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#1 Re: Puzzles and Games » 7x + 11y = N » Yesterday 21:59:52

Hi Amctest 123

Welcome to the forum.

There is an entire "family" of straight lines where 7x+ 11y = something.

They are all parallel and as N gets bigger the lines move away from the origin. Joining (11,0) to (0,7) will produce one.  What is N for this case?

That should give you a starting point.

Bob

#2 Re: Help Me ! » area of polygons review » 2017-10-19 22:44:56

Hi Taylor

Since you joined the forum you have started 6 threads  Each consists of 20 homework questions  I asked you to read the forum policy on help but you are still doing it several posts later. This is likely to result in you being banned.  Please reply with a proper request for help making it clear where your difficulty lies.

Bob

#3 Re: Help Me ! » Probabilty of a family's children » 2017-10-12 23:34:54

hi Abbas0000

You are correct that there are 8 possibilities:

BBB
BBG
BGB
BGG
GBB
GBG
GGB
GGG

We'll assume boy or girl is equally likely. (in fact it's not quite 50:50 but pretty close).

Each possibility has a probability of 1/8

Three of those listed have exactly two girls so P(two girls) = 1/8 + 1/8 + 1/8 = 3/8

There a formula for this:

p is the probability of an event; q = (1-p) is the probability of the event not happening; n is the total number of events; r is the number that we want.

You can find more on this here:

http://www.mathsisfun.com/data/binomial … ution.html

Bob

#4 Re: Puzzles and Games » New Puzzles 1 » 2017-10-11 21:01:06

hi RVave

Welcome to the forum.

Some puzzlers like to make a simple thing complicated just to make it hard.  The "As I was going to St Ives" puzzle is an example.

So is Lemonink.

The amount transferred is irrelevant.  Let's say we have a glass with 100mL lemonade and a second with 100mL of ink.

Start tipping liquid from one to the other in any way you like and as many transfers as you like. Stop when the first glass has exactly 100mL of liquid in some mixture.  There's no way we can know how much is lemonade so let's say it has x mL of lemonade.  As it has 100 altogether there must be 100 - x mL of ink.

The second glass must also have 100mL altogether as long as we didn't spill or drink (!) any.  How much is lemonade?  Well we know where x units is so the second glass must have 100 - x mL of lemonade.  And as there are 100mL altogether in that glass there must be x mL of ink.

So in the first glass lemonade : ink = x : 100-x and in the second ink : lemonade = x : 100 -x

I did read your explanation, but I got totally lost half way through.  Sorry.

Bob

#5 Re: Help Me ! » Need a beginner Algebra book that covers these ? » 2017-10-08 22:51:28

hi awholenumber

I don't know a good book for this but why not check out the Maths Is Fun teaching pages.  If you start at

http://www.mathsisfun.com/prime-factorization.html

you'll get a long way with what you want and there are many cross-links between pages so you can easily find other pages to give you more.  Or use the search with 'factors' or something similar.

I strongly recommend these pages because (1) they're free! and (2) the explanations are really clear.  Sometimes there are really helpful animations with interactive features and every page ends with some test yourself questions.

The site was made by the chief administrator, MathsIsFun, himself.

Bob

#6 Re: Help Me ! » Help with a proof » 2017-10-08 22:41:18

hi iamaditya

When I've got a proof like this I find it helpful to consider an example first.

Let's say a = 48 = 2 x 2 x 2 x 2 x 3
and        b = 56 = 2 x 2 x 2 x 7

The hcf = 2 x 2 x 2

and lcm = 2 x 2 x 2 x 2 x 3 x 7

so hcf x lcm = (2 x 2 x 2) x (2 x 2 x 2 x 2 x 3 x 7) = (2 x 2 x 2 x 2 x 3) x (2 x 2 x 2 x 7) = 48 x 56

The common factors occur in both the hcf and lcm and the not-common factors occur in the lcm.  So the re-arrangement allows us to pick out one set of common factors together with one set of not-common factors for the first number and what is left is the factors for the other.  Here's an attempt to make that rigorous:

Suppose a = hcf x N where N is the not-common factors, and similarly b = hcf x M

Then the lcm = all the common factors once and the not-common factors from both = hcf x N x M

So hcf x lcm = (hcf) x ( hcf x N x M) = (hcf x N) x ( hcf x M) = a x b

Hope that helps,

Bob

#8 Re: Maths Is Fun - Suggestions and Comments » Remainder Theorem » 2017-10-07 19:23:35

hi Arman Ansary

Welcome to the forum.

Start by writing

where Q is the quotient and R the remainder, both functions of x.

R must be a linear function (ax + b) because if it had higher power of x, you would be able to further divide it by x^2 -1

If you put x=1 you will get one equation with a and b; and putting x = -1 will give another.  Together you can solve for a and b.

Hope that helps,

Bob

#9 Re: Help Me ! » A simple question is not being able to solve by me » 2017-10-05 04:07:52

I think you're going to love this:

multiply the first term top and bottom by y and the third by xy:

smile

Bob

#10 Re: Help Me ! » How do you join the Forum? » 2017-10-04 22:09:30

She first logged in as a guest, but later decided to become a member.  The 'system' does not treat these as the same person so her early posts appear as a guest.

From memory, I think she used the same username.

