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 =
Answer: do the math
Step-by-step explanation:
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Set up a proportion:
3/4/24 = 1/x
Cross multiply:
3/4x = 24
Divide 3/4 to both sides or multiply by its reciprocal, 4/3:
x = 24 * 4/3
x = 96/3
x = 32
So you had a total of 32 muffins, now subtract this from the amount that blew away to find out how many you have left:
32 - 24 = 8
So you have 8 muffins left.
sin(x) = 2cos(x)
tan(x) = 2
tan⁻¹[tan(x)] = tan⁻¹(2)
x ≈ 63.4
sin(2x) = sin[2(63.4)]
sin(2x) = sin(126.8)
sin(2x) ≈ 0.801