Answer:
![\large\boxed{a(V)=\sqrt{\dfrac{3V}{8}}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7Ba%28V%29%3D%5Csqrt%7B%5Cdfrac%7B3V%7D%7B8%7D%7D%7D)
Step-by-step explanation:
The formula of a volume of a square pyramid:
![V=\dfrac{1}{3}a^2h](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B1%7D%7B3%7Da%5E2h)
a - base edge
h - height of a pyramid
We have H = 8in.
Substitute and solve for a:
![\dfrac{1}{3}a^2(8)=V\\\\\dfrac{8}{3}a^2=V\qquad\text{multiply both sides by}\ \dfrac{3}{8}\\\\\dfrac{3\!\!\!\!\diagup^1}{8\!\!\!\!\diagup_1}\cdot\dfrac{8\!\!\!\!\diagup^1}{3\!\!\!\!\diagup_1}a^2=\dfrac{3}{8}V\\\\a^2=\dfrac{3V}{8}\Rightarrow a=\sqrt{\dfrac{3V}{8}}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B3%7Da%5E2%288%29%3DV%5C%5C%5C%5C%5Cdfrac%7B8%7D%7B3%7Da%5E2%3DV%5Cqquad%5Ctext%7Bmultiply%20both%20sides%20by%7D%5C%20%5Cdfrac%7B3%7D%7B8%7D%5C%5C%5C%5C%5Cdfrac%7B3%5C%21%5C%21%5C%21%5C%21%5Cdiagup%5E1%7D%7B8%5C%21%5C%21%5C%21%5C%21%5Cdiagup_1%7D%5Ccdot%5Cdfrac%7B8%5C%21%5C%21%5C%21%5C%21%5Cdiagup%5E1%7D%7B3%5C%21%5C%21%5C%21%5C%21%5Cdiagup_1%7Da%5E2%3D%5Cdfrac%7B3%7D%7B8%7DV%5C%5C%5C%5Ca%5E2%3D%5Cdfrac%7B3V%7D%7B8%7D%5CRightarrow%20a%3D%5Csqrt%7B%5Cdfrac%7B3V%7D%7B8%7D%7D)
Answer:
LM = 14
Step-by-step explanation:
LM = AC/2
LM = 28/2
<u>LM = 14</u>
Hope this helps!
A=[p/4 ] the parenthesis is a ratio.
The most accurate "estimation" for 5+4 is um... 9