The given equation is:
a = 3b^2 + 2
For this case, what you should do is write carefully each term that belongs to the equation.
a is equal to three times b squared plus two.
answer:
a is equal to three times b squared plus two.
You can't take away 6 from 2 but you can switch them and get 6-2 and that's =4 10-8=2 7-5=2
![\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}](https://tex.z-dn.net/?f=%5Cdfrac%7Bsin%5Ctheta%20%2B%20cos%5Ctheta%7D%7Bsin%5Ctheta-cos%5Ctheta%7D%2B%5Cdfrac%7Bsin%5Ctheta-cos%5Ctheta%7D%7Bsin%5Ctheta%2Bcos%5Ctheta%7D%3D%5Cdfrac%7B2sec%5E2%5Ctheta%7D%7Btan%5E2%5Ctheta-1%7D)
From Left side:
![\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)](https://tex.z-dn.net/?f=%5Cdfrac%7Bsin%5Ctheta%20%2B%20cos%5Ctheta%7D%7Bsin%5Ctheta-cos%5Ctheta%7D%5Cbigg%28%5Cdfrac%7Bsin%5Ctheta%2Bcos%5Ctheta%7D%7Bsin%5Ctheta%2Bcos%5Ctheta%7D%5Cbigg%29%2B%5Cdfrac%7Bsin%5Ctheta-cos%5Ctheta%7D%7Bsin%5Ctheta%2Bcos%5Ctheta%7D%5Cbigg%28%5Cdfrac%7Bsin%5Ctheta-cos%5Ctheta%7D%7Bsin%5Ctheta-cos%5Ctheta%7D%5Cbigg%29)
![\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}](https://tex.z-dn.net/?f=%5Cdfrac%7Bsin%5E2%5Ctheta%2B2cos%5Cthetasin%5Ctheta%2Bcos%5E2%5Ctheta%7D%7Bsin%5E2%5Ctheta-cos%5E2%5Ctheta%7D%2B%5Cdfrac%7Bsin%5E2%5Ctheta-2cos%5Cthetasin%5Ctheta%2Bcos%5E2%5Ctheta%7D%7Bsin%5E2%5Ctheta-cos%5E2%5Ctheta%7D)
NOTE: sin²θ + cos²θ = 1
![\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%20%2B%202cos%5Ctheta%20sin%5Ctheta%7D%7Bsin%5E2%5Ctheta-cos%5E2%5Ctheta%7D%2B%5Cdfrac%7B1-2cos%5Ctheta%20sin%5Ctheta%7D%7Bsin%5E2%5Ctheta-cos%5E2%5Ctheta%7D)
![\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%20%2B%202cos%5Ctheta%20sin%5Ctheta%2B1-2cos%5Ctheta%20sin%5Ctheta%7D%7Bsin%5E2%5Ctheta-cos%5E2%5Ctheta%7D)
![\dfrac{2}{sin^2\theta-cos^2\theta}](https://tex.z-dn.net/?f=%5Cdfrac%7B2%7D%7Bsin%5E2%5Ctheta-cos%5E2%5Ctheta%7D)
![\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}](https://tex.z-dn.net/?f=%5Cdfrac%7B2%7D%7B%5Cbigg%28sin%5E2%5Ctheta-cos%5E2%5Ctheta%5Cbigg%29%5Cbigg%28%5Cdfrac%7Bcos%5E2%5Ctheta%7D%7Bcos%5E2%5Ctheta%7D%5Cbigg%29%7D)
![\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}](https://tex.z-dn.net/?f=%5Cdfrac%7B2sec%5E2%5Ctheta%7D%7B%5Cdfrac%7Bsin%5E2%5Ctheta%7D%7Bcos%5E2%5Ctheta%7D-%5Cdfrac%7Bcos%5E2%5Ctheta%7D%7Bcos%5E2%5Ctheta%7D%7D)
![\dfrac{2sec^2\theta}{tan^2\theta-1}](https://tex.z-dn.net/?f=%5Cdfrac%7B2sec%5E2%5Ctheta%7D%7Btan%5E2%5Ctheta-1%7D)
Left side = Right side <em>so proof is complete</em>
Answer:
yes, the student who has 8 pets.
Step-by-step explanation:
the majority of the class has between 0-4 pets leaving the student with 8 to be the "outlier"