answer:First, let's calculate the number of family members attending for each kid without considering the step-parents and grandparents: Each kid has 2 parents and 1 sibling attending, so that's 3 family members per kid. There are 10 girls and 12 boys, so there are a total of 10 + 12 = 22 kids in the play. The total number of family members attending for the kids is 22 kids * 3 family members per kid = 66 family members. Now, let's add the step-parents and grandparents: For the 5 kids with a step-parent attending, that's an additional 1 family member per kid, so 5 * 1 = 5 additional family members. For the 3 kids with 2 grandparents attending, that's an additional 2 family members per kid, so 3 * 2 = 6 additional family members. Adding these to the initial count, we get 66 + 5 + 6 = boxed{77} family members in the auditorium.

answer:1. **Given Information and Goal Setup**: - triangle XYZ is a right triangle at Y. - overline{WY} = overline{YW} means W is the midpoint of XY. - From overline{XY} = 26 cm and midpoint information, overline{YW} = overline{WY} = 13 cm. - WT perp XY, thus triangle WYT is a right triangle. 2. **Calculate overline{YZ}**: - Using the Pythagorean theorem in triangle XYZ, [ overline{YZ} = sqrt{overline{XY}^2 - overline{XZ}^2} = sqrt{26^2 - 10^2} = sqrt{676 - 100} = sqrt{576} = 24 text{ cm}. ] 3. **Determine the similarity ratio and area ratio of triangle WYT and triangle XYZ**: - Sharing the right angle and angle angle Y, triangle WYT sim triangle XYZ. - Ratio of corresponding sides: frac{overline{YW}}{overline{XY}} = frac{13}{26} = frac{1}{2}. - Area ratio of the triangles: left(frac{1}{2}right)^2 = frac{1}{4}. 4. **Calculate the areas**: - Area of triangle XYZ = frac{1}{2} times overline{XZ} times overline{YZ} = frac{1}{2} times 10 times 24 = 120 text{ cm}^2. - Area of triangle WYT = frac{1}{4} times 120 = 30 text{ cm}^2. 5. **Calculate the area of quadrilateral XWTZ**: - The area of XWTZ is area of triangle XYZ minus area of triangle WYT: [ [XWTZ] = [XYZ] - [WYT] = 120 - 30 = 90 text{ cm}^2. ] - [ 90 text{ cm^2} ] The final answer is C. boxed{90} cmboxed{^2}

answer:1. Given that the function [ f(x) = x^3 + a x^2 + 2 b x - 1 ] has three distinct real roots x_1, x_2, x_3, we can use Vieta's formulas to express the relationships between these roots and the coefficients of the polynomial. 2. From Vieta’s formulas, we have: [ begin{aligned} x_1 + x_2 + x_3 &= -a, x_1 x_2 + x_2 x_3 + x_3 x_1 &= 2b, x_1 x_2 x_3 &= 1. end{aligned} ] 3. To form the sum of the squares of the roots, we use the identity: [ x_1^2 + x_2^2 + x_3^2 = (x_1 + x_2 + x_3)^2 - 2(x_1 x_2 + x_2 x_3 + x_3 x_1). ] Substituting the expressions from Vieta’s formulas: [ begin{aligned} x_1^2 + x_2^2 + x_3^2 &= (-a)^2 - 2 cdot 2b &= a^2 - 4b. end{aligned} ] 4. By the Arithmetic Mean-Geometric Mean Inequality (AM-GM inequality), we know: [ frac{x_1^2 + x_2^2 + x_3^2}{3} geq sqrt[3]{x_1^2 x_2^2 x_3^2}. ] Since x_1 x_2 x_3 = 1, we have x_1^2 x_2^2 x_3^2 = 1. Therefore: [ frac{x_1^2 + x_2^2 + x_3^2}{3} geq sqrt[3]{1} = 1, ] which implies: [ x_1^2 + x_2^2 + x_3^2 geq 3. ] Hence, [ a^2 - 4b > 3. ] 5. Given that the quadratic function g(x) = 2x^2 + 2bx + a has no real roots, the discriminant must be less than zero: [ (2b)^2 - 4 cdot 2 cdot a < 0. ] Simplifying, we get: [ 4b^2 - 8a < 0 implies 2a > b^2. ] 6. Adding the inequalities (2a > b^2) and (a^2 - 4b > 3), we obtain: [ begin{aligned} (2a - b^2) + (a^2 - 4b) &> 0 + 3 Rightarrow 2a - b^2 + a^2 - 4b &> 3. end{aligned} ] 7. We can rearrange the above inequality as: [ (a+1)^2 - (b+2)^2 + 3 > 3, ] which further simplifies to: [ (a+1+b+2)(a+1-b-2) > 0. ] 8. Since (a) and (b) are real numbers, we have: [ a + b + 3 > 0. ] 9. Therefore: [ a - b - 1 > 0, ] which can be rewritten as: [ a - b > 1. ] Conclusion: [ boxed{a - b > 1} ]

answer:Given that x_{1} and x_{2} are the real roots of the equation 2x^{2}+kx-2=0, we can use Vieta's formulas to express the sum and product of the roots in terms of the coefficients of the equation. 1. The sum of the roots is given by x_{1} + x_{2} = -frac{b}{a} = -frac{k}{2}. 2. The product of the roots is given by x_{1} cdot x_{2} = frac{c}{a} = -1. Given the equation (x_{1}-2)(x_{2}-2)=10, we can expand and simplify it using the information from Vieta's formulas: [ begin{align*} (x_{1}-2)(x_{2}-2) &= x_{1}x_{2} - 2(x_{1} + x_{2}) + 4 &= -1 - 2left(-frac{k}{2}right) + 4 &= -1 + k + 4 &= k + 3. end{align*} ] Setting this equal to 10, we get: [ k + 3 = 10. ] Solving for k, we find: [ k = 10 - 3 = 7. ] Therefore, the value of k that satisfies the given conditions is boxed{7}.