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hi kubes

Welcome to the forum.

I've not met this before so I think you can call it kubes theorem. Well done for finding an algebraic proof. What about 3 or more digits ?

Bob

If you use http://www.mathsisfun.com/data/function-grapher-old.php

you can see the graph for yourself.

Differentiating gives 1 + sin(x) It's worth plotting this too. You'll see that maxima occur periodically.

Differentiating the slope function gives cos(x) and this is zero at π/2, 3π/2, 5π/2 etc. The first and every alternate one thereafter is a maximum. Plugging in π/2 for the slope function gives 1 + sin(π/2) = 2. I have zoomed in on the graph and used coordinates to get the gradient directly. Still get 2, not 1

It would be 1 for the function 1 - cos(x).

Bob

Hi,

If you differentiate the function, you have the slope at any value of x. So now you want to know when that is a maximum. So differentiate again and set equal to zero for stationary points. Investigate each to see which give a maximum.

These are only local maxima so you still need to consider which is the absolute one.

Bob

Hi Leila

Welcome to the forum.

I'm not doing your homework for you but I'll try to help a bit.

In logic p and q are statements and IMPLIES (sorry cannot do the symbol on my kindle) connects them in a complex statement. For example, "I've just won the lottery" IMPLIES "I'm going to be rich"

You can still make complex statements that aren't' true such as "It's raining" IMPLIES "it must be Tuesday"

Let's deal with those words next.

Counterpositive , contrary and counterintuitive aren't used in logic.

A converse statement would be the exact opposite, q IMPLIES p

A counterpositive negates both statements and reverses the order, not q IMPLIES not p. If p IMPLIES q is true then so is not q IMPLIES not p.

A counter example is an example that proves a statement is false. It's Saturday and it's raining proves the example statement above about it being Tuesday is false.

So, if we start with question 1, you need a complete statement (that rules out most of the possibilities) and it should be obvious which to choose from the rest.

Suggestion:

Try as many as you can and post your answers. I'll check them and we can move forward from there.

Bob

Have you read my post? I have answered this above and given a link to the MIF page on this.

Bob

Do you know how to 'reduce' a number to its prime factors? I'll do

12. √576

as an example.

576 = 2 x 288 = 2 x 2 x 144 = 2 x 2 x 2 x 72 = 2 x 2 x 2 x 2 x 36 = 2 x 2 x 2 x 2 x 2 x 18 = 2 x 2 x 2 x 2 x 2 x 2 x 9 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

You keep dividing by prime factors until the resulting multiplication is all primes. http://www.mathsisfun.com/prime-factorization.html

Now you can see if the number is a perfect square:

2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 = (2 x 2 x 2 x 3) x (2 x 2 x 2 x 3) = 24 x 24

Bob

#1 Correct, but how can you tell?

#2 Right, but again, how do you know?

#3 These ARE parallel... how do we know?

#7 Incorrect... show me how you would set these up to find the scale factors

#10 What is the sum of all angles? How many do you know? Subtract EVERYTHING you are given from the SUM of all... then simplify by combining like terms

If the angles are equal you have an example of 'corresponding angles' http://www.mathsisfun.com/geometry/corr … ngles.html

(7) is NOT incorrect! Your working is exactly what is needed. I'm guessing that your teacher is also expecting that you do (polygon B):polygon A, which would 'invert' the scale factor. So I would try x2 as the second answer.

(1) is also correct but you have an equals sign = where you need a minus sign - Looks like you just pressed the next door key on your computer.

I suspect that your answers are marked by a robot rather than a human. So, unless you get exactly what the robot has in its answer store, it gets marked wrong. I'd hope that the 'teacher' would sort this out and use some common sense, but it doesn't seem to have happened here. Maybe the teacher is not as good at math as you are

Hope this helps.

Bob

hi emmakatecumbo

Hurrah! At last you have given me a diagram for the shuffleboard. Now we can make some progress.

You have used 'postimage' for the diagram. I have a problem with that. (1) The image shows on a page that is full of adverts. MIF has to keep its pages suitable for children to view. Some of the adverts I saw are not. I use imgur for my images. The posters to imgur sometimes post unsuitable material too, but the link I provide on my posts is to the image only; you would have to go to the imgur site to see other material. That makes it better for MIF. So I have copied your diagram to my imgur account and replaced your image link with mine. I'll do the same with your other post when I get time. But I'd prefer that you do this yourself as it takes time.

Here's how to get a picture (on its own) into a thread:

http://www.mathisfunforum.com/viewtopic … 17631&p=23 post number 1686

Now to the problem.

3. What are the dimensions of the 1 rectangle?

4. What are the measurements of one of the 5 rectangles?

Both these can be worked out with some simple arithmetic.

You're told that the '1' is twice the width of the '2' so that's easy to calculate.

If you then add up all the widths and subtract from 108, you'll have the width of the open (non scoring) space.

The '4' space is 15 top to bottom, so subtracting from 30 will give you the remainder for the two '5s'. So divide that remainder by 2 to get the missing dimension of the '5'.

Now sketch the whole board and by length x width write in the area of every section. The overall area is 30 x 108.

