(2/1)^2*3.1415926*1.5=4.7m³什么意思

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/03 00:58:09
(2/1)^2*3.1415926*1.5=4.7m³什么意思

(2/1)^2*3.1415926*1.5=4.7m³什么意思
(2/1)^2*3.1415926*1.5=4.7m³什么意思

(2/1)^2*3.1415926*1.5=4.7m³什么意思
(2) =(4×2.1×2.5+9.7)÷(1.05÷0.5÷3+8.4÷0.7÷0.4...=(3/4+25/8/1.25)/(21/4-5/4*9/5)=(28-2.5)/(21-9

计算:3.1415926^2-2×3.1415926×13.1415926+13.1415926^2 2 * Sin(45 * 3.1415926 / 180) * Cos(45 * 3.1415926 / (2/1)^2*3.1415926*1.5=4.7m³什么意思 求极限 1.(1+cosπx)/ [(x-1)^2] x趋向于1 注:π是3.1415926的拍 2.[e^(2/x)-1]*x x趋向于正无穷大 求解matlab绘制参数函数y=50*cos(t)-40*cos(t+7.5*((1-cos(3.1415926*t/55))-0.25*(1-cos(2*3.1415926*t/55))+36.86);x=50*sin(t)-40*sin(t+7.5*((1-cos(3.1415926*t/55))-0.25*(1-cos(2*3.1415926*t/55))+36.86);就是绘制这个方程的函数图象,t是 已知函数f(x)=cosx分之1-(根号2)sin(2x-4分之派(就是圆周率3.1415926…). Fortran 程序错误 program gyyimplicit nonereal*8 t,d,l,c,k,f,f1write(*,*) 请输入周期T与水深d,中间用逗号连接.read(*,*)t,dl=(9.8*t**2)/(2*3.1415926)7 f=l-(9.8*(t**2)*TANH(2*3.1415926*d/l))/(2*3.1415926)f1=1+9.8*t**2/((l**2)*(COSH(2* 0 根号8 根号4 3.1415926 -2 根号3 根0 根号8 根号4 3.1415926 -2 根号3 根号3-1 22/7 0.1010010001.(相邻两个1之间依次增加1个0) 1.414 -0.020202. -根号7 -兀 - 语句“Circle(1000,1000),800,-3.1415926/3,-3.1415926/2”绘制的是A.弧 B.椭圆 C.扇形 D.同心圆 选择:一个圆柱体的侧面积展开后是正方形,这个圆柱体底面的直径与高的比是?A.1:2(3.1415926···派)B.1:(3.1415926···派)C.(3.1415926···派):1 关于正弦余弦诱导公式的一道化简题1+sin(x-2π)*sin(π+x)-2cos^2(-x)π就是3.1415926.x是一个角! lingo运行出错,Error Code 15 model:sets:points/1..24/:d;pp/1..23/:c;endsetsmax=@sum(points(i):@log(1+(d(i+1)-d(i)+(d(i)^2+d(i+1)^2-2*d(i)*d(i+1)*@cos(3.1415926/6))^(1/2))/(2*d(i))));@for(points(j):d(j)#lt#d(j+1));@for(points:d(1)#gt#1);@for(points 一个高度与底面直径相等的圆柱形容积为12pai(3.1415926.),这个容积的半径超过2米了吗?超过1. 3.1415926 3.1415926( ) lingo运行出错model:sets:points/1..24/:d;endsetsmax=@sum(points(i):@log(1+(d(i+1)-d(i)+(d(i)^2+d(i+1)^2-2*d(i)*d(i+1)*@cos(3.1415926/6))^(1/2))/(2*d(i)));@for(points(j):d(j)#lt#d(j+1));@for(d(1)#gt#1);@for(points(i):d(i)#gt#d(i+12)+2);end运行有 定积分计算,会的来,在线等2∫sin²t(1-sin²t)dt=?(上线是∏(3.1415926)/2,下线是0) lingo运行问题model:sets:points/1..24/:d;endsetsmax=@sum(points(i)|i#le#23:@log(1+(d(i+1)-d(i)+(d(i)^2+d(i+1)^2-2*d(i)*d(i+1)*@cos(3.1415926/6))^(1/2))/(2*d(i))));@for(points(j)|j#le#23:d(j)1);@for(points(i)|i#le#12:(d(i)+2)20);!该约束有问题