I'm not sure but I had it then I forgot
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
Hello there! I can help you! In order to answer those questions, we need to plug in the values, based off of the variable.
g. Okay. We are solving b - 10. b = -8. When you subtract something from a negative number, the number is even lower. Let's add the numbers first and then put in the negative symbol. 8 + 10 is 18. Put the negative sign and you get -18. The difference is -18.
h. Now, we solve a - b. a = 5 and b = -8. Because we are subtracting a negative number from a positive, we have to add both numbers, which means the number gets bigger. Ignore the negative sign and add. 5 + 8 is 13. There. The sum is 13.
i. The problem is c - a. c = -9 and a = 5. So as explained on problem G, for this problem, ignore the negative symbol and add. 9 + 5 is 14. Plug in the negative sign to get -14. There. The difference is -14.
Answer:
Step-by-step explanation:
for right-angled triangle: a^2 + b^2 = c^2
given a = 5 n b = 10
c^2 = 5^2 + 10^2
= 25 + 100
= 125
c = sqrt(125)
= 5*sqrt(5)
