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Depends on whether the sign means anything in the context of what the numbers represent.
E.g., if I owe you $10, then my balance with you is -10. Now I repay you $2, so +2 - 10 = -8, my new balance with you.
Other examples might be like comparing how far the two of us walk. Do we care about just the absolute value, or care whether you go North (+) and I go South (-) ?
Thanks, Phil.
No context was given in the original question. It was a proof question, about the difference of the squares of 2 consecutive integers. I used the difference between 2 and 10 for the sake of simplicity.
I get your points about absolute value etc, but what if nothing like that is stated. What if someone merely asks for the difference between 2 and 10?
Thanks, Bob.
“In mathematics or logic, an axiom is an unprovable rule or first principle accepted as true because it is self-evident or particularly useful. “Nothing can both be and not be at the same time and in the same respect” is an example of an axiom.”
Merriam-Webster
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Unprovable. That’s interesting. I’m guessing that for any given axiom, if it was provable someone would have proved it (or would be attempting to prove it).
So we can’t prove 2n = Even. It is either ‘self-evident’, or, ‘particularly useful’? Or both?
It does seem self-evident. And I’m guessing it’s useful, for those who apply that kind of thing.
How do they know something is unprovable?
And what do you think of that Merriam-Webster definition?
Having tried a few proof questions, I'm left wondering about the some of the assumptions made (axioms?) in the answers.
E.g,
Prove algebraically that the sum of any two consecutive integers is always an odd number
n+n+1
=2n+1
2n = even
+1 = odd
Therefore proven
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Is it just accepted in maths that 2n=even, and that +1 makes it odd? Have those been proven? Are they classed as axioms? Self-evident?
What is the difference between, say, 10 and 2?
Seems obvious to me. 10-2=8. The difference is 8.
What is the difference between 2 and 10?
Seems obvious to me, at first, i.e, 8 again. But then I’m wondering why I shouldn’t be calculating 2-10=-8 this time, given that the 2 appears first in the question this time.
Q. Can you have a difference of -8? Can you have a difference of negative anything?
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When I Googled it I was pointed to the number line.
When I then wrote out the 1 to 10 section of the number line, and circled 2 and 10, it showed that there were 7 numbers in between them (3 4 5 6 7 8 9).
This puzzled me further.
Any thoughts?
Thanks, Bob.
How would that be worded?
"Draw the locus of all points..."
I was asked to draw THE LOCUS OF ALL POINTS 3cm from a line.
I measured 3cm above and below the line, and drew two parallel lines there. Then drew two semi-circles with radii 3cm out from either end of the line to complete the locus, the shape(which Google tells me is a stadium).
But is the question imprecise?
Regards the lines above and below the line; shouldn’t it specify that the distance of 3cm is perpendicular?
There are many points 3cm from the line INSIDE THE STADIUM, no? (If you measure at an angle).
E.g. Draw a 3cm line perpendicular to the line, and north of the line. Then from the base of that line, draw another 3cm line at an angle of 45 degrees to the line. The point where this new line ends will be more than 1cm inside the stadium.
Doesn’t that mean that your original shape, your stadium, doesn’t actually cover THE LOCUS OF ALL POINTS 3CM FROM THE LINE?
Thanks, Bob.
Fascinating stuff (lines of infinite lengths, etc).
The very next question involved a ray!
It asked for the LOCUS OF ALL POINTS which are equidistant from lines CD and CE, which formed and angle of around 100 degrees.
I constructed my angular bisector and started drawing the ray, again thinking, this would extend infinitely if Euclid had anything to do with it.
So I drew it to the end of the page, but imagined it continuing across my desk, the rest of the room, through the wall, and, 'To infinity and beyond!'
P.S. I've done my GCSE Maths now (just waiting for the result), so I'm doing this for fun now (and with a view to maybe trying A Level).
I was asked to draw the LOCUS OF ALL POINTS which are equidistant from points A AND B.
A and B were two points 10 or so cm apart halfway down a page of A4.
I drew a perpendicular bisector connecting my compass markings to get the halfway mark between the points, which was a line of 10cm or so.
But then, as I’m new to this, I wondered, if the line is to be the locus of ALL points equidistant from A and B, should extend to the top and bottom of the page?
Then I wondered if it should extend infinitely in both directions, like one of those Euclidean theoretical (idealised?) lines of infinite length.
Am I on the right track in my thinking here?
P.S. The answer (this was a Corbettmaths question) showed a line extending beyond the compass markings, but not to the top or bottom of the page.
Thanks, Bob
h was for height (of tomato plants)
So, 140.000000001 etc makes sense.
My first thought was, "Greater than 140 therefore 141, 142..."
Doh!
What is the midpoint of the following class interval;
140 < h ≤ 150 ?
I’m used to seeing it given as 145 in this kind of example. But is it?
The possible values for h are, I think;
141,142,143,144,145,146,147,148,149,150.
The middle value of that group is, I think,145.5
So why is the midpoint of that class interval usually given as 145?
Thanks, Bob.
Not now though, although I will take up your offer later. Cheers.
Thanks, Bob.
Would that be AS, or A, level maths? Or further on than that?
Fearne invests £5600 in a savings account.
She gets 2% per annum compound interest.
After n years, Fearne has £6061.62 in her account.
Work out the value of n.
