Question
A ball is thrown straight up from 7 meters above the ground. The height y (in meters) of the ball after x seconds is modeled by the equation
y=−5x2+11x+7
𝑦
=
−
5
𝑥
2
+
11
𝑥
+
7
.
What is the height of the ball after 2 seconds?
(1 point)
meters
y=−5x2+11x+7
𝑦
=
−
5
𝑥
2
+
11
𝑥
+
7
.
What is the height of the ball after 2 seconds?
(1 point)
meters
Answers
GPT-4o mini
To find the height of the ball after 2 seconds, we can substitute \( x = 2 \) into the equation:
\[
y = -5x^2 + 11x + 7
\]
Substituting \( x = 2 \):
\[
y = -5(2)^2 + 11(2) + 7
\]
Calculating \( (2)^2 \):
\[
y = -5(4) + 11(2) + 7
\]
Calculating \( -5(4) \) and \( 11(2) \):
\[
y = -20 + 22 + 7
\]
Now, combine the values:
\[
y = -20 + 22 + 7 = 2 + 7 = 9
\]
Therefore, the height of the ball after 2 seconds is \( 9 \) meters.
\[
y = -5x^2 + 11x + 7
\]
Substituting \( x = 2 \):
\[
y = -5(2)^2 + 11(2) + 7
\]
Calculating \( (2)^2 \):
\[
y = -5(4) + 11(2) + 7
\]
Calculating \( -5(4) \) and \( 11(2) \):
\[
y = -20 + 22 + 7
\]
Now, combine the values:
\[
y = -20 + 22 + 7 = 2 + 7 = 9
\]
Therefore, the height of the ball after 2 seconds is \( 9 \) meters.