OCTA=8, an octagon has 8 sides, thus if one side is 6mm, then all sides, namely the perimeter, is 6+6+6+6+6+6+6+6, or 48.
now, we know what the apothem is.
Answer:
Step-by-step explanation:
The missing length is 13 because,
Lets say the top triangle is A and the bottom triangle is B.
Triangle A gives us the side GF, and Triangle B gives us the sides TU and ST. Since the triangles are similar(as stated in the problem), we can pair 2 sides GF(A) and TU(B) which is 11:22.(one way I usually figure which sides are similar is by first- matching the hypotenuse, then checking which of the remaining two is longer.. if that made any sense). You can see that their relationship is x2 or /2 (In another word, from A to B is multiplication- ex: 11 * 2 is 22, and from B to A is division- ex 22/2 is 11.) Since the missing number is the hypotenuse of triangle A and you know the Hypotenuse of triangle B all you have to do is divide side TS by 2 to get side SF. So the missing side is 13.
Answer: 13
Answer:
16. 200+3
17. 30+4+0.1+0.02+0.007
18. 200+70+6+0.1+0.03
19. 30000000+4000000+100000+20000+3000+6
Step-by-step explanation:
Answer: 
Step-by-step explanation:
Given the following inequality:

You need to solve for "x" in order to find the solution.
The steps are:
1. Add
to both sides of the inequality:

2. Add
to both sides:

3. Divide both sides by
:

Notice that "x" is less than 8. This indicates that 8 is not included in the solution and you must use parentheses.
The solution in Interval notation is:

Given the figure of a regular pyramid
The base of the pyramid is a hexagon with a side length = 6
The lateral area is 6 times the area of one of the side triangles
So, the side triangle has a base = 6
The height will be:
![\begin{gathered} h^2=6^2+(\frac{\sqrt[]{3}}{2}\cdot6)^2=36+27=63 \\ h=\sqrt[]{63} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20h%5E2%3D6%5E2%2B%28%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%5Ccdot6%29%5E2%3D36%2B27%3D63%20%5C%5C%20h%3D%5Csqrt%5B%5D%7B63%7D%20%5Cend%7Bgathered%7D)
so, the lateral area =