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

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Answer:

(4 Ø3) Ø (5 Ø2) =120

a Øb = [3b –(2a-b2)] + 10

Halla (4 Ø3) Ø (5 Ø2).

• (4 Ø3) =[3b –(2a-)] + 10

       (4 Ø3) =[3(3) –(2(4)-)] + 10

        (4 Ø3)=[9-(-1)]+10

         (4 Ø3)=10+10

          (4 Ø3)=10

  • (5 Ø2)=[3b –(2a-)] + 10

        (5 Ø2)=[3(2) –(2(5)-)] + 10

         (5 Ø2)=[6–6] + 10

          (5 Ø2)=10

(4 Ø3) Ø (5 Ø2)

10Ø 10 = [3b –(2a-)] + 10

10Ø 10 = [3(10) –(2(10)-)] + 10

10Ø 10 = [30 –(-80)] + 10

10Ø 10 =[30+80]+10

10Ø 10 =120

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