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  1. I

    Prelim 2016 Maths Help Thread

    With your username, of course you do. :D
  2. I

    Prelim 2016 Maths Help Thread

    Tried what? And I wasn't referring to Desmos.
  3. I

    Prelim 2016 Maths Help Thread

    You probably can, just need to learn how to use it.
  4. I

    Prelim 2016 Maths Help Thread

    Here's one: http://www.graphsketch.com/ . Or you can try WolframAlpha.
  5. I

    Prelim 2016 Maths Help Thread

    What will you do if no-one is available?
  6. I

    Prelim 2016 Maths Help Thread

    What do you think that reason is?
  7. I

    Prelim 2016 Maths Help Thread

    What do you mean mirroring of the graph? Graph of what? The inverse's graph is the reflection about the line y = x of the original graph.
  8. I

    Prelim 2016 Maths Help Thread

    How can I help you?
  9. I

    Prelim 2016 Maths Help Thread

    You probably just made a silly mistake in expanding something. Also make sure to take note of the range and domain of the inverse.
  10. I

    Leibniz's Rule

    $\noindent Let $h(a)=\int _{0} ^{1} \frac{x^a-1}{\ln{x}} \text{ d}x $. Applying Leibniz' Rule, $h^{\prime}\left(a\right) =\int _{0} ^{1} \frac{\partial}{\partial a} \left(\frac{x^a-1}{\ln{x}}\right) \text{ d}x = \int _0 ^1 x^{a}\text{ d}x = \frac{1}{a+1}$, since $\frac{\partial}{\partial a}...
  11. I

    First Year Mathematics A (Differentiation & Linear Algebra)

    Re: MATH1131 help thread Yes, it's right. You essentially had the same two vectors as it but put the "t" with the wrong one.
  12. I

    First Year Mathematics A (Differentiation & Linear Algebra)

    Re: MATH1131 help thread No, they're not the same. The easiest way to see this is to observe that the two answers' direction vectors aren't parallel, so they represent different lines. I think you basically mixed up where to put the parameter t / direction vector. The t is a scalar...
  13. I

    First Year Mathematics A (Differentiation & Linear Algebra)

    Re: MATH1131 help thread Yeah you can leave it out.
  14. I

    First Year Mathematics A (Differentiation & Linear Algebra)

    Re: MATH1131 help thread His one is fine. If it goes through the origin, don't need to include the (0,0) part (plus adding a zero vector to a vector doesn't change it, so you can leave it out).
  15. I

    HELP Binomial theorem exam

    Here are some hints. $\noindent Q.7. Recall that $\left(3+4x\right)^n = \sum _{k=0}^{n}\binom{n}{k}3^{n-k}4^k x^k$ from the Binomial Theorem. So the $x^2$ coefficient is (with $k=2$) $\binom{n}{2}3^{n-2}\times 16$ and the $x^3$ coefficient is (with $k=3$) $\binom{n}{3}3^{n-3}\times 64$. We are...
  16. I

    Prelim 2016 Maths Help Thread

    The Casio calculators have "log" as the button for log-base-10, but actually in mathematics log refers to base e.
  17. I

    First Year Mathematics A (Differentiation & Linear Algebra)

    Re: MATH1131 help thread $\noindent We have$ $$\begin{align*}\left(A^T A\right)^{-1} \left(A^T A\right)^T &= A^{-1} \underbrace{\left(A^{T}\right)^{-1} A^T}_{=I} \left(A^T\right)^T \\ &= A^{-1} A \\ &= I.\end{align*}$$ $\noindent We made use of the rules $\left(XY\right)^{-1} =...
  18. I

    First Year Mathematics A (Differentiation & Linear Algebra)

    Re: MATH1131 help thread $\noindent The vector $\overrightarrow{PA}$ is a vector connecting a point on the plane and the point whose distance from the plane we are finding. The vector $\bold{n}$ is a normal to the plane.$
  19. I

    First Year Mathematics A (Differentiation & Linear Algebra)

    Re: MATH1131 help thread It has applications in areas outside of mathematics, like physics (torque for instance is a cross product). Here's the section of the Wikipedia link on cross product about its applications: https://en.wikipedia.org/wiki/Cross_product#Applications .
  20. I

    First Year Mathematics A (Differentiation & Linear Algebra)

    Re: MATH1131 help thread Let f(t) := ln(1+t), for t ≥ 0. Let x > 0 be a fixed real number. Using MVT, as f is continuous on [0,x] and differentiable on (0,x), we know there is a c in (0,x) such that f'(c) = [f(x) - f(0)]/(x-0). But f'(c) = 1/(1+c), and f(0) = ln 1 = 0. So we get 1/(1+c) =...
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