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Thanks, Bob
When encountering vectors in maths, at the introductory level at least, is it appropriate to think of them in terms of displacement?
Take a triangle ABC, with AC being the base
Vectors AB and BC added together give the resultant vector AC, yes?
(imagine arrows going to the right above AB etc)
But this doesn’t mean that the lengths of AB and BC added together equal the length of BC (the added lengths will be longer than BC), yes?
What does it mean then?
Does it mean, or can it mean, the same as we’re taught in physics when dealing with displacement? The distance we end up from our point of origin A when we travel to C is the same regardless of whether we go directly to C, or via B?
So the distance covered going directly from A to C might be, say, 10m
And the distance going from A to C via B might be 12m
But the displacement is the same, i.e, 10m?
...the next calculation?
You perform a calculation
Your answer is 4.5... (long awkward number)
You now want to divide 5 by 4.5...
Is there a shortcut (rather than noting down the long awkward number, and typing in 5 divided by the long awkward number)?
Thanks, Bob.
So 1/12 = 42/N
1*42 = 42
12*42 = N
N=504?
Check;
504 ponies. Catch, tag and release 42.
Now 42/504, or 1/12, are tagged.
Catch 60 expecting 1/12 to be tagged.
5 are tagged which is what you would expect, on average?
Wyatt wants to work out an estimate for the total number of wild ponies in a forest
In July, Wyatt catches 42 ponies in the forest
He puts a tag on each of these ponies and releases them
In August, Wyatt catches 60 ponies in the forest
He finds that 5 of the 60 ponies are tagged
Work out an estimate for the total number of ponies in the forest
My question; Which branch of maths is this?
Also, can you give me a clue as to where to start regards solving this. Just a hint please, not a full explanation.
I'm struggling to even guess at this one. I did think, '60 ponies, 5 of which are tagged, so, 5/60 tagged, which is 1/12
1/12 of the 60 tagged...
How does this relate to the initial number of 42 caught and tagged, then released..?
Of the intial 42 caught, 100% were tagged...
Why did he only catch 42 in July, but caught 60 in August?
Thanks, guys.
I struggled to do this but managed a guess of 2^1
My thinking was √8 = 2, and another way of writing 2 is 2^1
The answer given was 2^(3/2)
Which I understood, eventually, as meaning (√2)^3
which equals (√2)(√2)(√2)
which equals (√4)(√2)
which equals 2(√2)
which equals 2^1*2^(1/2)
which equals 2^1+(1/2)
which equals 2^(3/2)
My question is; are both correct? My answer at the top, and the one I was given?
Thanks
Thanks, guys.
If we have small boxes and medium boxes in the ratio 2:5 does this exclude certain numbers of each box?
E.g, we could have;
2 small, 5 medium
or;
4 small, 10 medium
or;
200 small, 500 medium
Or any combination as long as there are 2.5 as many medium as there are small.
Does this mean we can't have, for example, 3 small (as that would mean 7.5 medium, and we can't have .5 of a box)?
Or, 1 small, 2.5 medium, or 5 small, 12.5 medium, etc, for the same reason?
Thanks, Bob.
I'll check out those links when I have time.
In the meantime, just a quick question.
Regards our saying that y=f(x), when looking for the inverse function; would it be accurate to say, in general (not just with functions), that, for example, y=x, or y=x+2, or whatever, that this is actually shorthand for y(x)=x, or y(x)=x+2, or whatever?
Thanks, everyone
Regards a straight line being somthing lik, Ax+By+C=0 ; would those kind of equations just be rearrangements of y=mx+c?
I can see that y=mx+c can be rearranged, for example, thus;
m=y-c/(x)
x=y-c/(m)
c=y-mx
Is the equation for a straight line graph always in the form y=mx+c?
If so, is an equation for a straight line such as y=4 shorthand for y=0x+4?
Thanks, guys.
One definition of function I came across online was;
"A function is a rule that maps a number to another unique number."
Q. Do you both agree with that defnition?
Q. Regards the 'unique number' part; I just came across a function question the answer to which was two numbers, either or, (a quadratic equation was involved; how does that fit with the 'unique number' part?
In a function is the part to the left hand side of the equals sign the OUTPUT?
For example, regards g(x) = x+2, is the following correct;
g(x) = OUTPUT
x = INPUT
x+2 = RELATIONSHIP
I know that if, regards the above, x=4, then we put 4 in and get 6 out, so 4 is the input and 6 is the output; but is g(x) also considered to be the OUTPUT? Or, in this case, g(4)?
