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hi iamaditya

There's a thread on the forum that includes the discussion and the link to the download. It was a while back and I've forgotten the details; sorry.

Bob

hi iamaditya

No Spam. Spam includes messages that have no relevance to the topic, that are annoying, repetitious or promotional in nature.

I remove adverts because they are (1) sometimes pornographic; (2) they 'clog' up the site with irrelevant rubbish; and (3) it's not fair to the paying sponsors to allow free advertising. If they stop supporting us the forum's very existence is at risk.

Mostly, these posters join the forum; create one or more posts which contain an advert, either in the post or the signature, then we never hear from them again.

Just once, we had a genuine mathematician who posted a link to the sale of his maths book. Bobbym had a word with him, and he removed the advert and replaced it with a link to a free download of the book. That seemed like a good compromise.

Bob

hi math9maniac

The 'simplification' was not valid. √ (A-B) is not generally equal to √ A - √ B

I squared both sides of the equation to get

I'm still not seeing any real solutions. See https://www.mathsisfun.com/data/functio … ^8 - x + 2

Bob

hi n0tjyxgsgs

Welcome to the forum.

This problem came up ages ago so I've mostly forgotten it. Some other posters provided quicker solutions than mine; that's fine; I never said mine was the only one nor that it was quick. The OP was happy to be taught some new maths along the way which I was happy to do.

PQRC is a sloping plane. So QR isn't a continuation of PQ. Not sure what you mean here.

Bob

hi math9maniac

I'm not seeing any real solutions.

https://www.mathsisfun.com/data/functio … t(x) - x^4

Bob

Hi Sterling94,

Welcome to the forum.

When two dice are thrown there are 36 possible outcomes but we can limit ourselves to just (5,x) 6 possibilities plus (x,5) another 6 possibilities less the one we've counted twice (5,5). So 11 in total. Of these two give a total of 8 so 2/11

Bob

hi joseph

Welcome to the forum.

Thanks for the supportive comments. MIF is funded by carefully selected advertisers. Please don't use the forum for free advertising.

Bob

It's very odd. You'll have to raise this with your teacher.

B

Arhh! Good old binary answer. Or maybe 1 + 1 = 1 as in one pile of sand added to one pile of sand = one pile of sand.

Bob

hi Mathegocart

Typos can creep into questions. In the UK the national exams are, nowadays, subject to a load of checking but question errors still get in sometimes. [eg. https://www.independent.co.uk/news/educ … 27141.html ]

Or maybe it was intended to test if you are aware that sometimes a calculation cannot be completed. If I was going to do this I would word my questions (have to be several) "Work out all the missing side lengths and angles in these triangles. If you think a triangle cannot exist explain why." Something like that would be reasonable if it was included like that as part of the assessment. If it was a single question on the topic, and you have copied it correctly**, then it looks like a serious error. What was the test?

Bob

** If the numbers are swapped around then it is possible for example.

hi Mathegocart

No error. No triangle can be made with those measurements. Try it yourself. Draw AB = 27 (may need to scale down) and angle B = 38. You don't know how long to make BC so just continue the line for some distance. Set a compass to a radius of AC = 15 and try to intersect BC. This fails. The shortest distance from A to BC is about 18.

Bob

hi Xepemu

Welcome to the forum.

That's a neat method which seems to overcome the objections some folk have about infinite digits.

How about joining the membership?

Bob

hi Sresta702

Welcome to the forum.

Which ans? I like to encourage posters to work things out for themselves.

Bob

hi Pablo,

Welcome to the forum.

Bob

The difficulty arises because it is a property of the real number system that between any two real numbers there is another real number. For all other reals this works ok but it fails whenever the decimal representation 'ends' with an infinite string of 9s. I have encountered two 'solutions' to the problem and both allow a consistent set of axioms.

(1) Allow that for example 0.99999999.... is the same as 1 I can understand why some people dislike this but, if you remember that the reals exist irrespective of the decimal representations, then why not? Lots of mathematical topics work perfectly using this solution. For example, you can convert an infinite, and recurring, decimal representation into a fraction.

(2) You can forbid the existence of such numbers from the real number system. That also works and is the basis of the approach used by Georgi E Shilov in his book Elementary Real and Complex Analysis.

It might help if you stop thinking that numbers have a concrete existence and accept that they are just convenient abstract ideas that help us to do certain types of mathematics. (You can hold 3 apples or even a wooden shaped '3' object but you cannot hold a 3.) So we can make numbers do what we wish according to the model we are making. Not everything obeys the rules for real numbers. For example add one pile of sand to another pile of sand and you have one pile of sand. You cannot have half a stick of chalk.

In summary, decide what mathematical model you are building; invent a consistent set of axioms for your model; and then use it to discover new things.

Bob

hi chinnu03

Welcome to the forum.

Good result. But I'm afraid it has been done before. This was, for a while, a GCSE maths project in the UK. You'd get a grade 'C' for what you have posted but would need to show the whole proof and try possible extensions to get an 'A'.

Bob

hi JoshNutter

Welcome to the forum.

Yes, you should be doing your chem engg project. How long did this take?

forum rules wrote:

No Swearing or Offensive Topics. Young people use these forums, and should not be exposed to crudeness.

My first reaction was to delete it completely but then we lose all your posts. I think you should do the right thing and edit it yourself to something more wholesome.

Thanks,

Bob

Not sure what you mean. My answer had 'x' in it. Isn't that algebra? Algebra doesn't exist independently from the maths topic it occurs in. This question requires trigonometry (in my opinion) and there's no getting away from that. You only need to pick one of the multi choice answers.

30:30 is impossible as the angles don't add up to 90.

45:45 won't work as the opposite and adjacent would be equal in this case.

30:60 won't work as the opposite and adjacent are in the ratio root 3 : 1

Bob

hi Sam_thing

I'm struggling a bit with your working. The octagon consists of 8 isosceles triangles. If 'a' is the height of one of these then I agree with your answer. So next I worked out the area of 8 of these triangles 8 times 2.5 times a. Where did 40 come from ?

Please re-post your working with some notes at each line to say what you are calculating.

Bob

In a right-angle triangle, perpendicular is 3 times the base.

So I made a triangle with B = 90, AC = 3 times BC.

I assumed that

What is ratio of their opposite angles?

means find the ratio of B:A

Bob

hi Monox D. I-Fly

I also get the x^3 coefficient as -1/3 Looks like the question has a typo as none of those answers works.

I get

and this checks out correctly with the given information.

I did this by substituting x = -3, 1, 0 and -2 into ax^3 + bx^2 + cx + d and then solving for a, b, c, and d.

When you put x = -2 , 3x - 2 becomes 3 times -2 -2 = -8.

Bob

hi Zeeshan 01

Let's call the two angles A and B. 3x and x are the opposite and adjacent, and the adjacent and opposite for these angles.

A/B = atan(3/1) / atan(1/3). This evaluates to about 3.6

Bob

Thanks Alg Num Theory. Hadn't thought of that.

But I think I can modify the plan to cover this. Draw a plane that intersects the three lines in A, B and C. Then choose D on that line as any point other than C. Then as before.

Bob

This has been asked before. See http://www.mathisfunforum.com/viewtopic.php?id=23304

What on-line course is this from?

Bob