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

Question

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

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Alice 1 month 2021-08-19T06:34:17+00:00 1 Answer 0 views 0

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    2021-08-19T06:35:55+00:00

    \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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