Bob

#11 Re: Help Me ! » How do you join the Forum? » 2017-10-04 18:52:53

hi Monox D. I-Fly

Sarah did join and had a lot of help with her studies.  She stopped posting once her work was complete.

Bob

#12 Re: Help Me ! » A simple question is not being able to solve by me » 2017-10-04 18:45:29

Yes, me too.

I managed to shorten the proof by re-writing like this

and so on.

I feel as though there ought to be a 'slick' way to do this easily, but I haven't found it yet.

Bob

#13 Re: Help Me ! » A simple question is not being able to solve by me » 2017-10-03 23:24:32

I've assumed that it's xyz = 1

Then I put everything over a common denominator, expanded all brackets and simplified.  The numerator and denominator came out the same so the expression = 1

It was a very complicated expansion, taking two sheets of A4, even with simplifications such as x.y^2.z = y so I won't post it here just yet.  I'll look for an easier way first.  Phew!

later edit: got it down to half a page of working but still the same method.

Bob

#15 Re: Help Me ! » A simple question is not being able to solve by me » 2017-10-03 21:10:24

I'm wondering what happens if you just expand the brackets.  Bit buzy right now.  I'll have a look later.

Bob

#16 Re: Help Me ! » geometry » 2017-10-03 20:33:57

hi iamaditya

This result is true for any prism and cone which have the same cross-section / base.  Curiously it's easier to prove in the general case than for a specific cross-section so I'll do that.

Sr4RzbO.gif

The first picture shows a prism which has a completely irregular, wiggly perimeter.  The second shows the same cross-section; this time forming the base of a pyramid.

Firstly: a general principle for calculating a volume. If you know the area of a shape and then extend it in a third dimension at right angles to the base then the volume of this solid will be area of shape times the height of the extension.  Imagine the area divided into square centimetres.  Each square makes a cuboid when extended so the result is true for these.  If the shape is irregular, make as many square centimetres as you can, and then fit square millimetres for the rest.  The result will be true for these square millimetres.  Continue to sub-divide any remaining area into smaller and smaller squares and the result continues to be true.  Hence it is true whatever the start area.

Now for the prism calculation:

Imagine the prism is divided into many small slices, each having the same cross-sectional area, A and 'height' delta x, a little bit in the x direction


Now for the pyramid.

The defining property for a pyramid is that the cross-section stays the same shape as the base but with a steadily diminishing size.  If the base has area A, then a cross-section at height x (measured from the vertex) will have an area that is proportional to the area of the base.  Say the base is at x = h and the vertex is at x = 0.  The cross-section at height x is a shape that is similar to the base.  The lengths are in the ratio h:x, so the areas will be in the ratio h^2 : x^2.

Hence


So the volume of the pyramid is one third the volume of the prism.  The result in post 1 is just a special case where the cross-section is a circle.

Bob

#17 Re: Help Me ! » Geomotry Work » 2017-10-03 20:14:25

Post your diagram on an image sharing site such as imgur.com and link to it using the BCCode.

Bob

#18 Re: Help Me ! » what is the the Radius of a Square in a circle? and how to calculate » 2017-10-03 20:12:09

hi Hannibal lecter

If the four corners of a square touch the circumference of a circle then the diagonal of the square must be the diameter of the circle.  So use Pythag. to calculate the diagonal and half it to get the radius.

If it's the other way round and the circle touches the sides of the square, then the length of the side is the diameter of the circle.

Bob

#20 Re: Help Me ! » Geomotry Work Help » 2017-10-03 20:07:42

???  See your 20 questions post.

Bob

#21 Re: Help Me ! » please help geometry work » 2017-10-03 20:06:42

hi emmakatecumbo

Please read this: http://www.mathisfunforum.com/viewtopic.php?id=14654

These look very much like a set of CompuHigh questions.  Posting them 'wholesale' like this is a breach of their copyright.  And it's unreasonable to post the lot anyway.  Further what chance does anyone have of helping without the diagrams?  Please modify this post or I will.  I suggest you post a single question with diagram; then get help; then try the rest yourself.

Best wishes,

Bob

#22 Re: Help Me ! » Trapezium » 2017-10-03 04:10:55

The non- parallel sides are 4 cms and 6 cms.

Bob

#23 Re: Help Me ! » geometry » 2017-10-02 23:57:52

hi Viki1177

Welcome to the forum.

If a cone and a cylinder have the same radius and height, you'd expect the volume of the cone to be less than the cylinder.  But why one third and not some other fraction, or maybe even an amount unconnected with the cylinder's volume?

I can show this if you know some calculus; specifically a little bit of integration. Post back if you want to see this.

Bob

#24 Re: Help Me ! » Trapezium » 2017-10-02 22:05:58

hi iamaditya

a + d = 4
b + c = 6

I only got one solution, same as thickhead.

Bob

#25 Re: Help Me ! » Help with solving the volume of a right triangular pyramid. » 2017-09-27 19:12:53

hi CIV,

Your answer and mine only differ by a factor of 2, so I think one of us is correct.  I've looked at mine several times to see if I've forgotten to half something, but I cannot see an error.  That's not to say I'm confident; I often make silly slipups like that.  shame

Please post when you've resolved this so I know where I've gone wrong.

Bob

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