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hi TheVoid87

**Welcome to the forum!**

Post back when you need help and someone will try for you.

Bob

hi math9maniac

That's odd. It used to be mathsisfun but today I found that mathisfun also works but, curiously, with an altered front page. Here's some things to try:

1. Clear your internet history and temporary files.

2. Try the alternative spelling.

3. Try this link:

http://www.mathsisfun.com/algebra/matri … ction.html

If still no joy, please report exactly the error message you receive.

Bob

hi sydbernard

1. I think to do this, you'd have to work through all the possibilities. geogebra as I mentioned before would help.

2. What is meant by a 'broken' line?

Bob

hi Abbas0000

Think of a vector as being a 'journey' ** from A to B. If you go A to B; then B to C; then C to A; then you end up where you started so your vector for the three journeys is zero.

Here's an example ( I'll use horizontal rather than the usual vertical to save time) :

(1,0) + (0,1) + (-1,0) + (0,-1) take you round the sides of a square and these add to (0,0).

What you describe in your post is a journey around the perimeter of a regular polygon. Once you've gone all the way round, you are back to the start.

http://www.mathsisfun.com/algebra/vectors.html

Bob

** vectors are used for other things than 'journeys', but it's a good place to start.

Hhhmm. A diagram would be useful here. I've given this some more thought and the main lesson we can claim from this question is the importance for the setter in making very clear what the question is. I've come up with three alternatives in addition to the one above:

In the first the vertex is A and the angles ABC and ACB are the base angles of a triangle and are bisected.

Let's suppose ABC = 2x and ACB = 2y, and that the bisectors cross at 90.

Then, in triangle BDE we have angles of x, 90 and y and so in triangle CDE angle BEC = x.

So angle BEA = 180 - x

and in triangle ABE angle A = 0.

So in this case the bisectors cannot cross at 90.

In my second diagram the vertex is at G and angles FGH and FGJ are bisected.

If FGH = 2x and FGJ = 2y, then 2x + 2y + 2x + 2y = 360, => x+y = 90, and the angles are supplementary.

For my third diagram the angles meet at a point and share a common line. If the angles are 2x and 2y and we know that x+y = 90, then 2x + 2y = 180 and the angles are supplementary.

??????

Maybe I should adapt the third so that the angles do not share a common line. Then the angles would not be supplementary. I leave this fourth diagram to the interested reader.

Bob

If you assume that the person setting the question did make a 'do-able' problem then a sign error making it impossible must be a typo. Over the years I've seen a few.

Bob

That's correct! Well done.

Bob

(1) I think that answer is correct.

(2) Did you get a quadratic? And if so, please post it.

Bob

hi DarkTangent

(1) Whoops, my mistake, sorry. I was trying to find tan BCD. For BDC work out tan BDA and then use the formula tan (180 - θ) = - tan (θ )

For (2) it looks like you may just have a simple error somewhere with the working as that result has features that suggest to me you are on the right lines. If you post your working I'll try to spot what went wrong. I'm confident about my answer for this one because when I substituted back the expression evaluated to 2.

Bob

hi sydbernard,

Welcome to the forum.

A is the common vertex, ABC is one of the angles; ADE the second; and the bisectors meet at F.

It didn't take long to construct this diagram; having some geometry software makes it easy. I use Geometer's Sketchpad which will cost you money. There's a similar program called Geogebra which is free. You'll find a download here:

Hope that helps,

Bob

hi

For + I'm getting two values; m=9 and m=18

What do you think?

Bob

Good morning zetafunc,

Are you assuming the operation is x; as in 9x9 = 81 ?

I had assumed the op is +

Bob

hi DarkTangent

1.) If you draw a line from D to BC parallel to AB, the point where this line crosses BC , say E, will be the midpoint of BC and ED will be half BA.

Use Pythag to calculate AC, work out DC and then tan BDC = ED/DC. (later edit: see post 4)

2.) I re-wrote this as:

Then re-write the sin as a cos:

Square this to remove the root and re-arrange as a quadratic.

(note: because this method has required a step with squaring it is possible the 'answer' won't fit the original question, so check by sunstituing back}

Bob

hi salaisuresh

And a welcome from me too.

Bob

hi iamaditya

Nice method

Bob

hi iamaditya

Don't worry; that's about three hour lessons for an A level class.

