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A lot of these would be by definition or convention.

Re: MATH1131 help thread $\noindent I'm assuming you mean the matrix one. Let $X=A^2$, so we are after $X^{-1}$. We can see that $A^6 = 64I$. Therefore, $A^4 A^2 = 64I$, as $A^4 A^2 = A^6$. So $A^4 X = 64 I \Rightarrow \left(\frac{1}{64}A^4\right)X=I$ (dividing through by $64$). This implies...

But I have ''no pure understanding of the sciences'', so it makes no difference for me. :lol:

1. Velocity is just the rate of change of displacement (by definition), which incorporates direction into it. 2. Convention (SI Units for distance is m, for time is s, and velocity is distance/time, so units become m/s). 3. Because m/s is like m s-1, recalling the definition of x-1 as 1/x. 4...

1. Because it's the vector that results from the addition. 2. By definition of addition of arrow representation of vectors. 2. By definition of addition of arrow representation of vectors. 4. By definition of addition of arrow representation of vectors. 5. Legitimately by inspection. 6. By...

Re: MATH1131 help thread I'm guessing you did inverse tan of (-1/-1). Since the real part of the complex number is negative, we need to add or subtract pi from what you got by doing that. Since the imaginary part is negative, we subtract pi. So the answer is pi/4 – pi = -3pi/4. It's...

Re: MATH1131 help thread The answer is -3pi/4. Did you get pi/4?

For what purpose should I have taken it into account?

You're welcome. :)

All the steps are trivial here apart from 3. And I already helped you with 3.

I pretty much did in that post. Just simplify that to show that it equals (k+1)*((k+1)+1)/2.

$\noindent These are all the same type of question (induction proving a sum). To do these, in the inductive step, write the $(k+1)$th partial sum as the $k$th partial sum plus the $(k+1)$th term, then invoke your inductive hypothesis. E.g., $1+2+3+\cdots + \left(k+1\right) = 1+2+3+\cdots +k+...

Re: MATH1131 help thread $\noindent You can think of this one as $\int f(x)g(x)\text{ d}x$, where $f(x) =1-x$ and $g(x)= \frac{1}{\left(1+x\right)^3}$.$

Re: MATH1131 help thread $\noindent We integrate it using the power rule of integration:$ $$\begin{align*}\int \frac{1}{\left(1+x\right)^3}\text{ d}x &= \int \left(1+x\right)^{-3}\text{ d}x \\ &= \frac{1}{-3+1}\left(1+x\right)^{-3+1} \\ &= \frac{1}{-2}...

$What progress have you made so far?$

Re: Multivariable Calculus $\noindent For (b), the vector notation $\bold{a}$ is just shorthand notation to say we are evaluating the partial derivative at the point $\left(r,a_0,a_1,a_2\right)$ or $\left(a_0,a_1,a_2\right)$. In other words, the notation $\bold{a} := \left(a_0,a_1,a_2\right)$...

Re: MATH1131 help thread Basically in integration by parts, we want to pick one factor that we'll differentiate and one that we'll integrate. Here, we'll want to integrate the 1/(1+x)^3 and differentiate the 1 – x. The reason for this is that when we differentiate the 1 – x, it'll just...

Re: MATH1131 help thread Yeah, then it'd definitely mean the log of: (sin x)^x (especially if it had y = (sin x)^x before that, which'd mean they're taking logs of both sides next line).

Re: MATH1131 help thread If it's written log(sin x)^x, from the context (I'm guessing), it probably means the log of: (sin x)^x. They didn't write (log(sin x))^x, did they? If they did, then that'd definitely be different.

It depends on the person.