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
The theater are relative to Sam 6.5 kilometers south, and 3 kilometers west.
We assume you intend
f(x) = 3x^-3
g(x) = 7x^-3
The y-value of g(x) will always be 7/3 times that of f(x). That is, g(x) > f(x).
Answer:
13) began with 34
Step-by-step explanation:
20 cards replaced and now have 37
37-20=17
17 is half of the baseball cards
17x2= 34 which is what you began with