answer:1. We are given that P is a point on the inscribed circle of the square (ABCD), and we need to find (tan^2 alpha + tan^2 beta) where (angle APC = alpha ) and (angle BPD = beta ). 2. Assume the coordinates of the center of the square are at the origin ((0,0)). Let the side length of the square (ABCD) be (2), and the square be oriented such that: - (A = (-1, -1), B = (1, -1), C = (1, 1), D = (-1, 1)). 3. Consider the point (P) on the inscribed circle with radius (1). Let the coordinates of (P) be: [ P = (1 + cos theta, 1 + sin theta) ] 4. Calculate vectors from (P) to the vertices: [ begin{aligned} overrightarrow{PA} &= (-1 - cos theta, -1 - sin theta), overrightarrow{PB} &= (1 - cos theta, -1 - sin theta), overrightarrow{PC} &= (1 - cos theta, 1 - sin theta), overrightarrow{PD} &= (-1 - cos theta, 1 - sin theta). end{aligned} ] 5. Compute the dot product: [ begin{aligned} overrightarrow{PA} cdot overrightarrow{PC} &= (-1 - cos theta)(1 - cos theta) + (-1 - sin theta)(1 - sin theta) &= -(1 + cos theta)(1 - cos theta) - (1 + sin theta)(1 - sin theta) &= - (1- cos^2 theta) - (1- sin^2 theta) &= - (1 - cos^2 theta + 1 - sin^2 theta) &= - (2 - (cos^2 theta + sin^2 theta)) &= - (2 - 1) &= -1. end{aligned} ] 6. Find magnitudes: [ begin{aligned} |overrightarrow{PA}|^2 &= (1 + cos theta)^2 + (1 + sin theta)^2 &= 1 + 2 cos theta + cos^2 theta + 1 + 2 sin theta + sin^2 theta &= 2 + (cos^2 theta + sin^2 theta) + 2 (cos theta + sin theta) &= 2 + 1 + 2 (cos theta + sin theta) &= 3 + 2 (cos theta + sin theta). end{aligned} ] [ begin{aligned} |overrightarrow{PC}|^2 &= (1 - cos theta)^2 + (1 - sin theta)^2 &= 1 - 2 cos theta + cos^2 theta + 1 - 2 sin theta + sin^2 theta &= 2 + (cos^2 theta + sin^2 theta) - 2 (cos theta + sin theta) &= 2 + 1 - 2 (cos theta + sin theta) &= 3 - 2 (cos theta + sin theta). end{aligned} ] 7. Now calculate the product of magnitudes: [ begin{aligned} |overrightarrow{PA}| cdot |overrightarrow{PC}| &= sqrt{(3 + 2 (cos theta + sin theta))(3 - 2 (cos theta + sin theta))} &= sqrt{9 - 4 (cos theta + sin theta)^2}. end{aligned} ] 8. Using the double angle identity (cos^2 theta + sin^2 theta = 1) and (sin 2theta = 2 cos theta sin theta): [ begin{aligned} sqrt{9 - 4 (cos theta + sin theta)^2} &= sqrt{9 - 4 (1 + sin 2theta)} &= sqrt{5 - 4 sin 2 theta}. end{aligned} ] 9. Now, the cosine of the angle (alpha) is then given by: [ cos alpha = frac{overrightarrow{PA} cdot overrightarrow{PC}}{|overrightarrow{PA}| cdot |overrightarrow{PC}|} = frac{-1}{sqrt{5 - 4 sin 2 theta}}. ] 10. Thus, [ tan^2 alpha = frac{1 - cos^2 alpha}{cos^2 alpha} = frac{(5 - 4 sin 2 theta - 1)}{-1^2} = 4 - 4 sin 2 theta. ] 12. Similarly, calculate for (beta): [ begin{aligned} overrightarrow{PB} cdot overrightarrow{PD} &= -(1 + cos theta)(1 - cos theta) - (1 + sin theta)(1 - sin theta) = -1, |overrightarrow{PB}|^2 &= 3 - 2 (cos theta - sin theta), |overrightarrow{PD}|^2 &= 3 + 2 (cos theta - sin theta). end{aligned} ] [ begin{aligned} |overrightarrow{PB}| cdot |overrightarrow{PD}| &= sqrt{9 - 4 (cos theta - sin theta)^2} &= sqrt{5 + 4 sin 2 theta}, end{aligned} ] [ cos beta = frac{-1}{sqrt{5 + 4 sin 2 theta}}, ] [ tan^2 beta = 4 + 4 sin 2 theta. ] 13. Summing up both terms: [ tan^2 alpha + tan^2 beta = 4 - 4 sin 2 theta + 4 + 4 sin 2 theta = 8. ] # Conclusion: [ boxed{8} ]

answer:First, let's find out how many Easter presents David received. Since the number of Easter presents is 10 fewer than half the number of Christmas presents, we calculate half of the Christmas presents and then subtract 10. Half of the Christmas presents is 60 / 2 = 30 presents. So, the number of Easter presents is 30 - 10 = 20 presents. Now, we know that the number of birthday presents is 3 times the number of Easter presents. So, David received 3 * 20 = 60 birthday presents. To find the total number of presents David got altogether, we add the number of Christmas presents, Easter presents, and birthday presents. Total presents = Christmas presents + Easter presents + Birthday presents Total presents = 60 + 20 + 60 Total presents = 140 David got boxed{140} presents altogether.

answer:First, calculate (1-i)^2: (1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i. Next, calculate frac{4+2i}{1-2i}: Multiply both the numerator and the denominator by the conjugate of the denominator: frac{4+2i}{1-2i} cdot frac{1+2i}{1+2i} = frac{4+2i+8i+4i^2}{1+2i-2i-4i^2} = frac{4+10i-4}{1+4} = frac{10i}{5} = 2i. Subtract the second result from the first: -2i - 2i = -4i. Therefore, the correct answer is boxed{text{C: } -4i}.

answer:To ensure the parabola y=x^2-3mx+m+n intersects the x-axis for all real numbers m, it means that for any value of m, the discriminant Delta = (-3m)^2-4 times 1 times (m+n) geq 0 must hold. Thus, 9m^2-4m-4n = 9left(m-frac{2}{9}right)^2-frac{4}{9}-4n geq 0, Therefore, -frac{4}{9}-4n geq 0. Solving this, we get: n leq -frac{1}{9}, Hence, the correct option is boxed{text{A}}.