zadania znajdują sie w załączniku, bardzo prosiłabym o poprawne rozwiązanie :)
4x^2 + 12x + 9 = x^2 +12
3x^2 + 12x -3
Δ = 144 + 36 = 180
√Δ = 6√5
x1= (-12 - 6√5) / 6 = -2 -√5
x2= (-12+6√5)/6 = -2 + √5
2;
Obliczymy podstawe za pomocą tgalfa
h= 6
tg30 = x/6
√3/3 = x/6
6√3 = 3x
x= 2√3
P = a*h/2
P = 2√3*6/2 = 12√3/2 = 6√3
3;
Log(√3)9 = 4 bo √3^4 = 9
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4x^2 + 12x + 9 = x^2 +12
3x^2 + 12x -3
Δ = 144 + 36 = 180
√Δ = 6√5
x1= (-12 - 6√5) / 6 = -2 -√5
x2= (-12+6√5)/6 = -2 + √5
2;
Obliczymy podstawe za pomocą tgalfa
h= 6
tg30 = x/6
√3/3 = x/6
6√3 = 3x
x= 2√3
P = a*h/2
P = 2√3*6/2 = 12√3/2 = 6√3
3;
Log(√3)9 = 4 bo √3^4 = 9