Answer:
the answer to the question is "C"
 
        
             
        
        
        
<u>211 - 87.40</u> equals 123.60. 
To see if your answer is correct just <u>add 123.60 and 87.40 together</u>. 
 
        
                    
             
        
        
        
Answer:
A) x = 20
Step-by-step explanation:
ABCD is a rhombus, AC and BD are diagonals which intersect each other at point E. 
Since, diagonals of a rhombus are perpendicular bisector. 

 
        
             
        
        
        
bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.
![\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}](https://tex.z-dn.net/?f=%5Cbf%20cos%5Cleft%5B%20sin%5E%7B-1%7D%5Cleft%28%20-%5Ccfrac%7B5%7D%7B13%7D%20%5Cright%29%20%5Cright%5D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bthen%20we%20can%20say%20that%7D~%5Chfill%20%7D%7Bsin%5E%7B-1%7D%5Cleft%28%20-%5Ccfrac%7B5%7D%7B13%7D%20%5Cright%29%5Cimplies%20%5Ctheta%20%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Btherefore%20then%7D~%5Chfill%20%7D%7Bsin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B-5%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B13%7D%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%7D%7D)
![\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%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%20%5Cpm%5Csqrt%7B13%5E2-%28-5%29%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B144%7D%3Da%5Cimplies%20%5Cpm%2012%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20cos%5Cleft%5B%20sin%5E%7B-1%7D%5Cleft%28%20-%5Ccfrac%7B5%7D%7B13%7D%20%5Cright%29%20%5Cright%5D%5Cimplies%20cos%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B%5Cpm%2012%7D%7D%7B13%7D)
le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.
 
        
             
        
        
        
Answer:
  x = -21/5 = -4.200
Step-by-step explanation: