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信息发布者:
解:
(1)设$y = \frac{k}{x},$由图可知点$(24,50)$在此图像上,
则$k = 50×24 = 1200.$
$\therefore y$与$x$之间的函数关系式为$y = \frac{1200}{x}.$
(2)由题意知,4台挖掘机每天能够开挖水渠$30×4 = 120$(米),当$x = 120$时,$y = \frac{1200}{x} = 10,$
$\therefore$该工程队需要用10天才能完成此项任务.
$\sqrt{3}$
$4$
解:
(1)$\because$双曲线$y = \frac{m}{x}$过点$A(1,4),$$B(4,n),$
$\therefore m = 1×4 = 4n.$
$\therefore m = 4,$$n = 1.$
$\therefore$反比例函数的表达式为$y = \frac{4}{x},$$B(4,1).$
$\because$直线$y = kx + b$过点$A,$$B,$
$\therefore\begin{cases}k + b = 4 \\4k + b = 1 \end{cases},$解得$\begin{cases}k = -1 \\b = 5 \end{cases}.$
$\therefore$一次函数的表达式为$y = -x + 5.$
(2)$0\lt x\leqslant1$或$x\geqslant4.$
(3)如答图,设直线$AB$与$x$轴交于点$C,$则$C(5,0).$

$\because\triangle ABP$的面积为6,
$\therefore\frac{1}{2}×PC×4 - \frac{1}{2}×PC×1 = 6.$
$\therefore PC = 4.$ $\therefore$点$P$的坐标为$(1,0)$或$(9,0).$