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A degree without Honours is referred to as a "pass" degree. If you don't do the Honours, you still get the pass degree. If you do Honours and fail it, you still get the pass degree - though third class honours is so vanishingly rare that it is effectively a fail.

If you are still around, one approach with inequalities are to consider LHs > RHS in the form of LHS - RHS... So, in this case: \begin{align*} \text{Our induction hypothesis:} \qquad k^2 + 1 &> k \qquad \qquad \text{. . . . (*)} \\ \\ \text{We seek to prove:} \qquad (k + 1)^2 + 1 &> k + 1 \\...

If you are going to make edits, might I suggest the cases where you put (1 + \cos\alpha + i\sin\alpha)^k + (1 + \cos\theta - i\sin\theta)^k on pages 1 and 3 have the \theta's changed to \alpha?

The right angle is definitely OAC as the tangent to a circle is always perpendicular to the radius at the point of contact. If OCA was a right angle, then the tangent at A would be parallel to OC and thus OA would not be a tangent, but rather would cross the circle at some point B between O and...

Note that this should say that \pi\ \text{radians} \equiv 180^{\circ}. Also, it is important to get comfortable with working in radians as soon as you can. Calculus of trig functions does not work in degrees, and radians are not something added to irritate / confuse, they are essential for...

There are a couple of issues with question 1: On page 2, you have |z| = \sqrt{4\cos^2{\frac{\alpha}{2}}} = 2\cos{\frac{\alpha}{2}} but this is not always true: \begin{align*} \text{Take $\alpha = 2\pi$:} \qquad \cos{\frac{\alpha}{2}} &= \cos{\pi} = -1 \\ |z| &= \sqrt{4\times(-1)^2} = 2 \\ \\...

I would refer to HCl (g) as hydrogen chloride or hydrogen chloride gas, whereas I would refer to HCl (aq), the acidic solution made by dissolving hydrogen chloride gas into water, as hydrochloric acid. Similarly, HCN (g) = hydrogen cyanide, HCN (aq) = hydrocyanic acid. For acidic substances...

You can prove the answer is wrong by working from it... \begin{align*} \text{Put $f(1) = -6$ into} \qquad f(x) &= f(x+1) + x^2 \qquad \text{taking $x = 1$} \\ f(1) &= f(2) + 1^2 \\ -6 &= f(2) + 1 \\ f(2) &= -7 \\ \\ \text{Put $f(2) = -7$ into} \qquad f(x) &= f(x+1) + x^2 \qquad \text{taking $x...

\bullet \ \text{1 Ind, 5 Libs, 1 Lab:} \quad \binom{5}{1} \binom{10}{5} \binom{8}{1} \bullet \ \text{1 Ind, 4 Libs, 2 Lab:} \quad \binom{5}{1} \binom{10}{4} \binom{8}{2} \bullet \ \text{1 Ind, 3 Libs, 3 Lab:} \quad \binom{5}{1} \binom{10}{3} \binom{8}{3} \begin{align*} \text{Total...

Suppose we had 10 balls of the five colours with no restrictions other than there must be at least 1 of each colour. Imagine the colours as different bins, into which we put some number of each colour - blue, green, pink, red yellow (BGPRY). A solution like B = 3, G = 1, P = 2, R = 3, Y = 1 is...

It would depend a lot on the question, and on whether it had bipartisan support. Howard demonstrated how a determined opponent can do a lot of damage to a "yes" campaign, and Dutton would likely campaign against it. Also, if the model became the central issue, some voters wanting a republic...

Some immediate thoughts: What did you get for 2(a)? Have you noticed the two pairs of differences of two squares on the RHS? And that the resulting \cos^2{\alpha} terms can be found by rewriting into \cos{2\alpha} terms? \cos{4\theta}\ \text{is the real part of}\ z^4\ \text{when}\ z =...

Carrying on from my earlier post, we had: Suppose we set the number or red balls, [text]r[/tex] to be at least 17, producing an invalid solution: \begin{align*} r + y + g + p + b &= 50 \\ (r'' + 16) + (y' + 3) + (g' + 3) + (p' + 3) + (b' + 3) &= 50 \qquad \text{where $r'' = r - 16$, so that}\...

As a general rule, unravelling the larger factorials makes for easier algebra: \begin{align*} \text{Required to Prove:} \qquad ^nP_r &=\ ^{n-2}P_r + 2r ^{n-2}P_{r-1} + r(r-1) ^{n-2}P_{r-2} \\ \\ \text{RHS} &= \frac{(n-2)!}{(n-r-2)!} + \frac{2r(n-2)!}{(n-r-1)!} + \frac{r(r-1)(n-2)!}{(n - r)!}...

Follow Up... how many permutations of the word PENCILS will have the "I", "C", and "E" occur in that order?

Not any more, but it was until the start of 2020. A parabola (x - h)^2 = \pm4a(y - k) has its vertex at (h,\,k) and a focal length of a (which is the vertical distance from the vertex to the focus). So, for example, (x - 2)^2 = 4(y - 1) has its vertex at (2,\,1) and a focal length of a = 1...

@ISAM77 's form is the general form: \begin{align*} y - h &= \frac{A}{x - k} \\ \\ y - 2 &= \frac{0.5}{x - 2} \end{align*} makes it easy to sketch... y - 2 = 0 \implies y = 2 is the horizontal asymptote (or, in the general case, y = h) x - 2 = 0 \implies x = 2 is the vertical asymptote (or...

The problem is finding the number of different solutions of the equation r + y + g + p + b = 50 where r is the number of red balls, etc, subject to the constraints that: r,\,y,\,g,\,p,\ \text{and}\ b \in \mathbb{Z}^+ r,\,y,\,g,\,p,\ \text{and}\ b \ge 4 r,\,y,\,g,\,p,\ \text{and}\ b <...

I wasn't aware of the other meetings, thanks for passing that info on.

Not surprising as the solutions are poorly expressed and contain errors. Issues that I note include: It is not clear that the "coordinate for the tangent" refers to the point T where the tangent touches the curve and not the point P that we seek, where the tangent meets the axis. The...