Answer:11
Step-by-step explanation:
let's recall that on the II Quadrant x/cosine is negative whilst y/sine is positive,
also let's recall that the hypotenuse is simply the radius distance and thus is never negative.
![\bf cot(\theta )=\cfrac{\stackrel{adjacent}{-2}}{\stackrel{opposite}{1}}\impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{(-2)^2+1^2}\implies c=\sqrt{5} \\\\[-0.35em] ~\dotfill\\\\ csc(\theta )=\cfrac{\stackrel{hypotenuse}{\sqrt{5}}}{\stackrel{opposite}{1}}\implies csc(\theta )=\sqrt{5}](https://tex.z-dn.net/?f=%5Cbf%20cot%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B-2%7D%7D%7B%5Cstackrel%7Bopposite%7D%7B1%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20hypotenuse%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20c%3D%5Csqrt%7B%28-2%29%5E2%2B1%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B5%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20csc%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bhypotenuse%7D%7B%5Csqrt%7B5%7D%7D%7D%7B%5Cstackrel%7Bopposite%7D%7B1%7D%7D%5Cimplies%20csc%28%5Ctheta%20%29%3D%5Csqrt%7B5%7D)
Answer:
SSS for Side Side Side
Step-by-step explanation:
To prove:
Triangle ABC is isosceles so AB and AC are congruent. BM and NC are congruent because it is given. Since Triangle AMN isosceles the sides AM and AN are congruent. Also the definition of the isosceles Triangle. Because of this the triangles BAM and CAN are congruent.
Answer:
18 2/3 miles. 4 x 2/3 =8/3 x 7 = 56/3 ÷3 = 18 2/3 miles.
4.47, if you put it in a calculator, not sure if teacher wants something different.
