Answer:
b. x>1
Step-by-step explanation:
2/3+x/3>1
Multiply the inequality by 3 to get rid of the fractions
3(2/3+x/3)>1*3
2 +x >3
Subtract 2 from each side
2+x-2 >3-2
x>1
Quien sera, sera que de noche sale y de dia se va?
V = (1/3) π r² t
= (1/3) π (10 cm)². 16 cm
= (1/3) π (100 cm²). 16 cm
= (1/3) π (1600 cm³)
= (1600π)÷3 cm³ (B)
Answer:
sample 1: 75
sample 2: 86
Step-by-step explanation:
The trick is to exploit the difference of squares formula,
![a^2-b^2=(a-b)(a+b)](https://tex.z-dn.net/?f=a%5E2-b%5E2%3D%28a-b%29%28a%2Bb%29)
Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:
![(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2](https://tex.z-dn.net/?f=%28%5Csqrt8%2B%5Csqrt6%29%28%5Csqrt8-%5Csqrt6%29%3D%28%5Csqrt8%29%5E2-%28%5Csqrt6%29%5E2%3D8-6%3D2)
Whatever you do to the denominator, you have to do to the numerator too. So
![\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt2%2B%5Csqrt6%7D%7B%5Csqrt8%2B%5Csqrt6%7D%3D%5Cdfrac%7B%28%5Csqrt2%2B%5Csqrt6%29%28%5Csqrt8-%5Csqrt6%29%7D%7B%28%5Csqrt8%2B%5Csqrt6%29%28%5Csqrt8-%5Csqrt6%29%7D%3D%5Cdfrac%7B%28%5Csqrt2%2B%5Csqrt6%29%28%5Csqrt8-%5Csqrt6%29%7D2)
Expand the numerator:
![(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2](https://tex.z-dn.net/?f=%28%5Csqrt2%2B%5Csqrt6%29%28%5Csqrt8-%5Csqrt6%29%3D%5Csqrt%7B2%5Ccdot8%7D%2B%5Csqrt%7B6%5Ccdot8%7D-%5Csqrt%7B2%5Ccdot6%7D-%28%5Csqrt6%29%5E2)
![(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6](https://tex.z-dn.net/?f=%28%5Csqrt2%2B%5Csqrt6%29%28%5Csqrt8-%5Csqrt6%29%3D%5Csqrt%7B16%7D%2B%5Csqrt%7B48%7D-%5Csqrt%7B12%7D-6)
![(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6](https://tex.z-dn.net/?f=%28%5Csqrt2%2B%5Csqrt6%29%28%5Csqrt8-%5Csqrt6%29%3D4%2B%5Csqrt%7B48%7D-%5Csqrt%7B12%7D-6)
![(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)](https://tex.z-dn.net/?f=%28%5Csqrt2%2B%5Csqrt6%29%28%5Csqrt8-%5Csqrt6%29%3D-2%2B%5Csqrt%7B12%7D%28%5Csqrt4-1%29)
![(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)](https://tex.z-dn.net/?f=%28%5Csqrt2%2B%5Csqrt6%29%28%5Csqrt8-%5Csqrt6%29%3D-2%2B%5Csqrt%7B12%7D%282-1%29)
![(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}](https://tex.z-dn.net/?f=%28%5Csqrt2%2B%5Csqrt6%29%28%5Csqrt8-%5Csqrt6%7D%3D-2-%5Csqrt%7B12%7D)
So we have
![\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt2%2B%5Csqrt6%7D%7B%5Csqrt8%2B%5Csqrt6%7D%3D-%5Cdfrac%7B2%2B%5Csqrt%7B12%7D%7D2)
But √12 = √(3•4) = 2√3, so
![\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt2%2B%5Csqrt6%7D%7B%5Csqrt8%2B%5Csqrt6%7D%3D-%5Cdfrac%7B2%2B2%5Csqrt3%7D2%3D%5Cboxed%7B-1-%5Csqrt3%7D)