[Mathematics from the beginning] No. 214 quarter with a computer to the college Entrance Examination (VI)

plot summary:
[Machine Xiao Wei] in [engineer Ah Wei] escorted into the [nine turn elixir] of the turn of the practice. Imagine a scenario:
If you were allowed to take an Internet-connected computer to the college entrance exam, would you give up choosing a hand-held calculator and a draft book?
？ Ah Wei decided and Xiao Wei to try to use the computer to calculate high questions will be how the feeling.

Drama Start:

Star Calendar May 20, 2016 11:40:58, the Milky Way Galaxy Earles the Chinese Empire Jiangnan Line province.
[Engineer Ah Wei] is working with [machine Xiao Wei] to do 2005 years of Jiangsu Province Mathematics college entrance exam questions].

Overall, the difficulty and the last year flat, are very kind,

But the last year a lot of questions are like a joke, this time slightly better, but the calculation is also very small.

Then the biggest change is to break the 22 pattern of the last n years, with the reduction of a subjective entitled cost, added two blank questions.

<span style= "FONT-SIZE:18PX;" >def tmp1 ():    A = Set ([1, 2]);    B = Set ([+ +]);    C = Set ([2,3,4]);    D = (a.intersection (B)). Union (C);    Print (D);    E = (A & B) | C;    Print (E);>>> {1, 2, 3, 4}{1, 2, 3, 4}</span>

<span style= "FONT-SIZE:18PX;"                    >//2if (1) {var r = 20;                      Config.setsector (1,1,1,1);                    Config.graphpaper2d (0, 0, R);                                      Config.axis2d (0, 0,180);                  Axis set var ScaleX = 2*r, ScaleY = 2*r;                   var SpaceX = 2, SpaceY = 2;                  var XS = -10, XE = 10;                  var YS = -20, YE = 20;                    Config.axisspacing (XS, XE, SpaceX, ScaleX, ' X ');                                        Config.axisspacing (YS, YE, SpaceY, ScaleY, ' Y ');                    var transform = new transform ();                                    Store the points on the function image var a = [], b = [], c = [], d = []; Function description//Greek alphabet to be displayed (save this for ctrl c/v//αβγδεζηθικλμνξ0πρστυφχψω//αβγδεζηθικλ                Μνξοπρστυφχψωvar f1 = ' y = 2^[1-x]+3 's inverse function ', F2 = ' y = 2^[1-x]+3 ', F3 = ', F4 = '; var y1 = y2 = 0;//(x+2) ^2) ^0.5for (var x = XS; x <= XE; x+=0.2) {y1 = Math.pow (2, 1-x) +3;a.push ([Y1, X]); B.push ([x, Y1]);                                  }//Store temporary array var tmp = []; Show transform if (A.length > 0) {a = Transform.scale (Transform.translate (A, 0, 0), scalex/s                     Pacex, Scaley/spacey);                        function 1 TMP = [].concat (a);                       Shape.pointdraw (TMP, ' red ');                        TMP = [].concat (a);                                            Shape.multilinedraw (tmp, ' Pink ');                    Plot.setfillstyle (' Red ');                  Plot.filltext (F1, 100,-90, 200); }//Show transform if (B.length > 0) {b = Transform.scale (transform.translate (b, 0, 0                     ), Scalex/spacex, Scaley/spacey);                        function 1 TMP = [].concat (b); Shape.pointdraw (TMP, ' BLue ');                        TMP = [].concat (b);                                            Shape.multilinedraw (tmp, ' #2288FF ');                    Plot.setfillstyle (' Blue ');                  Plot.filltext (F2, 100,-120, 200);                }} var f1 = ' Y = Math.log (2/(x-3))/math.log (2) ', F2 = ' y = 2^[1-x]+3 ', F3 = ', F4 = '; var y1 = y2 = 0;//(n (x+2) ^2) ^0.5for (var x = XS; x <= XE; x+=0.2) {y1 = Math.log (2/(x-3))/math.log (2) A.push ([x, Y1])        ; } </span>

