Math Is Fun Forum

Bob

16 is a power of 2 so it's worth re-writing this with powers of 2

You can extract 2x out of that first expression leaving (2x)^1/3 which becomes a common factor in both terms.

I started the second and got to wondering if that last term should be (8x^3y^3)^(1/3).

It simplfies nicely if that power is 1/3 rather than 1/2.

Bob

A long time ago, when I was preparing for my A level exams (age 18) I did two things for practice.

Starting with an A4 piece of paper I expanded that as a determinant into 27 terms each with three letters.

Then I reversed that using a different order to make, say, this:

which demonstrates a property of determinants.

Later edit: Actually, now I've tried it, one is the negative of the other.

It's hard work and you have to be carefully accurate not to miss a term.

Starting with a quadratic with integers coefficients but which won't factorise easily I would do the formula 'in my head'. ie. remember each calculation answer and carry it forward to the next stage in my head. It's hard work but these two things had two big advantages. (1) It trains the brain to do hard stuff and (2) it teaches you to be very careful with the algebra.

I think of it as similar to an athlete doing weight training.  They may not be using weights in their sport but it develops muscle, flexibilty and stamina. Now I'm reminded I'm going to give it a go again to try and keep away from dementia.

Bob

When you do numeric long division, you're more likely to succeed with errors if you keep the thousands, hundreds, tens and units neatly in columns.  The zero place holders (yes, that's the term) help with the algebraic version. I did this by hand first on a sheet of plain A4 paper. It took about a third of the page to complete.

Bob

Consider x^3 + 1.  The factor theorem can be used here. Set x = -1 and x^3 + 1 evaluates to zero. This means (x + 1) is a factor.

x^3 + 1 = (x+1)(x^2 -x +1)

Bob

I'm hoping you can do 'ordinary' long division with numbers. The method is similar. It's hard to show the whole process in a thread so I'll just try to get you started.

                                                   x^4      +ax^3
                              _________________________________________________
                              |
            x - a           |    x^5      + 0x^4     + 0x^3     +0x^2     + 0x      -a^5
                                   x^5       - ax^4
                                  ______________ -
                               
                                    0          +ax^4     + 0x^3
                                                +ax^4   -a^2x^3

etc

The method is easiest to use if you take great care to keep the powers of x in neat columns.  It looks ok on my laptop screen. Hope it stays that way for your screen.

Bob

hi Oculus8596

Welcome to the forum.

opposite integer :- the number that is the same distance from zero but with the opposite sign.

This restriction is unnecessary as the result follows even when m and n are any real numbers.

Start with  a^2 + b^2 = (m^2 - n^2)^2 + 4m^2n^2

Continue with the algebra until you reach c^2

Pythagoras theorem works 'in reverse' so a^2 + b^2 = c^2 is sufficent to prove the right angle.

Bob

From the (newtonian) equations of motion for a object  falling from rest with  an acceleration g (ok to take this as 10m/s/s) the final velocity is given by v = 0.5gt^2

So when t = 0.1   v = 0.05
              t = 0.2   v = 0.2
              t = 0.3   v = 0.45

................................................

              t = 1     v = 5

The v values aren't in arithmetic progression so that formula won't work.

Bob

Hmm. Good point. I've never thought about that before.  The nul vector is definitely a vector. If a vector had components x and y then the angle that vector makes with the positive x axis is arctan(y/x).

If x = y = 0 then y/x is indeterminate.  So I guess it has a direction but nobody knows, or will know, what it is. 

There's plenty of things you can work out so I think just file this away with Schrodinger's cat.

Bob

Yes. It follows from the following

If one of n events must occur then

P(event 1) + P(event 2) + ... + P(event n) = 1

Bob

Fair point. At the top of each post there's a list of symbols you can copy and paste. ≠  is available there. Does your phone allow you to copy any of those.

Whatever, I was happy with your answer as it had the essential bits.

Bob

As C complement is the elements not in C 6 doesn't feature whether it's there or not.  But If you change 6 to 0 for example it would change the answer.

Bob