Answer:
<h2>x = 0, y = 5, z = 3</h2>
Step-by-step explanation:



Answer:
Step-by-step explanation:
2X + 1 + 3X - 2 + 4X - 5 = 180
9X - 6 = 180
9X = 174
9 = 174
ـــــــــــــــــــــ
9 9
X = 19.333
Answer:
Step-by-step explanation:
Below is the pic of how this would be set up in order to determine what it is you are looking for. The angle is set in QI, and since csc A is the reciprocal of sin, the ratio is hypotenuse over side opposite. Solve for the missing side using Pythagorean's Theorem:
and
1369 = 144 + b² and
1225 = b² so
b = 35
The sec ratio is the reciprocal of cos, so if cos is adjacent over hypotenuse, the sec is hypotenuse over adjacent, which is 37/35
Answer:
i dont think you can simplify it any more
Step-by-step explanation:
Answer:
The answer to your question is AC = 9.9
Step-by-step explanation:
To solve this problem use trigonometric functions. The trigonometric function that relates the hypotenuse and the adjacent side is cosine.
cos α = 
Solve for hypotenuse

Substitution
hypotenuse = 
Simplification and result
hypotenuse = 9.85