answer:Analysis: From Basic Conclusion 6, we know that the projection of the apex of the cone on the base is the incenter of the base triangle. Then, given that the dihedral angle between the lateral face and the base is 30^{circ}, we only need to first find the inradius of the base triangle. Using plane geometry methods, it is easy to find that the inradius of the base triangle is 3, thus, the height of the pyramid is 3 tan 30^{circ}=sqrt{3}.

answer:6. C The distance, d, of the point (x, y) from the origin O,(0,0), is given by d=sqrt{x^{2}+y^{2}}. It is clear that the points (5,0),(4,3),(3,4),(0,5) are all at a distance of 5 units from O as sqrt{4^{2}+3^{2}}=sqrt{3^{2}+4^{2}}=5. So those four points all lie on a circle with centre O and radius 5 .However, the distance from the point (2,2) to O is sqrt{2^{2}+2^{2}}=sqrt{8} neq 5.Therefore the point (2,2) does not lie on the same circle as the other four points.(There is a unique circle through any three non-collinear points, so there can be no circle that passes through the point (2,2) and three of the other four given points.)

answer:Given that the contour line of the Dead Sea is labeled as -415m, we interpret this information as follows:- The negative sign before the number indicates that the altitude is below a reference point, which in this context is sea level.- The number 415 represents the magnitude of the difference in altitude from the reference point.Therefore, the interpretation of the contour line labeled as -415m is that the height at this point is below sea level by 415m. Hence, the correct way to fill in the blank is with the phrase "lower than", indicating that the height at this point is lower than sea level by 415m.Final Answer: boxed{text{lower than}}.

answer:(I) First, let's find out the propositions derived by replacing any two among a, b, and c with planes: If a and b are replaced with planes alpha and beta, respectively, the proposition becomes "alpha parallel beta, and alpha perp c Rightarrow beta perp c", according to the property of line-plane parallelism, this proposition is true; If a and c are replaced with planes alpha and gamma, respectively, the proposition becomes "alpha parallel b, and alpha perp gamma Rightarrow b perp gamma", b may intersect with gamma or lie within plane gamma, this proposition is false; If b and c are replaced with planes beta and gamma, respectively, the proposition becomes "a parallel beta, and a perp gamma Rightarrow beta perp gamma", according to the theorem for determining perpendicular planes, this proposition is true, thus, there are 2 true propositions; (II) Replacing all three among a, b, and c with planes, we get the proposition: "alpha parallel beta, and alpha perp gamma Rightarrow beta perp gamma", according to the theorem for determining perpendicular planes, this proposition is true, therefore, the correct choice is boxed{C}.

answer:The original equation is equivalent to the equation x^{2} z+y^{2} x+z^{2} y=m x y z, where x, y, z are non-zero integers, and pairwise coprime. Thus, yleft|x^{2} z, zright| y^{2} x, x mid z^{2} y. Also, (x, y)=1,(z, y)=1, so left(x^{2} zright., y)=1. From y mid x^{2} z we get y= pm 1, similarly we get z= pm 1, x= pm 1.If x, y, z are all positive or all negative, then from the original equation we get m=3. If two of x, y, z are positive and one is negative, or two are negative and one is positive, then from the original equation we get m as a negative number, which contradicts the problem statement.Therefore, the original equation, when m=3, has integer solutions x, y, z that are pairwise coprime, with only two sets of solutions: x=y=z=1, x=y=z=-1. When m neq 3, the original equation has no solutions that satisfy the problem conditions.

answer:3. B.According to the problem, we have a_{k+1}-a_{k}=1 or a_{k+1}-a_{k}=-1. If there are m ones, then there are 10-m negative ones. Therefore, 4=m-(10-m) Rightarrow m=7.Thus, the number of sequences we are looking for is mathrm{C}_{10}^{7}=120.