Let's say that the unknown value here is c. Although we are given two points on the line, it is only necessary to use one as there is only one unknown value, so let's say we choose the point (6, -12) and substitute this into the equation:
-12 = 1(6) - c
-12 - 6 = -c
-18 = -c
c = 18
Thus the equation for the line is y = 1x - 18
The answer is
2)40
3)y=3\2
You solve the problem and the answer is 982,810 :)
Step-by-step explanation:
S = ∫ 2π y ds
ds = √(1 + (dx/dy)²) dy
ds = √(1 + (8y)²) dy
ds = √(1 + 64y²) dy
S = ∫₁² 2π y √(1 + 64y²) dy
S = π/64 ∫₁² 128y √(1 + 64y²) dy
S = π/64 [⅔ (1 + 64y²)^(³/₂)] |₁²
S = π/96 (1 + 64y²)^(³/₂) |₁²
S = π/96 (1 + 256)^(³/₂) − π/96 (1 + 64)^(³/₂)
S = π/96 (257√257) − π/96 (65√65)
S = π/96 (257√257 − 65√65)
I found h max = 64 feet
Explanation: Ok...probably you can do this differently but I would try to find the vertex of the parabola describing the trajectory: 1) derive it: h ` ( t ) = 64 − 32 t 2) set derivative equal to zero: 64 − 32 t = 0 t = 64 32 = 2 sec 3) use this value of t into your trajectory: h ( 2 ) = h max 64 ⋅ 2 − 16 ⋅ 4 = 64 feet .
