var a, b, d: integer; n: real;
Begin Write(‗a= ‗); Readln(a); Write(‗b= ‗); ReadLn(b);
Write(‗d= ‗); ReadLn(d); n:=pi*sqr(d)+a*b; Write(‗n= ‗,n:0:2);
End. . . ,
.
M-5. Tomonlari a, b, с bo‗lgan uchburchakning yuzini hisoblash dasturini kiritish operatoridan
foydalanib tuzing va natijalar oling. a) a=5, b=7, c=4;
b) a=8, b=6, c=lO;
d)
a=3, b=4, c=5; e) a=10, b=8, c=10;
Yechim. Kiritish operatori deganda maTumotlami xotiraga mulo¬qot usulida kiritish, ya‘ni Read yoki
Readln operatorlaridan foydala¬nish, tushuniladi. Topshiriq shartidagi a, b, с o‗zgaravchilar butun va s
o‗zgaruvchi qiymati ildiz qatnashganligi sababli haqiqiy.
Vara, b, c: integer; p, §:real;
Begin Write(‗a ni kiriting= ‗); Readln(a); Write(‗b ni kiriting= ‗); Readln(b); Write(‗c ni kiriting= ‗);
Readln(c); p:=(a+b+c)/2; s:=sqrt(p*(p-a)*(p-b)*(p-c)); 1 Write(‗Uchburchk yuzi s= s:0:3); readln;
End.
M-6. y==23-x+l funksiyaning qiymatini x ning -5, ~4, -3, —2, -1, 0, 1, 2, 3, 4, 5 qiymatlanni xotiraga
qulay usulda kiritib hisoblash dasiu- rini tuzing va natijalar oling.
Yechim. MaTumotlami xotiraga kiritishni qulay usuli muloqot usulidir, ya‘ni Read yoki Readln
operatorlaridan foydalanishdir. X o‗zgaruvchining berilgan har bir qiymati uchun dastur qayta ishlatiladi
va natijalar olinadi.
Var x, у: integer;
Begin Write(‗x ning qiymatini kiriting= ‗); ReadLn(x);
y:=23:*!x+l; write(‗x= x, ‗-da ‗, ‗yf5 ‗,y ); readln; end.
M-7. y = 21-x2 +7 • x + 1963 funksiyaning qiymatini x ning -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 qiymatlarini
xotiraga qulay usulda kiritib hi¬soblash dasturini tuzing va natijalar oling.
Yechim. Avvalgi topshiriq dasturida y:=23*x+l; o‗miga y:= 21 *sqr(x)+7*x+1963; yoziladi.
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