Answer:
![\large\boxed{\ln\sqrt[3]{e^4}=\dfrac{4}{3}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B%5Cln%5Csqrt%5B3%5D%7Be%5E4%7D%3D%5Cdfrac%7B4%7D%7B3%7D%7D)
Step-by-step explanation:
![\text{Use}\\\\\sqrt[n]{a^m}=a^\frac{m}{n}\\\\\ln a^n=n\ln a\\\\\ln e=1\\-----------\\\\\ln\sqrt[3]{e^4}=\ln e^\frac{4}{3}=\dfrac{4}{3}\ln e=\dfrac{4}{3}\cdot1=\dfrac{4}{3}](https://tex.z-dn.net/?f=%5Ctext%7BUse%7D%5C%5C%5C%5C%5Csqrt%5Bn%5D%7Ba%5Em%7D%3Da%5E%5Cfrac%7Bm%7D%7Bn%7D%5C%5C%5C%5C%5Cln%20a%5En%3Dn%5Cln%20a%5C%5C%5C%5C%5Cln%20e%3D1%5C%5C-----------%5C%5C%5C%5C%5Cln%5Csqrt%5B3%5D%7Be%5E4%7D%3D%5Cln%20e%5E%5Cfrac%7B4%7D%7B3%7D%3D%5Cdfrac%7B4%7D%7B3%7D%5Cln%20e%3D%5Cdfrac%7B4%7D%7B3%7D%5Ccdot1%3D%5Cdfrac%7B4%7D%7B3%7D)
Line Q, you always want the line that has as many dots on it as possible!! :)
First you need to solve for angle F then you can use 1/2acSinB. If you haven't done this you can also use the Pythagoras theorem and solve for the unknown side once you do that you can just use 1/2 b*h
This inequality is true doll!!
A. 12x+y=6<span>B. −2x+y=6</span>
