3)
we know that θ is the IV quadrant, well, there the sine(y) is negative and the cosine(x) is positive, we have a tangent(sine/cosine) of -1/2, so one of those is the negative, in the IV Quadrant it has to be the sine, so the 1 is negative, so the fraction is (-1)/2.

![\bf c=\sqrt{2^2+(-1)^2}\implies c=\sqrt{5} \\\\\\ \stackrel{\textit{double-angle identities}}{cos(2\theta )=2cos^2(\theta )-1}\implies 2[cos(\theta )]^2-1\implies 2\left[ \cfrac{\stackrel{adjacent}{2}}{\stackrel{hypotenuse}{\sqrt{5}}} \right]^2-1 \\\\\\ 2\cdot \cfrac{2^2}{(\sqrt{5})^2}-1\implies 2\cdot \cfrac{4}{5}-1\implies \cfrac{8}{5}-1\implies \cfrac{3}{5}](https://tex.z-dn.net/?f=%20%5Cbf%20c%3D%5Csqrt%7B2%5E2%2B%28-1%29%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B5%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bdouble-angle%20identities%7D%7D%7Bcos%282%5Ctheta%20%29%3D2cos%5E2%28%5Ctheta%20%29-1%7D%5Cimplies%202%5Bcos%28%5Ctheta%20%29%5D%5E2-1%5Cimplies%202%5Cleft%5B%20%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B2%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B%5Csqrt%7B5%7D%7D%7D%20%5Cright%5D%5E2-1%20%5C%5C%5C%5C%5C%5C%202%5Ccdot%20%5Ccfrac%7B2%5E2%7D%7B%28%5Csqrt%7B5%7D%29%5E2%7D-1%5Cimplies%202%5Ccdot%20%5Ccfrac%7B4%7D%7B5%7D-1%5Cimplies%20%5Ccfrac%7B8%7D%7B5%7D-1%5Cimplies%20%5Ccfrac%7B3%7D%7B5%7D%20)
4)
recall that the hypotenuse is just a length unit, and is never negative, therefore on that sine ratio of opposite/hypotenuse, the negative is the 4, so the fraction is really (-4)/5,
also let's recall that the cosine(x) is also negative in the III Quadrant.


Answer:
Step-by-step explanation:
y / 4 = 8.....this is the same as (1/4)y = 8...so multiply both sides by 4...this will cancel out the 1/4 on the left side
4(1/4)y = 8(4)
y = 32
Answer:
2,698 mi
Step-by-step explanation:
3.8 hr × 1/2 = 1.9 hr
680 mi/hr × 1.9 hr = 1,292 mi
740 mi/hr × 1.9 hr = 1,406 mi
1,292 mi + 1,406 mi = 2,698 mi
Answer: A trapezoid is always a parallelogram.
Step-by-step explanation:
Answer: meter
Step-by-step explanation: