Draw a truth table & logic circuit corresponding to the following logic statement. X = 1 if ( ( M is 1 OR N is 1 ) AND ( M is 1 AND N is 1 ) ) OR ( O is NOT 1 ) I am gonna give to the first answer

You picked the corridor which led you here. if the guards find you, they're going to be really angry! what is the synonym of angry

Some of your friends have gotten into the burgeoning field of time-series data mining, in which one looks for patterns in sequences of events that occur over time. purchases at stock exchanges--what’s being bought-- are one source of data with a natural ordering in time. given a long sequence s of such events, your friends want an efficient way to detect certain "patterns" in them--for example, they may want to know if the four events buy yahoo, buy ebay, buy yahoo, buy oracle occur in this sequence s, in order but not necessarily consecutively. they begin with a collection of possible events (e.g., the possible’ transactions) and a sequence s of n of these events. a given event may occur multiple times in s (e.g., yahoo stock may be bought many times in a single sequence s). we will say that a sequence s’ is a subsequence of s if there is a way to delete certain of the events from s so that the remaining events, in order, are equal to the sequence s’. so, for example, the sequence of four events above is a subsequence of the sequence buy amazon, buy yahoo, buy ebay, buy yahoo, buy yahoo, buy oracle their goal is to be able to dream up short sequences and quickly detect whether they are subsequences of s. so this is the problem they pose to you: give an algorithm that takes two sequences of even~s--s’ of length m and s of length n, each possibly containing an event more than once--and decides in time o(m n) whether s’ is a subsequence of s

Write a program which asks you to enter a name in the form of first middle initial last. so you might enter for example samuel p. clemens. use getline to read in the string because it contains spaces. also, apparently the shift key on your keyboard doesn’t work, because you enter it all lower case. pass the string to a function which uses .find to locate the letters which need to be upper case and use toupper to convert those characters to uppercase. the revised string should then be returned to main in the form last, first mi where it will be displayed.