cumbersome

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cumbersome
cumbersome, a. (ˈkʌmbəsəm) [f. cumber v. + -some.] † 1. Of places or ways: Obstructing and impeding motion or progress; full of obstruction; troublesome to pass or get through. Obs.1375 Barbour Bruce xiii. 351 Bannokburne, that sa cummyrsum was Of slyk, and depnes for till pas. 1555 Fardle Facions i... Oxford English Dictionary
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Cumbersome
As of 2021, "Cumbersome" allegedly receives over 150 plays a week on popular radio. Track listings "Cumbersome" (LP version) "Cumbersome" (acoustic version) Charts References 1996 debut singles Seven Mary Three songs Post-grunge wikipedia.org
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Lee's Cumbersome Cavalry | American Battlefield Trust
J.E.B. Stuart commanded the cavalry wing of Robert E. Lee's Army of Northern Virginia and, as noted in a letter addressed to Stuart from Robert E. Lee, served as the "eyes and ears" of the army. By the time of the Gettysburg Campaign during the summer of 1863, J.E.B. Stuart had established himself as both a skilled cavalryman and, as ...
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cumbersome
cumbersome/ˈkʌmbəsəm; `kʌmbɚsəm/ adj1 heavy and difficult to carry, wear, etc (对於携带、 穿戴等)笨重的 a cumbersome parcel, overcoat 沉重的包裹、 大衣.2 slow and inefficient 迟缓而缺乏效率的 the university's cumbersome administrative procedures 这所大学拖拖拉拉的行政工作. 牛津英汉双解词典
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Less cumbersome notation for jacobian Is there a less cumbersome notation for the Jacobian of some multivariate functions? For instance, suppose I have $n$ multivariate functions of $n$ variables, $u_1(x_1,...,x_n),.....
I sometimes write $\mathbf{J}f(x)$ and leave it at that. That way I view the Jacobian as not just an object, but a sort of operator, akin to the derivative.
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Is there a conjunctive that's roughly equivalent to "in that case" or "if so"? Say for example I want to say something like > It feels cumbersome to me to have to repeat the verb in the second sentence. Is there a ...
"In that case" / "if so": * **** (colloquial) * **** (colloquial) * **** (politely colloquial) * **** (formal, bookish) Much like or which you may know, the conjugated forms of the copula can function as independent conjunctions (not conjunctive particles) at the beginning of a sentence, referring t...
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Using setminus notation with set elements The "correct" way to write a set without a specific element is as follows: $S \setminus \\{s\\}$ But in some contexts this is cumbersome to write/type or read, and it detract...
If you have to write this many times in a piece of text, and there is no possibility of confusion of $s$ being an actual set of elements you are removing, it is okay to first write one sentence reminding the reader of this abuse of notation ("In the following, we write $S\setminus s$ to mean $S \set...
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At first it was cumbersome, but now I feel safer to commute by train ...
I can understand what you're saying, but there are a few grammar issues: At first it was cumbersome, but now I feel safer to commute by train than by car. OR At first it was cumbersome, but now I feel safer to commute by train than to commute by car.
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Notation for the exclusive choice from the set I have the following formula: $$ \Omega \subset \mathbb{R}^d, $$ where $d=2$ or $d=3$. But I find this style of notation quite cumbersome. Is there any way, how to note...
You could write $\Omega\subset\mathbb{R}^2\lor \Omega\subset\mathbb{R}^3$, or $\Omega\in\mathcal{P}\left(\mathbb{R}^2\right)\bigcup\mathcal{P}\left(\mathbb{R}^3\right)$. Here $\mathcal{P}\left(S\right)$ denotes the power set of $S$, i.e. its set of subsets. (Note $S\in\mathcal{P}\left(S\right)$. If ...
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What will be the other vertex of the triangle? Two vertices of a triangles are $(5,-1)$ and $(-2,3)$. If the orthocenter of the triangle is the origin, what is the other vertex ? My approach was that since the three...
(I assume that you meant "the orthocenter of the _triangle_ is the origin.") Let $A$ be $(5,-1)$, $B$ be $(-2,3)$, $O$ be the origin, and your desired third vertex be $C$. Find the line through $O$ perpendicular to $\overline {AB}$: this will be the altitude of the triangle through $C$. (I get $-7x+...
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Evaluating the integral $\displaystyle \int_{0}^{\pi/2} (\cos^{5}x)\sin (7x)dx$? I tried to expand $\sin(7x)$ into $\sin x,\cos x $ terms but integral is getting pretty lengthy and cumbersome , how do i approach in a ...
Observe that $$\sin 7x = \sin 6x \cos x + \cos 6x \sin x,$$ hence $$\cos^5 x \sin 7x = \sin 6x \cos^6 x + \sin x \cos^5 x \cos 6x.$$ Now here is the tricky part: if we let $u = \cos^6 x$, $v = \cos 6x$, we get $$u' = -6 \cos^5 x \sin x,$$ and $$v' = -6 \sin 6x.$$ So what does this suggest?
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Function to create long rounded rectangles Is it possible to describe with a function the following shape or would that result in a too cumbersome approach? I am looking for something like this, because I would like t...
To me your shape looks like the convex hull of two circles. As such it can be parametrised as follows: Suppose the circles are centred at $\mathbf{x}_1$ and $\mathbf{x}_2$ and have radii $r_1$ and $r_2$ respectively. Let $$f\colon[0,\pi]\times[0,1]\to\mathbb{R}^2\colon(\theta,\lambda)\mapsto \lambda...
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Symmetric matrix under orthogonal transformation still symmetric? is there any way to see that a symmetric matrix is still symmetric after applying an orthogonal basis transformation to it? I would say that a proof th...
The important thing to note here is that the inverse of an orthogonal matrix is its transpose, so the basis change $S \leadsto O^{-1}SO$ is actually (also) $S \leadsto O^TSO$, in which form the symmetry-preserving nature of orthogonal basis changes is near-obvious: $$(O^TSO)^T = O^TS^T(O^T)^T = O^TS...
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When is a functor of bicategories part of an equivalence? Assuming the axiom of choice (I think), $F: \mathcal{C} \to \mathcal{D}$ is part of an equivalence of categories iff $F$ is fully faithful and essentially surj...
Weak equivalences of (either strict or weak) $n$-categories are defined inductively: > $F : \mathcal{C} \to \mathcal{D}$ is a **weak equivalence of $n$-categories** if it has the following properties: > > * For every object $Y$ in $\mathcal{D}$, there is an object $X$ in $\mathcal{C}$ and an equival...
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Solving a set of two polynomial equations For a given $a,b,c,d$ in $\mathbb{R}$, I want to prove that if $$ac-bd=0 \quad \text{and} \quad ad+bc=0$$ then $a=b=0$ or $c=d=0$. I am able to prove this in a long and cu...
Multiply $ac=bd$ by $d$ and $ad+bc=0$ by $c$ then $$bd^2+bc^2=b(c^2+d^2)=0$$ So, either $c=d=0$ or $b=0$. Take the second case then we get $ac=0$ and $ad=0$. So either $a=0$ or $c=d=0$. The second option brings us back to our first case.
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