Answer:
35 possible combinations would be there.
Step-by-step explanation:
We have total of seven students we have to choose three out of those seven
n is the total possibilities n =7 in the given case
And r is the chosen ones
Hence, r=3 in the given case.
Substituting the values of n and r in the formula we get
Using n!=n(n-1)----1
Cancel the common factor we get
possible combinations would be there.
Answer:
Option D
Step-by-step explanation:
Option A
x² - 3x² + 2x² = 0
Therefore, the given polynomial has no roots.
Option B
x² + 2x + 8
By quadratic formula,
x =
=
Therefore, roots are imaginary and for a polynomial with degree 1 or more than 1, both the roots can't be imaginary.
So the polynomial can't have exactly two roots.
Option C
99x³ - 33x + 1
Since the polynomial is of degree 3, so it will have three roots.
Therefore, the polynomial will not have exactly 2 roots.
Option D
√2x - 3x² + 7√2
By quadratic formula,
x =
x =
There are exactly two real roots.
Therefore Option D is the answer.
Option E
4x + 11x - 111
Since this polynomial is of a degree 1.
There will be only one root.