X=5, should be right correct me if i’m wrong.
We can only see one graph here
Answer:
x = 21/8
Step-by-step explanation:
So 8 is being multiplied by something to get to 21
So 8x = 21
Solve for x:
-Chetan k
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:
n = -1
Step-by-step explanation:
n^2 + 5n + 1 = 3n
N^2 +2n +1 = 0 (subtract 3n from both sides)
(n+1)(n+1) = 0 (factor
n+1 =0; n= -1
n+1 = 0 ; n=-1
