It’s 4 ik know this it’s four cixhdhdud didjdhdh djdjdhd djdjdhd
The surface area is the area of the outside of a 3D shape for example, let’s say you have a cube, a cube has 6 sides, so you have to find the area of each side and then add it up, since it’s a cube you would technically have to find the area of one side and then just multiply it by 6, that would be the easiest way to do it
Hope I helped!
So we have a 6 at the bottom, and the root is 3, so hmm how to take it out, simple enough, just let's get something to make the 6 a 6³, so it comes out of the root
so
![\bf \cfrac{2+\sqrt[3]{3}}{\sqrt[3]{6}}\cdot \cfrac{\sqrt[3]{6^2}}{\sqrt[3]{6^2}}\implies \cfrac{(2+\sqrt[3]{3})(\sqrt[3]{6^2})}{(\sqrt[3]{6})(\sqrt[3]{6^2})}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{3}\cdot \sqrt[3]{36}}{\sqrt[3]{6^3}} \\\\\\ \cfrac{2\sqrt[3]{36}+\sqrt[3]{3\cdot 36}}{6}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{108}}{6}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{3^3\cdot 4}}{6} \\\\\\ \cfrac{2\sqrt[3]{36}+3\sqrt[3]{ 4}}{6}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B2%2B%5Csqrt%5B3%5D%7B3%7D%7D%7B%5Csqrt%5B3%5D%7B6%7D%7D%5Ccdot%20%5Ccfrac%7B%5Csqrt%5B3%5D%7B6%5E2%7D%7D%7B%5Csqrt%5B3%5D%7B6%5E2%7D%7D%5Cimplies%20%5Ccfrac%7B%282%2B%5Csqrt%5B3%5D%7B3%7D%29%28%5Csqrt%5B3%5D%7B6%5E2%7D%29%7D%7B%28%5Csqrt%5B3%5D%7B6%7D%29%28%5Csqrt%5B3%5D%7B6%5E2%7D%29%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%5B3%5D%7B36%7D%2B%5Csqrt%5B3%5D%7B3%7D%5Ccdot%20%5Csqrt%5B3%5D%7B36%7D%7D%7B%5Csqrt%5B3%5D%7B6%5E3%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B2%5Csqrt%5B3%5D%7B36%7D%2B%5Csqrt%5B3%5D%7B3%5Ccdot%2036%7D%7D%7B6%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%5B3%5D%7B36%7D%2B%5Csqrt%5B3%5D%7B108%7D%7D%7B6%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%5B3%5D%7B36%7D%2B%5Csqrt%5B3%5D%7B3%5E3%5Ccdot%204%7D%7D%7B6%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B2%5Csqrt%5B3%5D%7B36%7D%2B3%5Csqrt%5B3%5D%7B%204%7D%7D%7B6%7D)
You would assume that in this figure, the number of colored sections with which are not colored with respect to a " touching " colored section, would be half of the total colored sections. However that is not the case, the sections are not alternating as they still meet at a common point. After all, it notes no two touching sections, not adjacent sections. Their is no equation to calculate this requirement with respect to the total number of sections.
Let's say that we take one triangle as the starting. This triangle will be the start of a chain of other triangles that have no two touching sections, specifically 7 triangles. If a square were to be this starting shape, there are 5 shapes that have no touching sections, 3 being a square, the other two triangles. This is presumably a lower value as a square occupies two times as much space, but it also depends on the positioning. Therefore, the least number of colored sections you can color in the sections meeting the given requirement, is 5 sections for this first figure.
Respectively the solution for this second figure is 5 sections as well.
