X +105 x+95 solve for x
1 answer:
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Answer:
t = 0 at the start of the projection
Step-by-step explanation:
To solve this we need to find the distance between the 2 positions at any given time, then solve for the least distance
Let t be the time of the 2nd ball, so t + 1 is the time of the first ball
Let g be the gravitational acceleration, v be the horizontal velocity
the y coordinates of the first and 2nd balls
The x coordinates of the 1st and 2nd balls:
The distance between the 2 balls is
As both v and g are constant and cannot be changed, d is minimum when (2t + 1) is minimum, which happens only when t is minimum = 0
Answer:
x = - 6
y = 4
Step-by-step explanation:
y = - x/3 + 2
y + 4 = - 4x/3 => y = - 4x/3 - 4
equality
- x/3 + ³⁾2 = - 4x/3 - ³⁾4
- x + 6 = - 4x - 12
-x + 4x = - 6 - 12
3x = - 18
x = - 6
y = - (-6)/3 + 2
y = 6/3 + 2
y = 2 + 2
y = 4
