Answer:
9 1/2 ÷20
9/1 +1/2 ÷20 ( convert to an improper frqction)
18/2 + 1/2 ÷20 ( Multiply 9/1 by 2/2)
19/2 ÷20 ( add 18/2 and 1/2)
19/2 ×1/20 (dividing by 20 is the same as multiplying by 1/20)
19/40
0.475
The answer would be 1. The commutative postulate for multiplication. Hope this helped! Good luck! :)
![\qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctextit%7Bperfect%20square%20trinomial%7D%20%5C%5C%5C%5C%20%28a%5Cpm%20b%29%5E2%5Cimplies%20a%5E2%5Cpm%20%5Cstackrel%7B%5Cstackrel%7B%5Ctext%7B%5Csmall%202%7D%5Ccdot%20%5Csqrt%7B%5Ctextit%7B%5Csmall%20a%7D%5E2%7D%5Ccdot%20%5Csqrt%7B%5Ctextit%7B%5Csmall%20b%7D%5E2%7D%7D%7B%5Cdownarrow%20%7D%7D%7B2ab%7D%20%2B%20b%5E2)
the idea behind the completion of the square is simply using a perfect square trinomial, hmmm usually we do that by using our very good friend Mr Zero, 0.
if we look at the 2nd step, we have a group as x² - x, hmmm so we need a third element, which will be squared.
keeping in mind that the middle term of the perfect square trinomial is simply the product of the roots of "a" and "b", so in this case the middle term is "-x", and the 1st term is x², so we can say that
![\stackrel{middle~term}{2(\sqrt{x^2})(\sqrt{b^2})~}~ = ~~\stackrel{middle~term}{-x}\implies 2xb~~ = ~~~~ = ~~-x \\\\\\ b=\cfrac{-x}{2x} \implies b=-\cfrac{1}{2}](https://tex.z-dn.net/?f=%5Cstackrel%7Bmiddle~term%7D%7B2%28%5Csqrt%7Bx%5E2%7D%29%28%5Csqrt%7Bb%5E2%7D%29~%7D~%20%3D%20~~%5Cstackrel%7Bmiddle~term%7D%7B-x%7D%5Cimplies%202xb~~%20%3D%20~~~~%20%3D%20~~-x%20%5C%5C%5C%5C%5C%5C%20b%3D%5Ccfrac%7B-x%7D%7B2x%7D%20%5Cimplies%20b%3D-%5Ccfrac%7B1%7D%7B2%7D)
so that means that our missing third term for a perfect square trinomial is simply 1/2, now we'll go to our good friend Mr Zero, if we add (1/2)², we have to also subtract (1/2)², because all we're really doing is borrowing from Zero, so we'll be including then +(1/2)² and -(1/2)², keeping in mind that 1/4 - 1/4 = 0, so let's do that.
![-3~~ = ~~-2\left[ x^2-x+\left( \cfrac{1}{2} \right)^2 ~~ - ~~\left( \cfrac{1}{2} \right)^2\right]\implies -3=-2\left(x^2-x+\cfrac{1}{4}-\cfrac{1}{4} \right) \\\\\\ -3=-2\left(x^2-x+\cfrac{1}{4} \right)+(-2)-\cfrac{1}{4}\implies -3=-2\left(x^2-x+\cfrac{1}{4} \right)+\cfrac{1}{2} \\\\\\ -3-\cfrac{1}{2}=-2\left(x^2-x+\cfrac{1}{4} \right)\implies -\cfrac{7}{2}=-2\left(x-\cfrac{1}{2} \right)^2\implies \cfrac{7}{4}=\left(x-\cfrac{1}{2} \right)^2](https://tex.z-dn.net/?f=-3~~%20%3D%20~~-2%5Cleft%5B%20x%5E2-x%2B%5Cleft%28%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%20~~%20-%20~~%5Cleft%28%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%5Cright%5D%5Cimplies%20-3%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D-%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20-3%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%2B%28-2%29-%5Ccfrac%7B1%7D%7B4%7D%5Cimplies%20-3%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%2B%5Ccfrac%7B1%7D%7B2%7D%20%5C%5C%5C%5C%5C%5C%20-3-%5Ccfrac%7B1%7D%7B2%7D%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%5Cimplies%20-%5Ccfrac%7B7%7D%7B2%7D%3D-2%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%5Cimplies%20%5Ccfrac%7B7%7D%7B4%7D%3D%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2)
![~\dotfill\\\\ \pm\sqrt{\cfrac{7}{4}}=x-\cfrac{1}{2}\implies \cfrac{\pm\sqrt{7}}{2}=x-\cfrac{1}{2}\implies \cfrac{\pm\sqrt{7}}{2}+\cfrac{1}{2}=x \implies \cfrac{\pm\sqrt{7}+1}{2}=x](https://tex.z-dn.net/?f=~%5Cdotfill%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B%5Ccfrac%7B7%7D%7B4%7D%7D%3Dx-%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20%5Ccfrac%7B%5Cpm%5Csqrt%7B7%7D%7D%7B2%7D%3Dx-%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20%5Ccfrac%7B%5Cpm%5Csqrt%7B7%7D%7D%7B2%7D%2B%5Ccfrac%7B1%7D%7B2%7D%3Dx%20%5Cimplies%20%5Ccfrac%7B%5Cpm%5Csqrt%7B7%7D%2B1%7D%7B2%7D%3Dx)