If you would like to solve - 3 * a^2 - b^3 + 3 * c^2 - 2 * b^3, if a = 2, b = -1, c = 3, you can calculate this using the following steps:
a = 2, b = -1, c = 3
- 3 * a^2 - b^3 + 3 * c^2 - 2 * b^3 = - 3 * 2^2 - (-1)^3 + 3 * 3^2 - 2 * (-1)^3 = - 3 * 4 - (-1) + 3 * 9 - 2 * (-1) = - 12 + 1 + 27 + 2 = 18
The correct result would be 18.
Answer: ![\bold{b)\quad \dfrac{2\pi}{3},\dfrac{4\pi}{3},0}](https://tex.z-dn.net/?f=%5Cbold%7Bb%29%5Cquad%20%5Cdfrac%7B2%5Cpi%7D%7B3%7D%2C%5Cdfrac%7B4%5Cpi%7D%7B3%7D%2C0%7D)
<u>Step-by-step explanation:</u>
Note the following identities: tan² x = sec²x - 1
![\sec x=\dfrac{1}{\cos}](https://tex.z-dn.net/?f=%5Csec%20x%3D%5Cdfrac%7B1%7D%7B%5Ccos%7D)
tan² x + sec x = 1
(sec² x -1) + sec x = 1
sec² x + sec x - 2 = 0
(sec x + 2)(sec x - 1) = 0
sec x + 2 = 0 sec x - 1 = 0
sec x = -2 sec x = 1
![\dfrac{1}{\cos x}=-2\qquad \dfrac{1}{\cos x}=1\\\\\cos x=-\dfrac{1}{2}\qquad \cos x=1\\\\x=\dfrac{2\pi}{3}, \dfrac{4\pi}{3}\qquad x=0](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5Ccos%20x%7D%3D-2%5Cqquad%20%5Cdfrac%7B1%7D%7B%5Ccos%20x%7D%3D1%5C%5C%5C%5C%5Ccos%20x%3D-%5Cdfrac%7B1%7D%7B2%7D%5Cqquad%20%5Ccos%20x%3D1%5C%5C%5C%5Cx%3D%5Cdfrac%7B2%5Cpi%7D%7B3%7D%2C%20%5Cdfrac%7B4%5Cpi%7D%7B3%7D%5Cqquad%20x%3D0)
Answer:
4x+3y=68
9x+2y=77
plain= 5 shiny=16
Step-by-step explanation:
plain=x shiny=y
The answer to your problem would be 0.05
There is a not so well-known theorem that solves this problem.
The theorem is stated as follows:
"Each angle bisector of a triangle divides the opposite side into segments proportional in length to the adjacent sides" (Coxeter & Greitzer)
This means that for a triangle ABC, where angle A has a bisector AD such that D is on the side BC, then
BD/DC=AB/AC
Here either
BD/DC=6/5=AB/AC, where AB=6.9,
then we solve for AC=AB*5/6=5.75,
or
BD/DC=6/5=AB/AC, where AC=6.9,
then we solve for AB=AC*6/5=8.28
Hence, the longest and shortest possible lengths of the third side are
8.28 and 5.75 units respectively.