答案：8.1或$-\frac{3}{4}$

答案：9.证明:(1)$\because CE$平分$∠ ACB,\therefore ∠ ACE=∠ ECB.$
$\because EG// BC,\therefore ∠ GEC=∠ ECB,$
$\therefore ∠ GEC=∠ ACE,\therefore EG=CG.$
(2)由(1)知$EG=CG,∠ GEC=∠ GCE.$
$\because G$是$EH$的中点,$\therefore EG=GH,$
$\therefore CG=GH,\therefore ∠ GCH=∠ GHC,$
$\therefore ∠ GCE+∠ GCH=90°,\therefore ∠ ECB+∠ HCD=90°.$
$\because ∠ GCE=∠ ECB,\therefore ∠ GCH=∠ HCD,$
$\therefore CF$平分$∠ ACD.$

答案：10.解:(1)$y_1=0.58x,y_2=0.28x+656,$
$\therefore y_1$与$x$之间的函数表达式为$y_1=0.58x$,$y_2$与$x$之间的函数表达式为$y_2=0.28x+656.$
(2)当$y_1=1160$时,$0.58x=1160$,解得$x=2000$,
当$y_2=1160$时,$0.28x+656=1160$,解得$x=1800$,
$\because 2000＞1800,\therefore$选择铁路运输方式运输牛奶多.
(3)当$y_1＜y_2$时,$0.58x＜0.28x+656$,解得$x＜2186\frac{2}{3}$,
当$y_1=y_2$时,$0.58x=0.28x+656$,解得$x=2186\frac{2}{3}$,
当$y_1＞y_2$时,$0.58x＞0.28x+656$,解得$x＞2186\frac{2}{3}$,
$\therefore$当运输牛奶少于$2186\frac{2}{3}$千克时,选择铁路运输所需费用较少;当运输牛奶等于$2186\frac{2}{3}$千克时,两种运输方式所需费用相等;当运输牛奶超过$2186\frac{2}{3}$千克时,选择公路运输所需费用较少.

答案：
11.解:(1)在$y=-x+6$中,令$x=0$,得$y=6$;令$y=0$,得$x=6$,故$A(6,0),B(0,6).$
(2)设$D(m,-m+6)$,由$S_{△ AOD}=\frac{1}{2}S_{△ ABO}$,得$\frac{1}{2}OA· |-m+6|=\frac{1}{2}×\frac{1}{2}OB· OA$,即$|-m+6|=3$,解得$m=3$或$m=9$,$\therefore$点$D$的坐标为$(3,3)$或$(9,-3).$
(3)不变.解法一:设$P(t,0)(t＞6)$,如答图,过点$Q$作$QM⊥ x$轴于点$M$,则$∠ QMP=∠ POB=90°$,
$\because △ BPQ$是等腰直角三角形,
$\therefore PB=PQ,∠ BPQ=90°$,
$\therefore ∠ OPB+∠ QPM=∠ PQM+∠ QPM=90°$,
$\therefore ∠ OPB=∠ PQM,\therefore △ POB≌ △ QMP(\mathrm{AAS})$,
$\therefore PM=OB=6,MQ=OP=t,\therefore Q(t+6,t).$
设直线$AQ$的函数表达式为$y=kx+b$,
则$\begin{cases}6k+b=0,\\(t+6)k+b=t,\end{cases}$解得$\begin{cases}k=1,\\b=-6.\end{cases}$
$\therefore$直线$AQ$的函数表达式为$y=x-6$,交$y$轴于点$(0,-6)$,$\therefore$点$K$的坐标为$(0,-6).$

解法二:由解法一可知$PM=OB=OA$,
$\therefore OA+AP=PM+AP$,即$OP=AM.$
又$\because QM=OP$,
$\therefore AM=MQ,\therefore △ AMQ$是等腰直角三角形.
$\therefore$无论点$P$如何运动,$QA$与$x$轴的正半轴的夹角都为$45°$,$\therefore △ AOK$是等腰直角三角形,
$\therefore$点$K$的坐标为$(0,-6).$