answer:- For r = sec(2theta) = frac{1}{cos(2theta)}, this polar equation represents the curve which includes the lines x = 1 and x = -1 because sec(theta) has vertical asymptotes at theta = pm frac{pi}{4}, pm frac{3pi}{4}, etc. However, in our region of interest (first quadrant), x = 1 is relevant. - For r = csc(2theta) = frac{1}{sin(2theta)}, this polar equation includes the lines y = 1 and y = -1 with vertical asymptotes at theta = 0, pm frac{pi}{2}, pm pi, etc. Similarly, only y = 1 affects the first quadrant. Now considering the region bounded by these, the x-axis, and the y-axis, the area of interest becomes a square formed by the intersection of x = 1 and y = 1 in the first quadrant. The vertices of this square are (0,0), (1,0), (1,1), and (0,1). The area of this square is 1 times 1 = boxed{1}.

answer:To find out how many more times a week Susannah swam than Camden, we first calculate how many times each of them went swimming per week by dividing the total number of times they went swimming in March by 4 (since March is divided into 4 weeks). For Camden: [ frac{16}{4} = 4 ] This means Camden went swimming 4 times a week. For Susannah: [ frac{24}{4} = 6 ] This means Susannah went swimming 6 times a week. To find out how many more times a week Susannah swam than Camden, we subtract the number of times Camden swam per week from the number of times Susannah swam per week: [ 6 - 4 = 2 ] Therefore, Susannah went swimming (boxed{2}) more times a week than Camden.

answer:Let’s denote the given triangle as ( triangle ABC ), where ( AB ) is the diameter of the circle. The circle intersects sides ( AC ) and ( BC ) at points ( P ) and ( Q ) respectively. Tangents to the circle at ( P ) and ( Q ) intersect at point ( F ). Our goal is to prove that the line ( CF ) is perpendicular to ( AB ). 1. **Auxiliary Lines**: - Draw the lines ( AQ ) and ( BP ). These lines serve as the altitudes of ( triangle ABC ), because ( AB ) is the diameter of the circle, thus ( angle AQB ) and ( angle APB ) are right angles by the Inscribed Angle Theorem and Thales's theorem. 2. **Intersecting Point ( H )**: - Let these altitudes intersect at point ( H ), which is the orthocenter of ( triangle ABC ). 3. **Angle Notations**: - Denote ( angle BAQ = alpha ) and ( angle ABP = beta ). 4. **Tangent-Chord Theorem Application**: - By the Tangent-Chord Theorem (The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment), we get: [ angle CQF = alpha quad text{and} quad angle CPF = beta ] 5. **Angles at Point ( H )**: - The angle at ( H ): ( angle AHB = 180^circ - alpha - beta = angle PHQ ). 6. **Angles at Points ( P ) and ( Q )**: - [ angle FPH = 90^circ - beta quad text{and} quad angle FQH = 90^circ - alpha ] 7. **Angle Calculations**: - Calculate the angle ( angle PFQ ): [ angle PFQ = 360^circ - angle FPH - angle FQH - angle PHQ = 360^circ - (90^circ - beta) - (90^circ - alpha) - (180^circ - alpha - beta) = 2(alpha + beta) ] 8. **Angle Measures in ( triangle ABC )**: - The triangle ( triangle ABC ) has its angles defined as follows: [ angle CAB = 90^circ - beta, quad angle CBA = 90^circ - alpha, quad angle ACB = alpha + beta ] 9. **Equality of Tangents**: - The segments ( FP ) and ( FQ ) are equal since they are tangents from a common external point ( F ) to the circle. - Construct a circle centered at ( F ) with radius ( FP = FQ ). 10. **Circumcenter Consideration**: - The central angle ( angle PFQ ) at point ( F ) is twice the inscribed angle ( angle PCQ ): [ angle PFQ = 2(alpha + beta) ] - Hence, point ( C ) must lie on this circle, making ( FP = FQ = CF ). 11. **( C ) Relative to the Circle**: - Since ( F ) is equidistant from ( P, Q ) and ( C ), the angle ( angle FCQ = alpha ) is identical to the angle at ( angle ACQ ) confirming ( C ) lies on the constructed circle centered at ( F ) with radius ( FP ). 12. **Conclusion**: - ( CF ) passes through ( H ) and coincides with the altitude from ( C ) to ( AB ), making ( CF ) perpendicular to ( AB ). Thus, it follows that ( CF perp AB ). [ boxed{CF perp AB} ]

answer:Since f′(x)=-3x^{2}+12=-3(x+2)(x-2), thus the function f(x)=-x^{3}+12x is decreasing on (-infty,-2), increasing on (-2,2), and decreasing on (2,+infty), therefore, the function f(x)=-x^{3}+12x reaches its maximum value at x=2, thus a=2. Hence, the correct choice is boxed{D}. First, find the derivative of the function, then solve the inequalities f′(x) > 0 and f′(x) < 0 to find the intervals of monotonicity of the function, and then by the definition of extreme values, find the points and values of the extremes. Using the derivative tool to find the extreme values of the function is key to solving this problem. It is necessary to first determine the real values of x for which the derivative equals zero, then discuss the intervals of monotonicity of the function, and according to the method of judging extreme values, find the extreme values of the function, which demonstrates the tool function of the derivative.