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nataly862011 [7]
3 years ago
14

Jay, Kay, and Ray found themselves far apart when they stopped for lunch while walking in a field. Jay could see Kay, then turn

through 75th, and could see Ray. Jay could see Ray, then turn through 50th amd see Jay. Ray could see Jay, the turn through 55th, and see Kay. Which two were the farthest apart?
A. Kay and Ray
B. Jay and Kay
C. Ray and Jay
D. Kay and Ray were the same distance apart as Ray and Jay
Mathematics
2 answers:
MariettaO [177]3 years ago
8 0
Hi There! :)

Jay, Kay, and Ray found themselves far apart when they stopped for lunch while walking in a field. Jay could see Kay, then turn through 75th, and couldsee Ray. Jay could see Ray, then turn through 50th amd see Jay. Ray could see Jay, the turn through 55th, and see Kay. Which two were the farthest apart? 

A. Kay and Ray
B. Jay and Kay
C. Ray and Jay
D. Kay and Ray were the same distance apart as Ray and Jay

I believe it's <span>A. Kay and Ray</span>
irakobra [83]3 years ago
3 0
Well, a Kay and Ray are the two farthest because Kay sees through 75th Ray. it couldn't be ,b Jay and Kay because Jay sees through 55th Kay.
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A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}
A=\tan^{-1}(\sin\theta)-\dfrac\pi4

B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt
B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}
B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2

Let's assume 0, so that |\csc\theta|=\csc\theta.

Now,

\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}
\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}
\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)
\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in A and B.
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m_a_m_a [10]

a^{p-1} \equiv 1 \pmod p where p is prime, a\in\mathbb{Z} and a is not divisible by p.

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