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3 years ago

O is the centroid. Find the value of x. (Sorry if u cant see the photo :/, but this is Geometry btw)

Mathematics

1 answer:

klemol [59]3 years ago

8 0

Answer: x = 6

==============================================

Explanation:

The medians of a triangle meet up at the centroid in such a way that they cut each other at a ratio of 2:1, meaning that one part of the median is twice as long as the other. In this case, MO is two times longer than OP, so,

OP = 2*MO

and

MP = MO + OP

MP = MO + 2*MO .... replace OP with 2*MO

MP = 3*MO

9x-24 = 3*(x+4) ... plug in the given expressions

9x-24 = 3x+12

9x-3x = 12+24

6x = 36

x = 36/6

x = 6

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2 years ago

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It's not difficult to compute the values of $A$ and $B$ directly:

$A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}$
$A=\tan^{-1}(\sin\theta)-\dfrac\pi4$

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