A few ways. you can enter it into a calculator for one, but the easiest way would be to reference a unit circle and look for an ordered pair where sin (the y value), is equal to -1/2
on a unit circle, the sin value is -1/2 at 7pi/6 and 11pi/6
because sin is the y value, it will additionally be in the quadrants III or IV based on the fact that (-1/2) as a sin value IS a negative and would have to be found in a quadrant in which sin is negative (the lower half of a coordinate plane, in this case)
-10/8 and + 15/8. Those should work.
Answer:
I believe this is correct
A is FALSE
B is TRUE
C is FALSE
tam olmasada yaklaşık 6.5 günde bitter
![\bf 1:1.618\iff \cfrac{1}{1.618}\cfrac{\leftarrow \textit{small side=width}}{\leftarrow \textit{longer side=length}}\\\\ -----------------------------\\\\ \textit{now, what if the length is 35?}\to \cfrac{1}{1.618}=\cfrac{w}{\boxed{35}}](https://tex.z-dn.net/?f=%5Cbf%201%3A1.618%5Ciff%20%5Ccfrac%7B1%7D%7B1.618%7D%5Ccfrac%7B%5Cleftarrow%20%5Ctextit%7Bsmall%20side%3Dwidth%7D%7D%7B%5Cleftarrow%20%5Ctextit%7Blonger%20side%3Dlength%7D%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Ctextit%7Bnow%2C%20what%20if%20the%20length%20is%2035%3F%7D%5Cto%20%5Ccfrac%7B1%7D%7B1.618%7D%3D%5Ccfrac%7Bw%7D%7B%5Cboxed%7B35%7D%7D)
solve for "w" to get the width
what's the area of a rectangle? length * width, or lw