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hi Zeeshan 01

I hope this diagram will help you.

If the lines are AB and CD and they are both in the same (2 dimensional) plane, then they must cross (when not parallel). When parallel the distance between them is constant but if not parallel then the distance between will vary and can be zero. This means they cross. In this case the answer is 1 point.

In 3 dimensions it is possible for the lines to be 'skew'. Skew lines are not parallel but do not cross. In the diagram AB and EF are a pair of skew lines. AB is confined to the top plane and EF cuts that plane once at F, so these lines never cross. Number of points is then zero.

If the 'two' lines are, in fact, just a single line then they 'cross' everywhere, so they would have an infinity of crossing points. But two lines that are the same line are considered to be parallel, so infinity will never be an answer.

Bob

hi Zeeshan 01

I thought your original question had been answered. If it hasn't please make clear: is this a question in 2 dimensions or 3? The answers is different in these two cases.

Bob

**bob bundy**- Replies: 0

hi Den,

It looks like this is a binomial probability question. http://www.mathsisfun.com/data/binomial … ution.html

If p is the probability of one network working, then the probability of exactly M working out of N is

where

But you would also be happy if more than M are working so you need to add up all Ps from M up to N:

Hope that's what you wanted.

Bob

hi coolcoder

Welcome to the forum.

Please don't post your problem in someone's else's thread. This will be confusing for that OP.

As it stands your problem doesn't make sense anyway. There is no fraction so "rationalise denominator" is not possible.

This is how you can display a fraction:

`[math]\frac{numerator}{denominator}[/math]`

will make this:

and square roots like this:

`[math]\sqrt{5}[/math]`

So start a new thread and try again.

Thanks,

Bob

hi !nval!d_us3rnam3

(Please don't just give me a link to somewhere else, work me through some of it here.)

I'm sorry, but I think that request is unreasonable. This question has been dealt with before and it took a lot of time to put together all the help in this post:

http://www.mathisfunforum.com/viewtopic.php?id=22435

Please take the time to read it and post back if you meet something that is not clear to you.

Bob

Sorry, my mistake. I should have worked with r = y/x, but I wrote r = x/y instead. So I got a similar quadratic and value for r, leading to the same cos(36) and cos(72). But not the same quadratic as you. That explains for me why your (d) answers were also mine. In re-assessing your results, I now think you are completely correct. Apologies.

Bob

hi alext

Welcome to the forum.

Your problem has been posted before (and answered).

http://www.mathisfunforum.com/viewtopic.php?id=23062

Bob

hi Mathegocart

In geometry you always have to be careful not to assume some property just because it looks like it's true. In that diagram you're not told that BDC = 90 or alternatively that BD is parallel to AE. If either is true then all the triangles are 30-60-90, but if you shift point D a little along CF then triangle BDC isn't 30-60-90 any longer.

Bob

hi !nval!d_us3rnam3

That's great! Glad you have got it sorted. Funny thing though. I've got some sign differences in the (c) answers. Same final result.

Bob

let's check you have my answers to those first. Please post.

Bob

hi taylorn5683

I don't mind helping, but your initial posts (on several threads) made it seem like you were just hoping someone would do your homework for you. Sorry to jump to the wrong conclusion.

Let's proceed with a few questions at a time.

Questions 1 to 6 are all about what possible triangles you can make. I assume you know that the angle sum is always 180 and that sets the limits. eg. You couldn't have two 90s because there's nothing left for the third angle. If you drew a base line and made angles of 90 at each end, the next two 'sides' would be parallel and so you'd never get to the vertex of the triangle.

If you try drawing some triangles with one, two and three acute angles, and also with one, two and three obtuse angles you'll soon see what is possible and what just will not make a sensible triangle.

The next questions are all about something called the triangle inequality: http://www.mathsisfun.com/geometry/tria … eorem.html

7E for example won't make a triangle because 1+3 = 4. If you made a base of 4 and then tried to get a line of 1 and a line of 3 to find a vertex, those lines would meet on the base so you'd just have a line with a point 1 from an end and 3 from the other. No triangle. The webpage is interactive; so you can try out different size triangles just by moving the points. The lengths are initially 2 or 3 digits and your questions only have relatively small numbers, but you could always scale everything up by a factor of 10. If you can make a x10 triangle, then you can make a x1 triangle with the same (proportionate) lengths.

