Recall your SOH CAH TOA, or
![\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad \qquad % cosine cos(\theta)=\cfrac{adjacent}{hypotenuse} \\ \quad \\\\ % tangent tan(\theta)=\cfrac{opposite}{adjacent}](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Bhypotenuse%7D%0A%5Cqquad%20%5Cqquad%20%0A%25%20cosine%0Acos%28%5Ctheta%29%3D%5Ccfrac%7Badjacent%7D%7Bhypotenuse%7D%0A%0A%5C%5C%20%5Cquad%20%5C%5C%5C%5C%0A%25%20tangent%0Atan%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Badjacent%7D)
now, bear in mind that, the opposite side, to an angle, is the side right "in front of it", that is, if you were to put your eye on that angle, "the wall" you'd see on the other end, is the opposite side
the adjacent side, adjacent = next to, is the side that's touching the angle itself
and the hypotenuse, is always the slanted and longest side of all three
for example on 16, tangent of Z
if you put one eye on Z, you'll see on the other end, the side of 30 units
the adjacent side is the one touching Z, or the 40 units side
![\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(Z)=\cfrac{30}{40} \\\\\\ \textit{which can be simplified to }\cfrac{3}{4} ](https://tex.z-dn.net/?f=%5Cbf%20tan%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Badjacent%7D%5Cimplies%20tan%28Z%29%3D%5Ccfrac%7B30%7D%7B40%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bwhich%20can%20be%20simplified%20to%20%7D%5Ccfrac%7B3%7D%7B4%7D%0A)
and you'd do all others, the same way, using those ratios
use
12ab+6ab
The answer is
= > 18ab
First add the whole numbers;
5 + 3 = 8
then add the numerators( you could add them cause they have like denominators);
8/12 + 5/12 = (8+5) = 13/12
13/12 = 1 1/2
so your final answer would be: 9 1/2 (I added the whole numbers)
Answer:
7
Step-by-step explanation:
Answer:
16.66months
Step-by-step explanation:
2500/150