answer:When (x leqslant 3), the inequality (|x-3|+|x-5|geqslant 4) becomes (3-x+5-xgeqslant 4), which simplifies to (x leqslant 2). Therefore, we have (x leqslant 2). When (3 < x < 5), the inequality (|x-3|+|x-5|geqslant 4) becomes (x-3+5-xgeqslant 4), which simplifies to (2 geqslant 4). This is impossible, so we have (x in varnothing). When (x geqslant 5), the inequality (|x-3|+|x-5|geqslant 4) becomes (x-3+x-5geqslant 4), which simplifies to (x geqslant 6). Therefore, we have (x geqslant 6). Combining all cases, we find that (x geqslant 6) or (x leqslant 2). Thus, the solution set is (boxed{[6,+infty) cup (-infty,2]}).

answer:We are given the quadratic expression x^2 - 20x + m and need to find m such that this expression is the square of a binomial. Let's assume the binomial is (x+a)^2. Then: [ (x+a)^2 = x^2 + 2ax + a^2 ] Matching the middle term 2ax with -20x, we find: [ 2a = -20 implies a = -10 ] Now, substituting a = -10 into a^2 gives: [ a^2 = (-10)^2 = 100 ] Thus, the expression (x-10)^2 expands to: [ (x-10)^2 = x^2 - 20x + 100 ] Therefore, the value of m is boxed{100}.

answer:If Nickel ate 3 chocolates and Robert ate 4 more chocolates than Nickel, then Robert ate 3 + 4 = boxed{7} chocolates.

answer:Solution: Given f(x) = sqrt{4x-3}, we can rewrite it as (4x-3)^{frac{1}{2}}. Therefore, f'(x) = frac{1}{2}(4x-3)^{-frac{1}{2}} cdot (4x-3)' = frac{2}{sqrt{4x-3}}. Hence, the answer is: boxed{frac{2}{sqrt{4x-3}}}. This problem involves applying the rules for differentiation and the chain rule for derivatives, making it a basic question.