Answer:
jump discontinuity at x = 0; point discontinuities at x = –2 and x = 8
Step-by-step explanation:
From the graph we can see that there is a whole in the graph at x=-2.
This is referred to as a point discontinuity.
Similarly, there is point discontinuity at x=8.
We can see that both one sided limits at these points are equal but the function is not defined at these points.
At x=0, there is a jump discontinuity. Both one-sided limits exist but are not equal.
Answer:
See problem 1 below.
Step-by-step explanation:
Problem 1.
1. <A, <D
Given
2. <1, <3
Given
3. BC is congruent to BC
Congruence of segments is reflexive.
4. ABC, DBC
AAS
1.4+2.7=4.1 is true the others are false
Yo sup??
To find velocity just differentiate the function r once wrt to t
v=dr/dt=(9t^2-1)i+(4t+2)j)
and to find acceleration differentiate v wrt to t
a=dv/dt=18ti+4j
Hope this helps.
