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I figured it couldn't be that easy.
Did you read my post carefully?
Do we also need to assume the opposite, i.e. that
If not the question is trivial: just let f(x) = c where c is any irrational number.
You cannot "prove" that 1 + 1 = 1, since it's not true. You can give a false proof that most people with limited mathematical knowledge would not catch, but it's nothing more than a magic trick.
x = y
x^2 = xy
x^2 - y^2 = xy - y^2
(x + y)(x - y) = y(x - y)
x + y = y
y + y = y
1 + 1 = 1
A good exercise would be to figure out where the mistake in this "proof" is.
I think what Bunno was asking is "How can f(x) not be continuous at x = 3, but if you factor the numerator and cancel it with the denominator then it is continuous at x = 3".
If that was your question Bunno, then the answer is that
If you are ever solving an equation and you get (x^2 - 9) / (x - 3), then you can never allow x to equal 3, even if you factor the numerator and cancel the denominator. You will still get x + 3 for the total expression, but you have to keep in mind that you cannot allow x to equal 3 even if the final answer you get would be continuous at that point.
I think substitution would be more useful here.
You can factor this into the form (2y^2 - a)(y^2 - b) = 0, from which you can easily solve for y and then x. a and b are irrational, which means x and y are too, and I'm too tired to deal with irrational numbers right now, but that should be good enough for you to take over.
Sorry, I edited to put the answer behind a hide tag.
With 1 try your chance of winning is 6/49 * 5/48 * 4/47 * 3/46 * 2/45 * 1/44 = 1/13983816. This means that your chances of not winning in 1 try are 13983815/13983816. Raise this number to the 100th power to find your chances of not winning in 100 tries, then subtract this number from 1 to find the chances of winning at least once in 100 tries. The final answer will be: 1 - (13983815/13983816)^100.
Actually, x = 61 is valid. The original problem doesn't restrict us to positive square roots, so you get 11 + (-8) = 3.
You have 2 simultaneous equations: ax + cy = 0 and bx + dy = 0. Try solving those 2 equations for x and y. Along the way you'll need to use the fact that ad - bc = 0.
You multiplied out the 4ac term wrong and forgot to substitute the c portion of b^2:
Yep, it's easy to pull LuisRodg's second answer from that:
On the integers it's easy to see that the RHS of the equation is a power of 10, which means you need to have the same number of 2's and 5's on the LHS. It just so happens that the value of x at which this occurs makes the entire equation true.
Non-integer solutions are more tricky. You need to do some fancy footwork with logarithms:
Now you can use the quadratic formula to solve for x.
The + half turns out to be x = 2. Whether or not the - half matches your TI-89's answer is beyond me, as I'm not sure how to simplify the expression further.
You have your definitions a little backwards. If you're able to find a pair (x, y) that satisfies your equation above then you've shown that they are linearly dependent. Since that's what you're asked to prove, just find that (x, y) pair and you're done.
To get from one term to the next in a geometric sequence you multiply each term by a fixed amount. In your example our first term is 27. The second term would be 27x, where x is the amount that every term will be multiplied by. The third term would be 27x^2, etc. until you reach the sixth term of 27x^5. You know this is equal to 519/9, so your equation is 27x^5 = 519/9. Solve for x, then multiply it by 519/9 for the seventh term.
All circles can be described by the equation (x - a)^2 + (y - b)^2 = r^2, where the center is located at (a, b) and has a radius of r. What you have to do is work with your equation to get it to fit into the general equation I just gave you.
Figuring out b is simple. Since there is no y term you know that b has to be 0.
Finding a is only slightly more difficult. Expanding the expression (x - a)^2 gives us x^2 - 2ax + a^2. Our x term is 8x, so we know that 8x = -2ax, which gives us a = -4.
We're not quite done with x yet, since (x + 4)^2 = x^2 + 8x + 16, but we don't have a term of 16. We do have an 8, so what we can do is add 8 to both sides. This gives us (x + 4)^2 + (y - 0)^2 = 8, and your work is done.
Instead of going term-by-term, use pairs. Show that |a - b| + |b - c| <= |1 - 0| + |1 - 1| = 1 for all 0 <= a, b, c <= 1.
I think leading 0's is the issue here. How about 9*8*7 = 504 4-digit numbers ending in 0 and 4*8*8*7 = 1792 4-digit even numbers that don't end or begin with 0, for a total of 2296?
I can agree with that, but I would argue in turn that the mathematical breakdown at the very specific distance of d = 0 is inconsequential because the model becomes wildly inaccurate before that point anyway. I'm more concerned about the fact that the answers that the model gives at very small values of d aren't even close to reality than I am by the fact that eventually it doesn't give an answer at all.
Yes, that's true, but it's not relevant to the specific example of Newtonian gravity because the model has been shown to be flawed. You can argue that we don't have sufficiently advanced knowledge of math to combine quantum mechanics and general relativity to handle situations like singularities (which, last I knew, is true), but in the very specific case of Newtonian gravity it is the model that is at fault, not math.
Alright, looks good now George.
How did you figure out that it needs a factor of n though? I tried following your work and thought it needed a factor of n + 1, which is obviously incorrect.
Your answer works for n = 1, but it doesn't appear to be correct for larger values of n. For example, for n = 2 we should get S(2) = 2q^2 + q = q(2q + 1). Here is what I get by using your formula:
It's pretty plain to see at this point that (q^3 - 2q^2 + 1) / (q^2 - 2q + 1) does not equal 2q + 1. You're close, but I think you've made a mistake in your algebra somewhere.
It seems you don't understand Newtonian gravity. The model doesn't break down; it was never intended to be used at d=0. Such a scenario cannot exist on the macro-scale for which Newton created his model. Of course that doesn't change the fact that mathematics, at our current level of understanding, breaks down when we attempt to divide by zero.
This is true, but it's also the reason I ask why you consider it to be a mathematical breakdown when the model isn't even intended to be used at such small distances. Yes, under certain situations that the model was never intended to be used it requires the impossible (division by 0), but I fail to see why you would blame it on the shortcomings of math rather than the shortcomings of the model.