Math Is Fun Forum

Good luck!

Bob

hi schoolrules1

Glad you're ok with the earlier parts because this last part is somewhat more abstract in nature.

Before giving a hint for that look again at

There's a little bit missing here.  Just a typo I'm sure! 

Now for the last part.

(i) Square (3n-2).

(ii) Add then subtract 2 from your expression so you get it in the form (expression) -2

(iii) You should find that the (expression) has 3 as a factor so you can write it as 3 x (another expression) - 2

(iv)  The (another expression) is a positive integer (why?) so the term squared is in the same form (3 x [pos. intg] - 2 so it's in the sequence.

Can you fill in the gaps?

Post again if you need more help.

Bob

Hi ihatecircles

Yes 57 is right!  You correctly remembered that the angle in a semi circle (PRQ)  is 90.

And PQRS is a cyclic quadrilateral (quad with 4 points all on a circle) so opposite angles add to 180.  That should take care of the second part.

edit: Sorry I've just read all of your post and you already knew this bit.  So you didn't really need help at all.  That's very good.  When are you sitting the exam?

Meanwhile ....  how have you got on with the rest of this paper?

Bob

hi ilovealgebra

I've been hoping someone else would pick this up and then I'd learn something.  But no one has so I'll throw in my small contribution just to show you're not being totally forgotten.

(i) 

I don't think this is ever going to simplify to

because

=

=

=

Why did you want this anyway?

(ii)  I need help following where your new limits came from.  Perhaps if you explain to me it will help you clarify what you have done so far.  I often find that helps.

Bob

Hi AU101,

You're welcome.

Here's a tip.  If you know it already, sorry to state the obvious;  but if you don't it will save you a lot of time.

The 2xn matrix can be any 'shape' so why not transform only the points (1,0) and (0,1)

The vector matrix for this is

But this is the identity matrix for multiplication
so if you want to find the transform for, say, a reflection in the x axis just consider what it does
to the points (1,0) --->  (1,0)   and (0,1)  ----->  (0,-1).....................line A

If the matrix is R and I is the identity ...............  R = R.I   (transformed shape equals reflection x identity)

so I can write down the right matrix straight away using the results on line A

I hope that makes sense.

Try this one:

I want to 'shear' shapes so that points on the x axis stay where they are and other points move parallel to that axis by an amount that is proportional to their distance from the  axis.  Here's a picture to show how (1,0) and (0,1) move.

(1,0) ---> (1,0)     and (0,1) ---->  (3,1)  that's a shear in the x direction with shear factor x3

The matrix for this will be

How about that?:)

Bob

ps.  to bobbym.  Thanks for your message on another post.  After tinkering with  'abstract algebra' I turned my hand to matrices and vector geometry because I liked the pictures better.:)

hi ampohmeow

In your version, you add a term on the right and then hop it across to become the middle term.  OK if you want to quote 'commutative' as your reason.

Then you have to use associativity on a new line.

What I did was added the same term but on the left.  Then it's in the right place to use only associativity to get the result.

As for 2a = 0, I'm stuck at the moment  as I don't know what axioms you started with, and I expect it is too much for you to post them all.  But I'm still thinking about it.

I'm painting my hall and landing at the moment which only takes a little bit of brain, so I'll devote the rest to the problem.

Bob

hi

Our posts crossed over I think.

I've edited post #8 to show how I think you can do the second part.

There's a bit missing where you say 'then' .....

I would say you should add then inverse of 'ab' to both sides and use associativity as necessary to 'move' ab from the LHS to -ab on the RHS.

Bob

hi again,

Windoze doesn't have such a helpful option, but I've got to it via the image properties so that's almost as good.

And it's bedtime here in Essex so I'll say night night for now.

Bob

Yes, that's what I want.

Please would you post the exact code (minus the first bracket I suppose so it shows as code rather than as a picture).

Bob

hi abhupathi

It could happen that none or one or  two or three or even all four have the disease.

This is an example of what is called the 'Binomial Distribution'.  It is worked out like this:

P(none have the disease) = 3/4  x  3/4  x 3/4  x 3/4  (As each must 'not get it')  = 81/256

P(a particular child has it) = 1/4 x 3/4  x 3/4  x 3/4    ( one has but three have not!)

but 'a particular child' means one of four so

P(any one child has it) = 4 x (1/4  x 3/4  x 3/4  x 3/4)  = 27/64

So now it is possible to answer the question.

P(at most one) = P(none) + P(any one has it) =  81/256  +  27/64 =   0.73828125...

I have added a diagram which may help.  Start at the left and follow a route going right.  I have shown one way in which a single child may get the disease.  You should be able to find three other routes for this outcome.  And I have shown the way in which no child gets the disease.

You can see that there are 16 posssible outcomes.  You could try to work out the probabilities for each route.  When you add up all 16 answers you should get exactly 1.0000000

Post back if you want to know more about the Binomial Distribution'.

Bob

Hi HanaAS

That looks possible, but I want to check some details.

You seem to have three lines, each perpendicular to the other two.  Correct?

Two lines are 600, the third is 550.  But do they cross at their midpoints?  Yes/No

Once I'm clear on that the rest is just trigonometry.  Do you know how to do SINE/COSINE/TANGENT ?

Bob

Hi Onyx,

Generally I feel that the first step in simplifying a fraction like this would be to remove the 'fractions within the main fraction'.

So multiply by the (common) denominator (provided it is not zero)  gives :

What I now wonder is 'Can this be further simplified?'

The answer has got to be 'no' because, if we replace the common elements with a single variable, say, x, we have

and that just won't simplify at all.

But, as bobbym says, it depends on what is meant by simplify.  If there's another bit to the question, this might provide a hint of where to go next.

Bob