comments » Possible to report spam members? » 2022-01-16 01:46:47

Bob

hi Mathegocart

Admin and Mods check the forum most days and deal with spammers and their posts. 

At the bottom of every post you should be seeing a 'Report' button that flags up any post of concern.  Admin and Mods get the report message when they log in and can jump to the post and do something about it.  Are you seeing that button?  Some forum features were lost when we moved to the new server and things that I can see are not necessarily visible to all. So please confirm if you have the report option and try it out by reporting this post for me. You should get a message space to give your reason, so please make it clear it's a test for me.

If you stumble on an edited post that isn't showing as a new post you should still be able to report it.

I've had a look through the options available to me on the forum.  I can switch off the information on the post side panel so that a members website doesn't show, but that would have apply to everyone, not just one member.  Similarly signatures can be turned off.  I'm very reluctant to do either of these as good members suffer along with the bad.

Bob

#5 Re: Exercises » Find the real solutions » 2022-01-15 08:04:19

Bob

Here we go:

All real log functions are undefined for x  ≤ 0 so x > 11 or the log powers don't work.

Replace x by X = x - 11

The equation becomes

For brevity I will use all logs in base 2 from now on.

case 1. log(X) = 0

case 2. if log(X)  ≠ 0 then it can be cancelled leaving

Raise 2 to the power of each side (you could also call this antilog}

The quadratic formula has one negative answer which contradicts the domain for X so

Bob

#6 Re: Introductions » Hello! » 2022-01-15 06:15:12

Bob

Thanks zetafunc,

These others had slipped past me so thanks for the info.  It does look like that word attracts them so I've cleaned up all, and removed a welcome from me that no longer applies. Would you mind editing out the 'e' word please.

Ganesh: I have left your posts but would prefer you take out the ones that now have no context. Thank you

Bob

#7 Re: Exercises » Find the real solutions » 2022-01-14 20:04:33

Bob

Hurray! I've got an answer for (2) that is confirmed by wolfram alpha.  Very busy today so I'll post it when I've got more time.

Bob

#8 Re: Help Me ! » Parallelism Need some serious help » 2022-01-14 20:02:07

Bob

hi Scuba_Joe

Welcome to the forum.

Compuhigh has asked us not to publish their copyright worksheets so I have removed the questions you posted.  They have better lawyers than we do so it's best not to upset them.

We will certainly help you with any topic in general but we won't do your homework for you.

For questions like these it is essential to include a diagram too.

Looking forward to helping once these matters are sorted out.

Bob

#9 Re: Maths Is Fun - Suggestions and Comments » Possible to report spam members? » 2022-01-14 19:57:14

Bob

hi

I used to ban the spammer, then search through for all their posts and remove these too.  Then I discovered that I have a delete member option which takes out the member and all their posts in one click.  So this has become my preferred option.

Mostly these posts are complete rubbish or blatant adverts but a new type is political posts.  These posters are putting links to propaganda and/or stories the poster thinks need to be in the public domain.  As we don't want to get embroiled in such arguments I delete these too.

A recent new member had put a website link to an essay writing service.  I'm wondering if it's really worth having such links as an option to make it less easy to create such advertising.

Bob

#10 Re: Introductions » Hello! » 2022-01-14 19:47:50

Bob

hi Mathegocart and Ganesh,

I noticed that keiraec had a link to an essay writing website.  I have deleted the link. Normally I would remove the poster too, but what do you think?

Bob

#11 Re: Coder's Corner » algorithm for tiles a floor , very important to move on » 2022-01-14 01:39:24

Bob

I had to look up the use of % here as I'd not met that notation before.  Here's what I found:

Modulo is a math operation that finds the remainder when one integer is divided by another. In writing, it is frequently abbreviated as mod, or represented by the symbol %.

So % finds the remainder after dividing by 2.  So what I said and what the book says are the same.  I note that the book is also applying the mod to the row and column values before the total is found.  mod 2 division will 'detect' even and odd numbers .

The use of % more than once is necessary.  If you just had

color =  (row % 2) + (column % 2)

There are the four cases are 0 + 0 = 0;  0 + 1 = 1; 1 + 0 = 1; and 1 + 1 = 2

The last answer would not give a 1 or 0 so would generate an error.

But I think this would work:

color = ( row + column ) % 2

Perhaps you could try it.

There's a short page about modulo arithmetic here: https://www.mathsisfun.com/numbers/modulo.html

Bob

#12 Re: Coder's Corner » algorithm for tiles a floor , very important to move on » 2022-01-13 20:51:25

Bob

You need to devise a way to tell if COL + ROW = an EVEN number or an ODD number.  EVEN means a BLACK square. ODD means a WHITE square.

This calculation will achieve that:

total = col + row

value = total/2 - INT(total/2) where the division is a normal division with possibly a remainder, and INT yields the integer value of the answer ie. discards the fraction.

Look what happens when total is even and when it is odd.

total even: value = <another integer> - <same integer> = zero.  This happens because there is no fractional part to discard.

total odd: value = <an integer + 0.5> - <same integer without the 0.5> = 0.5.

So the value calculation enables you to detect ODD and EVEN as needed.

Bob

#13 Re: Exercises » Find the real solutions » 2022-01-11 20:32:40

Bob

With one or both?

Please give more details.

LATER EDIT:

Ok I see what is wrong (I think) with each one, so it's back to the drawing board for me.  This may take a while.

Bob

#14 Re: Help Me ! » How would i start re learning mathematics ? » 2022-01-11 04:30:13

Bob

hi dummy39

Welcome top the forum.

I think this site https://www.mathsisfun.com/ is as good as any book and its FREE!

