Answer:
![f(x) = (x + 19)(x^2 + 1)](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%28x%20%2B%2019%29%28x%5E2%20%2B%201%29)
Step-by-step explanation:
Complex numbers:
The following relation is important for complex numbers:
![i^2 = -1](https://tex.z-dn.net/?f=i%5E2%20%3D%20-1)
Zeros of a function:
Given a polynomial f(x), this polynomial has roots
such that it can be written as:
, in which a is the leading coefficient.
Has zeros −19 and −i
If -i is a zero, its conjugate i is also a zero. So
![f(x) = a(x - (-19))(x - (-i))(x - i) = a(x+19)(x+i)(x-i) = a(x+19)(x^2 - i^2) = a(x + 19)(x^2 + 1)](https://tex.z-dn.net/?f=f%28x%29%20%3D%20a%28x%20-%20%28-19%29%29%28x%20-%20%28-i%29%29%28x%20-%20i%29%20%3D%20a%28x%2B19%29%28x%2Bi%29%28x-i%29%20%3D%20a%28x%2B19%29%28x%5E2%20-%20i%5E2%29%20%3D%20a%28x%20%2B%2019%29%28x%5E2%20%2B%201%29)
Output of 40 when x=1
This means that when
. We use this to find the leading coefficient a. So
![f(x) = a(x + 19)(x^2 + 1)](https://tex.z-dn.net/?f=f%28x%29%20%3D%20a%28x%20%2B%2019%29%28x%5E2%20%2B%201%29)
![40 = a(20)(2)](https://tex.z-dn.net/?f=40%20%3D%20a%2820%29%282%29)
![40a = 40](https://tex.z-dn.net/?f=40a%20%3D%2040)
![a = 1](https://tex.z-dn.net/?f=a%20%3D%201)
The polynomial is:
![f(x) = (x + 19)(x^2 + 1)](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%28x%20%2B%2019%29%28x%5E2%20%2B%201%29)
The other endpoint is (31, -4)
<em><u>Solution:</u></em>
Given that midpoint (23,- 10), endpoint (15.-16)
To find: The other endpoint
<em><u>The formula for midpoint is given as:</u></em>
![\text {For two points }\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right), \text { the midpoint }(x, y) \text { is given as: }](https://tex.z-dn.net/?f=%5Ctext%20%7BFor%20two%20points%20%7D%5Cleft%28x_%7B1%7D%2C%20y_%7B1%7D%5Cright%29%20%5Ctext%20%7B%20and%20%7D%5Cleft%28x_%7B2%7D%2C%20y_%7B2%7D%5Cright%29%2C%20%5Ctext%20%7B%20the%20midpoint%20%7D%28x%2C%20y%29%20%5Ctext%20%7B%20is%20given%20as%3A%20%7D)
![m(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)](https://tex.z-dn.net/?f=m%28x%2C%20y%29%3D%5Cleft%28%5Cfrac%7Bx_%7B1%7D%2Bx_%7B2%7D%7D%7B2%7D%2C%20%5Cfrac%7By_%7B1%7D%2By_%7B2%7D%7D%7B2%7D%5Cright%29)
Here in this problem,
![m(x, y) = (23, -10)](https://tex.z-dn.net/?f=m%28x%2C%20y%29%20%3D%20%2823%2C%20-10%29)
![(x_1, y_1) = (15, -16)](https://tex.z-dn.net/?f=%28x_1%2C%20y_1%29%20%3D%20%2815%2C%20-16%29)
![(x_2, y_2) = ?](https://tex.z-dn.net/?f=%28x_2%2C%20y_2%29%20%3D%20%3F)
<em><u>Substituting the values in formula we get,</u></em>
![(23,-10)=\left(\frac{15+x_{2}}{2}, \frac{-16+y_{2}}{2}\right)](https://tex.z-dn.net/?f=%2823%2C-10%29%3D%5Cleft%28%5Cfrac%7B15%2Bx_%7B2%7D%7D%7B2%7D%2C%20%5Cfrac%7B-16%2By_%7B2%7D%7D%7B2%7D%5Cright%29)
On comparing both sides we get,
![23=\frac{15+x_{2}}{2} \text { and }-10=\frac{-16+y_{2}}{2}](https://tex.z-dn.net/?f=23%3D%5Cfrac%7B15%2Bx_%7B2%7D%7D%7B2%7D%20%5Ctext%20%7B%20and%20%7D-10%3D%5Cfrac%7B-16%2By_%7B2%7D%7D%7B2%7D)
![\begin{array}{l}{46=15+x_{2} \text { and }-20=-16+y_{2}} \\\\ {x_{2}=46-15 \text { and } y_{2}=-20+16} \\\\ {x_{2}=31 \text { and } y_{2}=-4}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B46%3D15%2Bx_%7B2%7D%20%5Ctext%20%7B%20and%20%7D-20%3D-16%2By_%7B2%7D%7D%20%5C%5C%5C%5C%20%7Bx_%7B2%7D%3D46-15%20%5Ctext%20%7B%20and%20%7D%20y_%7B2%7D%3D-20%2B16%7D%20%5C%5C%5C%5C%20%7Bx_%7B2%7D%3D31%20%5Ctext%20%7B%20and%20%7D%20y_%7B2%7D%3D-4%7D%5Cend%7Barray%7D)
Thus the other endpoint is (31, - 4)
Answer: two
While you are dividing by a decimal you play a zero until it is the same place a the dividend
The points L(10,9)L(10,9), M(10,-5)M(10,-5), N(-1,-5)N(-1,-5), and O(-1,9)O(-1,9) form rectangle LMNOLMNO. Which point is halfwa
Inessa [10]
You are trying to find the halfway point between OO and NN.
OO: (-1,9) NN: (-1,5)
The x-coordinate does not change, because in both instances it is -1. The y-coordinate is (9-5)/2 AWAY from each point. AKA the number that is equidistant from 5 and 9 (7).