Olinguito

If [latex]a[/latex] is a repeated root of a polynomial [latex]p(x)[/latex] then [latex]p(x)=(x-a)^kq(x)[/latex] for some polynomial [latex]q(x)[/latex] and integer [latex]k\geqslant2[/latex]. Thus [latex]p(x)=(x-a)r(x)[/latex] where [latex]r(x)=(x-a)^{k-1}q(x)[/latex] and both [latex]x-a[/latex] and [latex]r(x)[/latex] are not units.

It depends of what your set of axioms is for. For example, your axioms might be for the real numbers as an ordered field. An field [latex]\langle F,+,\cdot\rangle[/latex] is said to be (totally) ordered by [latex]\leq[/latex] iff the following holds: [latex]\forall\,a,b,c\in F[/latex]: [latex]\text{either}\ a\leq b\ \text{or}\ b\leq a[/latex] [latex]a\leq b\ \text{and}\ b\leq a\ \Rightarrow\ a=b[/latex] [latex]a\leq b\ \text{and}\ b\leq c\ \Rightarrow\ a\leq c[/latex] [latex]a\leq b\ \Rightarrow\ a+c\leq b+c[/latex] [latex]0\leq a\ \text{and}\ 0\leq b\ \Rightarrow\ 0\leq ab[/latex] We first show that [latex]0\leq 1[/latex]. By axiom (1) either [latex]0\leq 1[/latex] or [latex]1\leq 0[/latex]. If [latex]1\leq 0[/latex] then [latex]1+(-1)\leq 0+(-1)[/latex] (axiom (4)) i.e. [latex]0\leq -1[/latex]. Then [latex]0\leq -1[/latex] and [latex]0\leq -1[/latex] imply (axiom (5)) [latex]0\leq (-1)(-1)=1[/latex]. (This is actually a contradiction because [latex]1\leq0[/latex] and [latex]0\leq 1[/latex] imply (axiom (2)) [latex]0=1[/latex].) So we can’t have [latex]1\leq 0[/latex]; hence we must have [latex]0\leq 1[/latex]. And now we are done, for [latex]0\leq 1[/latex] [latex]\implies[/latex] (axiom (4)) [latex]1=0+1\leq 1+1=2[/latex]. http://en.wikipedia.org/wiki/Ordered_field

This is an example of Pell’s equation.

That’s the assertion of Murphy’s law, which is clearly not true.

If something can go wrong, it means that the probability of it going wrong is greater than zero. If something will go wrong, it means that the probability of it going wrong is equal to one. [latex]\mathrm{P(go\ wrong)>0\ \not\Rightarrow\ P(go\ wrong)=1}[/latex].

It’s called aposiopesis.

[latex]\frac{(1+3)!}{1!+3!}=\frac{24}7\notin\mathbb N[/latex].

Suspend them and see which one always points north–south.

IMHO this is better: “I would be eating my lunch if I were not typing this response.” Example of a would have been + -ing construction: “If I had carried on waiting, I would have been waiting for two hours by the time she turned up.”

1 hour. The speed of the boat relative to the hat does not depend on the current.

For each [latex]x\in X[/latex], let [latex]x_1=x[/latex]. Then let [latex]x_2=f(1,x_1)[/latex]. Then let [latex]x_3=f(2,x_2)[/latex], [latex]x_4=f(3,x_3)[/latex], [latex]x_5=f(4,x_4)[/latex], … and so on. This is what’s called a recursively defined sequence. The existence of such a sequence is obvious; the corollary asserts that it is the only one possible. By the way, could you state the theorem or proposition to which this result is a corollary?

Basically you are looking at a triangle with side lengths 3, 5, and 7. The angle between the two shorter sides (call it [latex]\theta[/latex]) can be found by the cosine rule: [latex]c^2\ =\ a^2+b^2-2ab\cos\theta[/latex] Now you want to look at the triangle two of whose sides have lengths 3 and 5, the angle between them being [latex]\pi-\theta[/latex]. The length of the opposite side can again be found by the cosine rule.

It can be in any angular unit you like radians, degrees, grads, anything. You simply set your calculator to the angular unit you want and compute the inverse tangent. Note however that in some cases you have to use only radians. For example, if [latex]f(t)=\arctan t[/latex] and you compute [latex]f'(t)=\frac1{1+t^2}[/latex], then in order for the formula to work, [latex]t[/latex] must be in radians. You have to be aware of cases when you must use radians; otherwise you have a choice of radians or degrees.

