(2/2)h f() 三、Cars are a useful means of t().填空啊,
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(2/2)h f() 三、Cars are a useful means of t().填空啊,
(2/2)h f() 三、Cars are a useful means of t().填空啊,
(2/2)h f() 三、Cars are a useful means of t().填空啊,
transport运输
He can speak French fluently.
他能讲流利的法语。
Cars are a useful means of transportation.
汽车是一种有用的交通工具。
transportation
填写:transportation
翻译:汽车是一个有用的运输方式
绝对正确,希望帮到你,祝学习进步
在技术时代,音乐和美术仍有用的,因为… 2-3分钟 In the painting, class player for the music students in a relaxed, pleasant, and lively
(2/2)h f() 三、Cars are a useful means of t().填空啊,
(2/2)ch f() 三、Cars are a useful means of t().填空啊,帮帮忙啊
1,(O ,F ,Cl ) 2,(Na,Mg,Al) 3,(He,Ne,Ar) 三组化学元素的关系是?结论要多点的
读出下列单词,判断画线部分的读音是否相同1m[ay]----d[ay] 2t{all}---b[all} 3d[o]-----g[o] 4[wh]en---[wh]ere 5m[ee]t---b[ee] r6g[i]ve---f[i]ve 7h[a]ve---c[a]ke 8f[ar]m---p[ar]k
判断下列单词括号部分读音有几种A一种 B两种 C三种 D四种1.wh(e)n d(e)licious (e)vening cin(e)ma2.sn(ow) kn(ow) fl(ow)er n(ow)3.tr(u)e (u)se n(u)mber (u)mbrella4.w(ar)m c(ar) f(ar)m h(ar)d5.t(o)night c(o)ncert r(o)ck ag(o)6.r(ea)ll
化学上的基团 像这三个 =CH ,Ar-H ,三CH
lingo提示没有最优解,model:sets:k/1..10/:e,l,g;endsetsmin=(f/h+(a*b*c)/18000+(4/132.6)*@sum(k(j):g(j)));b=79.6;c=3;h=67;d=3.98;r=40;f/h=x/(a/2-d);a>150.06;ar/2-d;(h-f)^2+((a/2-d)-(a/2-d-x)*@cos(o))^2
lim [f(x0+h)-f(x0-h)]/h = 2 lim [f(x0+h)-f(x0-h)]/2h = 2 f'(x0)这个我还是不懂
f'(x)=2.则lim[f(x-h)-f(x+2h)]/2h
若f′(x0)=-2,则lim[f(x0+h)-f(x0-h)]/h=
变限积分求道问题对函数 f(t+h)-f(t-h) 在[-h,h]上的积分对h求导.F(h)=∫[-h,h]f(t+h)-f(t-h)dt (其中[-h,h]为积分区间,-h为下限,h为上限)参考答案中:∫[-h,h]f(t+h)dt = ∫[0,2h]f(u)du (做代换u=t+h) ∫[-h,h]f(t+h)dt
若f''(x)存在,证明:[f(x+2h)-2f(x+h)+f(x)]/(h^2)=f''(x)
f(x)在X0处二阶可导,证lim(h->0)[ f(x-h0)+f(x0+h)-2f(x0)]/h^2=f``(x0) 为什么不能这么做?原式=lim(h->0)[f(x0+h)-f(x0)]/h^2 — lim(h->0)[f(x0)-f(x0-h)]/h^2=f'(x)/h-f'(x)/h=0
f(x)在x=a处可导, lim(h→0) [f(a+h)-f(a-2h)]/h=
lim h趋于0时,(f(x0+h)-f(x0-h))/2h=f`(x0) 看不懂
如何证明lim[f(x+h)+f(x-h)-2f(x)]=f(x) 其中h趋向0
导数题 lim [f(a+h^2)-f(a)]/h=?(1)lim [f(a+h^2)-f(a)]/h=?(2)lim [f(a+3h)-f(a-h)]/2h=?
matlab编程中出现问题,说是矩阵维度不一致 ,M文件主函数如下:function f=myobj(x)arf=linspace(0.0001,0.5233,100);K=1162;M=600;L=1650;t=281-x(1).*cos(arf+1.134);h=sqrt(x(2).^2-(x(1).*sin(1.134+arf)-x(3)).^2);s=t-h;f=abs(acot(cot(ar