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hi lex,

Welcome to the forum.

You haven't given us any background to explain what these questions are about. The title is polygons, so I thought this might be the numbers of sides (N) and either the total angle sum, the exterior angle or interior angle if the polygon is regular. But I cannot make any of the answer choices fit with any of those possibilities, so I'm stuck with how to help. Please give the whole question and a diagram if possible. If you cannot show a diagram then describe in words.

Thanks,

Bob

hi Hannibal lecter

I think it is true. But say more about what led up to the question.

What is the mathematical topic ?

Do you have a definition for an 'expression' ?

Does aa mean 'a AND a' ?

Does a + b mean 'a OR b ?

Does a* mean 'NOT a' ?

Bob

hi laurenwest144

Yes, you'll definitely need a calculator for these. But there are plenty of on-line sites that will do the calculation as well.

When you make those triangles for a pentagon they are not equilateral. They are isosceles. 360/5 = 72 so not 60-60-60 but 72-54-54. When you split it in half the half triangle is 36-90-54 and you need to get the height by 1.5 x tan(54)

Bob

hi Monox D. I-Fly

Your example shows why no theoretical formula is possible. For example, if someone hadn't told you about a job, you'd never have applied. Some company bosses may have a preference for employing persons related to existing good workers because they think they'll get another good worker. Or a boss may be prejudiced against doing that because it was a mistake in an earlier appointment. Unless you know everything about everyone, you couldn't assess the probabilities and even if you had a super database capable of storing all the relevant information, it would be out of date by the time you got in all the data.

Best chance would be to try for an experimental result. ie. You collect lots of actual employment records and assess the probability from that. I don't think it would lead to consistent results if you repeated the collection with new (random) data.

Bob

hi laurenwest144

Thanks for the image. If you extend the 10 line until it reaches the 20 line you will see this splits the shape into a rectangle and a triangle. By subtracting 8 from 20 you'll get the size of one side of the triangle. Even though it points downwards in the image, let's call that the height. You also need the base of the triangle. Easiest way to get this is by using Pythagoras' theorem. With two sides you can calculate the third (the base).

Going back to the others, for an equilateral triangle and a hexagon divided into equilateral triangles, all sides are equal so you can use sin cos or tan. I think cos is the easiest but there's not much difference in the calculation. I've realised that it'll have to be tan for the pentagon because the distance to the centre of the pentagon is not the same as the side so cos is not possible. Sorry about that. I have edited my earlier post so it doesn't put any other reader off.

Q1. I agree with your answer for this.

Bob

hi laurenwest144

Once you are logged in, click for the index and then a subsection, most likely Help Me. There's a link for **post new topic** on the top right hand side.

Bob

hi laurenwest144

An equilateral triangle has angles of 60. If you split it in half with a line from the vertex to the midpoint of the base, you will create a right angled triangle 30-60-90.

So you can use trig. to work out the height of the triangle and then half base x height to get the area. eg with a hypotenuse of 1 the height will be 1 x cos(30)

A regular hexagon can be split into six equilateral triangles so you can do the above and then x by 6.

For a regular pentagon draw 5 lines from the centre to the points on the perimeter. This will divide it into five isosceles triangles. The angle at the centre is 360, so each triangle will have a top angle of 360/5. Split each triangle in half like you did for the equilateral triangle. **edit: Then work out the third angle of the right angled triangle and use tan to calculate the height.** Once you have one triangle you can times by 5.

Q15. That image is not loading for me. I expect CompuHigh have it password protected as it will be subject to copyright. When you post back with your answers to the others for me to check, try describing the diagram in words eg. There's a rectangle, L by W, and a ..........

Bob

hi iamaditya

According to my search I joined on 20th June 2010. My first post was on the 26th.

I think I was invited to become a moderator the next year. Because the user-status appears with the post and is taken from the profile, you cannot view early posts to see when I changed from a *member, and now all my posts show 'administrator'. I could probably work out the exact dates with some searches but it would take a while so I'd prefer if you didn't ask for this.

Bob

* I seem to remember I had reached my first 1000 and got 'promoted' to some category 'above' member but I forget now what it was called. And, actually, I don't think it important. As I said before, quality is better than quantity, so why chase after a high post count. I stopped counting after that 1000. But if you want to find post 1000 you'll see I was hoping to make a significant post with it. In the event the problem was not totally solved. It's in the back of my head to complete that problem one day but it would require me to learn about contour integration and I'm finding that tough.

hi iamaditya

I cannot remember or find out the exact date but it was just before March 2013 because that's when I started using imgur for images.

Bob

hi Monox D. I-Fly

Back then the membership was structured according to number of posts, with super member status being 'awarded automatically'. I think it helps explain some of the post-games that were introduced to boost people's totals. Then MIF switched to a new server and all that was no longer available so everybody just became a member. Personally I always thought quality rather than quantity more important. It was also at that time that it became no longer possible to up-load images directly to the server, so you may find old posts lack the correct images.

Bob

hi emmakatecumbo

This request for help has been posted 7 times by you. I consider this to be abuse of the forum. You have been shown how to add proper images and I have given a detailed answer to the shuffleboard question. The dartboard question cannot be answered because you have never provided a diagram and without this the answer could be anything. Please do not keep doing this or I shall have no choice but to delete your account. Please look at this post:

http://www.mathisfunforum.com/viewtopic … 20#p401220

and respond there. Either what I have said is helpful in which case you need to answer each of my questions so I can check your work or you need to ask something new relating to what I have said. Do not use postimg for your images. If you fail to do as I ask then your account will be removed.

Bob

hi emmakatecumbo

This request for help has been posted 7 times by you. I consider this to be abuse of the forum. You have been shown how to add proper images and I have given a detailed answer to the shuffleboard question. The dartboard question cannot be answered because you have never provided a diagram and without this the answer could be anything. Please do not keep doing this or I shall have no choice but to delete your account. Please look at this post:

http://www.mathisfunforum.com/viewtopic … 20#p401220

and respond there. Either what I have said is helpful in which case you need to answer each of my questions so I can check your work or you need to ask something new relating to what I have said. Do not use postimg for your images. If you fail to do as I ask then your account will be removed.

Bob

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.

later edit:

Robin knows that if he hits the white part of the target, he just slightly win, therefore not embarrassing John. What is the probability that Robin will hit the white part of the target (not the red and not the bull’s-eye)? Show your work.

3.14(2)2

= 12.56

3.14(6)2

= 113.04

3.14(9)2

= 254.34

There must have been a diagram for this which you haven't included. You could at least have described the target in words eg. : the target is a circular board with a circular centre, a red ring around this and a white outer. The radius values are ??? ??? and ???

This question (and the shuffle board question) rely on two unstated assumptions: "The player always hits somewhere on the board and it is equally likely which region the dart hits. This assumption is not true in real life. If I throw a dart at a target there is a high probability it will hit the wall and drop on the floor.

But, with the assumptions, what we need to know is the area of the regions involved. Larger areas are more likely to be hit than smaller areas.

So I'll assume you can work out:

(1) area of bull's eye (A1)

(2) red area (A2) = area of red circle minus A1

(3) white area (A3)= area of whole target minus red area.

Now the probability of landing a dart in a region is:

If you want further help on this it is essential] that you post back the radius of each circle and the calculations leading to the probabilities. I can then see if what you are doing is correct and either tell you the good news or tell you where you are going wrong.

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