Bob

#15 Re: Exercises » Find the real solutions » 2022-01-10 23:37:16

Bob

hi tony123,

I think these are valid:

Bob

#16 Re: Exercises » congruent » 2022-01-10 02:27:00

Bob

hi tony123

Thanks for another interesting puzzle. I think this is a proof but it's a bit messy.

Bob

#17 Re: Exercises » solve equation » 2022-01-07 22:57:39

Bob

hi zetafunc and tony123,

Thanks for providing me with ways to do this.

Bob

#18 Re: This is Cool » ‘Proof’ of 1=0 » 2022-01-07 03:36:17

Bob

Up to this point all is ok.  You can test this by evaluating the arithmetic.

This is ok because both sides evaluate to 1/4.

It's what happens next that is not correct.

If you just square root both sides you get

and that is not correct.  It's the careless square rooting that causes the false result.

Because a square root can be + or - the equation is valid if one of the signs is changed.

Take it as a warning not to square root without thinking about what is happening.

Calculators will only give a positive square root because they are programmed to give a single answer.  Some people regard the root sign as meaning this which is fine.  But it's always worth thinking about the negative case, because that might give the answer you want.

Bob

#19 Re: Help Me ! » Angles of a Parallelogram » 2022-01-07 02:08:16

Bob

hi

I've checked out the Latin. regula is the noun meaning rule.  In Latin, nouns take different endings depending on the status of the noun in the sentence (eg. cactus, cacti). regula is a 1st declension noun with ...am when it's in the accusative ie the object rather than the subject in a sentence.  There is no Latin word regulum but I have found internet sources where it is used with that spelling; so I suspect you picked it up from someone who just spelt it wrongly. ?? But do we really need a book of Latin commonly encountered in the workplace ?? Unless you're a doctor, lawyer or Latin teacher I'd have thought it unlikely you'd meet much Latin. 

probat comes from the verb probare meaning to test, prove, or approve.

exceptio is the Latin noun in the nominative (subject) case (3rd declension) meaning exception.

Thus we have a simple Latin sentence structured <subject> <verb> <object> and

my attempt at a translation is "The exception proves the rule"

If you truly want to make use of it in your signature then you need a 'rule' as well, hence my original question.

How about "MIF members are fun. Exceptio probat regulam!" But this would imply that those who are not members are not fun.  I'll leave you to decide if you want to go that far.  smile

Bob

#20 Re: Exercises » solve equation » 2022-01-07 01:45:24

Bob

hi tony123

Thanks for another great exercise.

I'm still stuck with

. Prove that

Easy to show it's a possible value for x+y and the graph suggests it's unique, but I still cannot prove it properly.  Hint please. smile

Bob

#21 Re: This is Cool » ‘Proof’ of 1=0 » 2022-01-06 22:16:43

Bob

A quicker route to a similar contradiction would be:

Now the error is obvious. Every number has two square roots, so you have to take care when using 'rooting' in an algebraic manipulation.

Bob

#22 Re: Help Me ! » Angles of a Parallelogram » 2022-01-06 21:51:41

Bob

hi CurlyBracket

That's the value of k that I got, well done!

I'd not seen your Latin phrase before so I googled and couldn't find 'regulum' but rather 'regulam'. 

It seems to be an abbreviated version of:

"exceptio probat regulam in casibus non exceptis" meaning the exception that proves the rule.

In maths that is bizarre because any exception (counterexample) means the rules hasn't been proved, so I investigated further.

Here's an example of what I think it really means:

A museum sign says "Entry free on Sundays".  It doesn't say what happens the rest of the week but most people would infer that there is a charge on the other 6 days.  So Sunday's rule is the exception that implies a rule that applies for the rest of the week.

A similar example would be "No parking here on market days".

Bob

#23 Re: Help Me ! » Angles of a Parallelogram » 2022-01-05 20:42:47

Bob

hi CurlyBracket

Welcome to the forum.

Here's my diagram for this.  I've used x and y to indicate the equal angles.

zE26toN.gif

I can give you a solution hint if you need it.

Exceptio probat regulum!

Which 'rule' do you have in mind? smile

Bob

#24 Re: This is Cool » Are these unique numbers? » 2022-01-03 01:25:47

Bob

Unless I've completely misunderstood, it has to be unique.  The number is 36 so the product begins 3 x 6 x <something> and has to result in 36.  There's only one solution to this equation:

3 x 6 times [letter x] = 36  => 18x = 36  => x = 36/18 = 2

Bob

#25 Re: This is Cool » Are these unique numbers? » 2022-01-02 02:47:15

Bob

hi Roger Austin

Welcome to the forum.

I think that there may be many such.  This is how I'm going to start looking.

You seem to be happy to multiply by 9 or 14 etc so I'll assume all you want is [number] = [any multiple of] [product of digits]   

So algebraically:

Find a b and n so that 10a + b = nab for whole numbers a, b and n.  {this is the two digit version}

I'll see if I can find some more .

On my spreadsheet allowing a and n to be from {2,3,4,5,6,7,8,9} I only got one more: b = 4 and 24 = 2 x 4 x 3

Three digits may take longer.

Actually not much longer as it was quick to add 100 to my 10a + b.

I found four in the one hundreds:

a = 7, b = 5 , n = 5                    175 = 1 x 7 x 5 x 5
a = 2, b = 8, n = 8                     128 = 1 x 2 x 8 x 8
a = 3, b = 5, n = 9                     135 = 1 x 3 x 5 x 9
a = 4, b = 4, n = 9                     144 = 1 x 4 x 4 x 9

Bob

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