[题目]将一张长方形纸片按图6所示的方式折-|||-叠,BC,BD为折痕,则 angle CBD 的度数为() ()-|||-B-|||-图-|||-A.60°-|||-B.75°-|||-C.90°-|||-D.95°

2.求曲面 =(x)^2+2(y)^2 及 =3-2(x)^2-(y)^2 所围成的立体的体积.

8.在无移除流行病模型中,人群是封闭的(总人数N不变),各人员间的接触是均衡的,易感-|||-者转变为感染者的变化率与当时易感人数与感染人数的乘积成正比,比例系数为b.假设在-|||-开始时只有一个感染者,求在时刻t的易感人数S(t)与感染人数I(t).

已知整式(a-1)x^3-2x-(a+3).(1)若它是关于x的一次式，求a的值并写出常数项；(2)若它是关于x的三次二项式，求a的值并写出最高次项；(3)在（2）的基础上，若x=-1，求这个整式的值.

11. 求 xy^2-e^x+y+x^3-25=0 所确定的隐函数的导数 (dy)/(dx)

2.求下列不定积分:-|||-(1) int (sin 2x-(e)^dfrac (x{3)})dx =-|||-(2) int ((2x+3))^99dx =-|||-(3) int dfrac (1-x)(sqrt {9-{x)^2}}dx :-|||-(4) int (e)^-xcos ((e)^-x)dx;-|||-(5) int dfrac (x)(sqrt [3]{2{x)^2+1}}dx :-|||-(6) int (x)^2(e)^-3(x^3+5)dx =-|||-(7) int dfrac (tan sqrt {x)}(sqrt {x)}dx =-|||-(8) int dfrac (1)(sqrt {x)sqrt (1-sqrt {x)}}dx =-|||-(9) int dfrac (ln x)(x({ln )^2x-1)}dx =-|||-(10) int dfrac (x)({sin )^2((x)^2+1)}dx :-|||-(11) int dfrac (1)(1+{e)^-x}dx =-|||-(12) int dfrac (1)({e)^x-(e)^-x}dx =-|||-(13) [tan^5xsec^3xdx;-|||-(14) int dfrac (1)(1+sin x)dx =-|||-(15) int dfrac (1)({x)^2+5x+6}dx =-|||-(16) int dfrac (x)({x)^2+3x+2}dx :-|||-(17) int dfrac (1)({x)^2+2x+3}dx =-|||-(18) int dfrac (1)(sqrt {1+x-{x)^2}}dx -

2阶行列式 |} -3(a)_(2) -3(b)_(2) . a1+2a2 |=6,则 |a1 a2 = __ 。-|||-|b1 b2-|||-B. -2 C.2 D. 6

[题目]如图,已知 AC=BC ,BD=BM ,ME=MF .-|||-angle C=(60)^circ ,则 angle F=-|||-C-|||-D-|||-E-|||-A B M F

2.设f(x)在x=0处可导，且f(0)=0，则lim_(x to 0) (f(x))/(x)等于（）A. f(0)B. f'(0)C. 0D. 不存在

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