(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>
Answer:
C) m∠H = 85°
Step-by-step explanation:
<em>KL is the mid-segment of the triangle</em>, and is correlated with line GH, therefore, they are parallel. This means that <em>m∠K = m∠G and m∠L = m∠H</em>, because they are corresponding and on parallel lines.
By substitution, m∠H = 85°
Answer:
Therefore total work done =2050 ft-lb
Step-by-step explanation:
Given that, a bucket are used to drawn water from a well that is 50 ft deep.
(1)
Work done = Force×displacement.
Work done to pull the bucket is
= Weight of the bucket × displacement
=(6×50) ft-lb
=300 ft-lb
(2)
The bucket fill with 40 lb of water and is pulled up at a rate of 2ft/s, but leaks out of a hole in the bucket at a rate of 0.2 lb/s.
Since the well is 50 ft deep.
It takes 
=25 s to pull the bucket at the top of the well.
After 25 s it lost (25×0.2)lb =5 lb water.
So the remaining water is = (40-5) lb= 35 lb
Let y is distance above the original depth of 50 ft.
When y=50 ft, weight of the bucket= 40 lb,when y=0 weight of the bucket= 35 lb
The slope of the of water leakage is
The bucket holds 
lb of water when it is y ft above the original depth.
Work done to pull the water in the bucket is
![=[40y- \frac1{10}y^2]_0^{50}](https://tex.z-dn.net/?f=%3D%5B40y-%20%5Cfrac1%7B10%7Dy%5E2%5D_0%5E%7B50%7D)
![=[(40\times 50)-\frac1{10}(50)^2]-[(40\times 0)-\frac1{10}(0)^2]](https://tex.z-dn.net/?f=%3D%5B%2840%5Ctimes%2050%29-%5Cfrac1%7B10%7D%2850%29%5E2%5D-%5B%2840%5Ctimes%200%29-%5Cfrac1%7B10%7D%280%29%5E2%5D)
=1750 ft-lb
Therefore total work done
=Work done to pull the bucket+Work done to pull the water in the bucket
=(300+1750) ft-lb
=2050 ft-lb
Answer:
Heyy
Step-by-step explanation:
