The quadrilaterals whose consecutive and opposite angles are always congruent are the square and the rectangle. All of the angles of the square and the rectangle are 90 degrees. The consecutive angles of the parallelogram and the rhombus are not equal.
now, if the denominator turns to 0, the fraction becomes undefined, and you get a "vertical asymptote" when that happens, so let's check when is that
![\bf sin\left(x-\frac{2\pi }{3} \right)=0\implies sin^{-1}\left[ sin\left(x-\frac{2\pi }{3} \right) \right]=sin^{-1}(0) \\\\\\ x-\frac{2\pi }{3}= \begin{cases} 0\\ \pi \end{cases}\implies \measuredangle x= \begin{cases} \frac{2\pi }{3}\\ \frac{5\pi }{3} \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20sin%5Cleft%28x-%5Cfrac%7B2%5Cpi%20%7D%7B3%7D%20%20%5Cright%29%3D0%5Cimplies%20sin%5E%7B-1%7D%5Cleft%5B%20sin%5Cleft%28x-%5Cfrac%7B2%5Cpi%20%7D%7B3%7D%20%20%5Cright%29%20%5Cright%5D%3Dsin%5E%7B-1%7D%280%29%0A%5C%5C%5C%5C%5C%5C%0Ax-%5Cfrac%7B2%5Cpi%20%7D%7B3%7D%3D%0A%5Cbegin%7Bcases%7D%0A0%5C%5C%0A%5Cpi%20%0A%5Cend%7Bcases%7D%5Cimplies%20%5Cmeasuredangle%20x%3D%0A%5Cbegin%7Bcases%7D%0A%5Cfrac%7B2%5Cpi%20%7D%7B3%7D%5C%5C%0A%5Cfrac%7B5%5Cpi%20%7D%7B3%7D%0A%5Cend%7Bcases%7D)
now, at those angles, the function is asymptotic, check the picture below
Answer:
4
Step-by-step explanation:
2+2 =4
Since we need a point that divides K-->J into two segments in a 1:3 ratio, let's flip it to a 3:1 ratio in the J-->K direction. That means that this point is 3/(3+1) = 3/4 of the way up line JK.
Let's take 3/4 of the x of K (plus the 1 for the x of J):
3/4×(9 - 1) + 1 (because the beginning ofvthe line, at J, has an x of 1)...
3/4×(8) + 1 = 6 + 1 = 7
Therefore the answer is C) 7
Answer:
the measure of the missing length is 3 mi
Step-by-step explanation:
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