pole trapezu przedstawionego na rysunku wynosi?prosze o wszystkie obliczenia.rysunek w zalaczniku
zad2.oblicz
6 6
___ = _______
x-3 2x+1
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Skoro mamy tam 45* to znaczy ze ten trójkat jets trójkątem prostokątnym i równoramiennym ,zatem
h = 3cm
b = 3cm
a = b + h = 3cm + 3cm = 6cm
P = 1/2 * (a + b) * h
P = 1/2 * (6 + 3) * 3
P = 1/2 * 9 * 3
P = 13,5cm^2
6 / (x - 3) = 6 / (2x +1) x≠ 3 i x ≠ -1/2
mnożę na "krzyż"
6 * (2x +1) = 6 * (x - 3)
12x + 6 = 6x - 18
12x - 6x = - 18 - 6
6x = - 24
x = -24 : 6
x = - 4
h = 3cm
b = 3cm
a = b + h = 3cm + 3cm = 6cm
P = 1/2 * (a + b) * h
P = 1/2 * (6 + 3) * 3
P = 1/2 * 9 * 3
P = 13,5cm^2
6 / (x - 3) = 6 / (2x +1) x≠ 3 i x ≠ -1/2
6 * (2x +1) = 6 * (x - 3)
12x + 6 = 6x - 18
12x - 6x = - 18 - 6
6x = - 24
x = -24 : 6
x = - 4
prosze ;)