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ascertained
ascertained, ppl. a. (æsəˈteɪnd) [f. ascertain v. + -ed.] † a. Determined, fixed (obs.). b. Discovered by investigation, known.1494 Fabyan ii. xxxiii. 26 So y{supt} no tyme asserteyned, is to her deputed or set. 1858 Sears Athan. xviii. 160 Professor Faraday considers it an ascertained fact.
Oxford English Dictionary
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Determined - Definition, Meaning & Synonyms | Vocabulary.com
determined: 1 adj having been learned or found or determined especially by investigation Synonyms: ascertained , discovered , observed discovered or determined by scientific observation Antonyms: undetermined not yet having been ascertained or determined unexplained not explained show more antonyms... adj devoting full strength and ...
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Sekong
It was created in 1984 after it was ascertained that Ban Phon's unexploded ordnance made it uninhabitable.
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Calculus in a discrete universe Suppose we ascertained that space and time are discrete and the units are Planck's. Would that affect calculus? I know that integration does not require a continuum, but about different...
"Would that affect calculus?" Of course not, and calculus in its present form would even remain a perfect tool for describing the macroscopic world. Note that, e.g,. we already _know_ that matter comes in the form of "little balls", but nevertheless we treat it as a homogeneous "glue" when we do ela...
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1991 Australian Endurance Championship
Results
Australian Endurance Championship
Note: Other placings in the Endurance Championship have not been ascertained. Australian Manufacturers Championship
Note: Other placings in the Manufacturers' Championship have not been ascertained.
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Let f(x) be a derivable function, $f'(x) > f(x)$ and $f(0) = 0$. What can be said about the sign of $f(x)?$ > Let $f(x)$ be a derivable function, $f'(x) > f(x)$ and $f(0) = 0$. Then > > (A) $f(x) > 0$ for all $x > 0...
Attempt: Let $x >0$; You got $f'(0)>0$. $\dfrac{f(x)-f(0)}{x}=\dfrac{f(x)}{x}$; $\lim_{x \rightarrow 0^+} f(x)/x= f'(0)>0$. For $x_0$ small enough $f(x_0)/x_0 >0$, i.e. $f(x_0)>0$. Assume there is an $x_1 >x_0$ s.t. $f(x_0)>f(x_1)$. The continuos function $f$ attains its maximum on $[x_0,x_1]$. $f'(...
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1964 Australian Tourist Trophy
Results
Race statistics
Race distance: 23 laps – 103½ miles
Number of starters: not yet ascertained
Number of finishers: not yet ascertained
Race
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How does the author of this probability book claim that a one-to-one function is invertible without showing it's onto? I know that the necessary conditions for a function to be invertible is that the function must be ...
When you have ascertained these for a one-to-one function, you can of course construct the inverse function from the range of the original function to
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1979 Australian Tourist Trophy
81.2 km
Pole Position: Paul Gibson
Race time of winning car: 42.30.0
Fastest lap: 61.1 - Paul Gibson & Stuart Kostera
Number of starters: Not yet ascertained Number of finishers: Not yet ascertained
References & notes
External links
Paul Gibson, Rennmax & David Richardson, Matich SR3a - Winton - 28 October
wikipedia.org
en.wikipedia.org
Changing co-ordinate systems quadratic forms A = $$\begin{bmatrix}1 & 4 \\\4 & -4\end{bmatrix}$$ and $$q(\mathbf{x})= \mathbf{x}\cdot A \mathbf{x}, $$ then what is? $$q(\mathbf{e_1})= $$ $$q(\mathbf{e_2})=$$ $$$$...
Why don't you just use the definition? $$ q(e_1)=e_1\cdot Ae_1=e_1\cdot\begin{pmatrix}1\\\4\end{pmatrix}=1 $$ and $$ q(e_2)=e_2\cdot Ae_2=e_2\cdot\begin{pmatrix}4\\\\-4\end{pmatrix}=-4. $$ You don't need the eigenvalues and eigenvectors of $A$ to compute the value of the quadratic form at some point...
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Evaluating probabilities with joint density. I have a density function $f(x,y)=\alpha x^2y^2$ for $y\in (0,1)$ and $x\in (-y,y)$. To find $\alpha$ I evaluated $$\alpha \int_{-y}^y\int_0^1 x^2y^2 \ dydx=1$$ and ascerta...
$$\alpha \int_0^1\int_{-y}^y x^2y^2 \ dxdy=1$$ Let's carry out the integration. $$\begin{align} \int_0^1\int_{-y}^y x^2y^2 \ dxdy&=\int_0^1y^2\left(\frac23 y^3\right)dy\\\\\\\ &=\frac19 \end{align}$$ Thus, setting $\alpha\left(\frac19\right)=1$ reveals that $\alpha = 9$. * * * To calculate the proba...
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Managing judges mathematically: an empirical study of the medical ...
The post-Mao China has been increasingly managed mathematically, not the least in its judicial system. In this paper, I looked into some of the mathematical indicators used to judge the performance of judges in this nation, and ascertained their …
www.springerprofessional.de
commercial-law-notes 1 .pdf - lOMoARcPSD|11303289...
(4) An agreement to sell becomes a sale when the time elapses or the conditions are fulfilled, subject to which the property in the goods is to be transferred. b. Transfer of property s 21 Goods must be ascertained Unascertained - not yet identified (may be existing or future) Ascertained - have become identified 就是"种类物—特定物"之分 ...
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Mussel-covered junk and mortar shell pulled out of Lake Geneva
"The mortar shell was found early this afternoon by a diver," said a spokesman for canton Vaud police on Saturday. "Demining experts arrived and ascertained that it didn't have a detonator."
www.swissinfo.ch
Proving that the $\lim_{n\to \infty} a_n = \infty$ where $\lim_{n \to \infty} \frac{a_n + 1}{a_n} = k > 1$ I am working on a proof as follows: Let $k > 1$ and $(a_n)_{n \in \Bbb{N}}$ be a sequence where $a_n > 0$ for...
> Herein, we present a standard approach that relies on the definition of the limit. To that end we proceed. * * * If $\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=k>1$, then for all $\epsilon>0$, there exists a number $N(\epsilon)$ such that whenever $n>N(\epsilon)$, $$k-\epsilonN\left(\frac{k-1}{2}\right...
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