C is the answer. By only looking at the domain (the stuff on the right in the functions), you can see that the first equation has a hole at x=2, meaning it does not have a value here at x=2. When x<2, there are values for the 1st function. When x>2, there are values for the second function, as well as when x=2
A ray is basically a line that has a "fixed point" on one end (this would be represented by a shaded circle on one end) and the opposite end goes to infinity (this would be represented by an arrow). The first image below is an example of a ray.
When writing rays out you must write it like in the third image below. Note that there is a "ray" on top of the "letters". The letter that signifies the end point (In the example below this would be "A") is under the side that has NO arrow symbol, and the letter that signifies the continues end is under the side of the ray that HAS the arrow. In other words the "formula" for writing out rays is:
End point; infinite end
We must find four rays with the endpoint E. This means that the "fixed point" must be E and the other end must continue indefinitely.
In the second image below I have colored each ray with the end point E in a different color.
First we have the ray EA (I don't know how to make the ray symbol on here so just assume that above EA the arrow looks like so: --->)
Next is ray EG (again the symbol above EG is --->)
Next is the ray ED (symbol above ED is --->)
Finally EF is a ray with the end point E (symbol above EF is --->)
When looking at the answer choices the only two that even use the correct symbol is B and C.
The issue with option C is that the end point (E) doesn't come first when they wrote it out.
That makes the answer B!
Hope this helped!
~Just a girl in love with Shawn Mendes
~Just a girl in love with Shawn Mendes
Answer: False
Step-by-step explanation:
Hope this helps
F(-8)=8^2-3(-8)
=64+24
=88
Hope this can help.
Answer:

Step-by-step explanation:
We have been an integral
. We are asked to find the general solution for the given indefinite integral.
We can rewrite our given integral as:


Now, we will apply the sum rule of integrals as:


Using common integral
, we will get:

Now, we will use power rule of integrals as:




We know that integral of a constant is equal to constant times x, so integral of 1 would be x.

Therefore, our required integral would be
.