x² + 9x + 1
let's start by grouping the ones with the same variable
(x² + 9x) + 1
(x² + 9x + [?]²) + 1
so, we seem to be missing a value there, to get a perfect square trinomial, let's recall a perfect square trinomial has a middle term that is 2 * "other two values", namely
so, the middle term on this group will be 9x.
we know 2*x*[?] = 9x, so then
so that's our mystery felllow.
now, let's bear in mind that we'll be borrowing from our very good friend Mr Zero, 0, so if we add (9/2)², we also have to subtract (9/2)².
![\bf \left[x^2+9x+\left( \cfrac{9}{2} \right)^2 - \left( \cfrac{9}{2} \right)^2 \right]+1\implies \left[x^2+9x+\left( \cfrac{9}{2} \right)^2 \right]+1- \left( \cfrac{9}{2} \right)^2 \\\\\\ \left( x+\cfrac{9}{2} \right)^2+1-\cfrac{81}{4}\implies \left( x+\cfrac{9}{2} \right)^2-\cfrac{77}{4}](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%5Bx%5E2%2B9x%2B%5Cleft%28%20%5Ccfrac%7B9%7D%7B2%7D%20%5Cright%29%5E2%20-%20%5Cleft%28%20%5Ccfrac%7B9%7D%7B2%7D%20%5Cright%29%5E2%20%5Cright%5D%2B1%5Cimplies%20%5Cleft%5Bx%5E2%2B9x%2B%5Cleft%28%20%5Ccfrac%7B9%7D%7B2%7D%20%5Cright%29%5E2%20%5Cright%5D%2B1-%20%5Cleft%28%20%5Ccfrac%7B9%7D%7B2%7D%20%5Cright%29%5E2%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%20x%2B%5Ccfrac%7B9%7D%7B2%7D%20%5Cright%29%5E2%2B1-%5Ccfrac%7B81%7D%7B4%7D%5Cimplies%20%5Cleft%28%20x%2B%5Ccfrac%7B9%7D%7B2%7D%20%5Cright%29%5E2-%5Ccfrac%7B77%7D%7B4%7D)
5) 19x - 6 = 8 + 9x +66°
(subtract 9x)
10x - 6 = 74°
(add 6)
10x = 80°
(divide by 10)
x = 8°
Then you put x into m< NMK
8 + 9(8°)
8 + 72°
m<NMK = 80°
6) because m<DER and m<REF are identical equations you can just divide m<DEF (96°) in half which is 48°
7) vertical pairs because they're right across from each other
Step-by-step explanation:
<h2>A set which is not finite is called an infinite set. Example: A set of all whole numbers. A = {0,1,2,3,4,5,6,7,8,9……}</h2>
It looks like you have the domain confused for the range! You can think of the domain as the set of all "inputs" for a function (all of the x values which are allowed). In the given function, we have no explicit restrictions on the domain, and no situations like division by 0 or taking the square root of a negative number that would otherwise put limits on it, so our domain would simply be the set of all real numbers, R. Inequality notation doesn't really use ∞, so you could just put an R to represent the set. In set notation, we'd write
and in interval notation,
The <em>range</em>, on the other hand, is the set of all possible <em>outputs</em> of a function - here, it's the set of all values f(x) can be. In the case of quadratic equations (equations with an x² term), there will always be some minimum or maximum value limiting the range. Here, we see on the graph that the maximum value for f(x) is 3. The range of the function then includes all values less than or equal to 3. As in inequality, we can say that
,
in set notation:
(this just means "f(x) is a real number less than or equal to 3")
and in interval notation:
![(-\infty,3]](https://tex.z-dn.net/?f=%20%28-%5Cinfty%2C3%5D%20)
(original - new)/original = discount
65-42.25)/65 = discount
22.75/65= discount
.35 = discount
multiply by 100 to put in percent form
35 percent
