A jump discontinuity occurs when the limits as x approaches a number from the left and right are not equal. Basically, the graph "jumps" from one number to another at that x value.
A point discontinuity occurs when limits as x approaches a number from the left and right are equal, but the actual value of f(x) at x is not equal to the limit. Basically, a point is missing and there is a "hole" in the graph at that x value.
Looking at your graph, you can see that at x=0, the graph "jumps" from a value of 2 as the graph approaches x=0 from the left to a value of 1 as the graph approaches x=0 from the right. That means there is a jump discontinuity at x=0.
You can also see that there is a "hole" in the graph at x=-2 and x=8 as seen by the open circle. There is no hole at x=3 because the circle is filled in. That means there is a point discontinuity at x=-2 and x=8.
Your answer is B) jump discontinuity at x=0; point discontinuities at x=-2 and x=8.
Answer:
The options 2x² + x² + x = 30 and 9x + 3x² = 14 + x + 1, and -x² + 4x + 7 = - x² - 9
Step-by-step explanation:
After being rearranged and simplified, which of the following equations could be solved using the quadratic formula?
The answers are:
2x² + x² + x = 30
9x + 3x² = 14 + x + 1
-x² + 4x + 7 = - x² - 9
Number 1 and 2 bc when y simplify you get x+5
Answer:
- 2 ≤ n < 8
Step-by-step explanation:
The closed circle at - 2 indicates that n can equal - 2
The open circle at 8 indicates that n cannot equal 8
Otherwise n can be all value between - 2 and 8 including - 2, thus
- 2 ≤ n < 8
Eugene can buy 19 oranges because 146.11 divided by 7.69 is 19
