The only games( )I play are football and tennis. A.whoB.whichC.whatD.when
The only games( )I play are football and tennis.
A.who
B.which
C.what
D.when
相关考题：
I _________ table tennis quite well, but I haven ’t had time to play since the New Year.A. will playB. have playedC. play
I() ping－pang quite well, but I haven t had time to play since the new year. A、will playB、have palyedC、playedD、play
根据下列程序的执行结果，可以断定划线部分的修饰符应为______。执行结果：i9a football is created.a football is created.i=10a football is destroyed.a football is created.i=20a football is destroyed.a football is destroyed.源程序：include＜iostream.h＞class Football{public:Football( ){cout＜＜"a football is created."＜＜endl;}～Football( ){tout＜＜"a football is destroyed."＜＜endl;}};void func(int i){______Football f1;Football f2;cout＜＜"i="＜＜i＜＜endl;}void main( ){func(10);func(20);}
The following statements have been made about life cycle costing:(i) It focuses on the short-term by identifying costs at the beginning of a product’s life cycle(ii) It identifies all costs which arise in relation to the product each year and then calculates the product’s profitability on an annual basis(iii) It accumulates a product’s costs over its whole life time and works out the overall profitability of a product(iv) It allocates costs to each stage of a product’s life cycle and writes them off at the end of each stageWhich of the above statements is/are correct?A.(i) and (iii)B.(iii) onlyC.(i) and (iv)D.(ii) only
The following information is available for a manufacturing company which produces multiple products:(i) The product mix ratio(ii) Contribution to sales ratio for each product(iii) General fixed costs(iv) Method of apportioning general fixed costsWhich of the above are required in order to calculate the break-even sales revenue for the company?A.All of the aboveB.(i), (ii) and (iii) onlyC.(i), (iii) and (iv) onlyD.(ii) and (iii) only
put computers to practical business use, but below the surface it is the heart of an(22) - the kernel - that provides the technical wizardry to juggle multiple programs, connect to networks and store(23).A traditional kernel provides all the functions for applications. The kernel(24) memory, I/O devices and parcels out processor time.The kernel also supports(25) and fault tolerance, which is the ability to recover automatically when parts of the system fail.A．DatabasesB．ApplicationsC．DriversD．Games
阅读下列说明、C++代码和运行结果，填补代码中的空缺，将解答填入答题纸的对应栏内。[说明]对部分乐器进行建模，其类图如下图所示，包括：乐器(Instrument)、管乐器(Wind)、打击乐器(Percussion)、弦乐器(Stringed)、木管乐器(Woodwind)、铜管乐器(Brass)。类图下面是实现上述设计的C++代码，其中音乐类(Music)使用各类乐器(Instrument)进行演奏和调音等操作。using namespace std; enum Note(/*枚举各种音调*/ MIDDLE_C,C_SHARP,B_FLAT }; classInstrument{/*抽象基类,乐器*/ public: ______; //play函数接口 virtual voidadjust()=0; //adjust函数接口 }; class Wind ______{ public: void play(Note n) { cout＜＜"Wind.play() "＜＜n＜＜endl; } void adjust(){cout＜＜"Wind.adjust()"＜＜endl; } ); /*类Percussion和Stringed实现代码略*/ class Brass ______{ public: void play(Note n) {cout＜＜"Brass.play() "＜＜n＜＜endl; } void adjUSt(){cout＜＜"Brass.adjust()"＜＜endl;) }; classWoodwind:public Wind{ public: void play(Note n) { cout＜＜"Woodwind.play()"＜＜n＜＜endl; } }; class MusiC { public: voidtune(Instrument*i) { i-＞play(MIDDLE_C.; } voidadjust(Instrument*i){ i-＞adjust(); } void tuneAll(______ e[],int numIns){ /*为每个乐器定调*/ for(int i=0; i＜numIns; i++){ this-＞tune(e[i]); this-＞adjust(e[i]); } } }; /*使用模板定义一个函数size,该函数将返回数组array的元素个数,实现代码略*/ int main(){ Music*music=______ Music(); Instrument*orchestra[]={new Wind(),new Woodwind() }; music-＞tuneAll(orchestra,size(orchestra));/*size返回数组orchestra的元素个数*/ for(int i=0;i＜size(orchestra);i++) deleteorchestra[i]; delete music; }本程序运行后的输出结果为：Wind.play()0 Wind.adjust() Woodwind.play()0 Wind.adjust()