Triangle LMN is similar to Triangle MKN.
If we look at the smaller triangle created by the line from angle M, we can see that if we flip the smaller triangle so angles M and L would match up, the triangles are similar.
Triangle LMN is similar to Triangle LKM.
If we look at the entire triangle as a whole, we can see that is is a right triangle as well. As long as we remember to match up the angles, we can say that the triangles are similar.
Hope this helps!! :)
Answer:
y = -½x + 4
Step-by-step explanation:
the line passes point of y-intercept (0, 4) ,
and another point (6, 1)
the slope = (1-4)/(6-0) = (-3)/6 = -½
so the Equation is y = -½x + 4
Answer:
132 i think
Step-by-step explanation:
Answer:
○![\displaystyle -\frac{2\sqrt{3}}{3}\:[or\:-\frac{2}{\sqrt{3}}]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20-%5Cfrac%7B2%5Csqrt%7B3%7D%7D%7B3%7D%5C%3A%5Bor%5C%3A-%5Cfrac%7B2%7D%7B%5Csqrt%7B3%7D%7D%5D)
Step-by-step explanation:
You have to know the Unit Circle on this one. So, to start off, in the degree-angle range from 180° - 270° [Quadrant III], you must figure out the cosine value that gives you −½, and according to the graph, <em>theta</em><em> </em>will be represented as 240°. So, now that we have our <em>θ</em>, looking at the <em>y-value</em><em> </em>240°, since we want the <em>cosecant</em><em> </em>function, all we have to do is take the multiplicative inverse of 
which gives you 
Extended Information
![\displaystyle -\frac{2\sqrt{3}}{3} = [-\frac{\sqrt{3}}{2}]^{-1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20-%5Cfrac%7B2%5Csqrt%7B3%7D%7D%7B3%7D%20%3D%20%5B-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5D%5E%7B-1%7D)
I am joyous to assist you anytime.
Answer:
in question one you will pick option 1
in second the 3rd option
in last question option 4
