Question

the value of p(t) =5t'2-6t +6at t = -1​

Answer 1

Answer:

$p(t) = 17$

Step-by-step explanation:

$given \: equation \: \\ p(t) = 5 {t}^{2} - 6t + 6 and \: t = - 1 \\ put \: t = - 1 \: in \: the \: given \: equation \: \\ p(t) = 5 {t}^{2} - 6t + 6 \: \\ p( - 1) = 5 {( - 1)}^{2} - 6( - 1) + 6 \\ = > \: \: 5 + 6 + 6 \\ = > \: \: 17$

Answer 2

QUESTION :

the value of p(t) =5t'2-6t +6at t = -1

$Answer: \\ </p><p></p><p>p(t) = 17 \\ </p><p></p><p>Step-by-step explanation: \\ </p><p></p><p>\begin{gathered}given \: equation \: \\ p(t) = 5 \\ {t}^{2} - 6t + 6 and \: t = - 1 \\ \\ put \: t = - 1 \: in \: the \: given \: equation \: \\ \\ p(t) = 5 {t}^{2} - 6t + 6 \\ \: \\ p( - 1) = 5 {( - 1)}^{2} - 6( - 1) + 6 \\ \\ = > \: \: 5 + 6 + 6 \\ \\ = > \: \: 17\end{gathered}</p><p></p><p>$

FINAL ANSWER : 17