The 2 Is at the Thousands Place for 2,783
Also the 2 is in the Hundreds place for 7,283
I hope this helps
In order to find the answer you have to first multiply 16 times 6. Then multiply that answer by 92. That will be your answer! Hope this helps!!
![\bf [cot(\theta )+csc(\theta )]^2=\cfrac{1+cos(\theta )}{1-cos(\theta )} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{doing the left-hand side}}{[cot(\theta )+csc(\theta )]^2}\implies cot^2(\theta )+2cot(\theta )csc(\theta )+csc^2(\theta ) \\\\\\ \cfrac{cos^2(\theta )}{sin^2(\theta )}+2\cdot \cfrac{cos(\theta )}{sin(\theta )}\cdot \cfrac{1}{sin(\theta )}+\cfrac{1}{sin^2(\theta )}\implies \cfrac{cos^2(\theta )}{sin^2(\theta )}+\cfrac{2cos(\theta )}{sin^2(\theta )}+\cfrac{1}{sin^2(\theta )}](https://tex.z-dn.net/?f=%5Cbf%20%5Bcot%28%5Ctheta%20%29%2Bcsc%28%5Ctheta%20%29%5D%5E2%3D%5Ccfrac%7B1%2Bcos%28%5Ctheta%20%29%7D%7B1-cos%28%5Ctheta%20%29%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bdoing%20the%20left-hand%20side%7D%7D%7B%5Bcot%28%5Ctheta%20%29%2Bcsc%28%5Ctheta%20%29%5D%5E2%7D%5Cimplies%20cot%5E2%28%5Ctheta%20%29%2B2cot%28%5Ctheta%20%29csc%28%5Ctheta%20%29%2Bcsc%5E2%28%5Ctheta%20%29%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7Bcos%5E2%28%5Ctheta%20%29%7D%7Bsin%5E2%28%5Ctheta%20%29%7D%2B2%5Ccdot%20%5Ccfrac%7Bcos%28%5Ctheta%20%29%7D%7Bsin%28%5Ctheta%20%29%7D%5Ccdot%20%5Ccfrac%7B1%7D%7Bsin%28%5Ctheta%20%29%7D%2B%5Ccfrac%7B1%7D%7Bsin%5E2%28%5Ctheta%20%29%7D%5Cimplies%20%5Ccfrac%7Bcos%5E2%28%5Ctheta%20%29%7D%7Bsin%5E2%28%5Ctheta%20%29%7D%2B%5Ccfrac%7B2cos%28%5Ctheta%20%29%7D%7Bsin%5E2%28%5Ctheta%20%29%7D%2B%5Ccfrac%7B1%7D%7Bsin%5E2%28%5Ctheta%20%29%7D)
![\bf \cfrac{\stackrel{\textit{perfect square trinomial}}{cos^2(\theta )+2cos(\theta )+1}}{sin^2(\theta )}\implies \boxed{\cfrac{[cos(\theta )+1]^2}{sin^2(\theta )}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{doing the right-hand-side}}{\cfrac{1+cos(\theta )}{1-cos(\theta )}}\implies \stackrel{\textit{multiplying by the denominator's conjugate}}{\cfrac{1+cos(\theta )}{1-cos(\theta )}\cdot \cfrac{1+cos(\theta )}{1+cos(\theta )}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Cstackrel%7B%5Ctextit%7Bperfect%20square%20trinomial%7D%7D%7Bcos%5E2%28%5Ctheta%20%29%2B2cos%28%5Ctheta%20%29%2B1%7D%7D%7Bsin%5E2%28%5Ctheta%20%29%7D%5Cimplies%20%5Cboxed%7B%5Ccfrac%7B%5Bcos%28%5Ctheta%20%29%2B1%5D%5E2%7D%7Bsin%5E2%28%5Ctheta%20%29%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bdoing%20the%20right-hand-side%7D%7D%7B%5Ccfrac%7B1%2Bcos%28%5Ctheta%20%29%7D%7B1-cos%28%5Ctheta%20%29%7D%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20by%20the%20denominator%27s%20conjugate%7D%7D%7B%5Ccfrac%7B1%2Bcos%28%5Ctheta%20%29%7D%7B1-cos%28%5Ctheta%20%29%7D%5Ccdot%20%5Ccfrac%7B1%2Bcos%28%5Ctheta%20%29%7D%7B1%2Bcos%28%5Ctheta%20%29%7D%7D)
![\bf \cfrac{[1+cos(\theta )]^2}{\underset{\textit{difference of squares}}{[1-cos(\theta )][1+cos(\theta )]}}\implies \cfrac{[cos(\theta )+1]^2}{1^2-cos^2(\theta )} \\\\\\ \cfrac{[cos(\theta )+1]^2}{1-cos^2(\theta )}\implies \boxed{\cfrac{[cos(\theta )+1]^2}{sin^2(\theta )}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5B1%2Bcos%28%5Ctheta%20%29%5D%5E2%7D%7B%5Cunderset%7B%5Ctextit%7Bdifference%20of%20squares%7D%7D%7B%5B1-cos%28%5Ctheta%20%29%5D%5B1%2Bcos%28%5Ctheta%20%29%5D%7D%7D%5Cimplies%20%5Ccfrac%7B%5Bcos%28%5Ctheta%20%29%2B1%5D%5E2%7D%7B1%5E2-cos%5E2%28%5Ctheta%20%29%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B%5Bcos%28%5Ctheta%20%29%2B1%5D%5E2%7D%7B1-cos%5E2%28%5Ctheta%20%29%7D%5Cimplies%20%5Cboxed%7B%5Ccfrac%7B%5Bcos%28%5Ctheta%20%29%2B1%5D%5E2%7D%7Bsin%5E2%28%5Ctheta%20%29%7D%7D)
recall that sin²(θ) + cos²(θ) = 1, thus sin²(θ) = 1 - cos²(θ).
Answer:
$25,600
Step-by-step explanation:
- Given: P = $50,000, r = 20% & t = 3 years
- Formula for depreciation is given as:
- Plugging the values of P, r and t in the above formula, we find:
- So, after third year it will be worth by $25,600.
Answer: 90 - 15-57 = 18 18 is your answer hope this helped
Step-by-step explanation: plz make brainly
