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well you need to change the form to get it in the derivative form - so f'(a) = (x+2) to the power of ½ so you multiply the ½ infront of (x+2) and substract 1 from the power which would be ½(x+2) and when u substract one from the power u get -½ and do the derivative of x would be 1 so f' = ½(x+2) to the -½ as the power *1 . how ever i am not sure what your limit is trying to say if u can resend the limit part clearly
ok this is what i have
p⇒q and q⇒r
is quivalent to (~p or q) and (~q or r) distribution of and
= ( (~p or q) and ~q) or ((~p or q) and r)
= ((~q and ~p) or (~q or q)) or ((~p and r) or( q and r)
= (((~q and ~p) or F or ((~p and r) or( q and r)
= ~p or (q and r)
= p implies (q and r)
i dont know if this is right but i tired please help me out
ok here is what i would on (( p imlies q) and (q implies r)) is equivalent to (p implies(q and r) prove this without doing truth tables
hello everyone, i'm new to the forum and i was wondering if anyone in here knows about logical equivalences for discrete maths since i have problem with them and im not getting by
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