We have x^2 + 2 · x · (11/2) + (11/2)^2 = - 24 + (11/2)^2;
Then, ( x + 11/2 )^2 = -24 + 121/4;
( x + 11/2 )^2 + 96/4 - 121/4 = 0;
( x + 11/2 )^2 - 25 / 4 = 0;
( x + 11/2 )^2 - (5/2)^2 = 0;
( x + 11/2 - 5/2)·( x + 11/2 + 5/2 ) = 0;
( x + 6/2 )·( x + 16/2 ) = 0;
( x + 3 )· ( x + 8 ) = 0;
x = - 3 or x = -8;
The first choice is the correct answer.
Only selections B and D give a maximum height of 13 at t=3. However, both of those functions have the height be -5 at t=0, meaning the ball was served from 5 ft below ground. This does not seem like an appropriate model.
We suspect ...
• the "correct" answers are probably B and D
• whoever wrote the problem wasn't paying attention.
You are given two points that could be plotted on a graph (1,-5) and (4,1). The formula for slope is y2-y1/x2 -x1
When plotted into the points it would look like 1-(-5)/4-1= 6/3
6/3=2. So, the slope for these two points is 2.
Answer:
a relation and a function
= A
Step-by-step explanation:
Move all terms not containing x to the right side of the equation.
Exact Form:
x=15/4
Decimal Form:
x=3.75
Mixed Number Form:
x=3 3/4
hope this helps