Artificial intelligent assistant

obversely

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1
obversely
obversely, adv. (əbˈvɜːslɪ) [f. prec. + -ly2. In Nat. Hist. after mod.L. obversè.] In an obverse form or manner; with an adj. of shape = ob- 2.1752 Sir J. Hill Hist. Anim. 565 The tailed Vespertilio, with a foliated and obversely cordated nose. 1753 Chambers Cycl. Supp. s.v. Leaf, Obversely ovated L... Oxford English Dictionary
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Arthrocereus
The spherical to obversely pear-shaped fruits are green and have white flesh. wikipedia.org
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ob-
ob-, prefix The Lat. prep. ob ‘in the direction of, towards, against, in the way of, in front of, in view of, on account of’. In combination with vbs. and their derivatives, the b is assimilated to certain consonants, becoming oc- before c-, of- before f-, op- before p-, and app. o- before m- (in om... Oxford English Dictionary
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vertically
ˈvertically, adv. [f. vertical a. + -ly2.] 1. a. In a vertical manner, direction, or position; so as to be vertical to the plane of the horizon, the earth's surface, or some other horizontal line or plane; perpendicularly, or approximately so; directly overhead or down below.1646 Sir T. Browne Pseud... Oxford English Dictionary
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obverse
obverse, a. and n. (see below) [ad. L. obversus, pa. pple. of obvertĕre to obvert. Single instances of the adj. and n. are known in 17th c.; otherwise the word is not exemplified till end of 18th c.; neither Johnson nor Todd has it, though giving obversant; under reverse, of coins, Chambers Cycl. (1... Oxford English Dictionary
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Love Medicine
If the purpose of telling Nanabozho stories is to challenge listeners and to obversely remind them of their roots, Owens argues, then the purpose of June wikipedia.org
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$f(x)$ are continues at x=0, "if $\exists \lim \frac{\left\{f(x)-f(-x)\right\}}{x}$ then $f^{\prime}(0)$ exist " are error. how to proof? $f(x)$ are continues at x=0, if $$\exists \lim _{x\to 0} \frac{f(x)-f(-x)}{x}$...
If $f$ is even, that means $f(x)=f(-x)$, then $\frac{f(x)-f(-x)}{x}=0$. Hence $\exists \lim _{x\to 0} \frac{f(x)-f(-x)}{x}$. Suppose that the statement is true. Then we would have: "each even function is differentiable at $0$". If $f$ is odd, that means $f(-x)=-f(x),$ then we have $\frac{f(x)-f(-x)}...
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