Hisha
Hisha Hisha
  • 04-02-2017
  • Mathematics
contestada

wнaт'ѕ тнe eхpreѕѕιon υѕιng eхponenтѕ м.м.м.p.p.p.p.p.p

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jdoe0001 jdoe0001
  • 04-02-2017
[tex]\bf m\cdot m\cdot m\cdot m\cdot p\cdot p\cdot p\cdot p\cdot p\cdot p \\ \quad \\ m^1\cdot m^1\cdot m^1\cdot m^1\cdot p^1\cdot p^1\cdot p^1\cdot p^1\cdot p^1\cdot p^1 \\ \quad \\ m^{1+1+1}p^{1+1+1+1+1+1}\implies m^{\boxed{?}}p^{\boxed{?}}[/tex]
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Shunterrica
Shunterrica Shunterrica
  • 04-02-2017
(m*m*m)*(p*p*p*p*p*p) is (m^1*m^1*m^1)*(p^1*p^1*p^1*p^1*p^1*p^1) so what I'm trying to say is m equals 1+1+1 and p equals 1+1+1+1+1+1.
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