An infinite series converges when abs(r) < 1.
Therefore, an infinite series converges for r = -2/3 and 3/4
Answer:
x = 3
Step-by-step explanation:
4(x+7)=9(x-1)+22
Begin solving for x by performing the indicated multiplication:
4x + 28 = 9x - 9 + 22
Combine the x terms:
28 = 9x - 4x - 9 + 22
28 = 5x + 13
Subtracting 13 from both sides isolates 5x:
15 = 5x. Then x = 3
![\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>