The probability of landing in any space is (area of that space) divided by (overall area)

Hope that helps.

Bob

hi EliTorres25

Welcome to the forum.

Individual albums first:

Any thing by the Beatles. (Whoops ... showing my age there!)

Pink Floyd's Dark side of the Moon*. And many others.

Dire Straits: Alchemy and many others.

So I'd say any of these bands is a strong contender.

Bob

* If I'm stranded on a desert island with only one album and a player**, this is the one I'd choose.

**Despite the opinion that vinyl sounds better, I'd prefer a CD player because that disc is going to get a lot of play.

Further thoughts:

Constructing a 30 degree angle is fairly straight forward but what if it's another angle, θ. Find F by (1) Drawing a line straight up from A; (2) Draw a line from B,

angle 90 - θ; (3) where these lines cross is F with AFB = θ

Some other ways of drawing the ellipse. We're used to using a compass to make a circle or an arc. There's a similar way to make an ellipse. Tie a piece of string in a loop exactly AB + BC + CA long. Put two drawing pins into the paper at the focal points. Loop the string around both pins and pull tight with a pencil. Allow the pencil to track around the foci and it will draw an ellipse.

There is also a construction which fixes points on the perimeter of the ellipse. Then you join them with a smooth curve. It's about 50 years since I last used this technique so I'll have to do some hard thinking if you want to see how it's done

Bob

I am not familiar with MATLAB but a quick google search gave me this:

http://hplgit.github.io/prog4comp/doc/p … ab025.html

Obviously, exact code will depend on what you're trying to solve.

B

hi sydbernard

It is a property of all ellipses that the distance from any point on the ellipse to one focus added to the distance to the other, is a constant. It is a property of all circles that if A, B and F are all points on the circumference then angle AFB = constant for all F on the same arc from A to B. My construction uses these two properties.

First some theory for ellipses.

The Cartesian equation for an ellipse is

Where 2a is the length of the major axis and 2b the length of the minor axis.

For this problem I'm labelling the foci (plural of focus) A and B. So AB will be the given length and AC + CB will be a specified constant amount.

I'll illustrate with AB = 8, AC + CB = 10 and ACB = 30. The method will work for any such measurements but these make the construction easier to follow.

On a coordinate diagram mark the centre of the ellipse as O, and put A and B equal distances (4 units) either side of O.

If point Y is chosen to be on the ellipse and directly above O then, as AO = 4 and AY = 5, this means that OY = 3 by Pythagoras.

So we know the minor axis has b = 3.

At the moment we don't know the major axis but it is easy to calculate using the constant distance property. Let the major axis be DE.

Then AD + DB = 10, so a = 5. So the equation of the ellipse is

So C has to be somewhere on that ellipse.

Now mark a point F above the line AB, so that angle AFB = 30. I want a circle to go through A, B and F. Clearly the centre has to be on the Y axis. I've bisected AF (point G) and drawn a line through G perpendicular to AF, to cross the Y axis at H. This is the centre of the required circle. I drew the circle and where it crosses the ellipse is the point C.

I used Geometers Sketchpad to make the construction but you can also use Geogebra.

Bob

hi sydbernard

(1) I think it's only one as well. If the altitudes are AD and BE and they meet at G, then DGB = 60 and GDB = 90 implies DBG = 30.

BEC = 90 and CBE = 30 implies BCE (BCA) = 60.

(2) Work in progress.

Later edit: I am unable to draw a diagram for this at all. Let D be where the bisectors cross.

A + B < 180 implies DAB + DBA < 90. Therefore ADB > 90

Do you have a diagram?

Other post: I'm working on a method involving ellipses and circles. If successful I'll post on that thread later. Later edit: Done in other thread.

Bob

hi Hannibal lecter

I don't know of a book just on this. I expect the book's title (if it exists) would be on a more general topic with, perhaps, a chapter on this. But I used 'google' to search for "brute force search" and got loads of useful hits, I suggest you try the same.

This link http://intelligence.worldofcomputing.ne … gQJpeSDPIV looks particularly detailed.

Bob

hi Hannibal lecter

Both cos(x) and x have continuous* graphs so cos(x) - x would too. You haven't said what value you want to achieve so I'll assume you want to solve cos(x) - x = 0

You could try a value of x and see what the function evaluates to. Then try a close by value of x. Do you get closer to zero or further away. Depending on the answer to that, continue trying values for x so you 'home in' on a solution. This is usually called trial and improvement but it is certainly also brute force because you are just trying lots and lots of values.

Because cos is a periodic function I thought there might be more than one solution but there isn't. It is close to 0.739... But you'll not get an exact solution for this one so you'd have to decide on the accuracy level you want.

Bob

If a function isn't continuous you might not be able to 'home in' like this.

Hi Gonzumzum

Welcome to the forum.

The tangent to two circles creates a pair of similar triangles. Thus

EOdash/AOdash = K J/AJ

If you put in r2 r1 r3 and d, you will find r3 in the required form straight away. The other is similar.

Bob

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

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

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

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

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

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

Yes but I'm just being helpful anyway.

Bob

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

I think you're going to love this:

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

Bob