herkes doğum tarihiyle 40 buluyo

17.12.1988

sin(1988) = 0.587648675
sin( 17π/12) =
sin( 3π/4 + 2π/3) =
sin( 3π/4) cos(2π/3) + sin( 2π/3) cos(3π/4) =
½√2 • -½ + ½√3 • -½√2 =
-¼√2 - ¼√6

(ie ½√3 = ¼√6)=136
sin(132) - 51 = -50.9469164
100(e^-ln(2)t)ise

e^lnx=x=1988
e^-ln2t=(e)^ln(2t)^-1=(2t)^-1=1/2t100.1/2t=der(50/t)=-50/(t)^2

olur

f(x)1988 bağıntısından

f(x) = x^4/5(x - 7)^2
f'(x) = [4x^3(5(-50/(t)^2 - 7)^2 - 10(x-7)*x^4]/(25(x-7)^4)
f"(x) = [20x^3(-50/(t)^2 - 7)^2 -10-50/(t)^2^4(-50/(t)^2 - 7)]/(25(-50/(t)^2-7)^4)
f'(x) = [20x^3(-50.9469164 - 7) -10-¼√2 - ¼√6^4]/(25(-¼√2 - ¼√6 - 7)^3)
10x^3[2(x - 7) - x] = 0
10x^3 (x - 14) = 0
0.587648675.20x^3(-50.9469164 - 7)=40 yapar