Math Is Fun Forum

Heyya Mathland,
I'm noticing a trend with your threads - you're splitting each individual problem(which has the same underlying theme) into its each own thread. Generally it is preferred here to keep problems with the same underlying topic in the same thread to reduce clutter.

With regards to a potential algebraic approach, we can either consider factoring the numerator and demoniator and cancelling out like terms, splitting the fraction, using L'Hopital's laws, or using the properties of the limit.

Try using the product rule. The limit of a product is the product of the limits. Do you have it?

But yes, your current approach does work. It is a little rudimentary though.

Do you know what to do from here?

EDIT 1:

> What if I select values to the left and right of -4 but not including -4? By doing this, I will then know if the interval (-00, -4) is positive or negative and if the interval (-4, 00) is positive or negative.

You're approaching from the right, so you don't need values to the "left" of four. Also, what is (-00, -4)? No such thing as negative 0 and why the two 00s? One 0 is fine. Besides, 0 is greater than -4, so it should appear as follows: (-4,0).

Hello mathland.

Firstly, I'm presuming that the "t" also represents theta.

(A graph of x(theta). x is the vertical axis in this case and theta the horizontal.)

It is evident that if we stay to the right and move left, the function is moving towards the value 10cos(pi/4)(pi/4). This is approximately equal to 5.553.

As for your second question with lim y(theta), we see that the function approaches -16(pi/4) + 10 sin(pi/4). This value, to the thousandths, is -2.799.

I'm not good ol' Bob, but I do have a lengthy history on this forum, so I do feel entitled to answer these questions:

1. Contributing regularly to the forum, asking out questions, answering inquiries when they pop up, and suggesting improvements to MIF.

2. Sure - I had proposed some relatively challenging math problems in the section in the past. That's what's the section's all about.

3. You can contribute and help correct apparent mistakes in answers - complex and convoluted answers will always contain a few errata.

4. We do receive an occasional contributor every now and then to the Help Me! section, but it's mainly Bob, phrohinster, irspow, zetafunc, and (sometimes) me.

5. Not necessarily. I'd characterize it as a collective process.

6. The MIF website specifically does contain a couple of rudimentary-to-intermediary linear algebra topics, like matrices, vectors, eigenvectors, etc... nothing too comprehensive but it's good to brush up on the basics. And what specifically do you mean by Calc III? Different schools and countries classify different aspects of Calc into Calc III... but MIF contains a myriad of Calc III related topics - see partial derivatives and diffeqs..

7. Bob is not the creator of the site - that honor would have to go to MathsIsFun. As for the specified divisions of mathematical subjects - I for one believe that an excessive amount of divisions would impede work done by the good people here..

The aforementioned linear system is how I determined the number of cases that total to an integer that did not total more than 40, yes.

The System
x = number of 1s
y = number of 2s
z = number of 3s

x + 2y + 3z = n (where n is a positive integer)

x + y + z = 20(i.e, 20 slots for 1, 2, and 3)

So for your first q, yes.

So to determine the number of combinations of 1s, 2s, and 3s that sum to 40, all we have to do is to replace n with 40(since the combo must sum to 40).

Solving for y and z in terms of x, we get that y=20-2x and that z=x.

I start finding said combos by manipulating the value of x - that is, the variable that represents the number of 1s in the combo. So it can range from 0(i.e, no 1s) to 10(not any higher as we would get a negative number for y - which represents 2.)

So there are 11 combos that sum to 40 that consist of 1s, 2s, and 3s.

Adding that to the 120 figure I reached earlier(forgot to add the number of combos for 40), we get 131.

Yeah - my method counts all of those as the same. You want each of them to count as individual permutations?

Hello, sorry for the delay - some life impediments proved to be quite tiresome and delayed me for a day. Nonetheless, here is what (I believe) to be a solution.

So I decided to tackle your intriguing problem by abstracting it to a system of equations (albeit underdetermined[i.e, there are less equations than variables]).

The System

x + 2y + 3z = n (where n is a positive integer)

x + y + z = 20(i.e, 20 slots for 1, 2, and 3)

__________________________________________________

Now, for obvious reasons, n can only range from 20 to 40 - and Denominator has solved the case for n=40, so really we can just try n=20 to n=39.

The most trivial case is obviously n=20 - there's only one set of slots that sums up to 20, and that's 20 1s. So there's one way to do that.

CASES
__________________________________________________
N = 21
Solving the system where n=21, we have that y=39-2x and z=x-19. Now from relatively obvious arithmetic reasoning(no negative amount of numbers), the only value that x can be 19. So that leaves us with z=0 and y=39-38=1. z=0, y=1, x=19. So 19 1s and 1 2s.
N=22
Solving the system where n=22, we have that y=38-2x and z=x-18. So x is constrained to {18, 19} for reasons above. So we have two solutions.
N=23
For the system n=23 we have that y=37-2x and z=x-17. x is constrained to {17,18}. Hence, two solutions.
N=24
For the system n=24, we have that y=36-2x and z=x-16. x is constrained to {16,17,18}. Hence, three solutions.
N=25
y=35-2x, z=x-15. x is constrained to {15,16,17}. Hence, three solutions.
N=26
y=34-2x, z=x-14. x is constrained to {14,15,16,17}. Hence, four solutions.
N=27
y=33-2x, z=x-13. x is constrained to {13,14,15,16}. Hence, four solutions.
N=28
y=32-2x, z=x-12. x is constrained to {12,13,14,15,16}. Hence, five solutions.
N=29
y=31-2x, z=x-11. x is constrained to {11,12,13,14,15}. Hence, five solutions.
N=30
y=30-2x, z=x-10. x is constrained to {10,11,12,13,14,15}. Hence, six solutions.
N=31
y=29-2x, z=x-9. x is constrained to {9,10,11,12,13,14}. Hence, six solutions.
N=32
y=28-2x, z=x-8. x is constrained to {8,9,10,11,12,13,14}. Hence, seven solutions.
N=33.
y=27-2x, z=x-7. x is constrained to {7,8,9,10,11,12,13}. 7 sols.
N=34, N=35, N=36, N=37, N=38, N=39
The pattern is relatively obvious so I'll just type up the number of solutions for each n.

[EDIT]: MISCOUNTED. ANSWER SHOULD BE 110 + 11 = 121
Respectively, they are: 8, 8, 9, 9, 10, 10. So adding them all up together.. we get 121 ways to fill 20 slots with 1s, 2s, and 3s that total no more than 40.

There is probably a more efficient way to determine the number of solutions, as my mathematical skills are a little rusty from years of inactivity, but 120 does seem to be the number you want.

I will have to say quite frankly that I do not understand the nature of your complaint against MIF. The animals are alive(and presumably enjoying their lives), so it doesn't offend any vegans taking a look. I presume that you are of a religious bent, which I must say rather respectfully that viewing images of live, real animals is not offensive.

Also, the website has been transformed a plethora of times over the past years of MIF's existence and quite obviously does not only serve UK viewers of MIF.

Woops. Didn't read too carefully.

Are you going to keep it a secret and profit off of it or are you going to enlighten our poor souls as to what it is?

Can you state this in clearer terms? It is... a little hard to comprehend.

With all due respect, can Knewlogik stop referring to irrelevant prayers?(I have no disrespect for religion - I am a Christian myself, but it is not necessary here.) Also, please learn how to use punctuation(commas, semicolons, you know... periods.)

Are you ESL?

Hmmm, I made a few mistakes there. Still no change, other than being more of a chess-obsessed maniac.