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adventitious
adventitious, a. (ædvənˈtɪʃəs) Also adventicious. [f. L. adventīci-us, in med.L. corruptly written adventiti-us, coming to us from abroad + -ous: see Advent, and -itious). The occas. adventicious is etymologically a better spelling.] 1. Of the nature of an addition from without; extrinsically added,... Oxford English Dictionary
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Adventitious Sweater pattern by Olga Putano Designs
4 days ago — While the short sleeved top is worked from the bottom up with one strand of lace yarn, the long sleeved sweater is worked from the top down ...
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Langley's Adventitious Angles
Langley's Adventitious Angles is a puzzle in which one must infer an angle in a geometric diagram from other given angles. Numerous adventitious quadrangles beyond the one appearing in Langley's puzzle have been constructed. wikipedia.org
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adventitious
adventitious/ˌædvenˈtɪʃəs; ˌædvɛn`tɪʃəs/ adj(fml 文) not planned; accidental 未经计划的; 偶然的 an adventitious occurrence 偶发事件. 牛津英汉双解词典
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Accidental
philosophical term See also Accidence (or inflection), a modification of a word to express different grammatical categories Accident (disambiguation) Adventitious , which is closely related to "accidental" as used in philosophy and in biology Random, which often is used incorrectly where accidental or adventitious wikipedia.org
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Adventive plant
Adventive plants or adventitious plants are plants that have established themselves in a place that does not correspond to their area of origin due to In natural and near-natural vegetation, adventitious plants are much rarer. Their share here is between zero and about 5%. wikipedia.org
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A question on the requirement of a quadrilateral being an adventitious quadrangle There is a special type of problem called Langley’s Adventitious Angles. See < The problem was solved and has the following generali...
It means that if $X$, $Y$, $Z$ are any three of $B,C,E,F$ then $\angle XYZ$ has measure (in radians) of the shape $r\pi$ for some rational number $r$. Equivalently, since $\pi$ radians is $180^\circ$, it means that every angle is a rational number of degrees. Thus angles like $\frac{99}{7}$ degrees ...
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Burrknot
Burrknot is a tree disorder caused by the formation of adventitious root primordia. First looks like a smooth orange bulge growing from the stem or a branch, later multiple adventitious roots form. wikipedia.org
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$20-80-80$ triangle, rhombus with orthocenter, circumcenter Let $ABC$ triangle such that $\angle A=20^{\circ}$ and $\angle B=\angle C=80^{\circ}$.Let $D,E$ be point on lines $AC,AB$ respectively such that $BD,CE$ are ...
![]( Let $\Gamma$ be circumcircle of $\triangle OHC$ and $S$ its circumcenter. Then $$\angle HSC = 2 \angle HOC = \angle BOC = 2\angle BAC = 40^\circ,$$ from which we get $$\angle SCH = \frac{180^\circ - \angle HSC}2=70^\circ = \angle ACH.$$ Thus $S$ lies on the line $AC$. Let $OB$ intersect $\Gamma...
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Austrocactus
of Austrocactus bertinii smaller stems (<25 cm) Austrocactus ferrarii Austrocactus longicarpus Austrocactus philippii stems prostrate stems with adventitious roots Austrocactus colloncurensis Austrocactus coxii Austrocactus gracilis – synonym of Austrocactus coxii Austrocactus hibernus stems without adventitious wikipedia.org
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Eowellerites
Most diagnostic character of its suture is its weakly developed adventitious lobe in the first lateral saddle. From E. quinni it distinguishes by more developed adventitious lobe of first lateral saddle. wikipedia.org
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Bud
In many plants buds appear in unexpected places: these are known as adventitious buds. buds may be lateral too); adventitious, when located elsewhere, for example on the trunk or roots (some adventitious buds may be former axillary ones wikipedia.org
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What exactly are narrower and wider conditions? (Polya's How To Solve It) In _How To Solve It_ , on page 56, Polya states that > "If we pass from a proposed condition to a new condition equivalent to it, we have the ...
If we only talk about transforming equations, then spurious solutions may appear whenever we apply a function $f$ to both sides that is not injective. In other words, we have $a=b\implies f(a)=f(b)$, but not vice versa. The case of squaring both sides is just a special case of this (with $f(x)=x^2$)...
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