新的开始的句子励志:一道中考题,快点回答~
(1)当 a=40时,求h的值;
(2)从 a=40开始,设螺旋装置顺时针方向旋转x圈,求h关于x的函数解析式;
(3)从 a=40开始,螺旋装置顺时针方向连续旋转2圈.设第1圈使“千斤顶”增高 ,第2圈使“千斤顶”增高 ,试判定s1 与 s2的大小,并说明理由.若将条件“从a=40 开始”改为“从某一时刻开始”,则结果如何?为什么?
从 a=40开始,螺旋装置顺时针方向连续旋转2圈.设第1圈使“千斤顶”增高s1 ,第2圈使“千斤顶”增高s2 ,试判定s1 与 s2的大小,并说明理由.若将条件“从a=40 开始”改为“从某一时刻开始”,则结果如何?为什么?
1. (h/2)^2+(40/2)^2=30^2
h=44.72
2.(h/2)^2+(40/2-x/2)^2=30^2
3..(h)^2=2000-x^2+80x
(h0)^2=2000-x^2+80x
(h1)^2=2000-(x+1)^2+80(x+1)
(h2)^2=2000-(x+2)^2+80(x+2)
(s1)^2=(h1)^2-.(h0)^2= 2x+81
(s2)^2=h2)^2-.(h1)^2=2x+83
所以s2>s1
1. (h/2)^2+(40/2)^2=30^2
h=44.72
2.(h/2)^2+(40/2-x/2)^2=30^2
3..(h)^2=2000-x^2+80x
(h0)^2=2000-x^2+80x
(h1)^2=2000-(x+1)^2+80(x+1)
(h2)^2=2000-(x+2)^2+80(x+2)
(s1)^2=(h1)^2-.(h0)^2= 2x+81
(s2)^2=h2)^2-.(h1)^2=2x+83
所以s2>s1
1. (h/2)^2+(40/2)^2=30^2
h=44.72
2.(h/2)^2+(40/2-x/2)^2=30^2
3..(h)^2=2000-x^2+80x
(h0)^2=2000-x^2+80x
(h1)^2=2000-(x+1)^2+80(x+1)
(h2)^2=2000-(x+2)^2+80(x+2)
(s1)^2=(h1)^2-.(h0)^2= 2x+81
(s2)^2=h2)^2-.(h1)^2=2x+83
所以s2>s1
1. (h/2)^2+(40/2)^2=30^2
h=44.72
2.(h/2)^2+(40/2-x/2)^2=30^2
3..(h)^2=2000-x^2+80x
(h0)^2=2000-x^2+80x
(h1)^2=2000-(x+1)^2+80(x+1)
(h2)^2=2000-(x+2)^2+80(x+2)
(s1)^2=(h1)^2-.(h0)^2= 2x+81
(s2)^2=h2)^2-.(h1)^2=2x+83
所以s2>s1
faint~