The shape above has the following

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

8 0

It’s A 44-1) and if not then it’s b

You might be interested in

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

ad-work [718]

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 $B$ .

3 0

3 years ago

Can somebody please help me?

klasskru [66]

Answer:

TRUE!!!!!!!!!!!!!!!!!! BRANLIST

Step-by-step explanation:

8 0

3 years ago

The sum of the interior angles of a regular pentangon

natima [27]

Answer:

540

Step-by-step explanation:

ARCHITECTURE The Pentagon in Washington, D.C. is shaped like a regular pentagon. Find the sum of the measures of the interior angles of the largest pentagon-shaped section of the Pentagon building. = 180(3) or 540 Simplify. The sum of the measures of the interior angles is 540.

4 0

2 years ago

Read 2 more answers

Simplest form<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B5%7D%20-%20%20%5Cfrac%7B3%7D%7B8%7D%20%20" id="TexFormula

raketka [301]