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Re: maths questions help $\noindent Set the discriminant $\Delta := (5m-1)^{2} -4(2m-3)(3m-2)$ to be greater than $0$ and solve for $m$. In other words, solve the quadratic inequation $(5m-1)^{2} -4(2m-3)(3m-2) > 0$ for $m$.$ Remember, if the given quadratic equation has two distinct (real)...

Re: University Statistics Discussion Marathon Note that the answer to the sum cannot exceed 0.5 (in general for X ~ Bin(n, 1/2), if k > n/2, then Pr(X ≥ k) ≤ 0.5).

Yes (to find the y'). E.g. if there were a point of intersection at x = 10, you would need to find the values of the derivatives at x = 10 (to get the slopes at x = 10).

Re: Several Variable Calculus $\noindent For $r\in\left\{1,2,3\right\}$, the $r$-th column of $A$ (which will be a $3\times 3$ matrix) will be equal to $A\mathbf{e}_{r} = \mathbf{a}\times \mathbf{e}_{r}$. So try and find an expression for $\mathbf{a}\times \mathbf{e}_{r}$ using the Einstein...

You'll need to find the points of intersection, and at each point, find the slopes of the tangents of the two curves. Then use the angle between two lines formula.

a) Yes. b) Pretty much. c) Correct! d) Have a play around with this: https://www.geogebra.org/m/W7dAdgqc (type in sech(y) - sech(1) for the f'(x,y) in the top left). Basically you need to sketch the direction field and draw some solution curves passing through various different points on...

In your original answer, did you mean cosh-1 (I'm guessing you did)? You wrote arccos (didn't put an "h" on the end).

$\noindent Did you manage to solve the ODE for $y$ as a function of $x$? Once we have this solution (noting the initial condition of $y = 0$ when $x = h$), we will get an equation relating $y\equiv \frac{1}{2}v^{2}$ to $x$, and will involve $h$. To get the impact speed, set $x = 0$ and solve for...

Your proposed answer isn't well-defined. For any h > 0, e^(h.sqrt(k/g)) > 1, so the arccos of this doesn't exist in the real numbers.

Re: MATH1131 help thread $\noindent The result $\lim_{\theta \to 0}\frac{\sin \theta}{\theta}$ is a standard result. Basically use the substitution $\theta = \frac{1}{x}$, so $\theta \to 0^{+}$ as $x \to \infty$ and the limit becomes $\lim_{\theta \to 0^{+}}\frac{\sin \theta}{\theta} = 1$.$

Re: MATH1131 help thread $\noindent Rewrite the function as $\frac{\sin \frac{1}{x}}{\frac{1}{x}}$. This tends to $1$ as $x \to \infty$, using the limit result $\lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1$. Hence the answer is $1$.$

i) We have dv/dt = (dv/dx)*(dx/dt) (chain rule) = (dv/dx)*v = v*(dv/dx) (as dx/dt = v). ii) We have dv/dt = kv, so from the first part, v*dv/dx = kv. It follows that dv/dx = k, so v = kx + C for some constant C. When x = 0, v = 7, so C = 7. So v = kx + 7. When x = 3, v = 13, so 13 = 3k + 7...

$\noindent b) The initial value problem (IVP) is$ $$\frac{\mathrm{d}v}{\mathrm{d}t} = g-kv^{2}, \quad v(0) = 0,$$ $\noindent where $k = \frac{1}{5}$.$ $\noindent Hence we have$ $$\begin{align*}\int_{0}^{v} \frac{\mathrm{d}\nu}{g - k\nu^{2}} &= \int_{0}^{t} \mathrm{d}\tau \quad (\text{since...

Re: MATH2601 Linear Algebra/Group Theory Questions Two different topics.

Re: MATH2601 Linear Algebra/Group Theory Questions $\noindent Take a map whose image is just $\mathrm{span}\left\{(1,2,3)^{\top}\right\}$. Then $T$ would map all non-zero vectors to multiples of $(1,2,3)^{\top}$, so wouldn't imply that those $\mathbf{v}$ and $\mathbf{w}$ are dependent. For...

Re: MATH2601 Linear Algebra/Group Theory Questions No.

Re: maths questions help Correct!

Re: maths questions help 1) Turning point at 0, local minimum. 2) Axis of symmetry is the y-axis. Method used: inspection.

Re: MATH2601 Linear Algebra/Group Theory Questions $\noindent It is clear that a necessary condition for $T$ to be invertible is that $a\neq c$ (otherwise the first and last entry of $T(p)$ are always the same, so $T$ cannot be onto). Note also with $a\neq c$, we also can't have $b =...

$\noindent Stretch the original graph vertically by a factor of $b$ and horizontally by a factor of $a^{-1}$, i.e. the graph becomes $y = b\left((ax)^{3} -ax \right) = ba^{3}x^{3} -ba x$. Comparing this to $x^{3} -3x$, we want $ba^{3} = 1$ and $ba = 3$. Dividing these equations shows $a^{2} =...