If the (n+1)th and lower figure is less than 1/2 of the 1unit for n-th figure 如何翻译啊 谢谢了
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振幅的扰动小于0.3倍
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仅供参考
lower figure is less than 1/2 of the 1unit for n-t
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莱婷康妇: 解:设a=2k;则b=3k,c=5k. 代入a*a+b*b+c*c=abc,得 4k*k+9k*k+25k*k=30k*k*k 化简,得:38k*k=30k*k*k 19k*k=15k*k*k 若k=0; 则a=0,b=0,c=0. 故a+b+c=0. 若k≠0; 则15k=19,k=19/15(十五分之十九). a=38/15,b=19/5,c=19/3. 故a+b+c...
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波密县19421759261: 写英语句子 .The+n.+谓语+倍数词+what+主语+谓语.. - ?
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莱婷康妇: #include <iostream> using namespace std;#define SIZE 10 int main() { int num; cout<<"Input the number of your figures:"; cin>>num; int array[SIZE]; for(int i=0;i<num;i++) { cout<<"Input the "<<i+1<<"th figure:"; cin>>array[i]; } int min = array[...
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波密县19421759261: C语言 n=1,表达式n+++n - --n结果是多少 - ?
莱婷康妇: 显然是1,原式可以看成(n++)+(n--)-n,++和--都在+和-运行后运行且影响n的值,不影响表达式值,所以表达式值为1+1-1=1