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Bob
hi Magicman89
Welcome to the forum.
A cube has 6 faces so I'm confused that your shape needs 12 cubes.
Unfortunately it is not possible to upload images directly to the forum server. Are you using a laptop or something similar? If so, you can join imgur.com, upload there, and then copy the bbcode into a post. Our server looks there for the image and it is displayed ok. Members who are trying to do this from a phone have had limited success with this. The Imgur app does not handle bbcode properly. I'm hopeful that their programmers are trying to add this feature soon, but don't hold your breath.
What has worked, but it's a bit clumsy, is to get the https address and post that. Then members can view it indirectly.
What software are you using for your image?
Bob
Having come up with a proof I was very keen to get it typed up for a post. Looking it over I decided it may be a bit 'heavy' for some members. Usually when I find the algebra is getting tough I ease the student in carefully with some number cases first. With this in mind I have edited my previous post by adding a table of formula values from 2 to 6. This shows that only second term prime gives a perfect number. Then I put numbers into my 'proof' to show it working for 6 and 28 and explained why it fails when p = 4 because 15 isn't prime.
Also I think it's easier to follow the proof if I start with the factors which are powers of two and then follow with the ones made by multiplying the prime by the powers of two. So I have switched these bits around in the proof.
In binary 2^p - 1 will always be a string of 1s and 2^(p-1) will always be a 1 followed by a string of zeros. So that would account for your binary observations. But, to get perfects, you need the string of 1s to be prime. There's no formula in decimals for generating primes so there cannot be one in binary either. So you're still stuck deciding which string of 1s is going to produce a prime.
Bob
Proof of Euclid's formula.
It is important to understand that this proof is of the form A => B. This does not mean that B => A
In other words, a number with this format can be shown to be perfect. But a number that is perfect may not have this format ie. there may be other perfect numbers that are not generated by this format. So far no mathematician has succeeded in finding such a number or proving the format produces the only perfect numbers.
Background for those who haven't met this before.
A number such a 6 is said to be perfect because its divisors (other than 6 itself) add uo to 6 (= 1+2+3)
Another such is 28 (= 1+2+4+7+14)
Euclid explored this and found a formula for generating perfect numbers.
If a number has the form
where (2^{p} - 1) is prime, will always be perfect.6 = 2 x 3 has this format. 28 = 2^2 x 7 has too.
This table with p from 2 to 6, shows this formula in action
Let's see why this happens for some of these and why it doesn't when the second term is not prime.
Without knowing what they are, we can say that 6 has 4 factors. Firstly there's the powers of 2 (2^0 and 2^1) and then the factors formed by multiplying the other term (3) by each power of 2 ( 2^0 times 3 and 2^1 times 3). This last factor is not counted for perfect number purposes. The factors we do count are 1, 2 and 3 and these add to 6.
28 = 2^2 times 7. From this we can say that 28 has 6 factors. Firstly the powers of 2 (2^0, 2^1 and 2^2) and then the factors formed by multiplying the other term (7) by each power of 2 (2^0 times 7, 2^1 times 7 and 2^2 times 7). Again the last of these isn't counted.
With numbers, this is what my proof does.
Add up
This is the formula with p=2.
Another example for the factors of 28
Next what happens when, say, p = 4. 2^p -1 is 15 which is not prime.
When the second term is prime there are always 2 times p factors and the highest is disgarded. When the second term isn't prime there are a lot more factors: the powers of two; the powers of two multiplied by (in this case) 3; the powers of two multiplied by 5; the powers of two multiplied by 15. Way too many factors to ever stand a chance of being perfect. Only second term prime keeps the total factors down to 2p altogether.
Proof
Consider a number with this format. You can make up factors in two ways (1) by multiplying the prime part by any of the powers of two from 2^0 up to 2^(p-2) and (2) by listing all the powers of two from 1 up to 2^(p-1). I'll add these two sets up.
Note: that first part stops at 2^(p-2) because the next term would generate the number itself and this is excluded when considering factors of the number.
Most of these powers of two cancel. Also the bracketed sum is 2^(p-1) less one so
which is the number I started with.
Bob
Good luck. It's been around a while and Euler spent a lot of time on it with some success.
Euclid's formula is
where the second term is prime.
