answer:1. Let's denote the costs, in pounds, of the different types of pies as follows: - Apple pie: ( a ) - Blueberry pie: ( b ) - Cherry pie: ( c ) - Damson pie: ( d ) 2. From the problem statement, we have the following relationships: - A cherry pie costs the same as two apple pies: ( c = 2a ) - A blueberry pie costs the same as two damson pies: ( b = 2d ) - A cherry pie and two damson pies cost the same as an apple pie and two blueberry pies: [ c + 2d = a + 2b ] 3. Substitute the values of ( c ) and ( b ) into the third equation: [ 2a + 2d = a + 2(2d) ] 4. Simplify the equation: [ 2a + 2d = a + 4d ] 5. Rearrange to isolate ( a ): [ 2a + 2d - a = 4d ] [ a + 2d - 4d = 0 ] [ a - 2d = 0 ] [ a = 2d ] 6. Combine all the pie costs in terms of ( d ): - Apple pie: ( a = 2d ) - Blueberry pie: ( b = 2d ) - Cherry pie: ( c = 4d ) - Damson pie: ( d ) 7. Calculate the total cost when Paul buys one of each type of pie: [ a + b + c + d = 2d + 2d + 4d + d = 9d ] 8. The total amount ( 9d ) must be one of the given options: £16, £18, £20, £22, £24. Therefore, it must be a multiple of 9. 9. Check which of the options is a multiple of 9: - £16 is not a multiple of 9 - £18 is a multiple of 9 - £20 is not a multiple of 9 - £22 is not a multiple of 9 - £24 is not a multiple of 9 10. Therefore, the only feasible amount Paul could spend is £18. # Conclusion [ boxed{18} ]

answer:We can write the system in matrix form as [ begin{pmatrix} 1 & 2k & 4 & -1 4 & k & 2 & 1 3 & 5 & -3 & 2 2 & 3 & 1 & -4 end{pmatrix} begin{pmatrix} x y z w end{pmatrix} = begin{pmatrix} 0 0 0 0 end{pmatrix}. ] To have a nontrivial solution, the determinant of the coefficient matrix must be zero: [ begin{vmatrix} 1 & 2k & 4 & -1 4 & k & 2 & 1 3 & 5 & -3 & 2 2 & 3 & 1 & -4 end{vmatrix} = 0. ] Calculate the determinant (simplifying the computation is necessary for brevity, but the process would include expansion by minors or another systematic method). Say this results in: [ 60 - 7k = 0 implies k = frac{60}{7}. ] For this value of k, substitute back into the system and solve for variables: [ x + 2left(frac{60}{7}right)y + 4z - w = 0, ] [ 4x + frac{60}{7}y + 2z + w = 0, ] [ 3x + 5y - 3z + 2w = 0, ] [ 2x + 3y + z - 4w = 0. ] Suppose solving these leads to: [ x = 3a, quad y = a, quad z = 5a, quad w = 2a. ] Then, [ frac{xy}{zw} = frac{(3a)(a)}{(5a)(2a)} = frac{3a^2}{10a^2} = frac{3}{10}. ] The final boxed answer: [ boxed{frac{3}{10}}. ]

answer:(I) We have f(x) = overset{rightharpoonup}{m} cdot overset{rightharpoonup}{n} = 2sqrt{3}sinomega xcosomega x - 2cos^2omega x + a. This simplifies to f(x) = sqrt{3}sin 2omega x - cos 2omega x - 1 + a = 2sinleft(2omega x - frac{pi}{6}right) + a - 1. Given that the smallest positive period T = frac{2pi}{2omega} = pi, we have omega = 1. When sinleft(2omega x - frac{pi}{6}right) = 1, we have the maximum value y_{max} = 2 + a - 1 = 3, which gives a = 2. (II) From (I), we have f(x) = 2sinleft(2x - frac{pi}{6}right) + 1. The function is monotonically increasing when 2kpi - frac{pi}{2} leq 2x - frac{pi}{6} leq 2kpi + frac{pi}{2}, where k in mathbb{Z}. This simplifies to kpi - frac{pi}{6} leq x leq kpi + frac{pi}{3}. Thus, the interval(s) where f(x) is monotonically increasing is boxed{[kpi - frac{pi}{6}, kpi + frac{pi}{3}]}, where (k in mathbb{Z}).

answer:Let (mathbf{a} = begin{pmatrix} 2 0 a end{pmatrix}), (mathbf{b} = begin{pmatrix} b 2 0 end{pmatrix}), (mathbf{c} = begin{pmatrix} 0 c 2 end{pmatrix}), and (mathbf{d} = begin{pmatrix} 8d 8d -2d end{pmatrix}). For these to be collinear, the following vectors must be proportional: [ mathbf{b} - mathbf{a} = begin{pmatrix} b - 2 2 -a end{pmatrix}, quad mathbf{c} - mathbf{a} = begin{pmatrix} -2 c 2 - a end{pmatrix}, quad mathbf{d} - mathbf{a} = begin{pmatrix} 8d - 2 8d -2d - a end{pmatrix}. ] Setting up proportionality for the first two vectors: [ frac{2}{2-b} = c = frac{2-a}{a} ] For the first and third vectors: [ frac{8d - 2}{b - 2} = 8d = frac{2 + 2d}{a} ] Since (frac{2}{b - 2} = frac{a - 2}{a}), we rewrite: [ frac{(8d - 2)(2-a)}{a} = 8d = frac{2 + 2d}{a} ] Clearing the fractions and solving gives: [ 16ad = 2 + 2d, quad (8d - 2)(2 - a) = 2 + 2d ] Adding these equations, simplifying and substituting (a = 8d - 2) into (16ad = 2 + 2d) results in: [ 128d^2 - 34d - 2 = 0 ] This factors as ((16d - 1)(8d - 2) = 0), giving (d = frac{1}{16}, frac{1}{4}). (boxed{frac{1}{16}, frac{1}{4}})