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KonstantinChe [14]
3 years ago
14

Que is on pic.i can't able to type in text.

Mathematics
1 answer:
ad-work [718]3 years ago
3 0
It's not difficult to compute the values of A and B directly:

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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The sum of three consecutive terms of an arithmetic sequence is 27, and the sum of their squares is 293. What is the absolute di
Ber [7]

The absolute difference between the greatest and the least of these three numbers in the arithmetic sequence is 10.

The sequence is an arithmetic sequence. Therefore,

d = common difference

let

a = centre term

Therefore, the 3 consecutive term will be as follows

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a - d +  a + a + d = 27

3a = 27

a = 27 / 3

a = 9

Therefore,

(a-d)² + (a)² + (a + d)² = 293

(a²-2ad+d²) + 9² + (a² + 2ad + d²) = 293

(81 - 18d + d²) + 81 + (81 + 18d + d²) = 293

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2d² = 50

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d = 5

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Therefore, the 3 numbers are as follows

9 - 5 , 9, 9 + 5 = 4, 9, 14

The difference between the greatest and the least of these 3 numbers are as follows:

14 - 4 = 10

learn more on Arithmetic progression: brainly.com/question/25749583?referrer=searchResults

7 0
2 years ago
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470

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hope this helped :)

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