Thanks, Bob.
What plotter was I using? I've never used one, nor seen one, until now
I was using some printed out exercise pages.
I tried your plotter; great stuff. I'll use that from now on
I’ve just completed a straight line graph for the equation y=4x-4.
y=mx+c tells me that the gradient(m)is 4 (and the y intercept is -4)
But the y values on the grid go up in 2s (2,4,6,8, etc) with 1 grid square for each value of 2. Whereas the x values on the grid go across in 1s (1,2,3) with two grid squares for each value of 1.
So the straight line graph on the grid looks like a y=x graph, with a gradient of 1, a line of 45 degrees, even though the gradient according to the values is 4, and would therefore be steeper, if drawn to scale.
I’m pretty sure the values on the grid were marked out this way due to the restriction of the page size (the y values went up to 16 and down to -12, so to give a whole square to the value of 1 would take up too much space vertically). But the x values only went from -2 to 3 so space wasn't an issue there.
So why don’t they do the grid to scale, where possible, which would make the line, and therefore the gradient, to scale? I think this would help the novice student grasp the steepnees of lines relative to various equations etc, and get the hang of graphs that appear frequently in maths.
I redrew the grid giving the x values 1 square for every value of two, like the y values to square ratio, and, for me, it looks better, and feels right. The other way just looks, and feels 'wrong'.
Also, just double checking on the difference between the grid and the graph. The grid is the thing we’re presented with on the page, the thing we have in front of us before we start drawing little dots or crosses on, before we draw a line through them; and the graph is the line itself, yeah? Or is the whole completed thing the graph?
Are the following two expressions equivalent;
(5+√5)/(5) and 1+(√5)/(5) ?
I've got to the end of a long surds problem and have to give the answer in the form a+(√5)/(b).
I got as far as (5+√5)/(5), then I'm not sure
P.s.
The problem was;
(√180-2√5) / (5√5 - 5)
I just noticed that in some questions for finding an angle instead of asking,
"Find angle x" the question is instead, "Find m angle x," m standing for measure of angle x.
This is interesting.
Q. Is it incorrect to say, for example, "Angle x is 45 degrees"?
Q. Would it be more correct to say, "The measure of angle x is 45 degrees"?
Again, we're back to how we define 'angle' (measure of turn/amount of space between/ figure created...)
Tricky one...
paulb203 wrote:Q. Do the lines (arms) in the MIFun illustration intersect? Or merely meet/join, at the vertex?
"intersect" doesn't have to mean "cross and continue"
It can mean "meet" or "just touch" like a repeated root of a quadratic just touches the x-axis
Ah, that's interesting. The Math is Fun dictionary gives the following defintition for intersect.
"To cross over (have some common point)"
The first part (to cross over) seems to imply cross and continue, yeah?
Whereas the second part (have some common point) suggests that to "meet" or "just touch" would satisfy the criteria, yeah?
Your remark is interesting.
As you know, a line is formed by dots (geometrical dots).
And by definition, a geometrical dot has no dimensions; actually, this means that it could be given any dimensions.
By 'geometrical dot' do you you mean a point? As in the point given by cartesian coordinates?
And why could it be given any dimensions?
paulb203 wrote:I’m looking for the proper method...for 3^c=1/ √3 ?
Remember that sqrt is 1/2 power & flipping is minus
1/sqrt{3}=1/(3^(1/2))=3^{-(1/2)}
so c=-1/2
Always try to convert sides to same base
Then set powers equal and solve - no guessing
Thanks, amnkb
3^c = 1 / √3
3^c = 1 / 3^(1/2)
3^c = 3^(-1/2)c = -1/2
Thanks, KerimF, you've been really helpful.
paulb203 wrote:Q. But presumably this only works if the numerator of the fraction on the right is a square of the integer value on the left (9 is a square of 3; 16 is a square of 4; and 4 is a square of 2)?
Not necessarily
You could have something like 4^n=8
so switch to (2^2)^n = 2^(2n) and 8 = 2^3
2n=3
n=3/2
But yeh it has to be the same base somehow (or else logs, i think)
Thanks, amnkb.
I thought I wasn't going to be able to follow that but managed with a bit of work. Cheers.
Thanks, Jai Ganesh