Math, asked by santoshisingh100585, 7 months ago


p(0) = 2 + t + 2t {}^{2}  - t {}^{2}

Answers

Answered by MrHyper
14

\huge\mathfrak\orange{∆nswer!}

\large\bold{Given:}

\bold{p(0)=2+t+2t^2-t^2}

\bold{=>p(0)=2+0+2(0)^2-(0)^2}

\bold{=>p(0)=2+0-0}

\bold{=>p(0)=2}

\huge\bold\purple{p(0)=2}

\huge\bold\blue{Hope \ this \ helps...}

\huge\bold\green{Brainliest \ Please!}

\huge\bold\red{★★★★★★★★★★}

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