If we had AB: BC: CD, we could easily have solved that problem. In order to combine these ratios, we need to have the same number for BC. We can create this number by finding LCM (Least Common Multiple) for 5 and 3, which is 15. Then, we can write ratios of AB:BC = 6:15 and BC:CD=15:20. Now, we can easily combine these ratios. AB:BC:CD = 6:15:20. Then, 6k+15k+20k = 82 and k=2 cm. And BC = 30 cm
Answer:
Step-by-step explanation:
(1)Identify the surface whose equation is r = 2cosθ by converting first to rectangular coordinates...(2)Identify the surface whose equation is r = 3sinθ by converting first to rectangular coordinates...(3)Find an equation of the plane that passes through the point (6, 0, −2) and contains the line x−4/−2 = y−3/5 = z−7/4...(4)Find an equation of the plane that passes through the point (−1,2,3) and contains the line x+1/2 = y+2/3 = z-3/-1...(5)Find a) the scalar projection of a onto b b) the vector projection of a onto b given = 〈2, −1,3〉 and = 〈1,2,2〉...(6)Find a) the scalar projection of a onto b b) the vector projection of a onto b given = 〈2,1,4〉 and = 〈3,0,1〉...(7)Find symmetric equations for the line of intersection of the planes x + 2 y + 3z = 1 and x − y + z = 1...(8)Find symmetric equations for the line of intersection of the planes x + y + z = 1 and x + 2y + 2z = 1...(9)Write inequalities to describe the region consisting of all points between, but not on, the spheres of radius 3 and 5 centered at the origin....(10)Write inequalities to describe the solid upper hemisphere of the sphere of radius 2 centered at the origin....(11)Find the distance between the point (4,1, −2) and the line x = 1 +t , y = 3 2−t , z = 4 3−t...(12)Find the distance between the point (0,1,3) and the line x = 2t , y = 6 2−t , z = 3 + t...(13)Find a vector equation for the line through the point (0,14, −10) and parallel to the line x=−1+2t, y=6-3t, z=3+9t<span>...</span>