In binary all powers of 2 are 1 followed by some zeros. For example 2^5 = 100000 If you subtract 1 the result, in binary is a string of 1s eg. 31 is 11111 in binary.
So when those terms are multipled together you get the string of 1s from the second term and a string of zeros after that because of the powers of 2.
So the format is always
But the question is how many 1s and how many 0s ?
Bob
Wow! I thought you had stumbled onto something here. But I did a bit of googling and found this
https://byjus.com/maths/perfect-numbers … 033550336.
If you look at how they are formed from that formula you'll see the connection with binary numbers. You just hadn't quite got the general term. Nice try
Bob
A, B, and C all correct.
D.sec_m = [2(x + h) + 5 - (2x + 5)]/h
Find the equation of the secant line at x = 1 with h = 0.01.
Sub in those values
sec_m = [2 times 1.01 + 5 - (2 + 5)]/0.01 = 0.02/0.01 = 2
So the line has gradient 2.
function is y = 2x + 5 so if x = 1, y = 7
So we have a line that has gradient 2 and goes through (1,7)
Because we're dealing with a straight line that is going to give y = 2x + 5 again.
Not sure why this question is even being asked.
I'd have chosen a function that isn't a line, eg y = x^2
Maybe that's the next question?
Bob
You can see the minimum. B is just the x value at that point and y is the cost at the minimum.
Bob
2. f(x) = -x^3 + 12x
f(-x) = -(-x)^3 + 12(-x)
f(-x) = -(-x^3) - 12x
f(-x) = x^3 - 12x
I say neither.
This has odd powers of x so it should be an odd function.
Oh I see. f(-x) = x^3 - 12x = - (-x^3 + 12x) = - f(x)
Bob
Because it's a odd function so you can use the rotational symmetry.
B
Calculate y for x = -3, -2, -1, 0, 1, 2 and plot the points.
I used the function grapher and there are max and min values but oddly not in that domain.
Spotted the problem. I copied your function as typed. I think there's a decimal point missing here 0.8x^2
Now the graph has max and min in the given domain.
Bob
Yes.
Where did you get -2 P_2 = (7, -2)
(3, -6) is right
Bob
Call the points A (0,0) B (a,0) and C (a/2, sqrt{3}a/2 )
AB = a No need to use pythag here, as the distance is straight along the x axis.
Exercise for you.
Work out BC similarly. (If you get 'a' again, you've done it.)
Bob
(x + 1)/2 = 2
x + 1 = 2
x = 1
line 2 should be x + 1 = 4
14 is right.
Bob
(3 + y_1)/2 = -4
That should be (-2 + y_1)/2 = -4
Bob
I agree.
B
The first two points tell us the length of that side is 'a'.
So work out the length of the lines joining point 1 to point 3 and also point 2 to point 3.
If all three answers are the same, then the triangle is equilateral.
Bob
Trig includes pythagoras. Any equilateral triangle can be split into two right angled triangles with a line through the midpoint of the base. You know the hypotenuse and half the base so you can work out the height, which is the x coordinate of the vertex.
Make a sketch and it should be fairly easy.
Bob
Distance would be hard because you don't know the direction. I've given two ways in the other post.
Bob
You don't have to know much. To 'travel' from P2 to M you need to go 2 left and 2 down. That's the vector.
So start at M and go another 2 left and 2 down.
The midpoint coords are [ (x1+x2)/2, (y1+y2)/2 ]
So you can make two equations, one for x and one for y. I'll give one here and leave the other as an exercise.
(x1 + 7)/2 = 5
Bob
Quite a bit of working here. You could find out the lengths of the three sides. For equilateral they should all be the same.
Then find the three midpoints.
Then do the equal length test again on those points.
Alternatively, you could use trig to get the angles.
There's a bit in Euclid that allows you to short cut the last bit by using parallel lines and similar triangles.
Bob
The third vertex will be directly across from the midpoint of the base (M) , and, as the y axis is that base, it will be on a horizontal line, drawn from M. That will fix its y coordinate. You'll have to do some trig. to get the x coordinate.
Bob
Use the same vector method in the previous post.
B
There's a quick way to get this using vectors.
The vector that takes you from P2 to M is (-2, -2)
So apply the same to M to get P1
Bob