Answer:
4(2x^2+5xy-3y)
Step-by-step explanation:
Part A
To graph this, we plot open holes at -2 and 3 on the number line. Shade between these open holes to represent values between -2 and 3, but we don't include the endpoints.
See figure 1 below.
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Part B
Solve the first inequality to get
Now solve the second inequality
We have 
The graph will have an open hole at -3 and a closed/filled in circle at 1. We shade everywhere but the region between these marked values. The left portion in blue represents stuff smaller than -3; the right portion in red represents values equal to 1 or larger.
See figure 2 below.
It would be (3,0) and (0,3) respectively.
Answer:
a) V_avg = 32 ft/s
b) V_avg = -30 ft/s
c) See explanation
Step-by-step explanation:
Given:
- The displacement function is given:
S(t) = -16t^2 + 64t + 80
Find:
a. Find the balls average velocity between the time it was thrown and 2 seconds later
b. Find the ball’s average velocity between 2 and 4 seconds after it was thrown
c. What do the signs in your answers to parts (a) and (b) mean in terms of the direction of the ball’s motion?
Solution:
- The average velocity of an object is the displacement over time, and it is expressed as follows:
V_avg = ( S_2 - S_1 ) / dt
- Evaluate S(2), S(0), and S(4):
S(2) = -16*2^2 +64*2 +80
S(2) = 144 ft
S(0) = -16*0^2 +64*0 +80
S(0) = 80 ft
S(4) = -16*4^2 +64*4 +80
S(4) = 80 ft
- For part a. Compute V_avg:
V_avg = ( S(2) - S(0) ) / 2
V_avg = (144 - 80) / 2
V_avg = 32 ft/s
- For part b. Compute V_avg:
V_avg = ( S(4) - S(2) ) / 2
V_avg = (80 - 140) / 2
V_avg = -30 ft/s
- We see that the signs of the Velocity have changed. The coordinate axis up against gravity was chosen to positive. While a negative access is in favor of the gravity. Hence, from the signs we can see that the ball is going down towards ground in motion from t = 2 s to t = 4 s.
