Just multiply 44 times 3 the you should get your answer
No, it is not a solution to the inequality.
The pro<span>bability of randomly pulling out a pineapple candy would be 5 over 8. Because there are 8 candies total (this is the denominator) and only 3 flavors, you just simply add the two given candies, which is 2 + 1 = 3, and subtract the sum (which is 3) by 8. Therefore, 8 - 3 = 5 and you put that as the numerator and the total number of candies, 8, as the denominator. </span>
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let's notice something, angles α and β are both in the I Quadrant, and on the first quadrant the x-coordinate/cosine and y-coordinate/sine are both positive.
![\bf \textit{Sum and Difference Identities} \\\\ cos(\alpha - \beta)= cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(\alpha)=\cfrac{\stackrel{opposite}{15}}{\stackrel{hypotenuse}{17}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BSum%20and%20Difference%20Identities%7D%20%5C%5C%5C%5C%20cos%28%5Calpha%20-%20%5Cbeta%29%3D%20cos%28%5Calpha%29cos%28%5Cbeta%29%20%2B%20sin%28%5Calpha%29sin%28%5Cbeta%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20sin%28%5Calpha%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B15%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B17%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%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%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D)
![\bf \pm\sqrt{17^2-15^2}=a\implies \pm\sqrt{64}=a\implies \pm 8 = a\implies \stackrel{I~Quadrant}{\boxed{+8=a}} \\\\[-0.35em] ~\dotfill\\\\ cos(\beta)=\cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cpm%5Csqrt%7B17%5E2-15%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B64%7D%3Da%5Cimplies%20%5Cpm%208%20%3D%20a%5Cimplies%20%5Cstackrel%7BI~Quadrant%7D%7B%5Cboxed%7B%2B8%3Da%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20cos%28%5Cbeta%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B3%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B5%7D%7D%5Cimpliedby%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%5Cpm%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)
![\bf \pm\sqrt{5^2-3^2}=b\implies \pm\sqrt{16}=b\implies \pm 4=b\implies \stackrel{\textit{I~Quadrant}}{\boxed{+4=b}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cpm%5Csqrt%7B5%5E2-3%5E2%7D%3Db%5Cimplies%20%5Cpm%5Csqrt%7B16%7D%3Db%5Cimplies%20%5Cpm%204%3Db%5Cimplies%20%5Cstackrel%7B%5Ctextit%7BI~Quadrant%7D%7D%7B%5Cboxed%7B%2B4%3Db%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf cos(\alpha - \beta)=\stackrel{cos(\alpha)}{\left( \cfrac{8}{17} \right)}\stackrel{cos(\beta)}{\left( \cfrac{3}{5} \right)}+\stackrel{sin(\alpha)}{\left( \cfrac{15}{17} \right)}\stackrel{sin(\beta)}{\left( \cfrac{4}{5} \right)}\implies cos(\alpha - \beta)=\cfrac{24}{85}+\cfrac{60}{85} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill cos(\alpha - \beta)=\cfrac{84}{85}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20cos%28%5Calpha%20-%20%5Cbeta%29%3D%5Cstackrel%7Bcos%28%5Calpha%29%7D%7B%5Cleft%28%20%5Ccfrac%7B8%7D%7B17%7D%20%5Cright%29%7D%5Cstackrel%7Bcos%28%5Cbeta%29%7D%7B%5Cleft%28%20%5Ccfrac%7B3%7D%7B5%7D%20%5Cright%29%7D%2B%5Cstackrel%7Bsin%28%5Calpha%29%7D%7B%5Cleft%28%20%5Ccfrac%7B15%7D%7B17%7D%20%5Cright%29%7D%5Cstackrel%7Bsin%28%5Cbeta%29%7D%7B%5Cleft%28%20%5Ccfrac%7B4%7D%7B5%7D%20%5Cright%29%7D%5Cimplies%20cos%28%5Calpha%20-%20%5Cbeta%29%3D%5Ccfrac%7B24%7D%7B85%7D%2B%5Ccfrac%7B60%7D%7B85%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20cos%28%5Calpha%20-%20%5Cbeta%29%3D%5Ccfrac%7B84%7D%7B85%7D~%5Chfill)
Answer: hi there! your answer should be d let me tell you why...
This function has no zeros.
This is because f(x)=12, meaning no matter the x value, the y value is always 12. If the y value is always 12, it is never 0, and therefore has no zeros.
hope this help have a great day! let me know if im right or wrong plz!:)
