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<h3>Answer:</h3><h3>D</h3><h3 /><h3>Step-by-step explanation:</h3><h3></h3><h3>In a function, an input (x) value should have only one output (y) value.</h3><h3 />
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<em>Example: (View attached image below). </em><em>Table A</em><em> is a function because each </em><em>x </em><em>value has only 1 </em><em>y</em><em> value. But </em><em>Table B</em><em> is not a function because the </em><em>x value</em><em> of </em><em>4 </em><em>has </em><em>2 y values</em><em>.</em>
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.
