Answer:
Part 1) 
Part 2) 
Part 3) 
Step-by-step explanation:
Part 1) Find AC
we know that
In the right triangle ABC of the figure
Applying the Pythagorean Theorem

substitute the given values


Part 2) Find the measure of angle A
we know that
In the right triangle ABC
----> by TOA (opposite side divided by the adjacent side)
substitute the values

using a calculator

Part 3) Find the measure of angle C
we know that
In the right triangle ABC
----> by complementary angles
substitute the given value



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:<span> 3/5</span>
Explanation:
<span>60/100</span> = <span>6/10</span> = <span>3<span>/5</span></span>
Answer:
False
Step-by-step explanation:
There are two concepts:
1. Row Echelon Form: There can be more than two <em>row echelon forms</em> of a single matrix, so different sequences of row operations can lead to different <em>row echelon forms</em> of a single matrix.
2. Reduced Row Echelon Form: It's unique for each matrix, so different sequences of row operations always lead to the same <em>reduced row echelon form</em> for the same matrix.