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kati45 [8]
3 years ago
7

Geometry help? How would I solve this?

Mathematics
1 answer:
Firlakuza [10]3 years ago
8 0
\bf \textit{sum of interior angles in a regular polygon}
\\\\\\
n\theta=180(n-2)\qquad 
\begin{cases}
n=\textit{number of sides}\\
\theta=\textit{angle in degrees}
\end{cases}
\\\\\\
\textit{notice, your two polygons, are just two regular octagons}\\
\textit{OCTAgons, thus OCTA = 8, 8 sides, thus}
\\\\\\
n\theta=180(n-2)\qquad n=8\implies 8\theta=180(8-2)
\\\\\\
\theta=\cfrac{180\cdot 6}{8}

so... that's how much an internal angle is
now, subtract that from 180 and you get the angle outside, in the picture

subtract that outside angle twice from 180, and you get angle "2"
because angle 2 is in the same triangle as those two outside angles, and all internal angles in a triangle is 180, thus 180 - (those two angles) is angle "2"

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7. A number of minor automobile accidents occur at various high-risk intersections in Teton County despite traffic lights. The T
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Answer:

Modification has not reduced the number of accidents.

Step-by-step explanation:

H_0: \mu\geq0\\H_a:\mu

                                 A  B  C   D   E   F  G   H

Before modification 5  7  6   4    8   9   8   10

After modification    3  7  7   0    4   6   8    2

Difference                -2 0  1   -4   -4   -3  0   -8

Mean of differences M_d = \frac{sum}{8}=\frac{2+0-1+4+4+3-0+8}{8}=2.5

Standard deviation of differences = \sqrt{\frac{\sum(x-\bar{x})^2}{n}}=2.9277

Formula of paired t test :

t=\frac{M_d}{\frac{s}{\sqrt{n}}}\\t=\frac{2.5}{\frac{2.9277}{\sqrt{8}}}\\t = 2.4152

Df = n-1 = 8-1 =7

t critical = t_{(df, \alpha)}=t_{7,0.01}=2.998

t critical> t calculated

So, We failed to reject null hypothesis

Hence modification has not reduced the number of accidents.

6 0
3 years ago
Quadratic equations and complex numbers PLEASE HELPPPP ASAP
kondor19780726 [428]

we are given

27x^3-1=0

we can also write it as

(3x)^3-(1)^3=0

now, we can use factor formula

a^3-b^3=(a-b)(a^2+ab+b^2)

we can use above formula

we get  

(3x)^3-(1)^3=(3x-1)((3x)^2+3x\times 1+(1)^2)

now, we can simplify it

27x^3-1=(3x-1)(9x^2+3x+1)

now, we can set it to 0

27x^3-1=(3x-1)(9x^2+3x+1)=0

and then we can solve for x

(3x-1)(9x^2+3x+1)=0

3x-1=0

x=\frac{1}{3}

9x^2+3x+1=0

now, we can use quadratic formula

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now, we can plug values

and we get

x=\frac{-3\pm \sqrt{3^2-4\cdot \:9\cdot \:1}}{2\cdot \:9}

x=-\frac{1}{6}+i\frac{\sqrt{3}}{6},\:x=-\frac{1}{6}-i\frac{\sqrt{3}}{6}

So, we will get solution as

x=\frac{1}{3},\:x=-\frac{1}{6}+i\frac{\sqrt{3}}{6},\:x=-\frac{1}{6}-i\frac{\sqrt{3}}{6}...............Answer

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nikklg [1K]

Answer:

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Then replace x with a

8 0
3 years ago
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