HELP:Sarah, who is 19 feet away from Justin, has a ball in her hand. Sarah throws the ball away from Justin at a speed of 5 feet
per second. It takes the ball 4 seconds to reach a third friend, Brandon. When the ball reaches Brandon, how far away is it from Justin?
1 answer:
Answer:
Justin is 1 feet far away from Brandon
Step-by-step explanation:
As shown in the diagram, we let the distance between Justin and Brandon be x
from the question
the total time taken by the ball to reach Brandson=4s
Also the total distance is travelled by the ball=19+x
while the average speed=5ft/s
By substitution we obtain
x=1 feet
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ANSWER
A) -1
EXPLANATION
The average rate of change of the given quadratic function on the interval 0 ≤ x ≤4 is the slope of the secant line connecting the points (0,f(0)) and (4,f(4)).
That is the average rate of change is:
From the graph, f(0) is 0 and f(4) is -4.
We plug in these values to obtain;
This simplifies to;
Hence the average rate of change for the given quadratic function whose graph is shown on 0≤x≤4 is -1
A) -1
EXPLANATION
The average rate of change of the given quadratic function on the interval 0 ≤ x ≤4 is the slope of the secant line connecting the points (0,f(0)) and (4,f(4)).
That is the average rate of change is:
From the graph, f(0) is 0 and f(4) is -4.
We plug in these values to obtain;
This simplifies to;
Hence the average rate of change for the given quadratic function whose graph is shown on 0≤x≤4 is -1
Answer is 3 seconds
When the bullet reaches the ground, ground being x in graph (and here its s which is = 0)
s = -16t^2 + 48t
s = 0, solve for t
0 = -16t^2 + 48t
0 = t ( -16t + 48)
0 = 16t ( - t + 3)
now you have two equation
0 = 16t and 0 = -t +3 ( you can look at the graph line touches x twice)
0 = 16 t
0 = t ( you know its false, because time = 0)
You are left with
0 = -t + 3
t = 3
It takes 3 seconds for the bullet to return to the ground.
// Hope this helps.
