How do you do this question?
1. A speed versus time graph shows the segment from (3, 4) to (8,4). The units for the vertical axis are m/s. If the object whose speed was being graphed is 8 m away from the point from which measurements are made at t= 3s, how far away is the object at t = 8s?
2. Determine the maximum and minimum values for y = 250(0.90)^x for 0<x<6
1. I can no see your graph but will assume that velocity remains 4 m/s from t = 3 to t = 8 seconds. In that case, the object moves 4 m/s *5 s = 20 m farther during that interval. That would put it at 28 m from the "point from which measurements are made", at t = 8 s.
2. y(x) is a monotonically decreasing function over that interval.
0.90^0 = 1
0.90^6 = 0.531441
y will be at its highest value at x = 0 and at its lowest value at x = 6.
1. To determine how far away the object is at t=8s, we need to find the distance traveled by the object during that time period.
- Given that the speed versus time graph shows a segment from (3, 4) to (8, 4), it means that the speed of the object remains constant at 4 m/s during this time interval.
- The formula for distance traveled is given by distance = speed × time. Considering that the speed is constant at 4 m/s and the time interval is 8s - 3s = 5s, we can calculate the distance as follows:
distance = 4 m/s × 5s = 20 meters.
Therefore, the object is 20 meters away from the point of measurement at t=8s.
2. To find the maximum and minimum values for y = 250(0.90)^x for 0 < x < 6, we can evaluate the expression for different values of x within the given range.
- Let's start by substituting x=0 into the equation:
y = 250(0.90)^0 = 250(1) = 250.
- Next, substitute x=6 into the equation:
y = 250(0.90)^6 ≈ 250(0.531441) ≈ 132.86.
Therefore, the maximum value of y is approximately 132.86, and the minimum value of y is 250.
To answer the given questions, we’ll go step by step:
1. A speed versus time graph shows the segment from (3, 4) to (8,4). The units for the vertical axis are m/s. If the object whose speed was being graphed is 8 m away from the point at t=3s, we can determine how far away it is at t=8s.
To find the distance, we need to calculate the area under the speed-time graph over the given time interval. Since the speed is constant at 4 m/s, the graph represents a rectangle.
The formula to find the area of a rectangle is:
Area = base × height
In this case, the base is the time interval, which is (8 - 3) = 5 seconds, and the height is the speed, which is 4 m/s.
So, the area is:
Area = 5 seconds × 4 m/s = 20 meters
Therefore, the object is 20 meters away from the starting point at t=8s.
2. To determine the maximum and minimum values for y = 250(0.90)^x for 0 < x < 6, we need to evaluate the function for different values of x within this range and observe the output.
Let's substitute various values of x into the equation and see what we get:
For x = 0:
y = 250(0.90)^0 = 250(1) = 250
For x = 1:
y = 250(0.90)^1 = 250(0.90) = 225
For x = 2:
y = 250(0.90)^2 = 250(0.90 * 0.90) = 202.5
For x = 3:
y = 250(0.90)^3 ≈ 182.25
For x = 4:
y = 250(0.90)^4 ≈ 164.03
For x = 5:
y = 250(0.90)^5 ≈ 147.63
So, the minimum value occurs at x = 5, where y ≈ 147.63, and the maximum value occurs at x = 0, where y = 250.
Therefore, the minimum value for y is approximately 147.63, and the maximum value for y is 250 within the given range 0 < x < 6.