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)
If you simply want to multiply the answers given to you I would say answer c. 6lbs peanuts, 3 pound almonds and 2lb raisins. If you multiply 6*1.50 you get 9, you multiply 3*3.00 would also equal 9, and 2*1.50 is 3. 9+9+3=21
Using order of operations, you square the 3 and get 9.
then divide 2 by 11, then multiply that by 9 and subtract 9. your answer is -7.363636364
If you reduce the 5 for 20, you will get 1 for $4. Therefore, 12×4 will be the answer ($48)
