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 · Then for all x, P(x) and Q(x) are both true, meaning for all x, P(x)∧Q(x) is true This proves that the first statement is always true The second statement is false Without going through the full process as above, we can simply show that the two propositions on either side of the ↔ do not always have the same truth value For example, suppose that for half of all possible x, P(x) isB ¥ V _ X ^ @ \ I G b#1= _ \>* }* ' L t S u _ c>* _>* Û#Õ _ P M Õ Ü \ 8 0i!l ?Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on

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X with a Modulus sign denotes the magnitude of x , ie value without sign So it will always give a positive number If x is positive then its mod will be x only as it is already positive If x is negative then mod x will give x as x is itself n° t ¤­ ` o _ Úb AU K } h Å É Ö ´ ç ` o ð ;Given f (x) = 2x, g(x) = x 4, and h(x) = 5 – x 3, find (f g)(2), (h – g)(2), (f × h)(2), and (h / g)(2) This exercise differs from the previous one in that I not only have to do the operations with the functions, but I also have to evaluate at a particular xvalue To find the answers, I can either work symbolically (like in the previous example) and then evaluate, or else I can find

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