Select the statement that best justifies the conclusion based on the given information.
1 answer:
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Explanation:
Each row in Pascal's triangle is a listing of the values of nCk = n!/(k!(n-k)!) for some fixed n and k in the range 0 to n. nCk is <em>the number of combinations of n things taken k at a time</em>.
If you consider what happens when you multiply out the product (a +b)^n, you can see where the coefficients nCk come from. For example, consider the cube ...
(a +b)^3 = (a +b)(a +b)(a +b)
The highest-degree "a" term will be a^3, the result of multiplying together the first terms of each of the binomials.
The term a^b will have a coefficient that reflects the sum of all the ways you can get a^b by multiplying different combinations of the terms. Here they are ...
- (a +_)(a +_)(_ +b) = a·a·b = a^2b
- (a +_)(_ +b)(a +_) = a·b·a = a^2b
- (_ +b)(a +_)(a +_) = b·a·a = a^2b
Adding these three products together gives 3a^2b, the second term of the expansion.
For this cubic, the third term of the expansion is the sum of the ways you can get ab^2. It is essentially what is shown above, but with "a" and "b" swapped. Hence, there are 3 combinations, and the total is 3ab^2.
Of course, there is only one way to get b^3.
So the expansion of the cube (a+b)^3 is ...
(a +b)^3 = a^3 + 3a^2b +3ab^2 +b^3 . . . . . with coefficients 1, 3, 3, 1 matching the 4th row of Pascal's triangle.
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In short, the values in Pascal's triangle are the values of the number of combinations of n things taken k at a time. The coefficients of a binomial expansion are also the number of combinations of n things taken k at a time. Each term of the expansion of (a+b)^n is of the form (nCk)·a^(n-k)·b^k for k =0 to n.
F(x) = 4x - 1
g(x) = 2x² + 3
1. (f + g)(x) = (4x - 1) + (2x² + 3)
(f + g)(x) = 2x² + 4x + (-1 + 3)
(f + g)(x) = 2x² + 4x + 2
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
2. (f - g)(x) = (4x + 1) - (2x² + 3)
(f - g)(x) = 4x + 1 - 2x² - 3
(f - g)(x) = -2x² + 4x + 1 - 3
(f - g)(x) = -2x² + 4x - 2
Domain: {x|-∞ < x < ∞}, (-∞, ∞)
3. (g - f)(x) = (2x² + 3) - (4x - 1)
(g - f)(x) = 2x² + 3 - 4x + 1
(g - f)(x) = 2x² - 4x + 3 + 1
(g - f)(x) = 2x² - 4x + 4
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
4. (f · g)(x) = (4x + 1)(2x² + 3)
(f · g)(x) = 4x(2x² + 3) + 1(2x² + 3)
(f · g)(x) = 4x(2x²) + 4x(3) + 1(2x²) + 1(3)
(f · g)(x) = 8x³ + 12x + 2x² + 3
(f · g)(x) = 8x³ + 2x² + 12x + 3
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
5.
Domain: 2x² + 3 ≠ 0
- 3 - 3
2x² ≠ 0
2 2
x² ≠ 0
x ≠ 0
(-∞, 0) ∨ (0, ∞)
6.
Domain: 4x - 1 ≠ 0
+ 1 + 1
4x ≠ 0
4 4
x ≠ 0
(-∞, 0) ∨ (0, ∞)
g(x) = 2x² + 3
1. (f + g)(x) = (4x - 1) + (2x² + 3)
(f + g)(x) = 2x² + 4x + (-1 + 3)
(f + g)(x) = 2x² + 4x + 2
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
2. (f - g)(x) = (4x + 1) - (2x² + 3)
(f - g)(x) = 4x + 1 - 2x² - 3
(f - g)(x) = -2x² + 4x + 1 - 3
(f - g)(x) = -2x² + 4x - 2
Domain: {x|-∞ < x < ∞}, (-∞, ∞)
3. (g - f)(x) = (2x² + 3) - (4x - 1)
(g - f)(x) = 2x² + 3 - 4x + 1
(g - f)(x) = 2x² - 4x + 3 + 1
(g - f)(x) = 2x² - 4x + 4
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
4. (f · g)(x) = (4x + 1)(2x² + 3)
(f · g)(x) = 4x(2x² + 3) + 1(2x² + 3)
(f · g)(x) = 4x(2x²) + 4x(3) + 1(2x²) + 1(3)
(f · g)(x) = 8x³ + 12x + 2x² + 3
(f · g)(x) = 8x³ + 2x² + 12x + 3
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
5.
Domain: 2x² + 3 ≠ 0
- 3 - 3
2x² ≠ 0
2 2
x² ≠ 0
x ≠ 0
(-∞, 0) ∨ (0, ∞)
6.
Domain: 4x - 1 ≠ 0
+ 1 + 1
4x ≠ 0
4 4
x ≠ 0
(-∞, 0) ∨ (0, ∞)