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:
Translation down 8 units, and left 9 units
(I hope this was the answer you were looking for if not im sorry.)
Answer:
Volume of pyramid with square base is 1057.54 m³
Step-by-step explanation:
We are given:
Length of base of pyramid = 14.4 m
Height of pyramid = 15.3 m
We need to find:
Volume of pyramid with square base
The formula used is: ![Volume=\frac{1}{3}\times a^2 \times h](https://tex.z-dn.net/?f=Volume%3D%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20a%5E2%20%5Ctimes%20h)
Where a = base of pyramid and h is height
Putting values and finding volume
![Volume=\frac{1}{3}\times a^2 \times h\\Volume=\frac{1}{3}\times (14.4)^2 \times (15.3)\\Volume=\frac{1}{3}\times 207.36 \times 15.3\\Volume=1057.54 \ m^3](https://tex.z-dn.net/?f=Volume%3D%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20a%5E2%20%5Ctimes%20h%5C%5CVolume%3D%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%2814.4%29%5E2%20%5Ctimes%20%2815.3%29%5C%5CVolume%3D%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20207.36%20%5Ctimes%2015.3%5C%5CVolume%3D1057.54%20%5C%20m%5E3)
So, Volume of pyramid with square base is 1057.54 m³
Answer:
x = 5, y = 20
Step-by-step explanation:
Since AB and CD are parallel, then
∠ AOC and ∠ OCD are Alternate angles and congruent, thus
12x + 8 = 68 ( subtract 8 from both sides )
12x = 60 ( divide both sides by 12 )
x = 5
-------------------------------
∠ OCD and 5y + 12 are adjacent angles and supplementary, thus
5y + 12 + 68 = 180 , that is
5y + 80 = 180 ( subtract 80 from both sides )
5y = 100 ( divide both sides by 5 )
y = 20
Your answer is <span>132^2
......</span>