[latex]2\left[\sum_{i=0}^{n}{i}\right]+(n+1)\ =\ 2\left[\frac{n(n+1)}2\right]+(n+1)\ =\ (n+1)^2[/latex] Also [latex]\left(\sum_{i=0}^n10^i\right)^2\ =\ \sum_{i=0}^n10^{2i}+2\sum_{0\leqslant i<j\leqslant n}^n10^{i+j}[/latex] using the formula [latex]\left(a_n+a_{n-1}+\cdots+a_1+a_0\right)^2\ =\ a_n^2+\cdots+a_0^2+2\sum_{0\leqslant i<j\leqslant n}^na_ia_j[/latex] Given [latex]1\leqslant r\leqslant n[/latex], how many pairs [latex]i,j[/latex] are there with [latex]0\leqslant i<j\leqslant n[/latex] such that [latex]i+j=r[/latex]? Suppose [latex]r=2k[/latex] is even. Then [latex]r=0+2k=1+(2k-1)=\cdots=(k-1)+(k+1)[/latex] so there are [latex]k[/latex] such [latex]i,j[/latex] pairs. Thus the coefficent of [latex]10^r[/latex] is [latex]2k+1=r+1[/latex]. Suppose [latex]r=2k-1[/latex] is odd. Then [latex]r=0+(2k-1)=1+(2k-2)=\cdots=(k-1)+k[/latex] so again there are [latex]k[/latex] such [latex]i,j[/latex] pairs. Then the coefficent of [latex]10^r[/latex] is [latex]2k=r+1[/latex] again. For [latex]n+1\leqslant r\leqslant2n-1[/latex], we let [latex]s=r-n[/latex]; then [latex]r=s+n=(s+1)+(n-1)=\cdots=(s+k)+(n-k)[/latex] giving [latex]k+1[/latex] pairs. Since [latex]s-k<n-k[/latex] we have [latex]n-s>2k[/latex]; therefore [latex]n-s=2k+1[/latex] or [latex]n-s=2k+2[/latex]. If [latex]r[/latex] is even, then so is [latex]n-s=r-2s[/latex]; then [latex]n-s=2k+2[/latex] and the coefficient of [latex]10^r[/latex] is [latex]2(k+1)+1=n-s+1=2n-r+1[/latex]. If [latex]r[/latex] is odd, then so is [latex]n-s[/latex]; so [latex]n-s=2k+1[/latex] and the coefficient of [latex]10^r[/latex] is [latex]2(k+1)=n-s+1=2n-r+1[/latex] again. Hence [latex]\left(\sum_{i=0}^n10^i\right)^2\ =\ 10^{2n}+1+\sum_{r=1}^n(r+1)10^r+\sum_{r=n+1}^{2n-1}(2n-r+1)10^r\ =\ \sum_{r=0}^n(r+1)10^r+\sum_{r=n+1}^{2n}(2n-r+1)10^r[/latex].

Logarithms and square roots of negative numbers do exist. They are complex rather than real.

Draw a diagram of vectors. The velocity of the plane relative to ground is the vector sum of the two given velocities. You’ve calculated the magnitude; the angle can be found by trigonometry.

I don’t see why that step is not valid. Let me make be more clear. To say that a series [latex]\sum_n a_n[/latex] converges means the sequence of partial sums [latex]\sum_{i=1}^n a_i[/latex] tends to a limit as [latex]n[/latex] tends to infinity. In fact I’m splitting hairs here: a series is rigorously defined as nothing more or less than the sequence of its partial sums. The limit of a convergent series is denoted [latex]\sum_{n=1}^\infty a_n[/latex]. This notation is often used to denote the series itself but for this post I’ll write [latex]\sum_n a_n[/latex] for a series and [latex]\sum_{n=1}^\infty a_n[/latex] for the limit of a (convergent) series. My contention is that if the series [latex]\sum_n a_n[/latex] is (absolutely) convergent and [latex]k[/latex] is a fixed real number, then the series [latex]\sum_n ka_n[/latex] is (absolutely) onvergent and [latex]\sum_{n=1}^\infty ka_n=k\sum_{n=1}^\infty a_n[/latex].

Oops! I beg your pardon. I didn’t think clearly. You don’t need to subtract (that would be finding the death rate, not survival rate). You just divide the final count by the original count and then multiply by 100. The forumula is: [latex]\text{percentage survival rate}\ =\ \frac{\text{final count}}{\text{original count}}\times100\%[/latex]

As long as the series is absolutely convergent (as this one is) then such an operation is perfectly valid.

I’d say this was correct.

Try plotting the graph with the % survival on the vertical axis and time on horizontal axis. In such questions time is usually the independent variable and so is plotted on the horizontal axis. Also in such questions, if you have to calculate anything, there is usually a formula you have to use. What’s the formula relating % survival and time?

The statement [latex]a\ne b\Rightarrow ag\ne bg[/latex] is the contrapositive of [latex]ag=bg\Rightarrow a=b[/latex] (which is what you proved in the opening statement of your post). There is no need to state it as an extra condition.