Answer:
Step-by-step explanation:
3/16= 0.1875
Though if you mean 16/3 it is 5.333.....
we know that the cos(θ) is -(3/5), however θ is in the II Quadrant, where the cosine is negative whilst the sine is positive, meaning the fraction is really (-3)/5, so
![cos(\theta )=\cfrac{\stackrel{adjacent}{-3}}{\underset{hypotenuse}{5}}\qquad \qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}](https://tex.z-dn.net/?f=cos%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B-3%7D%7D%7B%5Cunderset%7Bhypotenuse%7D%7B5%7D%7D%5Cqquad%20%5Cqquad%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Bopposite%20side%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Csqrt%7Bc%5E2-a%5E2%7D%3Db%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D)
![\pm\sqrt{5^2-(-3)^2}=b\implies \pm\sqrt{25-9}=b\implies \pm 4=b\implies \stackrel{II~Quadrant}{+4=b} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill csc(\theta )=\cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{4}}~\hfill](https://tex.z-dn.net/?f=%5Cpm%5Csqrt%7B5%5E2-%28-3%29%5E2%7D%3Db%5Cimplies%20%5Cpm%5Csqrt%7B25-9%7D%3Db%5Cimplies%20%5Cpm%204%3Db%5Cimplies%20%5Cstackrel%7BII~Quadrant%7D%7B%2B4%3Db%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20csc%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bhypotenuse%7D%7B5%7D%7D%7B%5Cunderset%7Bopposite%7D%7B4%7D%7D~%5Chfill)
Answer:
a = -3/2
Step-by-step explanation:
You can set x=0 and solve for y to find the y-intercept in each case.
first equation: y-intercept = -3/2
second equation: y-intercept = a
To make these equal, we must have ...
a = -3/2
They repeat by adding that same number again and again