If you would like me to check your answers, I will, but please post back like this:

Question followed by your answer, so I don't have to keep scrolling back to the first post.

Once we have sorted this worksheet, I'll look at the others.

Bob

hi !nval!d_us3rnam3

(b) Let x = BC and let y = CD. Using similar triangles ABC and BCD, write an equation relating x and y.

(c) Write the equation from Part (b) in terms of r=y/x and find r.

(d) Compute cos 36 degrees and cos 72 degrees using Parts a-c.

So identify two isosceles triangles, with 36-72-72 angles.

Write the letters for these, in order, one below the other. This makes it easier to construct the ratios you'll need.

Fill in every length on the diagram with x and y as appropriate.

Now write the ratio r = x/y in all ways possible using the similar triangles.

From these you should be able to eliminate x and y and so get an equation for r.

If you split an isosceles triangle down its line of symmetry you'll get right angled triangles with 36 and 72 in them, so you can then use A/H to get the cosines, initially with x and y, but then convert to 'r' and use the known value from above. You can check your answer by evaluating it and seeing if your calculator agrees using the cos button.

Bob

hi Abu Ahmad

Welcome to the forum.

Plenty on fractions here:

http://www.mathsisfun.com/fractions-menu.html

Bob

hi Geomtry.202

Welcome to the forum.

When you split an isosceles triangle along its line of symmetry, you create a right angled triangle. So you can determine cos(36) using A/H.

Bob

hi markosheehan

Q, of mass 2m kg, moving towards each other with speed 2u m/s.

Strange wording. Does this mean a relative velocity of -2u, or an absolute velocity of -2u ?

I've tried both and cannot get your final velocities with either. Here's a start with the interpretation that leads to the book answer:

momentum along line of centres:

restitution law:

It's easy to show that Vq is greater than zero so the condition on Vp leads to the required condition on e.

Incidentally, you should have realised there's something wrong with your solution as you have

solving these i get e<3 and e>3/2

e can never be greater than 1.

Hope that helps,

Bob

hi Monox D. I-Fly

Yes, that's the puzzle. Words like 'four times', and 'greater than' and 'negative four' all have well defined mathematical meanings, but 'hotter' doesn't. Rather it's a colloquial term that doesn't seem to go with the rest of the question. It's like asking 'Which is bigger, a triangle or a cube?' In colloquial terms most people would regard -4 degrees C as cold. You could easily say what is 4 degrees hotter than the given amount; it's the '4x as hot' that messes the question up and I agree with thickhead that it is meaningless as stated.

Where did it come from?

Bob

hi Monox D. I-Fly

I understand why you're puzzled. That negative makes the question silly. My son suggests:

(1) Convert to the absolute scale, kelvins, 269

(2) times by 4, 1076

(3) convert back, 803 degrees celcius.

Ho hum. Sometimes questions have no answer.

Bob

Well it is. I don't know why you're surprised. Am I missing something? If you have that software then give it a try. If you don't then stop worrying.

Bob

Not exactly. Microsoft Excel uses it to mean:

You will find a lot of helpful maths on this site:

http://www.mathsisfun.com/algebra/expon … ithms.html

When you want a new topic, use the search engine at the top of every page, or the cross-links between pages.

Bob

to the power of e some say exp

Bob

We have met this before. ln and e^ are inverse functions.

If y = e^x then x = ln(y)

So if you raise ln(something) to the power of e, the ln and e cancel out, and you're left with the 'something' ... in this case just y

It's like doubling one half of something. Doubling 'undoes' the effect of halving so you get back to where you started.

One way to define a log is to say it has this property.

eg If y = a^x then we define the log of y (in base a) to be x.

Or you can say, if I have a number y, what power of 'a' will give me y. This power of 'a 'is called the log of y.

eg. What power of 10 will give me 1000 ? Answer 3. So we say 3 = log(1000) in base 10.

So you can always re-write a power expression as a log expression or the other way round.

Bob

Is it possible y=e^(alnx)???

You can differentiate this:

Replacing the first part by y brings you back to my earlier answer.

My version is the one that you will find in lists of standard differentials such as:

http://www.math.wustl.edu/~freiwald/131 … etable.pdf

Bob

are different ways of writing the same thing. ln was the quickest for me. (But not as it turned out, as I've had to add this post )

Bob