For this case we have the following quadratic equation:
![2x ^ 2-4x + 9 = 0](https://tex.z-dn.net/?f=2x%20%5E%202-4x%20%2B%209%20%3D%200)
The solutions are given by:
![x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}](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%7B2a%7D)
We have to:
![a = 2\\b = -4\\c = 9](https://tex.z-dn.net/?f=a%20%3D%202%5C%5Cb%20%3D%20-4%5C%5Cc%20%3D%209)
Substituting:
![x = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (2) (9)}} {2 (2)}\\x = \frac {4 \pm \sqrt {16-72}} {4}\\x = \frac {4 \pm \sqrt {-56}} {4}\\x = \frac {4 \pm \sqrt {-1 * 56}} {4}\\x = \frac {4 \pmi \sqrt {2 ^ 2 * 14}} {4}\\x = \frac {4 \pm2i \sqrt {14}} {4}\\x = \frac {2 \pm i\sqrt {14}} {2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%20%7B-%20%28-%204%29%20%5Cpm%20%5Csqrt%20%7B%28-%204%29%20%5E%202-4%20%282%29%20%289%29%7D%7D%20%7B2%20%282%29%7D%5C%5Cx%20%3D%20%5Cfrac%20%7B4%20%5Cpm%20%5Csqrt%20%7B16-72%7D%7D%20%7B4%7D%5C%5Cx%20%3D%20%5Cfrac%20%7B4%20%5Cpm%20%5Csqrt%20%7B-56%7D%7D%20%7B4%7D%5C%5Cx%20%3D%20%5Cfrac%20%7B4%20%5Cpm%20%5Csqrt%20%7B-1%20%2A%2056%7D%7D%20%7B4%7D%5C%5Cx%20%3D%20%5Cfrac%20%7B4%20%5Cpmi%20%5Csqrt%20%7B2%20%5E%202%20%2A%2014%7D%7D%20%7B4%7D%5C%5Cx%20%3D%20%5Cfrac%20%7B4%20%5Cpm2i%20%5Csqrt%20%7B14%7D%7D%20%7B4%7D%5C%5Cx%20%3D%20%5Cfrac%20%7B2%20%5Cpm%20i%5Csqrt%20%7B14%7D%7D%20%7B2%7D)
Answer:
![x_ {1} = \frac {2 + i \sqrt {14}} {2}\\x_ {2} = \frac {2-i \sqrt {14}} {2}](https://tex.z-dn.net/?f=x_%20%7B1%7D%20%3D%20%5Cfrac%20%7B2%20%2B%20i%20%5Csqrt%20%7B14%7D%7D%20%7B2%7D%5C%5Cx_%20%7B2%7D%20%3D%20%5Cfrac%20%7B2-i%20%5Csqrt%20%7B14%7D%7D%20%7B2%7D)
Answer:
![BD = 30](https://tex.z-dn.net/?f=BD%20%3D%2030)
Step-by-step explanation:
Given
![AB = 17](https://tex.z-dn.net/?f=AB%20%3D%2017)
![AC = 16](https://tex.z-dn.net/?f=AC%20%3D%2016)
See attachment
Required
Find BD
If AC = 16, then:
i.e. half the diagonal AC
![AO = 8](https://tex.z-dn.net/?f=AO%20%3D%208)
The diagonals of a rhombus are perpendicular.
This implies that we can apply Pythagoras theorem.
Using Pythagoras theorem on triangle AOB, we have:
![AB^2 = AO^2 + OB^2](https://tex.z-dn.net/?f=AB%5E2%20%3D%20AO%5E2%20%2B%20OB%5E2)
![17^2 = 8^2 + OB^2](https://tex.z-dn.net/?f=17%5E2%20%3D%208%5E2%20%2B%20OB%5E2)
![289 = 64 + OB^2](https://tex.z-dn.net/?f=289%20%3D%2064%20%2B%20OB%5E2)
Collect like terms
![OB^2 = 289 - 64](https://tex.z-dn.net/?f=OB%5E2%20%3D%20289%20-%2064)
![OB^2 = 225](https://tex.z-dn.net/?f=OB%5E2%20%3D%20225)
Take positive square roots of both sides
![OB = 15](https://tex.z-dn.net/?f=OB%20%3D%2015)
To solve for BD, we use:
--- i.e. half the diagonal BD
![BD = 2 * OB](https://tex.z-dn.net/?f=BD%20%3D%202%20%2A%20OB)
![BD = 2 * 15](https://tex.z-dn.net/?f=BD%20%3D%202%20%2A%2015)
![BD = 30](https://tex.z-dn.net/?f=BD%20%3D%2030)
Answer
480
Step-by-step explanation:
Congrats on your first question!
The investment is 480 because
(80 / 100) x 1000 = 80 x 6 = 480
The total is 1480.
I think it is 740 inches because 6ft and 2 inches is 74 inches in total and you multiply that by how many times the dog walked the fence (10 times) and you get 740 inches