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I found finding the value of n one way straightforward. I put 5600 into my calculator, pressed equal then pressed ANS to store it. Then put the multiplier (1.02, for the 2% interest) next to ANS in brackets. Then pressed = repeatedly until I got my answer of 6061.62. As I had to press it 4 times this told me n=4.
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But before doing that I wrote out the following equation;
5600(1.02)^n=6061.62
(1.02)^n=6061.62/5600
(1.02)^n=1.082432143
1.02=nth root of 1.082432143
Nb; I wasn’t sure of this last step but it seemed to check out re;
2^2=4
Therefore;
Sqrt4=2
And then I substituted the 4 (from my first answer using the calculator) for the n and that checked out;
1.02=4th root of 1.082432143
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My question;
How do I get the nth root from the second method, once I’ve got to the stage;
1.02=nth root of 1.082432143 ?
Is there a way, other than putting the 5600 into the calculator and multiplying it by 1.02 repeatedly?
Thanks, Bob.
Thanks, Bob.
Why do we equate the two values of x?
I managed to follow your algebra for finding the value of x, thanks; great practice for me!
And the co-ordinates for the centre matched up with my graph, where the rays met (1,0), which was satisfying.
But why, once we established that x=1, do we conclude that y=0?
It's all in the name. HCF is highest common factor. The common factors must be in the intersection because that's what the intersection means. We want the highest so multiply them together.
Splitting the factors into their primes components helps us to identify the common ones. You could put {1,2,3,6,7,14,21,42} all in one circle and {1,2,4,7,8,14,28,56} in the other, and then you could spot that 14 is the highest in both sets but it's a lot more work.
Bob
Thanks, Bob.
I got that the common factors were in the intersection. It was the bit where multiplying them together produced the highest common factor that I didn't get. I think the penny is starting to drop now.
Thanks, Bob.
"Find where they cross. that's the centre."
Can that be worked out from the two equations, without drawing the two rays?
I've worked out the following two equations from an example;
y=2x-2
y=(1/2)x-(1/2)
To make a Venn diagram, draw one circle for each number. Put the prime factors inside the circles so that common primes are in the intersecting part of the diagram.
https://i.imgur.com/O2ganzy.gif
The union gives the LCM.
Bob
Thanks, Bob.
But why does intersection give us the products of the HCF? I get how to obtain the HCF from that method, but I don't know why it works.
Why does the prime factors/Venn diagram system work for this?
Example.
Find the HCF of 42 and 56
Prime factors of 42; 2,3,7.
Prime factors of 56; 2,2,2,7.
The intersection of the Venn diagram contains 2 and 7, so the HCF is 14, which checks out when you use the simple method of listing all factors of 42 and 56 and look for the HCF that way.
But why does this work?
All I can glean from the 2 and the 7 in the intersection (imagining I didn’t know what the 2 original numbers were, that I didn’t know they were 42 and 56) is that both numbers must be at least 14; and that one of them must be at least 28.
They must be at least 14 because both have 2 and 7 as factors.
One must be at least 28 because its circle in the diagram must contain at least one other prime factor (beyond the intersection) and the lowest of these is 2.
But I can’t work out why the 2 and the 7 produce the HCF.
Is there a way to find the centre of enlargement without drawing rays?
I see once I have drawn rays, and found the centre, that there is a pattern in the co-ordinates of the respective shapes, related to the scale factor, but is there a kind of reverse process or similar; use the co-ordinates of the shape given, and the scale factor, to find the centre (without the rays)?
Thanks, KerimF, thanks, Bob.
You're talking about a regular hexagon I think; so the answer is yes. You can divide the hexagon into 6 equilateral triangles each with side = the same as for the hexagon itself. Let's say side = a. Then perimeter = 6a and diagonal = 2a.
Bob
Thanks, Bob, but someone elsewhere has pointed out to me an error in my thinking.
I've been thinking about this in relation to my sqi ar ^2 idea.
But with sqi ar ^2 I was using the centre of the square to the centre of a side for the 'radius'. To be consistent I should be using that for the 'radius' of the regular hexagon
I found out that's called the apothem (which means I can dispense with the 'ar' for 'radius' in my formula and now use 'a' for apothem.
So,
sqi(a^2)
And,
hxi(a^2)
respectively, are the formulae
But I got 2 different answers for the value of hxi
Because I got 2 different answers for the area of the regular hexagon (sides 6cm)
I got 2 different formulae for the area online;
(3√ 3/(2))*a^2 where a=length of sides
and
0.5(a)(P) where a=apothem and P= perimeter
apothem seems to equal 5.5cm (?)
Thanks, Bob.
I was annoyed at this one. I got the answer and thought, 'That can't be right...' Then puzzled over it for ages. So when I saw that was the answer...
Doesn't seem like a sensible approach to to me; 'Here's the answer, but it doesn't make sense.'
A shop sells small chocolate bars and large chocolate bars
The small bars are sold in packs of 4
The large bars are sold in packs of 9
On one day;
packs of small sold ; packs of large sold = 5:2
A total of 95 bars were sold
How many small bars were sold?
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5(4) ; 2(9) = 20;18
20:18=10:9
10:9 = 19 parts
95/19 = 5
10x5=50 and 9x5=45
50 small bars and 45 large bars
But the small bars come in packs of 4
50/4=12.5
How can you buy 12.5 packs?