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We always have F = ma where F is the net force. The fact that the force is opposing the motion is reflected by the fact that F and a are negative.

Re: Several Variable Calculus $\noindent It's an operator that acts on scalar functions. Let $\textbf{F} = \left(F_{x}, F_{y}, F_{z}\right)$, then as $\nabla = \left(\partial_{x}, \partial_{y}, \partial_{z}\right)$, we have $\textbf{F} \cdot \nabla =...

Re: Statistics $\noindent Basically, the pdf of $\mathrm{Unif}(0,\theta)$ ($\theta > 0$) isn't just $f(x) = \frac{1}{\theta}$, it's $f(x) = \frac{1}{\theta}I_{[0,\theta]}(x)$. Therefore, given non-negative data $x_{1},x_{2},\ldots, x_{n}$, we have $f(x_{i}; \theta) =...

Re: Discrete Maths Sem 2 2016 The empty set.

Re: Discrete Maths Sem 2 2016 Once you change the summation index, you should see a lot of terms are in common so cancel. Only some of the terms are the start and end won't cancel. The terms at the end will leave you with something involving N.

Generally speaking, no.

Whole graph should be reflected about x = 0 (the y-axis).

Basically the first line – it should be a = F/m. (Since F = ma.)

$\noindent The $r$ should just be $1-y$ and the $h$ should be $4\sqrt{1-y^{2}}$. So $2\pi r h = 2\pi (1-y)\times 4\sqrt{1 - y^{2}} = 8\pi(1-y)\sqrt{1 - y^{2}}$.$

$\noindent The volumes one is (B). Note that the last two integrals can be immediately ruled out as they are $0$ (integrating an odd function over a symmetric-about-0 interval). So if you had no idea and had to guess, you could narrow it down to between (A) and (B). Anyway, one way to see that...

$\noindent For the mechanics one, using $v^{2} = u^{2} + 2as$ with $v=0$, $u = V_{0}$, $s = l$, we can solve for $a$ to find that the required (constant) acceleration is $a = -\frac{V_{0}^{2}}{2l}$. The force is thus $ma = -\frac{mV_{0}^{2}}{2l}$, so the answer is (C).$

Re: Statistics $\noindent The MLE is that $\widehat{\theta}$ (a function of $X_{1},\ldots,X_{n}$) that maximises that likelihood function (as a function of $\theta$). Due to what the $L$ is you've written down, the way to maximise this is to make $\widehat{\theta}$ as small as possible (since...

Re: MATH1131 help thread $\noindent Right idea, but the chain rule calculation is $\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2}F'\left(x^{3}\right)$ (rather than $F'(x)$).$ $\noindent Since $F'(u) = \cos \left(u^{2}\right)$ (by the FTC), we have $F'\left(x^{3}\right) = \cos...

$\noindent Because we're told the graph of the cubic passes through the point $(0, -6)$, so if we sub. $x = 0$ into the cubic, we get $-6$. Also, if we sub. $x = 0$, we get $d$. Hence $d = -6$.$ In general, the constant term of a polynomial is the y-intercept (value when x = 0).

First find f(-1). Then plug this into the equation for g(x) to get the value of g(f(-1)).

Re: MATH1131 help thread $\noindent In other words, let $F(u) = \int_{0}^{u} \cos \left(t^{2}\right)\, \mathrm{d}t$. Then we are asked to find $\frac{\mathrm{d}y}{\mathrm{d}x}$, where $y = \int_{0}^{x^{3}} \cos \left(t^{2}\right) \, \mathrm{d}t = F\left(x^{3}\right)$. You should be able to get...

Re: MATH1131 help thread We're multiplying it by the derivative of the x^3 in the integral upper limit, because of the chain rule.

If that were the case, the sides of vertical cross-sections taken from the centre of the base would essentially be parabolic arcs. It's because the sides of vertical cross-sections from the centre of the base are straight lines that the w is linearly related to h.

Re: Statistics Converging in distribution to a constant means the limiting random variable X that we are converging in distribution to is a constant random variable (see https://en.wikipedia.org/wiki/Degenerate_distribution ). It does not mean the CDF of the limiting random variable X is...

Basically logb(x) is the inverse function to bx when b > 0 (and b ≠ 1), but if b < 0, then bx isn't generally well-defined in the real numbers for general real x (like (-1)1/2 doesn't exist in the real numbers (and in fact is multivalued in complex numbers)), so we can't define an inverse...