Answer:
x₁ = -4
x₂ = 3
Step-by-step explanation:
x²+ x + 12 = 0
x = {-1±√((1²)-(4*1*-12))} / (2*1)
x = {-1±√(1+48)} / 2
x = {-1±√49} / 2
x = {-1±7} / 2
x₁ = {-1-7} / 2 = -8/2 = -4
x₂ = {-1+7} / 2 = 6/2 = 3
Check:
x₁
-4² + (-4) - 12 = 0
16 - 4 - 12 = 0
x₂
3² + 3 - 12 = 0
9 + 3 - 12 = 0
Answer:
1+i
Step-by-step explanation:
To find the 8th roots of unity, you have to find the trigonometric form of unity.
1. Since
then

and

This gives you 
Thus,

2. The 8th roots can be calculated using following formula:
![\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7Bz%7D%3D%5C%7B%5Csqrt%5B8%5D%7B%7Cz%7C%7D%20%28%5Ccos%5Cdfrac%7B%5Cvarphi%2B2%5Cpi%20k%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B%5Cvarphi%2B2%5Cpi%20k%7D%7B8%7D%29%2C%20k%3D0%2C%5C%201%2C%5Cdots%2C7%5C%7D.)
Now
at k=0, ![z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;](https://tex.z-dn.net/?f=z_0%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%200%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%200%7D%7B8%7D%29%3D1%5Ccdot%20%281%2B0%5Ccdot%20i%29%3D1%3B)
at k=1, ![z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_1%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%201%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%201%7D%7B8%7D%29%3D1%5Ccdot%20%28%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
at k=2, ![z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;](https://tex.z-dn.net/?f=z_2%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%202%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%202%7D%7B8%7D%29%3D1%5Ccdot%20%280%2B1%5Ccdot%20i%29%3Di%3B)
at k=3, ![z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_3%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%203%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%203%7D%7B8%7D%29%3D1%5Ccdot%20%28-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
at k=4, ![z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;](https://tex.z-dn.net/?f=z_4%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%204%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%204%7D%7B8%7D%29%3D1%5Ccdot%20%28-1%2B0%5Ccdot%20i%29%3D-1%3B)
at k=5, ![z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_5%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%205%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%205%7D%7B8%7D%29%3D1%5Ccdot%20%28-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
at k=6, ![z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;](https://tex.z-dn.net/?f=z_6%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%206%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%206%7D%7B8%7D%29%3D1%5Ccdot%20%280-1%5Ccdot%20i%29%3D-i%3B)
at k=7, ![z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_7%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%207%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%207%7D%7B8%7D%29%3D1%5Ccdot%20%28%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
The 8th roots are

Option C is icncorrect.
The base of its triangular face of the triangular prism is found as 17.7 cm.
<h3>What is defined as the triangular prism?</h3>
- A polyhedron with two triangular bases as well as three rectangular sides is known as a triangular prism.
- It is a three-dimensional shape with three side faces but also two base faces that are connected by the edges.
The formula for the volume of triangular prism is;
Volume = base area x length
The base area of the triangular base.
The area of the triangle = 1/2 base×height
The given values are-
- volume = 2,354. 1 cubic centimeters.
- length = 19 cm
- height = 14 cm.
Thus, volume = base area x length
volume = 1/2 base × height × length
base = 2v/(height × length)
Put the values;
base = 2×2,354.1/(14 × 19)
base = 4708.2/(14 × 19)
base = 17.7 cm
Thus, the base of the triangular prism is found as 17.7 cm.
To know more about the triangular prism, here
brainly.com/question/16128664
#SPJ9
Answer:
Nico
Step-by-step explanation:
In order to compare who had the best performance on the aptitude test, we can use Z score. Z score indicates how many standard deviation a result is over the mean. For this case the higher the result is, the best is the performance on the aptitude test.
Z score is expressed with the following formula:
Z=(X-μ)/σ
Where:
X is the test score
µ is the mean
σ is the standard deviation
Replacing in the formula we obtained
For Nico:
X = 88.3
µ = 66.2
σ = 13
Z=(88.3-66.2)/13=1.7
For Justin:
X = 226.2
µ = 216
σ = 17
Z=(226.2-216)/17=0.6
For Tera:
X = 7.97
µ = 7.2
σ = 0.7
Z=(7.97-7.2)/0.7=1.1
Nico had the higher Z score which means he had the better performance on the test and he should be offered the job.
8n. 8 represents the number of hours she sleeps, and n represents the number of night she sleeps.