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List all common factors of 6x^3 and 15x^4.

I think: { 1, 3, 1x, 3x, 1x^2, 3x^2, 1x^3, 3x^3}

Bob

List the factors of 3x^3.

Note: Do not consider factors that are evenly divided by 3x^3. For example, 3x should not be on your list.

I have no idea what 'evenly divided' means in this question so I'll list them all.

I think {1, 3, 1x, 3x, 1x^2, 3x^2, 1x^3, 3x^3}

The underlined 1s could be omitted.

Bob

GCF is 1 tick.

Relatively prime tick apparently. I'd not met this term but Wolfram Alpha agrees so it must be true..

Bob

I'm happy with your working and final answer.

Bob

I'm assuming AI questions is a 'robot'. Robots don't always get things right because 'their' answers depend on the programming by humans.

Recommended background reading: I Robot series by Isaac Azimov.

How to test if something is a factor: Does it divide exactly?

The AI doesn't list x^2 but (12x^2)/x^2 = 12 so it is a factor.

If you write the expression like this:

then a factor can be 'constructed' by taking some of the above factors multiplying them together.

Thus: leave out all the xs and you have the usual number factors {1, 2, 3, 4, 6, 12}

Now include one x and you have {1x, 2x, 3x, 4x, 6x, 12x}

Now include both xs and you have {1x^2, 2x^2, 3x^2, 4x^2, 6x^2, 12x^2}

I'm claiming that a systematic approach like this makes sure none are left out.

Bob

The factors of x^2 are {1, x, x^2}

The factors of x^4 are {1, x, x^2, x^3, x^4}

1 is always a factor of anything because it always divides exactly. So it is always a common factor.

Bob

1 is also a common factor.

Bob

hi FelizNYC

Welcome to the forum.

I think there are more than this.

eg. 2x is a factor; 2x^2 is a factor.

If you treat x as 1x and x^2 as 1x^2 then there must be

6 factors where no x is present

6 factors where x and a number is present.

6 factors where x^2 and a number is present.

Bob

hi Oscar,

Welcome to the forum.

Shouldn't it be

Chinese-proverb wrote:

teacher open the door, you enter by yourself

Do you want the length of the diagonal or the area of the rectangle?

Bob

hi Opichem

Welcome to the forum.

Bob

This can (and does) happen in other branches of mathematics. There's no absolute authority that determines definitions and anyone can write a book developing an area of maths using whatever definitions they like.

eg. Is 0.999999 recurring the same as 1 ?

Many say yes and use this as a way to convert recurring decimals into fractions for example. But I have a maths text that avoids the question completely by declaring that decimals 'ending' with a recurring set of 9s as being invalid real numbers. So 1 exists but 0.99999 recurring doesn't.

There is also a question about how to define parallel lines: two lines that make the same angle with a transversal or two lines that never meet when extended indefinitely. This arose recently in a post about Euclidean geometry.

The Edexcel exam board avoids this issue completely by only requiring a knowledge of integers. I assume this means that 'natural numbers' and 'whole numbers' are never used in exam questions.

Bob

?? Surface area is the same as area for a circle. How did this question arise?

Bob

Go back to your sketch in post 1. The line marked with an angle of 100 is not correctly drawn. As 100 > 90 this angle should be obtuse so the line should slope the other way. I chose to make an accurate diagram hoping you'd realise this. I added a protractor so you'd be able to see how 100 and 80 are related on the diagram.

Bob

The method works like this.

Suppose the number is N and G1 is your first guess. Let G2 = N / G1

N = square root x square root. Also N = G1 x G2

If G1 is lower than the root then G2 must be higher; and if G1 is higher then G2 must be lower.

So, either way, root N lies between G1 and G2.

So, if you work out G3 = (G1+G2)/2 you get a number that is closer to the root than G1.

Using G3 as a 'better' guess leads to G4 and once again the true root must be between them.

So continuing the procedure leads to Gs that are closer and closer to the actual root. When Gx and Gx+1 appear the same (ie differ by an amount so small as to be less that the displayed accuracy, you've got a root that is close enough.

Bob

If it said express d in terms of F then you'd need d = .... and would have to have a square root. But it doesn't.

Bob

Yes, they're the same thing. If you look at your vectors you'll see you've got the same ones; just in a different order so only the diagram is different. I was imagining my square on a coordinate graph so x across and y up; but one of the useful things about vectors is it'll come out the same on any grid.

You can even have a grid based on axes at, say, 60 degrees so the usual squares become parallelograms. All the usual results come out the same. (except maybe dot and cross products although you could probably find a way to include them)

Bob

Yes, that's right. If you have vectors around a square then the sum is zero. If you write the vectors as 2D components it is more obvious:

Bob

That's it. Just remember it's an estimate. There is a statistical test that can be used to say how good an estimate it is, but I cannot remember it at the moment.

Bob

You may have a reciprocal button (1/x)

If you enter 2 then press the button you'll get 0.5 If you do it again you'll get back to 2.

So do the long calc (say it's y) ; press reciprocal; times by 5. That's equivalent to doing 5/y

If you haven't got that button then an alternative is store y in memory, then do 5 divide by memory recall.

Bob

I think this is an application of probability theory.

Say there are N ponies.

He tags 42, so the probability of getting a tagged one if you pick one at random is 42/N

Later you find 5 are tagged out of 60 so assume that gives the correct experimental probability value (5/60) and equate that with 42/N so giving an equation for N.

Bob

The key to problems like this is

So start with

then

......(i)So we have an answer, but it would be good to simplify it.

Substituting into (i) we get

Bob

I have made an accurate diagram and put a protractor image over point D.

When you measure an angle using a protractor you will see there is a choice of angles.

If angle BDE is measured the zero on the protractor is on the inner scale. Going round anticlockwise when we reach the line DE the scale says 100.

If angle ADE is measured the zero is on the outer scale. Going round clockwise when we reach the line DE the scale says 80.

As you can see 80 + 100 makes a straight line and is equal in total to 180. Lines DE and BC are parallel.

Bob

In your diagram the line made by 100 degrees should slope away from the other line as 100 is obtuse. These lines are parallel.

Bob

Parallel lines are cut by a transversal in equal angles arises first in book 1 proposition 28. That is the theorem that is being used here.

Bob

hi AlachimUI

Welcome to the forum.

From the graph it looks like (2,1) and (1,2) are two solutions.

That makes x3 = 3 x4 = 1 and x3 = 1 and x4 = 3

What happens if you start with x3 and x4 taking those values?

Bob