im not great at this im learning it to but im going to say b from what ive learned if you can find a better answer find it
It’s D that is the answer
Hello!
So, this is quite the complex question, and here are the following steps:
What is the quotient of ![\frac{6-3\sqrt[3]{6}}{\sqrt[3]{9}}](https://tex.z-dn.net/?f=%20%5Cfrac%7B6-3%5Csqrt%5B3%5D%7B6%7D%7D%7B%5Csqrt%5B3%5D%7B9%7D%7D%20%20)
?
![\frac{(6 - 3\sqrt[3]{6})\sqrt[3]{9^{2}}}{9}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%286%20-%203%5Csqrt%5B3%5D%7B6%7D%29%5Csqrt%5B3%5D%7B9%5E%7B2%7D%7D%7D%7B9%7D%20%20%20%20%20)
(rationalize the denominator)
![\frac{3(2 -\sqrt[3]{6})\sqrt[3]{9^{2}}}{9}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%282%20-%5Csqrt%5B3%5D%7B6%7D%29%5Csqrt%5B3%5D%7B9%5E%7B2%7D%7D%7D%7B9%7D%20%20%20%20%20)
(factor 3 from the expression)
![\frac{(2-\sqrt[3]{6})\sqrt[3]{9^{2}}}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%282-%5Csqrt%5B3%5D%7B6%7D%29%5Csqrt%5B3%5D%7B9%5E%7B2%7D%7D%7D%7B3%7D%20%20%20%20%20)
(reduce the fraction with 3)
![\frac{2\sqrt[3]{9^{2}}-\sqrt[3]{9^{2}}}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%5Csqrt%5B3%5D%7B9%5E%7B2%7D%7D-%5Csqrt%5B3%5D%7B9%5E%7B2%7D%7D%7D%7B3%7D%20%20%20%20%20%20%20%20)
(distributive property)
![\frac{2\sqrt[3]{81}-\sqrt[3]{486}}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%5Csqrt%5B3%5D%7B81%7D-%5Csqrt%5B3%5D%7B486%7D%7D%7B3%7D%20)
(simplify 3 · 9²)
![\frac{6\sqrt[3]{3}-3\sqrt[3]{18}}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B6%5Csqrt%5B3%5D%7B3%7D-3%5Csqrt%5B3%5D%7B18%7D%7D%7B3%7D%20%20%20%20)
(simplify the radical)
![\frac{3(2\sqrt[3]{3}-3\sqrt[3]{18})}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%282%5Csqrt%5B3%5D%7B3%7D-3%5Csqrt%5B3%5D%7B18%7D%29%7D%7B3%7D%20%20%20%20)
(factor 3 from the expression)
![2\sqrt[3]{3}-\sqrt[3]{18}](https://tex.z-dn.net/?f=%202%5Csqrt%5B3%5D%7B3%7D-%5Csqrt%5B3%5D%7B18%7D%20%20%20%20)
(reduce the fraction)
The answer, is simply, choice A, ![2\sqrt[3]{3} -\sqrt[3]{18}](https://tex.z-dn.net/?f=%202%5Csqrt%5B3%5D%7B3%7D%20-%5Csqrt%5B3%5D%7B18%7D%20%20)
≈ 0.263758.
From the question, we know that we will be looking at <3, <4, and angles TKL and TLK. That being said, since <3 is congruent to <4, that means that angles TKL and TLK, which are each supplementary to either angle 3 or 4, are congruent because, since angles 3 and 4 are congruent, they are congruent because the supplements of congruent angles are congruent.
3 ln 3 + 4 ln b => ln 3^3 + ln b^4 => ln (3^3)*(b^4)
