Answer: x = 3/2
<u>Step-by-step explanation:</u>
log (2) + log (x) = log (3) <em>Given</em>
log(2x) = log(3) <em>Condensing log rules</em>
2x = 3 <em>Cancellation log rule</em>
<em>Division Property</em>
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Answer:
15, -7 x=15 y=-7
Step-by-step explanation:
x+y=8, we'll come back to this
x-y=22, move y to the other side, x=y+22
now insert the value of x =(y+22) to the other equation like so, (y+22)+y=8, simplify
2y+22=8 2y=(-14), y=(-7)
Insert the value of y into the first equation again like so, x+(-7)=8, simplify
x=15
15, -7
The answer is
y^2-4x+4y-4=y^2+4y-4x-4=(y+2)²-4x - 8= 0
it is the same of x + 2 = (1/4 ) (y+2)²
the main formula is x-h= a (y-k)² so h=-2, k= - 2 and a=1/4
Answer:
36
Step-by-step explanation:
90/5 because CBE is 90 so CBA would be 90 too.
90/5 is 18 so do 2 times 18.
x=18
2(18)=36
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)
