(-2,-5) is the answer for this by doing -6-4 and -8–2
Complete question:
A student throws a heavy ball downward off the top of a building with a speed of 18m/s. The ball reaches a speed of 41m/s just before striking the ground. Neglect drag, find the height of the building.
Answer:
The height of the building is 69.235 m
Step-by-step explanation:
Given;
initial velocity of the ball, u = 18 m/s
final velocity of the ball, v = 41 m/s
The height of the building is equal to distance traveled by the ball downward.
Apply the following kinematic equation;
v² = u² + 2gh
where;
g is acceleration due to gravity
h is height of the building
41² = 18² + 2(9.8)h
1681 = 324 + 19.6h
19.6h = 1681 - 324
19.6h = 1357
h = 1357 / 19.6
h = 69.235 m
Therefore, the height of the building is 69.235 m
Answer:
Step-by-step explanation:
By the polygon exterior angle sum theorem, all the exterior angles of a polygon add up to equal 360 regardless of the number of sides it has. That means that 82 + 150 + b = 360 so
b + 232 = 360 and
b = 128
Answer:
698 cm²
Step-by-step explanation:
The volume is given by ...
V = LWH
Filling in the given values, we have ...
1020 = (17)(5)y . . . . . . . using L=17, W=5, H=y
y = 1020/(17·5) = 12
The surface area is given by ...
A = 2(LW +H(L+W))
A = 2(17·5 +12(17+5)) = 2(85 +264) . . . . . . . using L=17, W=5, H=y=12
A = 698 . . . . square centimeters
