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hi Monox D. I-Fly

I have only 'scratched the surface' of this topic. You could start by looking at this article:

https://en.wikipedia.org/wiki/Key_(cryptography)

Bob

hi Cliff

Welcome to the forum.

I think the answer to (a) is r=1

Here's my logic:

Certainly 1 is a possibility.

eg. a, a + π, a + 2π + a + 3π, …. where a is an integer and π = pi, is such a sequence. 'a' doesn't have to be the first term; you could have many terms leading up to 'a' and then as shown.

So what about r > 1 ?

Let's say a and b are two integer terms in a sequence, m terms apart

so b - a = md where d id the common difference. As a and b are integers, so is md.

so another integer term is c = b + md.

So if there are r such integer terms then it is always possible to find one more by adding md to the last. So if r > 1 there will be an infinite number of such terms.

I'll post again if I find an answer to (b)

LATER EDIT:

I think r = 1 or r = 2 are the only possibilities, but I'm still working on a proof.

Bob

hi Joe walsh,

Welcome to the forum.

Sounds to me like you also need a general way to do this for any regular polygon.

Here's a diagram for a heptagon (7 sides). You can easily adapt it for n sides.

(1) Provided the polygon is regular you can calculate the central angle (AOB). Divide 360 by 7 ( or n).

(2) Half this to get AOH and then subtract from 90 to get OAH.

(3) In the green triangle AOH, AH is adjacent to angle OAH and OH is opposite so OH = AH x tan(OAH)

(4) The yellow area is then 1/2 AB x OH

(5) Multiply by 7 (n) to get the area of the whole polygon.

Hope that is useful for you,

Bob

hi Amartyanil

First you could calculate the number of arrangements where there is no restriction.

Then imagine EA locked together as a single 'letter' and calculate how many ways you can arrange these five letters. Do a similar thing with AE etc*. Then subtract.

Bob

*You'll need to take care to avoid repeats caused by all three vowels together.

hi Amartyanil

What exactly do you mean by "it's not working properly" ? This series converges only very slowly. I needed over 1200 terms before I got 3.14....

Bob

OK. Q15 is straight forward. An angle = "To define an angle, you need to be able to define both rays, and they need to have the same endpoint."

So let's say the ray is ry_AB and C is a point not on AB so that <BAC is the angle.

Extend the ray to make an line, extending beyond point A. Then we have a line and a point not on the line. This satisfies "a line and a point not lying on the line" so we have a plane.

Q14 is more problematic. Let's call the lines l, m and n. Since we have l and m parallel this satisfies "two lines which intersect in a single point or are parallel"

The reason I think this problem is problematic is that the third line, n, may or may not be in the same plane as the other two. Does this matter? Strictly, no. We can say yes I have a plane, in fact I have three so I certainly have one. Since we are 'in court' it seems odd not to reveal the whole truth and say we have one or three planes. That's why I'm curious about the course you are doing. There are two on-line courses that I know of. Sometimes they set silly questions and I wondered if I could add this question to the list. Ha ha!

Bob

hi riaaaa

If someone makes a statement and you can see it is untrue because you can think of an example that proves it's wrong this example is called a 'counter example'. In English we could say this example runs counter to the statement, meaning it is against the statement.

eg. Statement: "All swans are white" I go to a zoo and see an Australian black swan. It is not white. This swan is a counter example to the statement.

Bob

hi riaaaa

For number 7, I'm thinking it could be counterpositive?

There's a mathematical term 'contrapositive'. Is that what you mean? What course are these from?

The contrapositive of x^2 > 10 => x >0 would be ' not x>0 => not x^2 > 10

I tried to help with this one in my last post. Have another look at it.

As for number 13, I'm thinking E.

Correct!

And for number 3, I'm thinking F.

Correct!

Bob

hi riaaaa

Welcome to the forum.

statement wrote:

So I'll choose a value for x^2 that makes x^2 > 10.

Let's try x^2 = 16.

We know that positive numbers have two roots; in this case x = 4 and x = -4.

So it is not true that x must be > 0.

So what is it called when you find a value that proves a conclusion is wrong ?

Bob

hi Doggone

Welcome to the forum.

So

That has to be true for all values of x, y and z, so I chose a set of values that makes life easy:

If x = y = 1 and z = -2, then x + y + z = 0 and the equation becomes

From here you can get k.

Bob

hi Loki,

Re-arrange the equation to make y the subject. As x is an integer so is X+7 so replace x by X+7 and simplify that equation for y.

It should be fairly easy to work out which values of X (I can only see three and three more negative that will work) so you can then convert back to get values of x.

Bob

hi mrpace,

So what answer are you getting?

Note: x^2 is a common factor of 'top' and 'bottom' and so can be cancelled.

