Answer:
The number of different combinations of three students that are possible is 35.
Step-by-step explanation:
Given that three out of seven students in the cafeteria line are chosen to answer a survey question.
The number of different combinations of three students that are possible is given as:
7C3 (read as 7 Combination 3)
xCy (x Combination y) is defines as
x!/(x-y)!y!
Where x! is read as x - factorial or factorial-x, and is defined as
x(x-1)(x-2)(x-3)...2×1.
Now,
7C3 = 7!/(7 - 3)!3!
= 7!/4!3!
= (7×6×5×4×3×2×1)/(4×3×2×1)(3×2×1)
= (7×6×5)/(3×2×1)
= 7×5
= 35
Therefore, the number of different combinations of three students that are possible is 35.
Answer:
thw answer is -3/2 and in decimal form its -1.5
Step-by-step explanation:
hoped I helped:)
X^3-7x^2-5x+35
= x^2(x-7) - 5 (x-7) now by definition
=(x^2-5)(x-7)
hope this helps
Answer:
and 
Step-by-step explanation:
We are given the term 
.
<em>It is known that, 'A monomial is an algebraic expression consisting of one term'.</em>
So, we add the given options to .
1. 
2. 
3. 
4. 
5. 
6. 
7. 
As, we can see that only option 2 and 6 results in an expression having one term.
Thus, adding and to results in a monomial.
