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 =
![6\cdot\frac{1}{2}\cdot6\cdot\sqrt[]{63}=18\sqrt[]{63}](https://tex.z-dn.net/?f=6%5Ccdot%5Cfrac%7B1%7D%7B2%7D%5Ccdot6%5Ccdot%5Csqrt%5B%5D%7B63%7D%3D18%5Csqrt%5B%5D%7B63%7D)
Answer:
This is the alternate exterior angle theorem
Step-by-step explanation:
This is because angle 2 and angle 7 are in the <u>outer</u> part of each side of opposite lines
Answer: the box contained 9 square chocolates and 15 round chocolates.
Step-by-step explanation:
Let x represent the number of square chocolates contained in the box.
Let y represent the number of round chocolates contained in the box.
The box of chocolates contains square chocolates, which weigh 10g each and round chocolates which weigh 8g each. The combined weight of all the chocolates is 210g. It means that
10x + 8y = 210- - - - - - - - - - -1
The number of round chocolates is 3 less than twice the number of square chocolates. It means that
y = 2x - 3
Substituting y = 2x - 3 into equation 1, it becomes
10x + 8(2x - 3) = 210
10x + 16x - 24 = 210
26x = 210 + 24
26x = 234
x = 234/26
x = 9
y = 2x - 3 = 2 × 9 - 3
y = 18 - 3
y = 15
Answer:
9d you need to take the 3 to the power outside the (). I can't read it. 4???
10c x needs to be raised to the power also.
10e simplify the fraction
Step-by-step explanation:
