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電子課本網 第113頁

第113頁

信息發(fā)布者:
??$8\sqrt{2}$??
解??$:(2)$??當??$0<t≤4$??時,??$P$??在線段??$AB$??上,此時??$CQ=2t,$????$PB=8-2t$??
??$∴s=\frac {1}{2}×2t×(8-2t)=2t^2+8t.$??
當??$t>4$??秒時,??$P$??在線段??$AB$??得延長線上,此時??$CQ=2t,$????$PB=2t-8$??
??$∴s=\frac {1}{2}×2t×(2t-{8})=2t^2-8t,$??
??$∵S_{△ABC}=\frac {1}{2}AB·AC=32$??
∴當??$t≤4$??時,??$S_{△PCQ}=-2t^2+8t=32$??
整理得??$t^2-4t+16=0$??無解,
當??$t>4$??時,??$S_{△PCQ}=2t^2-8t=32$??
整理得??$t^2-4t-16=0$??解得??$t=2±2\sqrt{5}($??舍去負值)
∴當點??$P_{運動}(2+2\sqrt{5})$??秒時,??$S_{△PCQ}=S_{△ABC}.$??
解:??$(1)$??設??$3$??月份再生紙的產量為??$x$??噸,則??$4$??月份再生紙的產量為??$(2x-100)$??噸,
依題意得:??$x+2x-100=800,$??
解得:??$x=300,$??
??$∴2x-100=2×300-100=500.$??
答:??$4$??月份再生紙的產量為??$500$??噸.
??$(2)$??依題意得:??$1000(1+\frac {m}{2}\%)×500(1+m\%)=660000,$??
整理得:??$\ \mathrm {m^2}+300m-6400=0,$??
解得:??$m_1=20,$????$m_2=-320($??不合題意,舍去).
答:??$m $??的值為??$20.$??
??$(3)$??設??$4$??至??$6$??月每噸再生紙利潤的月平均增長率為??$y,5$??月份再生紙的產量為??$a$??噸,
依題意得:??$1200(1+y)^2?a(1+y)=(1+25\%)×1200(1+y)?a,$??
??$∴1200(1+y)^2=1500.$??
答:??$6$??月份每噸再生紙的利潤是??$1500$??元.