Answer:
First off, we look for which circles are open or closed.
We start with an open interval since the circle on the left is open and end with a closed interval since the circle on the right is closed.
Domain is all x values, Range is all y values
The graph shows the continous function going from -3 to 1 on the x axis.
According to the circles, this means our domain will be (-3,1].
Now, the range doesn't care about if its closed or not. So we can say the graph is on the y axis from -4 and 0. This means the range is -4<y<0
I used different notations for both just incase you need to represent your answer differently :)
-3<x<1 & (-3,1] . Range is [-4,0]. 0>y>-4 looks correct as-well.
<u>Answer:</u>
-2
<u>Step-by-step explanation:</u>
We have been given a function f(x)=\frac{-2x}{x+1} and we are asked to find the horizontal asymptote of our given function.
Recalling the rules for a horizontal asymptote:
1. If the numerator and denominator have equal degree, the horizontal asymptote will be the ratio of the leading coefficients.
2. If the polynomial of denominator has larger degree than the numerator, then the horizontal asymptote will be the x-axis or y=0.
3. If the polynomial of numerator has larger degree than denominator, then the function has no horizontal asymptote.
Here, the numerator and denominator are of the same degree. So the horizontal asymptote will be the ratio of the coefficients.
Horizontal asymptote = 
= -2
Here we must see in how many different ways we can select 2 students from the 3 clubs, such that the students <em>do not belong to the same club. </em>We will see that there are 110 different ways in which 2 students from different clubs can be selected.
So there are 3 clubs:
- Club A, with 10 students.
- Club B, with 4 students.
- Club C, with 5 students.
The possible combinations of 2 students from different clubs are
- Club A with club B
- Club A with club C
- Club B with club C.
The number of combinations for each of these is given by the product between the number of students in the club, so we get:
- Club A with club B: 10*4 = 40
- Club A with club C: 10*5 = 50
- Club B with club C. 4*5 = 20
For a total of 40 + 50 + 20 = 110 different combinations.
This means that there are 110 different ways in which 2 students from different clubs can be selected.
If you want to learn more about combination and selections, you can read:
brainly.com/question/251701
Answer:
Mizuki is here to help! The answer is 8!
Step-by-step explanation:
5 + 30 ÷ 10 =
5 + 3 =
8
Remember PEMDAS!
