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Let P(x) + ax^7 + bx^3 +cx -5 where a,b,c are constants. Given that P(-7) = 7, find the value of P(7).
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What course are you taking, this is very perplexing to me at least. Hopefully robtobey can solve, it is above my paygrade.
\[ P(x) =ax^7 + bx^3 +cx -5\]?
note that all the exponents are odd numbers
\[-P(7)=-a7^7-b7^3-7c+5=a(-7)^7+b(-7)^3+c(-7)-5+10=P(-7)+10\]does that help?
this is Algebra Grade 9
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This helps mulishly
very well :)
what that word means? mulishly?
sorry..I typed your name and the computer recognized it differently . when I clicked "return" it sent that name...mukushla
oh...np..:)
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