answer:1. First, introduce variables to represent the number of occurrences of the values -2, 0, and 1 among x_1, x_2, ldots, x_n. Let: - p be the number of occurrences of 1. - q be the number of occurrences of -2. - The remaining values take on 0. 2. We are given two conditions: [ x_1 + x_2 + cdots + x_n = -17 ] and [ x_1^2 + x_2^2 + cdots + x_n^2 = 37. ] 3. Calculate the sum and the sum of squares using p and q: - The sum is given by: [ p cdot 1 + q cdot (-2) + (n - p - q) cdot 0 = p - 2q ] Therefore, we have: [ p - 2q = -17 ] - The sum of squares is given by: [ p cdot 1^2 + q cdot (-2)^2 + (n - p - q) cdot 0^2 = p + 4q ] Therefore, we have: [ p + 4q = 37 ] 4. We now solve the system of linear equations: [ begin{cases} p - 2q = -17 p + 4q = 37 end{cases} ] 5. To eliminate p, add both equations: [ (p - 2q) + (p + 4q) = -17 + 37 ] Simplify: [ 2p + 2q = 20 ] [ p + q = 10 quad text{(Equation A)} ] 6. To find q, substitute p from Equation A into one of the original equations: [ p - 2q = -17 ] Using p = 10 - q from Equation A: [ (10 - q) - 2q = -17 ] Simplify: [ 10 - 3q = -17 ] [ -3q = -27 ] [ q = 9 ] 7. Substitute q = 9 back into Equation A: [ p + 9 = 10 ] Solving for p: [ p = 1 ] So, we have p = 1 and q = 9. 8. Calculate the sum of cubes: [ left(x_1^3 + x_2^3 + cdots + x_n^3 right)^2 ] Each x_i is either 1, -2 or 0. Therefore, [ left( p cdot 1^3 + q cdot (-2)^3 + (n - p - q) cdot 0^3 right)^2 ] Substitute p = 1 and q = 9: [ left( 1 cdot 1^3 + 9 cdot (-2)^3 right)^2 ] Evaluate each term inside: [ 1 cdot 1 + 9 cdot (-8) = 1 - 72 = -71 ] Finally, square the result: [ (-71)^2 = 5041 ] Conclusion: [ boxed{5041} ]

answer:The function given is ( y = -5 cos left( x + frac{pi}{4} right) ). 1. **Amplitude Identification**: The coefficient of cos, which is -5, indicates the amplitude. The amplitude is the absolute value, hence ( text{Amplitude} = |-5| = 5 ). 2. **Phase Shift Identification**: The phase shift of the cosine function (cos left( x + frac{pi}{4} right)) is determined by the argument (x + frac{pi}{4}). Since the addition of (frac{pi}{4}) shifts the graph to the left, the phase shift is ( -frac{pi}{4} ). Thus, the amplitude is ( 5 ) and the phase shift is ( -frac{pi}{4} ), so the final answers are the amplitude is boxed{5} and the phase shift is boxed{-frac{pi}{4}}.

answer:The problem is focused on understanding the domain and range of a trigonometric function and the periodic nature of the function. As the function is given by y=2sin x, and we know the sine function has a period of 2pi, we need to deduce the possible value of b-a. Since the range of y=2sin x is [-2,1] which includes the minimum value but does not include the maximum value, it follows that the span of the domain, b-a, must be less than a full period. The sine function reaches its maximum amplitude at sin x = 1. However, for 2sin x, the maximum value achievable before scaling is still sin x = 1, which would correspond to y = 2. Within any interval smaller than the full period, for every point in the range to be covered, the domain must at least span from the point where sin x = -1 to just before sin x goes back to 1. The values of x for which sin x = -1 occur at odd multiples of pi, so at x = (2k+1)pi, with k being any integer. The values for sin x = 0.5 (leading to y=1), a sub-maximum on the original sine curve, occur at x = frac{pi}{6} + 2kpi and x = frac{5pi}{6} + 2kpi. Given the constraint that the range does not include the maximum value 2, the domain in [a,b] must end before 2sin x hits its peak at y=2. The closest we can approach this value from the left (taking the negative value of the sine wave first) is at x = frac{5pi}{6} + 2kpi, not exceeding the point where sin x = 1. Thus, the domain covered from the negative peak to just before returning to the positive peak (without reaching the full amplitude of the sine function) is less than the full periodic length, 2pi. Option D, frac{3pi}{2}, is less than the full period of the sine function and could potentially be the length of the domain. However, the period is symmetric about the x-axis, and to include one extreme value and exclude the other, the domain needs to be asymmetric about the midline, implying that the correct answer must exclude the full half-period, which is pi. Options A (frac{5pi}{6}) and C (frac{7pi}{6}) are both possible since they are asymmetric about the x-axis and are smaller than the full period. This is because they allow us to include the lower bound of the range while excluding the upper bound. Option B (pi) cannot be correct because a stretch of pi starting from a negative peak would include the next positive peak due to the symmetry of the sine wave. Consequently, the value of b-a cannot be boxed{frac{3pi}{2}}.

answer:To find the molecular weight of BaF2, we first need to calculate the molecular weight of one mole of BaF2. Since we know the molecular weight of 6 moles of BaF2 is 1050 grams, we can divide this by 6 to find the molecular weight of one mole. Molecular weight of one mole of BaF2 = 1050 grams / 6 moles = 175 grams/mole So, the molecular weight of BaF2 is boxed{175} grams/mole. This means that regardless of the number of moles, the molecular weight of BaF2 will always be 175 grams/mole.