40º
7) In this problem, we can see that both tangent lines to that circle come from the same point O.
So, we can write out the following considering that there is one secant line DO and one tangent line to the circle AO
2 .............:)
Good luck
Answer:
<em>The first step is to determine the average
</em>
<em>The exercise says it’s a normal distribution: (n=8)</em>
<em>According to the exercise, the mean is equal to 0,5 then the value of t of the distribution can be obtained
</em>
<em />
<em>The variable t has 7 grade to liberty, we calculate the p-value as:
</em>
This value is very high, therefore the hypothesis is not rejected
Answer:
A, B, C, D
Step-by-step explanation:
They All add up to 360
Step-by-step explanation:
With reference to the regular hexagon, from the image above we can see that it is formed by six triangles whose sides are two circle's radii and the hexagon's side. The angle of each of these triangles' vertex that is in the circle center is equal to 360∘6=60∘ and so must be the two other angles formed with the triangle's base to each one of the radii: so these triangles are equilateral.
The apothem divides equally each one of the equilateral triangles in two right triangles whose sides are circle's radius, apothem and half of the hexagon's side. Since the apothem forms a right angle with the hexagon's side and since the hexagon's side forms 60∘ with a circle's radius with an endpoint in common with the hexagon's side, we can determine the side in this fashion:
tan60∘=opposed cathetusadjacent cathetus => √3=Apothemside2 => side=(2√3)Apothem
As already mentioned the area of the regular hexagon is formed by the area of 6 equilateral triangles (for each of these triangle's the base is a hexagon's side and the apothem functions as height) or:
Shexagon=6⋅S△=6(base)(height)2=3(2√3)Apothem⋅Apothem=(6√3)(Apothem)2
=> Shexagon=6×62√3=216
