I need help with this question
1 answer:
You might be interested in
Answer:
1) Please find attached, the graph of the function
The domain of the function is 0 ≤ x ≤ 30
The range of the function is 330 ≥ x ≥ 0
2) The slope of the given equation, shows that the elevation decreases, by (-)11 feet for each increase in time by one second
3) The y-intercept gives the initial elevation of the balloon at time t = 0 seconds
The x-intercept gives the final elevation of the balloon when it lands at time t = 30 seconds
Step-by-step explanation:
1) The function modelling the height of the balloon is h(x) = -11x + 330
Where;
x = The time of elevation of the balloon
The data for the graphing derived from Microsoft Excel are as follows
x h(x)
0 330
5 275
10 220
15 165
20 110
25 55
30 0
Please find attached, the graph of the function
The domain of the function is 0 ≤ x ≤ 30
The range of the function is 330 ≥ x ≥ 0
Comparing the given function to the general equation of a straight line, y = m·x + c, we have that the slope, m = -11
2) The slope of a straight line graph gives the rate of change of the dependent variable, per unit change in the independent variable
Therefore, the slope of the given equation, -11 gives the rate of change of the function per unit increase in the independent variable x
Therefore, the elevation decreases, by (-)11 feet for each increase in time by one second
3) The y-intercept gives the initial elevation of the balloon at time t = 0 seconds
The x-intercept gives the final elevation of the balloon when it lands at time t = 30 seconds
