Answer:
24
6
1/12
Step-by-step explanation:
a) This is a counting problem 4*3*2*1 = 24 different ways.
b) 1 * 2 * 1 * 1 = 2 ways
This is one you should try.
Jim Kat Larry Kim
Jim Larry Kat Kim
You can't get another way.
c)
There are 2 ways that Jim can arrive first and Kim Last shown in b
2/24 = 1/12
Answer: {(x + 2), (x - 1), (x - 3)}
Step-by-step explanation:
Presented symbolically, we have:
x^3 - 2x^2 - 5x + 6
Synthetic division is very useful for determining roots of polynomials. Once we have roots, we can easily write the corresponding factors.
Write out possible factors of 6: {±1, ±2, ±3, ±6}
Let's determine whether or not -2 is a root. Set up synthetic division as follows:
-2 / 1 -2 -5 6
-2 8 -6
-----------------------
1 -4 3 0
since the remainder is zero, we know for sure that -2 is a root and (x + 2) is a factor of the given polynomial. The coefficients of the product of the remaining two factors are {1, -4, 3}. This trinomial factors easily into {(x -1), (x - 3)}.
Thus, the three factors of the given polynomial are {(x + 2), (x - 1), (x - 3)}
There's nothing preventing us from computing one integral at a time:



Expand the integrand completely:

Then

Answer:
adenosine triphosphate.
Step-by-step explanation: