Did you mean

Have a look here:

http://www.mathsisfun.com/exponent.html

Bob

]]>I'm sorry you didn't get what you wanted in time for your exam. What you have to realise is that MIF is a world wide, and free to use, forum. While you are studying, the person helping you may be sleeping, or at work, or just digging the garden.

When you know you have an exam approaching you need to allow plenty of time for other members to respond. I'm surprised that you knew in advance what question would be asked. This doesn't happen in the UK; candidates can practise similar questions but the examiners set new questions for the exam itself. Had I known this I would have been even more reluctant to provide the actual answer since it goes against what I believe is the purpose of the exam in the first place. If you give an answer having no idea why it is correct but just copying someone else then you give the examiners a false impression of your understanding.

This is why I try to provide general help on the topic rather than just a model answer. It says in the rules: "This is a Forum, it is not instant chat. Leave your message, and come back later (hours or days) to see what responses you got." And in the advice when asking for help: "We are happy to help! But we don't do your homework for you." This could equally have said: "We won't do your exam for you."

Bob

]]>It all started when I proved .

Let ,

where

Now I would like to show that ,

where .

So I proved the following inequality for positive integers :

.

Expanding

gives.

As n grows towards infinity, the binomial coefficients go to . Now taking the absolute values, that sequence is bounded above by .

But now I got stuck. One idea that I have is the sandwich theorem -- i.e. to 'squeeze' the sequence of partial sums of the cosine series between that sequence above and another sequence

The trouble is the altering signs, so I just can't add up inequalities ...

Any ideas?

Regards,

zahlenspieler

I chose D for my answer. I chose this since we are unaware of what A and B are, so that means we it cannot be determined without any other calculations, which is answer D. I will let you know if this is correct.

Kayla

]]>FP = PH = 2.5 by Pythag. PO = 5 and angle OPH is 90 degrees so OH is given by

Angle PHO = PBO = alpha

Bob

]]>Welcome to the forum.

Yes, you can use the rational root theorem. You'll find it here:

https://en.wikipedia.org/wiki/Rational_root_theorem

If that leaves your head spinning let's simplify by looking at a quartic equation.

If p/q is a rational solution, ie p and q are integers

and if we times by q^4

If we assume p/q is in its lowest terms then => q divides a and p divides e.

But you are told that the solutions are all integers; as all integers are rationals this means that q = 1 and p divides e.

So could p be 3 ? Yes as 3 divides 6

Could p be 9 (ie a repeated root of 3) ? No because 9 doesn't divide 6.

Hope that helps,

Bob

]]>Bob

]]>Q12. I see why you thought supplementary at first but supplementary only applies to a pair of angles and this is about a triangle. D is correct.

Q17' I agree.

Bob

]]>Bob

]]>The space to the left of the red may be any of 3. We'll assume we use one of these. The space to the right of the red may be any from the remaining two. Now to place the rest. Two ways to do that.So that's three x two x two = 12 ways.

Now to consider the total number of possibilities.

We've placed one red. Left may be one from four and right one from three and once again two ways to place the rest. That's 24 ways.

So P = 12/24

Bob

Good luck with the exam.

]]>