The answer is 3.16
explanation:
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Can you give me the problem
The equation to calculate how many ways that "r" things can be chosen from a larger set of "n" things is:n! / [r! * (n-r)!] =
12! / (5! * 7!) =
12 * 11 * 10 * 9 * 8 * 7! / (5! * 7!) =12 * 11 * 10 * 9 * 8 / 5 * 4 * 3 * 2 =11 * 9 * 8 =792 ways
http://www.1728.org/combinat.htm
Answer:
- zeros: x = 1, x = 3
- y = (x -1)(x -3)
Step-by-step explanation:
The zeros are the x-values where the graph crosses the x-axis (y=0; That's why it is called a "zero.") The graph crosses y=0 at x=1 and x=3.
A factor of the polynomial is zero at each zero. Hence the factorization is ...
y = (x -1)(x -3)
The first factor is zero at x=1; the second factor is zero at x=3.
If you would like to know what is f(x), you can find this using the following steps:
f(x) = ?
2 * f(x) + f(1-x) = x^2
2 * f(x) = x^2 - f(1-x) /2
f(x) = (x^2 - f(1-x)) / 2
f(x) = x^2 / 2 - f(1-x) / 2
The correct result would be f(x) = x^2 / 2 - f(1-x) / 2.
