@leslietownes Thanks. Yeah I did this brute force method. Usually when I brute force things the mark scheme shows a one liner solution to the problem and that is often gut wrenching. But this time the mark scheme did the same thing
When two planes intersect, the intersection is a line. The cross product of the normal vectors of both planes will lie on that line of intersection. I understand that there is several ways in which 3 planes can intersect. When 3 planes intersect at a point, will the direction vector of the line (when two planes intersect) be parallel to all 3 planes?
@o.9 How are they different? The way that I was taught combinations was with the letters of the alphabet. So say you had ABCD. We could permutate this to find all the ways in which the characters ABCD could be arranged. However, this would only be 1 combination. So how does that relate to this?
Would like to ask a small question about binomial distribution. I am not getting the concept. Suppose we flip a coin 4 times. We would like to find the probability of getting exactly 2 heads in those 4 flips. If we think in terms of tree diagrams, then there will be many ways in which we can get 2 heads. I.e. HHTT, THTH, THHT, etc. So what we are doing is finding all the ways we can permutate HHTT right? So why do we use combinations for this?
Did Stack change it's user profile interfaces or did I click on something that I shouldn't have? It seems I can no longer stalk people and find out when they were last online... Can I revert back?
@TedShifrin I'm not cheating on a test or anything haha. it's a maths admission test. a university admission test only runs on the 5th november each year. im just practicing old past papers
@leslietownes Oh yes, the derivative was a better way to approach this question. thanks. The value of the coefficients is the thing that's confusing me. Because if you use $B = a^2+b^2$ you get the correct coefficient for the $x^2$ term, but you get the wrong coefficent for the $x$ term.
Greetings all. I need some assistance on (ii). In order to show that there is no value of $b$ for which $x=1$ is a repeated root of the cubic. I attempted to factorise the cubic. If $x=1$ is a solution, then the cubic may be expressed as follows with no remainder. $x^3+2bx^2-a^2x-b^2 \equiv (x-1)(Ax^2+Bx+C)$. Comparing coefficients, $A = 1, C = b^2, B = a^2+b^2$ $\therefore (x-1)(x^2+(a^2+b^2)x+b^2)$. Finally, if we sub $x=1$ into the resulting quadratic, we get $1+a^2+b^2+b^2 > 0$
Nevermind, figured it out. Seems instead of expanding out $(\alpha + \beta + \gamma)^n$ to get an identity, you were supposed to make use of the cubic equation. I.e. $\alpha^3 = 2\alpha^2 - k$ and from there it's easy to find $\alpha^3+\beta^3+ \gamma^3$, $\alpha^4+\beta^4+ \gamma^4$ and $\alpha^5+\beta^5+ \gamma^5$
I understand what the question wants me to do. However, I would like to find out if there a clever way or a pattern that can be observed to find the sum the of $\sum \alpha^n$? I brute forced the $\sum \alpha^2$ and $\sum \alpha^3$. But I think the question wants me to observe a pattern or something, but I don't see it. By $\sum \alpha^3$, I mean $\alpha^3 + \beta^3 + \gamma^3 = (\alpha+\beta+\gamma)^3-3(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\alpha\gamma)+3\alpha\beta\gamma $
Any hints on how to go about part (c)? The roots $z$ are the following $\tan \theta$ where $\theta = \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8} and \frac{7\pi}{8}$
What programs support the graphing of the loci of complex numbers? I would like to find out what this sort of thing would trace out $$arg(\frac{z-a}{z+b}) = \theta$$. It doesn't seem Desmos can do it.
if $ \lim_{n \rightarrow \infty} a_n = 0$, wouldn't that be sufficient to conclude that $\sum a_n$ converges? If not, doesn't the ratio test say about the same thing?
However, the ratio test says something that is identical to this in my opinion. That is, if $\lim_{n\rightarrow\infty} \frac{a_{n+1}}{{a_n}} = L$ then $\sum a_n$ will converge if $L<1$. That is because for each successive term in the $a_n$ sequence, it approaches $0$. And if $L>1$ each successive term is getting larger hence it will diverge. So, the way I see it, the divergence test says the same thing but more concisely so what can't we use that to prove convergence?
Doesn't the divergence and ratio test say about the same thing? But for some reason we are told we cannot use the divergence test to prove convergence, but we can use the ratio test to prove convergence. The divergence test states that if $ \lim_{n \rightarrow \infty} a_n \not= 0$ then clearly $\sum a_n$ will not converge. But we are told the converse of this statement is not true.That is, if $\lim_{n \rightarrow \infty} a_n = 0$ then we cannot infer from this that $\sum a_n$ is convergent.
Wouldn't the solution set be all the points on the line $y = x$? Clearly it isn't, but why? Surely you can just inverse sine both sides and arrive at $y = x$ ?
Since the object is moving up and incline and the weight is opposing the motion of the object. Will the weight of the object also be taking away energy from the object?
Hello all. In this example question, I understand that this question can be solved with the kinematics formulae. However, I am trying to solve it with an energy approach. I understand that if an object is in motion, then that object has energy. I also understand that if an object is moving along a surface, then friction will oppose the motion and take away energy from the object.
