Answer: the last one, 31.6
For this case we must find the roots of the following equation:
![x ^ 2-7x-4 = 0](https://tex.z-dn.net/?f=x%20%5E%202-7x-4%20%3D%200)
We have to:
![x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%20%7B-b%20%5Cpm%20%5Csqrt%20%7Bb%20%5E%202-4%20%28a%29%20%28c%29%7D%7D%20%7B2%20%28a%29%7D)
Where:
![a = 1\\b = -7\\c = -4](https://tex.z-dn.net/?f=a%20%3D%201%5C%5Cb%20%3D%20-7%5C%5Cc%20%3D%20-4)
Substituting the values:
![x = \frac {- (- 7) \pm \sqrt {(- 7) ^ 2-4 (1) (- 4)}} {2 (1)}\\x = \frac {7 \pm \sqrt {49 + 16}} {2}\\x = \frac {7\pm\sqrt {65}} {2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%20%7B-%20%28-%207%29%20%5Cpm%20%5Csqrt%20%7B%28-%207%29%20%5E%202-4%20%281%29%20%28-%204%29%7D%7D%20%7B2%20%281%29%7D%5C%5Cx%20%3D%20%5Cfrac%20%7B7%20%5Cpm%20%5Csqrt%20%7B49%20%2B%2016%7D%7D%20%7B2%7D%5C%5Cx%20%3D%20%5Cfrac%20%7B7%5Cpm%5Csqrt%20%7B65%7D%7D%20%7B2%7D)
We have two roots:
![x_ {1} = \frac {7+ \sqrt {65}} {2} = 7.53\\x_ {2} = \frac {7- \sqrt {65}} {2} = - 0.53](https://tex.z-dn.net/?f=x_%20%7B1%7D%20%3D%20%5Cfrac%20%7B7%2B%20%5Csqrt%20%7B65%7D%7D%20%7B2%7D%20%3D%207.53%5C%5Cx_%20%7B2%7D%20%3D%20%5Cfrac%20%7B7-%20%5Csqrt%20%7B65%7D%7D%20%7B2%7D%20%3D%20-%200.53)
Answer:
![x_ {1} = \frac {7+ \sqrt {65}} {2} = 7.53\\x_ {2} = \frac {7- \sqrt {65}} {2} = - 0.53](https://tex.z-dn.net/?f=x_%20%7B1%7D%20%3D%20%5Cfrac%20%7B7%2B%20%5Csqrt%20%7B65%7D%7D%20%7B2%7D%20%3D%207.53%5C%5Cx_%20%7B2%7D%20%3D%20%5Cfrac%20%7B7-%20%5Csqrt%20%7B65%7D%7D%20%7B2%7D%20%3D%20-%200.53)
Given:
Two sides of a triangle are 9 and 5.
To find:
The range of possible measures for the third side.
Solution:
In a triangle, the sum of smaller sides must be greater than the largest side.
Let x be the measure of third side.
Case 1: When x is the largest side, then
![9+5>x](https://tex.z-dn.net/?f=9%2B5%3Ex)
...(i)
Case 2: When 9 is the largest side, then
![x+5>9](https://tex.z-dn.net/?f=x%2B5%3E9)
![x>9-5](https://tex.z-dn.net/?f=x%3E9-5)
...(ii)
Using (i) and (ii), we get
![4](https://tex.z-dn.net/?f=4%3Cx%3C14)
Therefore, the range of possible measures for the third side is
.
<u>Answer:</u>
![y = - \frac{1}{24} (x + 5) + 1](https://tex.z-dn.net/?f=y%20%3D%20%20-%20%20%5Cfrac%7B1%7D%7B24%7D%20%28x%20%2B%205%29%20%2B%201)
<u>Explanation</u>
The directrix y=7, is above the y-value of the focus. The parabola must will open downwards.
Such parabola has equation of the form,
![{(x - h)}^{2} = - 4p(y - k)](https://tex.z-dn.net/?f=%20%7B%28x%20-%20h%29%7D%5E%7B2%7D%20%20%3D%20-%20%204p%28y%20-%20k%29)
where (h,k) is the vertex.
The vertex is the midway from the focus to the directrix
The x-value of the vertex is x=-5 because it is on a vertical line that goes through (-5,-5).
The y-value of the vertex is
![y = \frac{ 7 + - 5}{2}](https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B%207%20%2B%20%20-%205%7D%7B2%7D%20)
![y = \frac{ 2}{2} = 1](https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B%202%7D%7B2%7D%20%20%3D%20%201)
The equation of the parabola now becomes
![{(x + 5)}^{2} = - 4p(y - 1)](https://tex.z-dn.net/?f=%7B%28x%20%20%2B%205%29%7D%5E%7B2%7D%20%20%3D%20-%20%204p%28y%20%20%20-%201%29)
p is the distance from the focus to the vertex which is p=|7-1|=6
Substitute the value of p to get:
![{(x + 5)}^{2} = - 4 \times 6(y - 1)](https://tex.z-dn.net/?f=%7B%28x%20%20%2B%205%29%7D%5E%7B2%7D%20%20%3D%20-%20%204%20%5Ctimes%206%28y%20%20%20-%201%29)
![{(x + 5)}^{2} = - 24(y - 1)](https://tex.z-dn.net/?f=%7B%28x%20%20%2B%205%29%7D%5E%7B2%7D%20%20%3D%20-%2024%28y%20%20%20-%201%29)
We solve for y to get:
![y = - \frac{1}{24} (x + 5) + 1](https://tex.z-dn.net/?f=y%20%3D%20%20-%20%20%5Cfrac%7B1%7D%7B24%7D%20%28x%20%2B%205%29%20%2B%201)