answer:Let's break down the problem step by step: 1. Rene received 600. 2. Florence received twice as much as Rene, so Florence received 2 * 600 = 1200. 3. Isha gave half of the squared amount to Florence. Let's denote the amount Isha received from Maria as x. Then, (x^2) / 2 = 1200. To find x, we solve for x: x^2 = 1200 * 2 x^2 = 2400 x = sqrt(2400) x = 49 (approximately, since the square root of 2400 is not a whole number, but we'll use the approximate value for the sake of the calculation). 4. Maria gave a quarter of her money to Isha, so the total amount Maria had was 4 * x. Let's denote Maria's total money as M: M = 4 * x M = 4 * 49 M = 196 (approximately). 5. John received one-third of what Florence received, so John got 1200 / 3 = 400. 6. John then halved it, getting 400 / 2 = 200, and gave half to Emma, so Emma received 200 / 2 = 100. Now, let's add up the amounts received by each friend to find out how much money Maria distributed: - Isha received approximately 49. - Florence received 1200. - Rene received 600. - John received 200 (before giving half to Emma). - Emma received 100. Total distributed = Isha + Florence + Rene + John + Emma Total distributed ≈ 49 + 1200 + 600 + 200 + 100 Total distributed ≈ 2149 So, Maria distributed approximately boxed{2149} among her five friends.

answer:First, calculate the binomial coefficient dbinom{30}{3}: [ dbinom{30}{3} = dfrac{30!}{27!3!} = dfrac{30 times 29 times 28}{3 times 2 times 1} = dfrac{24360}{6} = 4060. ] Next, square this result to find (dbinom{30}{3})^2: [ (dbinom{30}{3})^2 = 4060^2 = 16483600. ] Thus, the solution to the problem is: [ boxed{16483600}. ]

answer:Since the function f(x)=sin ωx+acos ωx=sqrt{1+a^2}left(frac{1}{sqrt{1+a^2}}sin ωx+frac{a}{sqrt{1+a^2}}cos ωxright)=sqrt{1+a^2}sin(ωx+φ) (tan φ=a) is symmetric about the point M(frac{π}{3},0), we have frac{π}{3}ω+φ=kπ (k∈mathbb{Z}), (①) Also, the function has a minimum value at x=frac{π}{6}, so frac{π}{6}ω+φ=frac{3π}{2}+2kπ (k∈mathbb{Z}), (②) Solving the system of equations (①) and (②), we obtain ω=-9-6k and φ=kπ+3π (k∈mathbb{Z}). Thus, a=tan(kπ+3π)=0, And a+ω=-9-6k (k∈mathbb{Z}). Given that a+ω∈[0,10], we have k=-2 and a+ω=3. Therefore, the answer is boxed{3}. By analyzing the function and utilizing its symmetry, we can derive ω=-9-6k and φ=kπ+3π (k∈mathbb{Z}). From this, we can calculate a and subsequently a+ω=-9-6k (k∈mathbb{Z}), which enables us to find the answer. This problem primarily evaluates the understanding of the symmetry point, symmetry axis, and period of a sine and cosine function, making it a moderately difficult question.

answer:To solve the given problem, we start by analyzing the given inequality involving a complex number: m-3+(m^{2}-9)igeqslant 0 This inequality involves both real and imaginary parts of a complex number. For a complex number to satisfy this inequality, both the real part and the imaginary part must satisfy certain conditions. First, let's separate the real and imaginary parts: 1. Real part: m-3 2. Imaginary part: m^{2}-9 Given that the inequality involves a complex number, we interpret the conditions for the real and imaginary parts separately: 1. For the imaginary part m^{2}-9 to not affect the inequality (since an inequality cannot directly apply to complex numbers as it does to real numbers), it must be equal to 0. This gives us the equation: m^{2}-9=0 Solving this equation for m gives: m^{2}=9 m=pm3 2. For the real part m-3 to satisfy the inequality, it must be greater than or equal to 0. This gives us the inequality: m-3geqslant 0 Solving this inequality for m gives: mgeqslant 3 Combining these two conditions, we find that m must satisfy both m=pm3 from the imaginary part condition and mgeqslant 3 from the real part condition. The only value of m that satisfies both conditions is m=3. Therefore, the value of the real number m is boxed{3}.