Where is this triangle you speak of
Answer:
Solve this equation. Enter your answer in the box. 75 â€"" 3. 5y â€"" 4y = 4y 6.
f(x) = 1 - ²/ₓ₃
y = 1 - ²/ₓ₃
y = 1 - ²/ₓ₃
y - 1 = ⁻²/ₓ₃
x - 1 = -2/y³
y³(x - 1) = -2
y³ = ⁻²/ₓ₋₁
y = ∛⁻²/ₓ₋₁
y = -∛(2x² - 4x + 2)/x - 1
f⁻¹(x) = -∛(2x² - 4x + 2)/x - 1
Now, we know that 90°< θ <180°, that simply means the angle θ is in the II quadrant, where sine is positive and cosine is negative.
![\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{13}}\impliedby \textit{now let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B5%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B13%7D%7D%5Cimpliedby%20%5Ctextit%7Bnow%20let%27s%20find%20the%20%5Cunderline%7Badjacent%20side%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%5C%5C%5C%5C%0Ac%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ac%3Dhypotenuse%5C%5C%0Aa%3Dadjacent%5C%5C%0Ab%3Dopposite%5C%5C%0A%5Cend%7Bcases%7D)
Answer:
y= -1x+7
Step-by-step explanation:
I believe this is the correct answer. The graph starts at 7 which is the y-intercept. And starts declining my -1