answer:If Josh had 9 marbles and lost 5, then he would have: 9 - 5 = boxed{4} marbles left.

answer:1. **Decomposition of n**: - When n is divided by 50, n = 50q + r. - q represents a number consisting of the digits before the last two when n is expressed in base 50. - r is the remainder, thus 0 leq r < 50. 2. **Range of q**: - As n is a 5-digit number, 10000 leq 50q + r leq 99999, leading to 200 leq q leq 1999 when r is maximized at 49. 3. **Condition for divisibility by 7**: - We need q + r to be divisible by 7, hence q + r equiv 0 pmod{7}. 4. **Counting valid r for each q**: - For each q, r needs to satisfy r equiv -q pmod{7}. - Since 0 leq r < 50, there are leftlfloor frac{50}{7} rightrfloor + 1 = 7 + 1 = 8 values of r for each residue class modulo 7. 5. **Total valid pairs (q, r)**: - Each q (from 200 to 1999, total 1800 values) has 8 corresponding rs. - Total valid (q, r) pairs is 1800 times 8 = 14400. 6. **Conclusion**: Thus, there are 14400 valid 5-digit numbers n such that q + r is divisible by 7. The final answer is boxed{B}

answer:1. **Start by solving the first equation**: [ 2x - y = 5 implies y = 2x - 5 ] Substitute y in the second equation: [ kx^2 + (2x - 5) = 4 implies kx^2 + 2x - 9 = 0 ] 2. **Solve the quadratic equation for x**: We solve kx^2 + 2x - 9 = 0. This is a quadratic equation, and we apply the quadratic formula: [ x = frac{-2 pm sqrt{4 + 36k}}{2k} = frac{-2 pm sqrt{4(1 + 9k)}}{2k} ] To ensure real solutions (thus valid points in the first quadrant), the discriminant must be non-negative: [ 1 + 9k geq 0 implies k geq -frac{1}{9} ] 3. **Conditions for x > 0 and y > 0**: - From the solution for x, we choose the positive root for x (assuming k neq 0): [ x = frac{-2 + sqrt{4(1 + 9k)}}{2k} ] We need -2 + sqrt{4(1 + 9k)} > 0 implies sqrt{4(1 + 9k)} > 2 implies 1 + 9k > 1 implies k > 0. - Ensure y > 0: [ y = 2x - 5 > 0 implies x > frac{5}{2} ] We substitute x = frac{-2 + sqrt{4 + 36k}}{2k}: [ frac{-2 + sqrt{4 + 36k}}{2k} > frac{5}{2} implies -4 + sqrt{4 + 36k} > 5k implies sqrt{4 + 36k} > 4 + 5k. ] Solving this inequality provides another condition on k. 4. **Combine conditions**: We need both k geq -frac{1}{9} and k > 0. The more restrictive condition is k > 0. [ k > 0 ] The final answer is C) boxed{k > 0}

answer:1. Since overrightarrow{a}=(1+sin 2x,sin x-cos x) and overrightarrow{b}=(1,sin x+cos x), we have f(x)=1+sin 2x+sin^2 x-cos^2 x = 1+sin 2x-cos 2x = sqrt {2}sin (2x- frac {pi}{4})+1. Thus, f(x) attains its maximum value of sqrt {2}+1 when 2x- frac {pi}{4}=2kpi+ frac {pi}{2}, i.e., x=kpi+ frac {3}{8}pi (kinmathbb{Z}). 2. From f(theta)=1+sin 2theta-cos 2theta and f(theta)= frac {8}{5}, we get sin 2theta-cos 2theta= frac {3}{5}. Squaring both sides gives 1-sin 4theta= frac {9}{25}, which simplifies to sin 4theta= frac {16}{25}. Hence, the value of cos 2( frac {pi}{4}-2theta) is cos ( frac {pi}{2}-4theta)=sin 4theta=boxed{frac{16}{25}}.