Asked by s
An object is launched from a platform. Its height, y (in meters), x seconds after the launch is modeled by the function y=−5(x−4)2+180
𝑦
=
−
5
(
𝑥
−
4
)
2
+
180
.
What is the object’s height after 10 seconds?
What is the rate of change for the interval 5 seconds to 8 seconds?
Show your work. Please make it simple.
𝑦
=
−
5
(
𝑥
−
4
)
2
+
180
.
What is the object’s height after 10 seconds?
What is the rate of change for the interval 5 seconds to 8 seconds?
Show your work. Please make it simple.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To find the object's height after 10 seconds, we can substitute \( x = 10 \) into the height function \( y = -5(x - 4)^2 + 180 \).
### Step 1: Calculate the height after 10 seconds
Substitute \( x = 10 \):
\[
y = -5(10 - 4)^2 + 180
\]
\[
y = -5(6)^2 + 180
\]
\[
y = -5(36) + 180
\]
\[
y = -180 + 180
\]
\[
y = 0
\]
So, the object's height after 10 seconds is **0 meters**.
### Step 2: Calculate the rate of change from 5 seconds to 8 seconds
To find the rate of change over the interval from 5 seconds to 8 seconds, we first need to find the heights at these two times.
#### Height at 5 seconds:
Substitute \( x = 5 \):
\[
y = -5(5 - 4)^2 + 180
\]
\[
y = -5(1)^2 + 180
\]
\[
y = -5(1) + 180
\]
\[
y = -5 + 180
\]
\[
y = 175
\]
#### Height at 8 seconds:
Substitute \( x = 8 \):
\[
y = -5(8 - 4)^2 + 180
\]
\[
y = -5(4)^2 + 180
\]
\[
y = -5(16) + 180
\]
\[
y = -80 + 180
\]
\[
y = 100
\]
#### Step 3: Calculate the rate of change
The rate of change between 5 seconds and 8 seconds can be calculated using the formula:
\[
\text{Rate of Change} = \frac{y(8) - y(5)}{8 - 5}
\]
Where \( y(5) = 175 \) and \( y(8) = 100 \):
\[
\text{Rate of Change} = \frac{100 - 175}{8 - 5}
\]
\[
\text{Rate of Change} = \frac{-75}{3}
\]
\[
\text{Rate of Change} = -25
\]
Thus, the rate of change of the object's height from 5 seconds to 8 seconds is **-25 meters per second**. This indicates that the object is descending at a rate of 25 meters per second during this interval.
### Step 1: Calculate the height after 10 seconds
Substitute \( x = 10 \):
\[
y = -5(10 - 4)^2 + 180
\]
\[
y = -5(6)^2 + 180
\]
\[
y = -5(36) + 180
\]
\[
y = -180 + 180
\]
\[
y = 0
\]
So, the object's height after 10 seconds is **0 meters**.
### Step 2: Calculate the rate of change from 5 seconds to 8 seconds
To find the rate of change over the interval from 5 seconds to 8 seconds, we first need to find the heights at these two times.
#### Height at 5 seconds:
Substitute \( x = 5 \):
\[
y = -5(5 - 4)^2 + 180
\]
\[
y = -5(1)^2 + 180
\]
\[
y = -5(1) + 180
\]
\[
y = -5 + 180
\]
\[
y = 175
\]
#### Height at 8 seconds:
Substitute \( x = 8 \):
\[
y = -5(8 - 4)^2 + 180
\]
\[
y = -5(4)^2 + 180
\]
\[
y = -5(16) + 180
\]
\[
y = -80 + 180
\]
\[
y = 100
\]
#### Step 3: Calculate the rate of change
The rate of change between 5 seconds and 8 seconds can be calculated using the formula:
\[
\text{Rate of Change} = \frac{y(8) - y(5)}{8 - 5}
\]
Where \( y(5) = 175 \) and \( y(8) = 100 \):
\[
\text{Rate of Change} = \frac{100 - 175}{8 - 5}
\]
\[
\text{Rate of Change} = \frac{-75}{3}
\]
\[
\text{Rate of Change} = -25
\]
Thus, the rate of change of the object's height from 5 seconds to 8 seconds is **-25 meters per second**. This indicates that the object is descending at a rate of 25 meters per second during this interval.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.