Answer:
There are more boys than girls. The red chips are girls are the boy are the yellow chips. If more yellow chips were pulled out then there are more boys. 11 + 9 = 2. 10 + 9 = 19 + 1 = 20. The fraction of how many boys there are would be 11/21 and the fraction for girls would be 9/21. Also 11 x 9 = 99. There would be 99 people if you times the number of girls and boys. But it's best to add because it would be in the smaller form if you were to times it.
Answer:
Step-by-step explanation:
![x^2+3x=10^{1\frac{1}{2}}\\\\x^2+3x=10^{1+\frac{1}{2}}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\x^2+3x=10\cdot10^\frac{1}{2}\qquad\text{use}\ \sqrt[n]{a}=a^\frac{1}{n}\\\\x^2+3x=10\sqrt{10}\qquad\text{subtract}\ 10\sqrt{10}\ \text{from both sides}\\\\x^2+3x-10\sqrt{10}=0\\\\\text{Use the quadratic formula}\\\\ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\a=1,\ b=3,\ c=-10\sqrt{10}\\\\b^2-4ac=3^2-4(1)(-10\sqrt{10})=9+40\sqrt{10}\\\\x=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2(1)}=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2}\\\\x=\dfrac{-3-\sqrt{10+10\sqrt{10}}}{2}\notin D](https://tex.z-dn.net/?f=x%5E2%2B3x%3D10%5E%7B1%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C%5C%5Cx%5E2%2B3x%3D10%5E%7B1%2B%5Cfrac%7B1%7D%7B2%7D%7D%5Cqquad%5Ctext%7Buse%7D%5C%20a%5En%5Ccdot%20a%5Em%3Da%5E%7Bn%2Bm%7D%5C%5C%5C%5Cx%5E2%2B3x%3D10%5Ccdot10%5E%5Cfrac%7B1%7D%7B2%7D%5Cqquad%5Ctext%7Buse%7D%5C%20%5Csqrt%5Bn%5D%7Ba%7D%3Da%5E%5Cfrac%7B1%7D%7Bn%7D%5C%5C%5C%5Cx%5E2%2B3x%3D10%5Csqrt%7B10%7D%5Cqquad%5Ctext%7Bsubtract%7D%5C%2010%5Csqrt%7B10%7D%5C%20%5Ctext%7Bfrom%20both%20sides%7D%5C%5C%5C%5Cx%5E2%2B3x-10%5Csqrt%7B10%7D%3D0%5C%5C%5C%5C%5Ctext%7BUse%20the%20quadratic%20formula%7D%5C%5C%5C%5Cax%5E2%2Bbx%2Bc%3D0%5C%5C%5C%5Cx%3D%5Cdfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D%5C%5C%5C%5Ca%3D1%2C%5C%20b%3D3%2C%5C%20c%3D-10%5Csqrt%7B10%7D%5C%5C%5C%5Cb%5E2-4ac%3D3%5E2-4%281%29%28-10%5Csqrt%7B10%7D%29%3D9%2B40%5Csqrt%7B10%7D%5C%5C%5C%5Cx%3D%5Cdfrac%7B-3%5Cpm%5Csqrt%7B40%2B10%5Csqrt%7B10%7D%7D%7D%7B2%281%29%7D%3D%5Cdfrac%7B-3%5Cpm%5Csqrt%7B40%2B10%5Csqrt%7B10%7D%7D%7D%7B2%7D%5C%5C%5C%5Cx%3D%5Cdfrac%7B-3-%5Csqrt%7B10%2B10%5Csqrt%7B10%7D%7D%7D%7B2%7D%5Cnotin%20D)
It equal 12 have a nice day
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.
Your answer would be B.
The triangles would be congruent, because the lines GH and FI are congruent and the angle H and I are the same, and since the line HI is shared by the two, it would then be SAS.
