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Topic: I need help with Latin roots!?!?**Question:**
What is the root for naive, portrayal, secondary, excecutive, portrait, and pursuit? Please? My teacher won't tell me and my dictionary doesn't say. Latin roots are hard...

July 20, 2019 / By Ciera

Most of these are French, not Latin! The etymologies of all English words, if known, are given in the American Heritage Dictionary, which is freely available online. You can start there. If you still further help after that, please repost, and I will certainly help. I just don't want to do all your homework.

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Did you like the answer? Flavus comes from the root meaning yellow, as in riboflavin http://dictionary.reference.com/search?q... I believe hisutus should be hirsutus, which means hairy, or bristly, rough, or shaggy http://dictionary.reference.com/browse/h... scandens is probably climber. http://www.archives.nd.edu/cgi-bin/looku... One Latin dictionary says quarrelsome. But that doesn't sound like an animal name. Another says scand means climb, and I don't have anything better (scansorial means adalpted for climbing).Scandalum relates to scandal and moral failing. That also seems like a strange choice for an animal. Scandinavia is a dead end

naive comes from nativus, innate, natural portrayal and portrait come from protrahere, to depict pursuit comes from prosecuta, to follow executive comes from executus, to perform secondary comes from secundus, following You need to look in a bigger dictionary. You are probably going to get more assignments like this. Go to the big dictionary in your school library. After the word is defined, the dictionary will list something like this: ML, or ME, or some other set of letters. There is a key in the front of the dictionary that explains these. If the word came to English from Latin through French, you can look further to find the Latin root where the French got the word, so that you can answer the question completely.

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Dictionaries do tell roots. Look the words up on Merriam Webster.com. I just looked at naive and it gives the latin root.

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Quite a number of the 'nuber' prefixed phrases have latin roots: pentatonic (scale) approach a five be aware (tonus) scale in tune pentadactyl approach having five hands (sorry can not recollect the latin for the dactyl bit) quadrilateral - quad, four, lateral from latin for 'facet' bisect - bi, two, sect from the latin for reduce unison - uni from unus that means one million quotidien - historic phrase for day-to-day - from the latin quotidies that means day-to-day tripod - tri from latin for three, pod from pes, latin for foot uniform - uni, one million, sort from formus (I suppose) that means style/type acquainted - from latin familia that means household equine - from latin equus that means a horse. prejudice - from latin pre, earlier than, and judicare, to pass judgement on disciple - from discipulus - a disciple Could move on however have not the time! Use a well etymological dictionary - one that offers you the origins of phrases - and you'll be able to uncover hundreds of thousands. The reference component to your neighborhood library will undoubtedly have one and the librarian will typically be simplest too inclined to support!

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Your teacher won't tell you because he or she wants you to be resourceful and find out for yourself. Try: http://www.dictionary.com Good luck!

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x^3 = 1 x^3 - 1 = 0 We can see by observation that x = 1 is a root of this equation, so it follows that x - 1 is a factor of x^3 - 1. Dividing x^3 - 1 by x - 1 synthetically gives us a quotient of x^2 + x + 1, so: (x - 1)(x^2 + x + 1) = 0 x = 1 is one root of the initial equation, and the other two cube roots of unity will be the roots of x^2 + x + 1. But x^2 + x + 1 = 0 can be solved using the quadratic formula, which tells us that: x = (-1 +/- i*sqrt(3))/2, which are exactly the answers you gave. This is the method to find the cube roots of unity; i.e. the three distinct complex numbers such that when you cube them you get one. If you know a little bit about complex numbers, a better way to think of the cube roots of unity is: cis(0) = e^(i*0) = 1 cis(2pi/3) = e^[(2pi*i)/3] = -1/2 + i*(sqrt(3)/2) cis(4pi/3) = e^[(4pi*i)/3] = -1/2 - i*(sqrt(3)/2) The reason this helps is that, for the general problem of finding the nth roots of unity, or the n complex solutions to the equation x^n = 1, we can immediately say that the nth roots of unity are: cis[(2pi*k)/n] = e^[(2pi*k*i)/n], where k = 0, 1, ..., n - 1. Check out Wikipedia: http://en.wikipedia.org/wiki/Root_of_uni... and Wolfram MathWorld: http://mathworld.wolfram.com/RootofUnity... for more info on the subject.

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