Question

Se define el operador Ø como a Ø b = [3b –(2a-b2)] + 10 Halla
(4 Ø 3) Ø (5 Ø2).

Answer 1

Respuesta:

(4 Ø3) Ø (5 Ø2) =120

Explicación paso a paso:

a Øb = [3b –(2a-b2)] + 10

Halla (4 Ø3) Ø (5 Ø2).

• (4 Ø3) =[3b –(2a-$b^{2}$)] + 10

       (4 Ø3) =[3(3) –(2(4)-$3^{2}$)] + 10

        (4 Ø3)=[9-(-1)]+10

         (4 Ø3)=10+10

          (4 Ø3)=10

• (5 Ø2)=[3b –(2a-$b^{2}$)] + 10

        (5 Ø2)=[3(2) –(2(5)-$2^{2}$)] + 10

         (5 Ø2)=[6–6] + 10

          (5 Ø2)=10

(4 Ø3) Ø (5 Ø2)

10Ø 10 = [3b –(2a-$b^{2}$)] + 10

10Ø 10 = [3(10) –(2(10)-$10^{2}$)] + 10

10Ø 10 = [30 –(-80)] + 10

10Ø 10 =[30+80]+10

10Ø 10 =120