A) 0.5 would be the greater number out of them all.
Answer:
Step-by-step explanation:
This problem is just really long- it's trying to confuse you with complicated wording. Here's it demystified:
a)
t = 0 degrees Celsius
d = 1500 meters
Find the temperature, T.
b)
d = 300 meters
T = 26 degrees Celsius
Find the ground temperature, t.
NOTE: The temperatures are different! T is the final temperature and t is the ground temperature.
Saturn is 10 times farther from the sun than Venus
Every function is a relation because the numbers have relationships but not every relation is a function because for it to be a function there has to be one y fo every x if any x(input) has more than one y(output) it's not a function
well, for both angles A and B we're on the IV Quadrant, meaning, the sine is negative, the cosine is positive, likewise, the opposite side is negative and the adjacent side for the angle is positive.
![\bf cos(A)=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{5^2-3^2}}\implies b = \pm 4 \\\\\\ \stackrel{IV~Quadrant}{b = -4}\qquad \qquad sin(A)=\cfrac{\stackrel{opposite}{-4}}{\underset{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill\\\\ cos(B)=\cfrac{\stackrel{adjacent}{12}}{\underset{hypotenuse}{13}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{13^2-12^2}}\implies b = \pm 5](https://tex.z-dn.net/?f=%5Cbf%20cos%28A%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B3%7D%7D%7B%5Cunderset%7Bhypotenuse%7D%7B5%7D%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bgetting%20the%20opposite%20side%7D%7D%7Bb%3D%5Cpm%5Csqrt%7B5%5E2-3%5E2%7D%7D%5Cimplies%20b%20%3D%20%5Cpm%204%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7BIV~Quadrant%7D%7Bb%20%3D%20-4%7D%5Cqquad%20%5Cqquad%20sin%28A%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B-4%7D%7D%7B%5Cunderset%7Bhypotenuse%7D%7B5%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20cos%28B%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B12%7D%7D%7B%5Cunderset%7Bhypotenuse%7D%7B13%7D%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bgetting%20the%20opposite%20side%7D%7D%7Bb%3D%5Cpm%5Csqrt%7B13%5E2-12%5E2%7D%7D%5Cimplies%20b%20%3D%20%5Cpm%205)
![\bf \stackrel{IV~Quadrant}{b = -5}\qquad \qquad sin(B)=\cfrac{\stackrel{opposite}{-5}}{\underset{hypotenuse}{13}} \\\\[-0.35em] ~\dotfill\\\\ sin(A-B)=\cfrac{-4}{5}\cdot \cfrac{12}{13}-\left( \cfrac{3}{5}\cdot \cfrac{-5}{13} \right)\implies sin(A-B)=\cfrac{-48}{65} - \left( \cfrac{-15}{65} \right) \\\\\\ sin(A-B)=\cfrac{-48}{65} + \cfrac{15}{65}\implies sin(A-B)=\cfrac{-33}{65}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7BIV~Quadrant%7D%7Bb%20%3D%20-5%7D%5Cqquad%20%5Cqquad%20sin%28B%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B-5%7D%7D%7B%5Cunderset%7Bhypotenuse%7D%7B13%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20sin%28A-B%29%3D%5Ccfrac%7B-4%7D%7B5%7D%5Ccdot%20%5Ccfrac%7B12%7D%7B13%7D-%5Cleft%28%20%5Ccfrac%7B3%7D%7B5%7D%5Ccdot%20%5Ccfrac%7B-5%7D%7B13%7D%20%5Cright%29%5Cimplies%20sin%28A-B%29%3D%5Ccfrac%7B-48%7D%7B65%7D%20-%20%5Cleft%28%20%5Ccfrac%7B-15%7D%7B65%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20sin%28A-B%29%3D%5Ccfrac%7B-48%7D%7B65%7D%20%2B%20%5Ccfrac%7B15%7D%7B65%7D%5Cimplies%20sin%28A-B%29%3D%5Ccfrac%7B-33%7D%7B65%7D)
