answer:According to the given information, forming a parallelogram with overrightarrow{a} and overrightarrow{b} as adjacent sides, the vector corresponding to the shorter diagonal is overrightarrow{a} - overrightarrow{b}. Since |overrightarrow{a}|=1, |overrightarrow{b}|=2, and the angle between overrightarrow{a} and overrightarrow{b} is frac{pi}{3}, we have the following relationship for the length of the vector overrightarrow{a} - overrightarrow{b}: begin{align*} |overrightarrow{a} - overrightarrow{b}|^2 &= (overrightarrow{a} - overrightarrow{b}) cdot (overrightarrow{a} - overrightarrow{b}) &= overrightarrow{a} cdot overrightarrow{a} + overrightarrow{b} cdot overrightarrow{b} - 2 overrightarrow{a} cdot overrightarrow{b} &= |overrightarrow{a}|^2 + |overrightarrow{b}|^2 - 2 |overrightarrow{a}| |overrightarrow{b}| cos left( frac{pi}{3} right) &= 1^2 + 2^2 - 2(1)(2)left(frac{1}{2}right) &= 1 + 4 - 2 &= 3. end{align*} Therefore, the length of the shorter diagonal is |overrightarrow{a} - overrightarrow{b}| = sqrt{3}. The answer is boxed{B}.

answer:Grayson spends 2 hours answering questions, and he takes 2 minutes for each answer. First, let's convert the 2 hours into minutes: 2 hours * 60 minutes/hour = 120 minutes Now, let's find out how many questions he answers in those 120 minutes: 120 minutes / 2 minutes/question = 60 questions Grayson answers 60 questions in 2 hours. Therefore, the number of questions he leaves unanswered when he submits his test is: 100 questions - 60 questions = 40 questions Grayson leaves boxed{40} questions unanswered.

answer:Solution: (1) Since overrightarrow{m}=(sin x,-1) and overrightarrow{n}=(sqrt{3}cos x,-frac{1}{2}), the function f(x)=(overrightarrow{m}+overrightarrow{n})cdotoverrightarrow{m}=(sin x+sqrt{3}cos x,-frac{3}{2})cdot(sin x,-1) =sin x(sin x+sqrt{3}cos x)+frac{3}{2}=sin^2x+sqrt{3}sin xcos x+frac{3}{2} =frac{1-cos 2x}{2}+frac{sqrt{3}}{2}sin 2x+frac{3}{2}=frac{sqrt{3}}{2}sin 2x-frac{1}{2}cos 2x+2 =sin(2x-frac{pi}{6})+2. From -frac{pi}{2}+2kpileqslant 2x-frac{pi}{6}leqslant frac{pi}{2}+2kpi, kinmathbb{Z}. We get -frac{pi}{6}+kpileqslant xleqslant frac{pi}{3}+kpi, kinmathbb{Z}. Therefore, the interval of monotonic increase for f(x) is left[-frac{pi}{6}+kpi, frac{pi}{3}+kpiright]; (2) Since xinleft[0, frac{pi}{2}right], 2x-frac{pi}{6}inleft[-frac{pi}{6}, frac{5pi}{6}right]. Then, the maximum value of f(x) in left[0, frac{pi}{2}right] is 3. That is, f(A)=3, sin(2A-frac{pi}{6})+2=3, 2A-frac{pi}{6}=frac{pi}{2}, we get A=frac{pi}{3}. Also, a=2sqrt{3}, c=4, by frac{2sqrt{3}}{sinfrac{pi}{3}}=frac{4}{sin C}, we get sin C=1, thus C=frac{pi}{2}. Then, B=pi-frac{pi}{2}-frac{pi}{3}=frac{pi}{6}. Therefore, the area of triangle ABC is S=frac{1}{2}accdotsin B=frac{1}{2}times2sqrt{3}times4timesfrac{1}{2}=2sqrt{3}. So, the final answers are A=boxed{frac{pi}{3}}, b is not directly calculated but implied, and S=boxed{2sqrt{3}}.

answer:Neither 0.8 nor 1/2 (which is boxed{0.5} when converted to a decimal) is greater than 3. Therefore, there are 0 numbers greater than 3 in the given set.