I'll assume that's an exponent at the end there:
The first derivative gives the velocity. The second derivative gives the acceleration; increasing velocity is the same as positive acceleration. So we want to find when the second derivative is positive.
Let's see if we can use 
to avoid multiplying this out.
That worked; let's do it again:
That's a nice parabola. It's zero or negative for 
so positive everywhere else:
Answer: Increasing velocity when t < 1 or t > 3
Answer:
Example Quadrant 3 coordinate value (-2, -2)
Step-by-step explanation:
Coordinates are written (x, y)
Quadrant 1: both x and y are positive values. Example (2, 2)
Quadrant 2: x is negative and y is positive. Example (-2, 2)
Quadrant 3: both x and y are negative values. Example (-2, -2)
Quadrant 4: x is positive and y is negative. Example (2, -2)
Answer:
64
Step-by-step explanation:
5ft x 4ft = 20
4ft x 6ft = 24
5ft x 4ft =20
24+20+20=64
Hope this helps u
Answer:
$1.60 per pen
Step-by-step explanation:
60-30=30
30-13= 17
17-9=8
8/5= 1.6
Answer:
True.
We can see that the line goes through (0, 0), however, the line is not dotted, therefore that point is included. (0,0) also coincides with the other inequality, which means that (0, 0) is a solution.
