Answer:
The surface area of the right regular hexagonal pyramid is 50.78 cm².
Step-by-step explanation:
Given:
A right regular hexagonal pyramid with sides(s) 2 cm and slant height(h) 5 cm.
Now, to find the surface area(SA) of the right regular hexagonal pyramid.
So, we find the area of the base(b) first:
Area of the base = ![\sqrt[3]{3}\times s^{2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B3%7D%5Ctimes%20s%5E%7B2%7D)
= ![\sqrt[3]{3}\times 2^{2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B3%7D%5Ctimes%202%5E%7B2%7D)
On solving we get:
Area of the base(b) = 
Then, we find the perimeter(p) :
Perimeter = s × 6

Now, putting the formula for getting the surface area:
Surface area = perimeter × height/2 + Area of the base.




As, <em>the surface area is 50.784 and rounding to nearest hundredth becomes 50.78 because in hundredth place it is 8 and in thousandth place it is 4 so rounding to it become 50.78.</em>
Therefore, the surface area of the right regular hexagonal pyramid is 50.78 cm².
The height of the doorway is 252 inches
Answer:
![\large\boxed{\ln\sqrt[3]{e^4}=\dfrac{4}{3}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B%5Cln%5Csqrt%5B3%5D%7Be%5E4%7D%3D%5Cdfrac%7B4%7D%7B3%7D%7D)
Step-by-step explanation:
![\text{Use}\\\\\sqrt[n]{a^m}=a^\frac{m}{n}\\\\\ln a^n=n\ln a\\\\\ln e=1\\-----------\\\\\ln\sqrt[3]{e^4}=\ln e^\frac{4}{3}=\dfrac{4}{3}\ln e=\dfrac{4}{3}\cdot1=\dfrac{4}{3}](https://tex.z-dn.net/?f=%5Ctext%7BUse%7D%5C%5C%5C%5C%5Csqrt%5Bn%5D%7Ba%5Em%7D%3Da%5E%5Cfrac%7Bm%7D%7Bn%7D%5C%5C%5C%5C%5Cln%20a%5En%3Dn%5Cln%20a%5C%5C%5C%5C%5Cln%20e%3D1%5C%5C-----------%5C%5C%5C%5C%5Cln%5Csqrt%5B3%5D%7Be%5E4%7D%3D%5Cln%20e%5E%5Cfrac%7B4%7D%7B3%7D%3D%5Cdfrac%7B4%7D%7B3%7D%5Cln%20e%3D%5Cdfrac%7B4%7D%7B3%7D%5Ccdot1%3D%5Cdfrac%7B4%7D%7B3%7D)
Combine 6.1a and 3a
9.1a + 5.7b
Diameter is 2 times the radius. Radius is 6.
Circumference is 2(pi)(radius)
Circumference is approximately 37.7cm