Bob

hi iamaditya

If there are no restrictions on a, b, and c then there are an infinite numbers of solutions. Re-write the equation with a as the subject and then choose any b and any c to generate a value for a.

So I'm assuming you want positive integer values.

Good puzzle, thanks.

Bob

iamaditya wrote:

and i will post the solution.

Not yet please. I've just been busy with other stuff. Give me a couple of weeks more.

Bob

hi math9maniac

Sorry to have to break this news to you:

http://www.mathisfunforum.com/viewtopic.php?id=23912

Bob

hi iamaditya

Firstly note that

(whoops ... cannot find the Latex for lim as delta x tends to zero

lim as delta x tends to zero

is the gradient of e^x at (0,1)

What do you know about e^x ?

If you know it's a power series then you can get the result easily by differentiating that.

I usually start a class with what follows where the class no nothing yet about e^x

Consider y = a^x for some real a

At x = 0 y = 1. As x gets bigger y gets bigger and as x gets more negative y get smaller (but still positive)

Thus we have a 'family' of curves all with similar properties. If y = b^x is another where b is slightly bigger than a, then b^x is bigger than a^x for positive x and smaller for negative x.

As a may be any real number the 'family' all go through (0,1) all have a fixed but not yet known gradient at (0,1) ; let's say that gradient is k ; and dy/dx = k.a^x

Thus dy/dx = ky for all a in the family.

[note: if a is negative then y = -a^x is the familiar member of the family and just the mirror image of y = |a|^x in the y axis. So we only need consider a to be zero or positive]

If a = 0 the 'curve' is just the line y=1.

As a gets bigger from 0, curves have steadily increasing gradient at (0,1) so choose the one where this gradient, k,

is 1 as special and call that value of a, the special letter e.

So dy/dx = y for this value e.

So the function has the property that it differentiates to give itself.

Are there other functions with this property?

Let's suppose there are two, df/dx = f and dg/dx = g.

Consider f/g and differentiate using the quotient rule

So if we integrate this we get

So all functions with the property are multiples of e^x

Now consider the power series

This will differentiate to give itself so it must be a multiple of e^x.

But the series has value 1 at x = 0 so it must be e^x itself.

That's as far as I'll go for now. I can continue to show that integral of 1/x is ln(x) and hence obtain a value for k for all values of a.

Bob

hi Bryan,

Sounds like an interesting idea. Beyond my abilities to create what is needed though. It would have to be by Rod, the originator of MIF, I think.

Bob

hi iamaditya

In differential calculus, we are trying to develop a rule for how the gradient of a function varies with x. The usual method for working this out is this:

(1) Pick a point on the curve for the function, say ( x, f(x) )

(2) Move to a point close to ( x, f(x) ). The way this is written is ( x + Δx, f(x+Δx) ) meaning ("x plus a little bit more in the x direction", "value of the function at this new value of x")

note: Δx should be regarded as a single variable, not as two variables multiplied together. Thus Δy/Δx must not be simplified to y/x by cancelling! This should be considered as "a little bit in the y direction"/"a little bit in the x direction"

(3) The gradient of the chord joining these points is then considered, simplifying where possible.

(4) Then Δx is reduced towards zero, to see if the gradient found in step (3) tends towards a recognisable gradient function.

(5) This gradient function is usually written dy/dx. Again these two, dy and dx, should be treated as two single variables so the 'd's cannot be cancelled.

Some texts avoid this confusion by using x + h and f(x+h) where 'h' is a small amount in the x direction.

eg https://diversity.umn.edu/multicultural … nRules.pdf

Hope that helps,

Bob

hi Dark Tangent

I get this:

Q (2,7) rotates to Q' ((0,5) and then reflects to (2,5)

thickhead wrote:

So the coordinates of various points will be N(0,-4) ,P( 2,-4) and Q(2,0).

Is this correct? The rotation is 270 clockwise.

Bob

hi DarkTangent

First draw a diagram. Then make use of some tracing paper. In the UK in GCSE exams, tracing paper is allowed.

Bob

hi samuel.bradley.99

I've been away so I've only just seen this problem. I had a vague memory that the volume of a tetrahedron could be calculated using vector cross and dot products. A quick 'google' took me to this page:

https://math.stackexchange.com/question … ot-product

Your tetrahedron is a regular one by the way which fixes the directions of the vectors IA, IB and IC. So it looks like you can use this formula to get a result. But post back if you need a quick lesson in cross and dot products.

Bob