運算式 20: "g" equals StartRoot, "x" squared plus "y" squared minus "r" squared , EndRootg=x20+y20−r2
equals=
2.4 1 3 5 8 6 5 4 2 8 92.41358654289
20
運算式 21: alpha Subscript, "C" "e" "n" "t" "e" "r" , Baseline equals tangent to the negative 1st power left parenthesis, StartFraction, "y" Subscript, 0 , Baseline Over "x" Subscript, 0 , Baseline , EndFraction , right parenthesisαCenter=tan−1y0x0
運算式 23: "b" Subscript, "l" "e" "n" , Baseline equals StartRoot, 1 squared plus "b" squared , EndRootblen=12+b22
equals=
2.5 3 1 9 7 8 2 6 4 22.5319782642
23
運算式 24: "c" Subscript, 1 , Baseline equals "x" Subscript, 0 , Baseline minus "a" Subscript, 1 , Baselinec1=x0−a1
equals=
negative 2.2 9 6 7 5 8 5 8 5 1 7−2.29675858517
24
運算式 25: "c" Subscript, 2 , Baseline equals "y" Subscript, 0 , Baseline minus "a" Subscript, 2 , Baselinec2=y0−a2
equals=
negative 0.9 8 7 3 7 0 2 4 5 3 7 1−0.987370245371
25
運算式 26: "d" equals 1 times "x" Subscript, 0 , Baseline plus "b" Subscript, 2 , Baseline times "y" Subscript, 0 , Baselined=1·x0+b2·y0
equals=
negative 6.1 1 1 1 4 8 6 6 5 3 6−6.11114866536
26
運算式 27: theta Subscript, 1 , Baseline equals tangent left parenthesis, alpha Subscript, "C" "e" "n" "t" "e" "r" , Baseline minus tan to the negative 1st power StartFraction, "r" Over "g" , EndFraction , right parenthesisθ1=tanαCenter−tan−1rg
equals=
negative 2.3 2 6 1 3 7 1 2 6 3 1−2.32613712631
27
運算式 28: theta Subscript, 2 , Baseline equals tangent left parenthesis, alpha Subscript, "C" "e" "n" "t" "e" "r" , Baseline plus tan to the negative 1st power StartFraction, "r" Over "g" , EndFraction , right parenthesisθ2=tanαCenter+tan−1rg
equals=
0.4 7 2 2 2 4 0 8 2 8 3 20.472224082832
28
Punkter
對學生隱藏此資料夾。
29
36
技術支援
技術支援
Start black, left parenthesis, "x" Subscript, 0 , Baseline ; "y" Subscript, 0 , Baseline , right parenthesis equals ( negative 3.2 5 ; 1.2 3 ) End blackx0;y0=(−3.25;1.23)
Start black, left parenthesis, "a" Subscript, 1 , Baseline ; "a" Subscript, 2 , Baseline , right parenthesis equals ( negative 0.9 5 ; 2.2 2 ) End blacka1;a2=(−0.95;2.22)