<span style= "FONT-SIZE:18PX;"    > #题4def tmp4 (): panel = panel ();    #正三棱柱ABC-a1b1c1 A, B, C = [0, 0, 0], [2,0,0], [1, 3**0.5, 0];    A1, B1, C1 = [0, 0, 1], [2,0,1], [1, 3**0.5, 1];    #平面ABC的法向量 norm_abc = Panel.normal ([A, B, C]);    Print (NORM_ABC);    #平面ABC的一般式方程 genre_abc = Panel.genreform ([A, B, C]);    Print (GENRE_ABC);    #占A到平面A1BC的距离 DIS_A_A1BC = Panel.pt2panel (A, [A1, B, C]); Print (DIS_A_A1BC);>>> [0.0, 0, 3.4641016151377544][0.0, 0, 3.4641016151377544, -0.0]0.8660254037844386#### @ The plane class in the analytical geometry of the usage plane, containing a set of related calculation methods # @author mw# @date May 20, 2016 Friday 09:02:31 # @param # @return # # # # # # # # # # # # # #涉及到平面, all coordinate points are spatial coordinates system [x, Y, z] form class Panel (): #平面的法向量, determined by [p_[1], p_[2], p_[3]] unique flat def normal (self, plain): #平面上的三个点 poi        Nt_1, point_2, point_3 = Plain[0], plain[1], plain[2];        X_1, y_1, z_1 = Point_1[0], point_1[1], point_1[2];        X_2, y_2, z_2 = Point_2[0], point_2[1], point_2[2]; X_3, y_3, z_3 = Point_3[0], point_3[1], point_3[2];        #中间变量 a = (y_2-y_1) * (z_3-z_1)-(y_3-y_1) * (z_2-z_1);        b = (z_2-z_1) * (x_3-x_1)-(z_3-z_1) * (x_2-x_1);        c = (x_2-x_1) * (y_3-y_1)-(x_3-x_1) * (y_2-y_1);    return [A, b, c];        #平面一般式方程, the four parameters a,b,c,d def genreform (self, plain) are obtained from three flat pastry ax+by+cz+d=0: Norm = self.normal (plain);        A, b, C = norm[0], norm[1], norm[2];        #平面上的三个点 point_1, point_2, point_3 = Plain[0], plain[1], plain[2];        X_1, y_1, z_1 = Point_1[0], point_1[1], point_1[2];        X_2, y_2, z_2 = Point_2[0], point_2[1], point_2[2];                                          X_3, y_3, z_3 = Point_3[0], point_3[1], point_3[2];        D =-(a*x_1+b*y_1+c*z_1);    return [A, B, C, d];        #点到平面的距离 def pt2panel (self, point, Panel): #平面一般式的参数 A, B, c, d = self.genreform (panel);        X, y, z = point[0], point[1], point[2];        Distance = ABS ((a*x+b*y+c*z+d)/(a**2+b**2+c**2) **0.5); Return distance;</span>

<span style= "FONT-SIZE:18PX;" > #题5def tmp5 ():    #三角形ABC, give one more side trial calculation    TRI_ABC = Geo.solvetriangle ([3, 2, '? ', '? ');    Print (TRI_ABC);    BC, AC, AB = Tri_abc[0], tri_abc[1], tri_abc[2];    #周长    perimeter = bc+ac+ab;    print (perimeter);    #角B, converted to radians    B = Tri_abc[4]/180*math.pi;    #选项, used to compare    choice = [4*1.732*math.sin (B+MATH.PI/3) +3,              4*1.732*math.sin (B+MATH.PI/6) +3,              6*math.sin (B +MATH.PI/3) +3,              6*math.sin (B+MATH.PI/6) +3              ];    Answer = ' ABCD ';    For I in range (4):        If ABS (choice[i]-perimeter) < 1e-6:            print (answer[i]);>>> [3, 2, 3.4494897427831783, 60.0, 35.264389682754654, 84.73561031724536]8.449489742783179d</span>

