The height of the ball 2 seconds after being thrown is 20.8m
In order to get the required quadratic function, we need to write the given information as coordinate.
- If the time the ball was initially thrown in the air, it was 7. 5 meters off the ground, this is written in a coordinate form as (0, 7.5)
- If after 1. 25 seconds, the ball was at its maximum height of 15.3125 meters in a coordinate form as (1.25, 15.3125)
Get the slope of the equation;
Get the y-intercept "b"
Recall that y = mx + b
Get the required equation:
Next is to get the height of the ball 2 seconds after being thrown
Hence the height of the ball 2 seconds after being thrown is 20.8m
Learn more on functions here: brainly.com/question/1214333
Answer:
slope = - 
Step-by-step explanation:
calculate the slope m using the slope formula
m = 
with (x₁, y₁ ) = (3, 2 ) and (x₂, y₂ ) = (- 3, 4 )
m = 
= 
= -
Answer:
I don't know that is hard
Step-by-step explanation:
You said this was an easy math question.
Divide 2.65 by 0.25
So 2.65/0.25= 10.6
So 10.6 ounces
Answer:
Step-by-step explanation:
If the position function is
and we are looking for time when the height is 0, we sub in a 0 for h(t) and solve for t:
and the easiest way to do this is to factor by taking the GCF of -40t:
0 = -40t(t - 5) and by the Zero Product Property,
-40t = 0 or t - 5 = 0. Solving for t, we get
t = 0 (which is before the object is launched) and
t = 5 (which is how long it takes the object to go from the ground, up to its max height, and then back to the ground again).
Your choice is c) 5
