Answer:
by combining like terms.
Step-by-step explanation:
so you look at what you are given and you see that you have 1x and no other x's. you see that you are given 1y and 2 more y's therefore having 3y's since theyre both positive so you add them. you then see that you have a 4 and a 1 left over, which add to make 5. hope this helped!!!
To solve for
![\frac{1}{2}- \frac{1}{10}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D-%20%5Cfrac%7B1%7D%7B10%7D%20)
You first need to find a common denominator.
To do so, you need to make both denominators 10 by multiplying the top and bottom of
![\frac{1}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20)
by 5
![\frac{5}{10}- \frac{1}{10}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7B10%7D-%20%5Cfrac%7B1%7D%7B10%7D%20)
=
![\frac{4}{10}](https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7B10%7D%20)
Reduce by dividing both the top and bottom by 2
Your answer is
Answer: A
Step-by-step explanation:
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)