Answer:
Step-by-step explanation:
The position function is
and if we are looking for the time(s) that the ball is 10 feet above the surface of the moon, we sub in a 10 for s(t) and solve for t:
and
and factor that however you are currently factoring quadratics in class to get
t = .07 sec and t = 18.45 sec
There are 2 times that the ball passes 10 feet above the surface of the moon, once going up (.07 sec) and then again coming down (18.45 sec).
For part B, we are looking for the time that the ball lands on the surface of the moon. Set the height equal to 0 because the height of something ON the ground is 0:
and factor that to get
t = -.129 sec and t = 18.65 sec
Since time can NEVER be negative, we know that it takes 18.65 seconds after launch for the ball to land on the surface of the moon.
Based on the given parameters about old cone, the height of the new cone is 12 inches
<h3>Equivalent ratio</h3>
Old cone:
- Radius, r = 2 inches
- Height, h = 6 inches
New cone:
- Radius, r = 4 inches
- Height, h = h
equate ratio of radius to height in old and new cone
2 : 6 = 4 : h
2/6 = 4/h
cross product
2 × h = 6 × 4
2h = 24
h = 24/2
h = 12 inches
Learn more about ratio:
brainly.com/question/2328454
#SPJ1
Its B trust i took the test and got it right
Part A. y = -3x - 2
Part B. = y = -3x - 11
Hope this helps!!
~Kiwi