<span style= "FONT-SIZE:18PX;" >//7if (1) {analyze = new dataanalyze (); a = [9.4, 9.4, 9.4, 9.6, 9.7];hint (Analyze.average (a)); Hint (analyze.variance ( a));}        9.50.015999999999999886# 7def TMP7 (): Analyze = Dataanalyze ();    Array = [9.4, 8.4, 9.4, 9.9, 9.6, 9.4, 9.7];    Array = sorted (array);    Array = array[1:-1];    print (array);    Array = Analyze.format (array);    Sum_ = analyze.sum (array);    Print (SUM_);    Aver_ = analyze.average (array);    Print (' mean ', aver_);    Variance = analyze.variance (array);    Print (' Variance ', variance);    #numpy数组的属性 print (Array.ndim);    Print (Array.shape);    Print (array.size);    Print (Array.dtype);    Print (array.itemsize); Print (Array.data);>>> [9.4, 9.4, 9.4, 9.6, 9.7]47.5 mean 9.5 variance 0.0161 (5,) 5float648<memory at 0x0205ad50>### # @usage Centralized Analysis class # @author mw# @date May 20, 2016 Friday 10:06:47 # @param # @return # # # #class Dataanalyze (): #由于nump The method interface of Y is only open to Narray, so the array should be formatted first #对于自己的数组而言, this method must be called first, in order to use the NumPy method DEF format (self, array): Return Numpy.array (array);    #求和 def sum (self, array): Return Array.sum ();            #均值 def average (self, array): Return Self.sum (Array)/len (array);        #方差 def variance (self, array): Array_ = Array*array;        Sum_ = Array_.sum ();        Aver_ = self.average (array);        result = Sum_/len (array)-aver_**2;        return result; #标准差 def RMS (self, Array): Return (self.variance) **0.5;</span>

<span style= "FONT-SIZE:18PX;" > #题9def tmp9 ():    a = [' X ', ' 2 '];    A = Alg.strformat (a);    A = Alg.strpow_n (A, 5);    A = Alg.strcombine (a);    Print (a);>>> [' (1) *x^[5] ', ' (Ten) *x^[4] ', ' (+) *x^[3] ', ' (+) *x^[2] ', ' () *x^[1] ', ' (+) ']</span>

<span style= "FONT-SIZE:18PX;" > #题16def tmp16 ():    print (Math.log (0.618)/math.log (3));</span>

<span style= "FONT-SIZE:18PX;" > #题20def tmp20 ():    #甲, B hit    p_1 = 2/3;    p_2 = 3/4;    # (1)    print (1-p_1**4);    # (2) There is a combination number multiplied    by a = Alg.combination (4, 2) *p_1**2* (1-p_1) **2;    b = Alg.combination (4, 3) *p_2**3* (1-p_2);    Print (a*b);    # (3)    A12 = N (1-p_2) **2; #前两次, as long as not all do not hit    a3 = p_2; #第三次, you must hit    A45 = (1-p_2) **2; #第四五次, you must not    print (a12* A3*A45);>>> 0.80246913580246910.125000000000000030.0439453125</span>

<span style= "FONT-SIZE:18PX;" > #题21def tmp21 ():    #已知边 as    = AB = AE = 2;    BC = DE = 1.732;    #已知角    A = C = D = +;    #解三角形, the array is    tri_abs = Geo.solvetriangle (['? ', as, AB, '? '] in the opposite side of the diagonal);    BS = tri_abs[0];    ES = BS;    Tri_abe = Geo.solvetriangle (['? ', AE, AB, A, '? ', '? ']);    be = tri_abe[0];    Tri_bes = Geo.solvetriangle ([ES, BS, be, '? ', '? ', '? ']);    Print (' CD with AB angle: ', tri_bes[3]);>>> CD with AB angle: 52.23875609296496>>> math.acos (6**0.5/4) *180/ 3.14252.23198356916861>>> Math.acos (6**0.5/4) *180/math.pi52.23875609296496    # (2)    #角B    B = ( 540-a-c-d)/2;    #tri_BCD = Geo.solvetriangle ([CD, '? ', BC, '? ', C, '? ']);    Print (B);    "    " >>>     CD with AB angle: 52.23875609296496    90.0    ' </span>

The end of this section, to know how to funeral, please see tell.

[Mathematics from the beginning] No. 214 quarter with a computer to the college Entrance Examination (VI)