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Text 3 This line of inquiry did not begin until earlier this month--more than three months after the accident--because there were "too many emotions, too many egos," said retired Adm. Harold Gehman, chairman of the Columbia Accident Investigation Board. Testifying before the Senate Commerce Committee, Gehman said this part of his inquiry Was in its earliest stages, starting just 10 days ago. But Gehman said he already has concluded it is "inconceivable" that NASA would have been unable or unwilling to attempt a rescue for astronaut, s in orbit if senior shuttle managers and administrators had known there was fatal damage to Columbia’s left wing. Gehman told reporters after the hearing that answers to these important questions could have enormous impact, since they could place in a different context NASA’s decisions against more aggressively checking possible wing damage in the days before Columbia’s fatal return. Investigators believe breakaway insulating foam damaged part of Columbia’s wing Shortly after liftoff, allowing superheated air to penetrate the wing during its fiery re-entry on Feb. 1 and melt it from the inside. Among those decisions was the choice by NASA’s senior shuttle managers and administrators to reject offers of satellite images of possible damage to Columbia’s left wing before the accident. The subject dominated the early part of Wednesday’s hearing. Gehman complained that managers and administrators "missed signals" when they rejected those offers for images, a pointedly harsh assessment of the space agency’s inaction during the 16 day shuttle mission. "We will attempt to pin this issue down in our report, but there were a number of bureaucratic and administrative missed signals here," Gehman told senators. "We’re not quite so happy with the process." The investigative board already had recommended that NASA push for better coordination between the space agency and military offices in charge of satellites and telescopes. The U. S. National Imagery and Mapping Agency in March agreed to regularly capture detailed satellite images of space shuttles in orbit. Still, Gehman said it was unclear whether even images from America’s most sophisticated spy satellites might have detected on Columbia’s wing any damage, which Gehman said could have been as small as two inches square. The precise capabilities of such satellites was a sensitive topic during the Senate hearing. Gehman believed that had its administrators known the damage to Columbia, NASA would ______.

A. have managed the rescue of the astronauts orbiting the earth
B. still have been unable to conceive of any way to save the astronauts
C. have attempted to help the astronauts out of danger unwillingly
D. have succeeded in repairing the damaged left wing of the shuttle

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使用“答题”菜单或从VC6中打开考生文件夹proj2下的工程proj2。此工程包含一个程序文件main.cpp,其中有类Quadritic、类Root以及主函数main的定义。一个Quadritic对象表示一个ax2+bx+c的一元二次多项式。一个Root对象用于表示方程ax2+bx+c=0的一组根,它的数据成员num_of_roots有3种可能的值0、1和2,分别表示根的3种情况:无实根、有两个相同的实根和有两个不同的实根。请在程序中的画线处填写适当的代码,然后删除横线,以实现上述类定义。此程序的正确输出结果应为(注:输出中的X^2表示x2): 3X^2+4X+5=0.0 无实根 4.5X^2+6X+2=0.0 有两个相同的实根: -0.666667和-0.666667 1.5X^2+2X-3=0.0 有两个不同的实根: 0.896805和-2.23014 //源程序 #include<iostream> #include<iomanip> #include<cmath> using namespace std; class Root //一元二次方程的根 public: const double x1; //第一个根 const double x2; //第二个根 const int num_of_roots; //不同根的数量:0、1或2 Root():x1(0.0),x2(0.0),num_of_roots(0) //创建一个“无实根”的Root对象 Root(double root) //************found************ :______ //创建一个“有两个相同的实根”的Root对象 Root(double root1,double root2) :x1(root1),x2(root2),num_of_roots(2) //创建一个有两个不同的实根的Root对象 void show() const //显示根的信息 cout<<"\t\t": switch(num_of_roots) case 0: //************found************ case 1: cout<<"有两个相同的实根:"<<x1<<"和"<<x2;break; default: cout<<"有两个不同的实根:"<<x1<<"和"<<x2;break; ; class Quadratic //二次多项式 public: const double a,b,c; //分别表示二次项、一次项和常数项3个系数 Quadratic(double a,double b,double c) //构造函数 //************found************ :______ Quadratic(Quadratic&x) //复制构造函数 :a(x.a),b(x.b),c(x.c) Quadratic add(Quadratic x)const //求两个多项式的和 return Quadratic(a+x.a,b+x.b,c+x.c); Quadratic sub(Quadratic x)const //求两个多项式的差 //************found************ double value(double x)const //求二次多项式的值 return a *x*x+b*x+c: Root root() const //求一元二次方程的根 double delta=b*b-4*a*e; //计算判别式 if(delta<0.0)return Root(); if(delta==0.0)return Root(-b/(2 *a)); double sq=sqrt(delta); return Root((-b+sq)/(2*a),(-b-sq)/(2*a)); void show() const //显示多项式 cout<<endl<<a<<"X^2"<<showpos<<b<<"X"<<c<<noshowpos; void showFunction() //显示一元二次方程 show(); cout<<"=0.0": ; int main() Quadratic q1(3.0,4.0, 5.0), q2 (4.5,6.O,2.O),q3(q2.sub(ql)); q1.showFunction(); q1.root().show(); q2.showFunction(); q2.root().show(); q3.showFunction(); q3.root().show(); cout<<endl; return 0:

案例六:张小姐是某外企的市场部经理,年轻有为,收入客观。最近张小姐打算给自己添置一辆新车,但她前不久刚刚买了房,手中资金有限,一次付清车款有一定困难。针对这种情况,张小姐的理财规划师建议其通过贷款方式购车。 根据案例六,回答30~39题: 若张小姐在北京工作,但没有北京户口,她选择( )会更容易获得买车贷款。

A. 通过银行进行汽车消费贷款
B. 通过汽车金融公司进行汽车消费贷款
C. 通过投资银行进行汽车金融贷款
D. 私人借贷

关于中度脱水的临床表现,以下哪项不对

A. 失水量为体重的5%~10%
B. 皮肤弹性较差
C. 眼窝、前囟明显凹陷
D. 尿量明显减少
E. 四肢厥冷

下列哪项不是失代偿性代谢性酸中毒的血气分析和PH改变

A. HCO3-/H2CO3<20%
B. PH↓
C. HCO3-↓
D. BE-↑
E. PaCO2↑

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