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Diffstat (limited to 'Master/texmf-dist/doc/support/ketcindy/source/ketmanual/Fig/mxtex02.tex')
-rw-r--r-- | Master/texmf-dist/doc/support/ketcindy/source/ketmanual/Fig/mxtex02.tex | 12 |
1 files changed, 6 insertions, 6 deletions
diff --git a/Master/texmf-dist/doc/support/ketcindy/source/ketmanual/Fig/mxtex02.tex b/Master/texmf-dist/doc/support/ketcindy/source/ketmanual/Fig/mxtex02.tex index ca72d2d3660..7059840ea97 100644 --- a/Master/texmf-dist/doc/support/ketcindy/source/ketmanual/Fig/mxtex02.tex +++ b/Master/texmf-dist/doc/support/ketcindy/source/ketmanual/Fig/mxtex02.tex @@ -5,14 +5,14 @@ ( 10.22000, 3.87000)( -0.61000, 2.03000)% \special{pn 8}% % -\settowidth{\Width}{部分分数への分解 $\frac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\frac{8}{x+2}-\frac{1}{x+1}+x-3$}\setlength{\Width}{0\Width}% -\settoheight{\Height}{部分分数への分解 $\frac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\frac{8}{x+2}-\frac{1}{x+1}+x-3$}\settodepth{\Depth}{部分分数への分解 $\frac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\frac{8}{x+2}-\frac{1}{x+1}+x-3$}\setlength{\Height}{-0.5\Height}\setlength{\Depth}{0.5\Depth}\addtolength{\Height}{\Depth}% -\put(0.0500,5.0000){\hspace*{\Width}\raisebox{\Height}{部分分数への分解 $\frac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\frac{8}{x+2}-\frac{1}{x+1}+x-3$}}% +\settowidth{\Width}{Decomposition into partial fractions}\setlength{\Width}{0\Width}% +\settoheight{\Height}{Decomposition into partial fractions}\settodepth{\Depth}{Decomposition into partial fractions}\setlength{\Height}{-0.5\Height}\setlength{\Depth}{0.5\Depth}\addtolength{\Height}{\Depth}% +\put(0.0500,5.0000){\hspace*{\Width}\raisebox{\Height}{Decomposition into partial fractions}}% % % -\settowidth{\Width}{部分分数への分解 $\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\setlength{\Width}{0\Width}% -\settoheight{\Height}{部分分数への分解 $\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\settodepth{\Depth}{部分分数への分解 $\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\setlength{\Height}{-0.5\Height}\setlength{\Depth}{0.5\Depth}\addtolength{\Height}{\Depth}% -\put(0.0500,3.0000){\hspace*{\Width}\raisebox{\Height}{部分分数への分解 $\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}}% +\settowidth{\Width}{$\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\setlength{\Width}{0\Width}% +\settoheight{\Height}{$\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\settodepth{\Depth}{$\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\setlength{\Height}{-0.5\Height}\setlength{\Depth}{0.5\Depth}\addtolength{\Height}{\Depth}% +\put(0.0500,3.5000){\hspace*{\Width}\raisebox{\Height}{$\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}}% % % \end{picture}}%
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