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gulaghasi [49]
3 years ago
10

Attached is a screenshot of another problem I am having issues with. please please please help me

Mathematics
1 answer:
anyanavicka [17]3 years ago
4 0
\bf cos(\alpha+\beta)cos(\alpha-\beta)\\\\
-----------------------------\\\\
\textit{using the sum identities}\\\\\

\begin{array}{ccccccll}
[cos(\alpha)cos(\beta)&-&sin(\alpha)sin(\beta)]&[cos(\alpha)cos(\beta)&+&sin(\alpha)sin(\beta)]\\
a&-&b&a&+&b
\end{array}\\\\
\textit{notice above, is just a difference of squares, thus}
\\\\\\\
[cos(\alpha)cos(\beta)]^2-[sin(\alpha)sin(\beta)]^2\\\\
\textit{now let us distribute the exponents}\\\\

\bf cos^2(\alpha)cos^2(\beta)-sin^2(\alpha)sin^2(\beta)\\\\
-----------------------------\\\\
\textit{now, let us use the pythagorean identity of }\\\\
sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
-----------------------------\\\\\
[ 1-sin^2(\alpha)][ 1-sin^2(\beta)]-sin^2(\alpha)sin^2(\beta)
\\\\\\\
1-sin^2(\beta)-sin^2(\alpha)\underline{+sin^2(\alpha)sin^2(\beta)-sin^2(\alpha)sin^2(\beta)}
\\\\\\
\underline{1-sin^2(\beta)}-sin^2(\alpha)\implies cos^2(\beta)-sin^2(\alpha)
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Now, the number the freshmen in public schools is considered as 100% or 1.

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5 0
4 years ago
Helppppppppppppppppp
valina [46]
This is a simple "solve for x" equation.
5k-2k=12
3k=12
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You should try doing some more on your own, as this will be coming up a lot in the future.
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