Bob

hi Oran2009

The horizontal component of velocity will be a linear expression and the vertical a quadratic so those equations look right. But the values will depend on your initial throw (magnitude and direction) so I'll have to assume you have those right.

Your initial velocity calculations are then correct.

I do not use Desmos but I'm sure any graph plotter will allow you to enter those equations to get a graph of velocity against time.

Bob

hi liza2020

Welcome to the forum.

Pplease also post the definitions you have been given. That way I can try to get from the definitions to the statements given.

Bob

hi George,Y

Good to see you back here!!!

Bob

OK, here we go:

For one 'round' of advantage there are three outcomes: A WINS (P = 0.49), REPLAY FROM 40/40 (P = 0.42), B WINS (P = 0.09)

If the REPLAY occurs, there are another three outcomes with the same probabilities. Then again, then again and so on.

The REPLAYs could, in theory, go on forever so As chance of winning is the sum of an infinite number of probabilities.

If a GP has first term a and ratio r I will write it as GP(a,r)

Using the sum to infinity formula

Hope that helps,

Bob

hi mrpace

Give me a moment to make a 'tree diagram' for this and I'll show you how to do this.

Bob

That's very ingenious! Why a quartic?

Bob

hi Siva882

Welcome to the forum.

At the moment I cannot see any obvious answer to this. The differences started nicely as 1, 8 , 27 and then that didn't continue. Sequence puzzles can be tough. For any set of numbers you can always fit a formula that will generate the sequence but this is not usually what the puzzle requires. Sometimes they use a single digit from each term to get the next; or some complicated sum using all the digits. We have been asked before for next numbers in sequences and the best efforts of several members have not found an answer. Check you've got the right numbers, please.

Bob

hi Monox D. I-Fly

I notice no one has answered this yet. Maybe this post will 'bump it' to someone's attention. In case not:

I don't know about using annuities for borrowing purposes so I cannot at the moment understand what you are doing here. But it's still possible we can work together to a solution. I have done this in the past. The poster explains more about what they are doing and then I can understand enough to make a contribution. Together we may get there

Here's what I do understand.

(1) If you borrow money you have to pay it back with interest. I cannot see what the rate for this would be in your question.

(2) If you invest money (usually for retirement) in an annuity, you pay regular amounts and the value accumulates because of this and interest earned. (at 10% ??)

(3) You want the ten year annuity value to be equal to the ten year amount owed for the loan so it can be paid off.

Please let me know if I'm correct so far.

Bob

Thanks! I think I deserve an award for getting all those brackets right!

Bob

hi Grantingriver

The sum can be written as a series of GPs. Simplifying, it becomes a single GP with a sum of 1.5

If GP(a,r) means a GP with first term a and ratio r:

If |r| < 1 then sum to infinity is a/(1-r)

Bob

hi mrpace

Imagine there is a field of infinite daisy chains.

The probability that a given chain contains n daisies is given by the following.

P = 2 / (3^n)

What is the average length of the daisy chains in the field?

Is this the exact wording of the question? If the field has infinite daisy chains, to me that means every chain is infinite in length. So I guess that isn't what is intended.

If there are an infinite number of chains, each of which has some finite number of daisies, then the answer cannot be determined. Say nearly every chain has 3 daisies and just a tiny number have a different number of daisies then the expected length would be close to 3. Substitute x for 3 and you can see that any answer is possible because we don't know how many chains there are of each length.

The only way I can make sense of this is if the question reads "A field has an infinite number of daisy chains. One chain has length 1, one chain has length 2, one chain has length 3 and so on. This I can work out and I get the expected length = 1.5

If this interpretation is what you want then post back and I'll complete the proof.

Bob

hi Monox D. I-Fly

I thought 'who is this examiner to set the exam for a Saturday'? but then I noticed the post date. Let's hope he did well.

This is what I did for number 4:

Note: It's easy to eliminate x and y to get an equation in z.

2 .... gives x = -z/k and 3 ... gives y = -kz so substitute these into 1 to get an equation for z.

It has a quadratic denominator so the roots will lead to impossible to solve.

I used the function plotter https://www.mathsisfun.com/data/function-grapher.php? to see what values of z are possible. There are three asymptotes, two vertical at the roots mentioned above and one horizontal.

As this is a valid function, any value of k that leads to a (unique) value of z means a single set of values for z, x and y. That's most values of k. There are two values at approximately - 0.6 and +0.8 (if my algebra is correct) where z tends to infinity so no solutions. I cannot find any values of k where more than one solution exists. At one point in my algebra I multiplied through by k to remove the fraction so I separately tested k=0 and that leads to a single solution too.

Bob

hi Mathegocart

I don't see this field in either my own profile or yours. Please explain what that acronym stands for.

Bob