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.
<span>First note that y cannot equal -7 or -1 as division by zero is undefined. So that if it comes up in our solution we know that it is not a real solution.
[(y+4)/(y+7)]+1=(y+3)/(y+1) make the left side have a common denominator
(y+4+y+7)/(y+7)=(y+3)/(y+1) combine like terms on left side
(2y+11)/(y+7)=(y+3)/(y+1) cross multiply
(2y+11)(y+1)=(y+3)(y+7) expand
2y^2+13y+11=y^2+10y+21 subtract y^2 from both sides
y^2+13y+11=10y+21 subtract 10y from both sides
y^2+3y+11=21 subtract 21 from both sides
y^2+3y-10=0 expand so factoring is easier
y^2-2y+5y-10=0 factor
y(y-2)+5(y-2)=0 which is equal to
(y+5)(y-2)=0
y=-5 or 2</span>
Answer:
y=2/3x+2
Step-by-step explanation:
Answer:
D
Step-by-step explanation: :)
