He should first multiply both sides by 8, to get rid of the denominator.
5/8 m = 25
8(5/8 m) = 8(25)
5m = 200
He should then divide both sides by 5, to get rid of the coefficient
5m/5 = 200/5
m = 40
So m = 40 by these steps
The answer should
A. 8 shells because Jerrell collected 2 more than half as many shells as Paula.
Answer:
![\frac{4}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B3%7D)
Step-by-step explanation:
The area of a regular octahedron is given by:
area =
. Let a is the length of the edge (diagonal).
area = ![2\sqrt{3}\ *a^2](https://tex.z-dn.net/?f=2%5Csqrt%7B3%7D%5C%20%2Aa%5E2)
Given that the diagonal of the octahedron is equal to height (h) of the tetrahedron i.e.
a = h, where h is the height of the tetrahedron and a is the diagonal of the octahedron. Let the edge of the tetrrahedron be e. To find the edge of the tetrahedron, we use:
![h=\sqrt{\frac{2}{3} } e\\but\ h=a\\a=\sqrt{\frac{2}{3} } e\\e=\sqrt{\frac{3}{2} }a](https://tex.z-dn.net/?f=h%3D%5Csqrt%7B%5Cfrac%7B2%7D%7B3%7D%20%7D%20e%5C%5Cbut%5C%20h%3Da%5C%5Ca%3D%5Csqrt%7B%5Cfrac%7B2%7D%7B3%7D%20%7D%20e%5C%5Ce%3D%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%20%7Da)
The area of a tetrahedron is given by:
area =
=
The ratio of area of regular octahedron to area tetrahedron regular is given as:
Ratio = ![\frac{2\sqrt{3}\ *a^2}{\frac{3}{2} \sqrt{3}*a^2} =\frac{4}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2%5Csqrt%7B3%7D%5C%20%2Aa%5E2%7D%7B%5Cfrac%7B3%7D%7B2%7D%20%5Csqrt%7B3%7D%2Aa%5E2%7D%20%3D%5Cfrac%7B4%7D%7B3%7D)