answer:1. The side length of the square ( KLMN ) is given as 1. 2. Let us denote ( KX ), ( XL ), ( KY ), ( YN ), ( NZ ), and ( ZM ) using their respective ratios: [ frac{KX}{XL}= frac{3}{2}, quad frac{KY}{YN} = frac{4}{1}, quad frac{NZ}{ZM} = frac{2}{3} ] Using these ratios and the side length of the square, we can write: [ KX = frac{3}{3+2} = frac{3}{5}, quad XL = frac{2}{3+2} = frac{2}{5} ] [ KY = frac{4}{4+1} = frac{4}{5}, quad YN = frac{1}{4+1} = frac{1}{5} ] [ NZ = frac{2}{2+3} = frac{2}{5}, quad ZM = frac{3}{2+3} = frac{3}{5} ] 3. Now, compute the area of the white triangles and the trapezium inside the square: - Area of triangle ( KXL ): [ text{Area} = frac{1}{2} cdot KX cdot XL = frac{1}{2} cdot frac{3}{5} cdot frac{2}{5} = frac{3}{25} ] - Area of triangle ( KYZ ): [ text{Area} = frac{1}{2} cdot KY cdot YN = frac{1}{2} cdot frac{4}{5} cdot frac{1}{5} = frac{2}{25} ] - Area of trapezium ( NZMZ' ): Since ( Z' ) is point on ( LZ ) such that ( ZM = ZM ), the height is ( 1 ): [ text{Area} = frac{1}{2} cdot (NZ + ZM) cdot 1 = frac{1}{2} cdot left( frac{2}{5} + frac{3}{5} right) cdot 1 = frac{1}{2} cdot 1 = frac{1}{2} ] 4. Summing up the areas of the two white triangles and the trapezium: [ text{Total area} = frac{3}{25} + frac{2}{25} + frac{1}{2} = frac{3}{25} + frac{2}{25} + frac{12.5}{25} = frac{17.5}{25} ] 5. The area of triangle ( XYZ ) is: [ text{Area of } triangle XYZ = text{Area of the square} - text{Area of the white portions} = 1 - frac{17.5}{25} = 1 - 0.7 = 0.3 ] Conclusion: [ boxed{frac{3}{10} text{ units}^2} ]

answer:First, expand and simplify the given equation: [ (x+3)(x-4) = x^2 - 4x + 3x - 12 = x^2 - x - 12 ] Now substitute the right side expression: [ 2(x+1) = 2x + 2 ] Setting them equal gives: [ x^2 - x - 12 = 2x + 2 Rightarrow x^2 - 3x - 14 = 0 ] For a quadratic equation in the form ax^2 + bx + c = 0, the product of the roots is given by frac{c}{a}. Thus, for x^2 - 3x - 14 = 0, the product of the roots is: [ frac{-14}{1} = boxed{-14} ]

answer:Start by simplifying the expression: [ 0.08 div 0.002 div 0.04 ] Convert the decimals to fractions and simplify the expression: [ frac{0.08}{0.002} div 0.04 = frac{0.08}{0.002} cdot frac{1}{0.04} = left(frac{0.08}{0.002}cdotfrac{1000}{1000}right) cdot left(frac{1}{0.04}cdotfrac{100}{100}right) ] [ = left(frac{80}{2}right) cdot left(frac{1}{0.04}right) = 40 cdot 25 = boxed{1000}. ]

answer:To solve this problem, we consider all students except for students A and B, which leaves us with 36 students. Our task is to choose 4 of these 36 students to be included in the group of 5, as we will always include student A to meet the requirement of the problem. We can calculate the number of combinations of 4 students from 36 using the combination formula C(n, k) = frac{n!}{k!(n-k)!}, where n is the total number of items to choose from, and k is the number of items to choose. So, the number of combinations where 4 students are chosen from the remaining 36 is C(36, 4). Then we will always include student A in the group, which we can consider as C(1, 1) because we are choosing 1 student from 1 (student A himself). Therefore, the number of possible selections where student A is chosen but student B is not is the product of these two combinations: begin{align*} text{Total number of possible selections} &= C(36, 4) times C(1, 1) &= frac{36!}{4!(36-4)!} times 1 &= frac{36!}{4! times 32!} &= frac{36 times 35 times 34 times 33}{4 times 3 times 2 times 1} &= 58905. end{align*} So, there are boxed{58905} possible selections where student A is included but student B is not.