Search results

$\noindent Alternatively, just note that $\alpha$ is the acute angle the line makes with the $y$-axis, so $|m|=\cot \alpha$, where $m\in \mathbb{R}$ is the slope of the line, which we can find from its equation (note $m\neq 0$). Hence $\tan \alpha = \frac{1}{|m|}$, so $\alpha = \tan ^{-1}...

$\noindent Had a look at the Q. now. That line is vertical (i.e. its slope is undefined, or `infinite'). So to find the angle between the lines, let this angle be $\alpha$. Let the other line (the non-vertical one) have slope $m$ (which you can find easily from rearranging its equation). Let the...

The inverse tan of that should be 67.619864.... deg.

Use the (acute) angle between two lines formula. Have you learnt this yet? It is: $$\tan \theta = \left |\frac{m_1 - m_2}{1+m_1 m_2} \right|$$ provided that m1m2 is not equal to -1. (m1 and m2 refer to the slopes of the two lines. Obviously if m1m2 were equal to -1, the lines would be...

Inverse tan of a positive number will always be an acute angle. You may need to add or subtract a multiple of 180 deg. if you want the answer to be in some other range.

The fourth one is only to avoid losing cheap marks from the markers (in other words, "because the HSC says so"). The first three are the ones needed for a mathematically valid induction proof of that kind.

Re: HSC 2016 2U Marathon $\noindent Find $\frac{\mathrm{d}}{\mathrm{d}x}\big{(}\ln |x|\big{)}$.$

In the end, they actually will come down to inspection mostly. By this I mean, scientists observe ("inspect") the world via experiments and then may try and propose models explaining the 'why'. If you keep asking why though it'll probably come down to 'by inspection', since we'll arrive at the...

Most of your Q's can be answered in two words: "by inspection". :p

Because they represent two different things. If you use the same symbol for them, it's easy to get confused. It's like in a maths word problem if there are two unknown quantities and you let both of them be x.

$\noindent Otherwise, we can do it how you seem to know how to do Q's like this if it just said do it for $x>0$. Fix $x>a>0$. Let $g(s):=\ln \left(\frac{s}{a}\right)$ for $s\in \left[a,x\right]$. Then $g$ is continuous on $\left[a,x\right]$ and differentiable on $\left(a,x\right)$. It follows by...

$\noindent For that particular question, we can turn it into a one parameter problem if we like. Note that for $x>a>0$, $\frac{x-a}{x}=1-\frac{a}{x}$ and $\frac{x-a}{a}=\frac{x}{a}-1$. As $x$ and $a$ vary with $x>a>1$, the ratio $t\equiv \frac{x}{a}$ covers all real numbers greater than $1$...

Here's how to write the inequalities normally and avoid the text disappearing: (x-a)/x < ln(x/a) < (x-a)/a. I.e. put spaces between the inequality signs and nearby text.

Not sure what the Q. is. Can you provide an example?

Given a point z, the point z+i is one unit up from this in the complex plane, because the real part is unchanged but the imaginary part has increased by 1 due to adding i. Meanwhile, for arg(z+i), this is arg(z – (-i)). Recall that to sketch arg(z - z0) = theta (where z0 is a fixed complex...

It's easier to just think about it intuitively. It's obvious from an intuitive point of view that increasing the distance will lead to intensity going to down (e.g. if you take a heat source away from you, you feel less heat) and increasing the power of the source leads to intensity going up...

$\noindent What do you mean by $2I = \frac{1}{2^2}$? We should carefully define our variables. Here is one way to do it.$ $\noindent Let the original intensity be $I_0$ and the new one be $I_1$ (i.e. after doubling the distance and doubling the power of the source). Let $I^{\star}$ be the...

Are you saying the final intensity is half the original one? That is the answer that I gave.

For 1., if we first double the power of the source, then the intensity doubles. Then if we double the distance from the source, this has the effect of 'quartering' the intensity. Since 2*(1/4) = 1/2, the end result is that the intensity is halved.

Re: First Year Uni Calculus Marathon This is basically a hint to one of seanieg89's hint exercises.