题目 #183410

(1)根据表格，气体温度升高1，气体体积 C 2L，则y = 596 + 2(− = 2x + 546,frac增大y fracxy frac25)x 与的函数关系式为 = 2 +\(\frac{x}{y} 546\). 当 = 700时，得2x + 546 = 700，解得\(\frac{(2)}{x} = 77．\)答：停止加热时的气体温度为77 ． C (1)当0„  150时，设y =\(kx_{1}\)+\(bk_{1}\)( 0)，\(\frac{x}{图象经过点}\) (0,60),(150,35)，    = −\(\frac{1}{6}\)，  = 60, \(\frac{b_{1}}{150}\)解得 \(\frac{k_{1}}{b_{1}}\)\(k_{1}\)+\(b_{1}\)= 35,   = 60，   y = −\(\frac{1}{6}\)x + 60. 200时，设y =\(kx_{2}\)+\(b_{2}\)(\(k_{2}\) 0)， 当150剟\(\frac{x}{图像经过点} (150,35),(200,10)\)  + = 35,  = −\(\frac{1}{2}\)， \(\frac{150}{200}\frac{k_{2}}{k_{2}} \frac{b_{2}}{b_{2}}\)解得\(\frac{k_{2}}{b_{2}}\) + = 10,   = 110，  y = −\(\frac{1}{2}\)x + 110. 综上，当0„ x  150时，y = −\(\frac{1}{6}\)x + 60； 当150„ x  200时，y = −\(\frac{1}{2}\)x + 110. (2)当x = 180时，y = −\(\frac{1}{2}\)180 +110 = 20. 故此时蓄电池的剩余电量是20kW h