Question

find P (2/3) for P (y)=2y^3-y^2-13y-6​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

$2 {y}^{3} - {y}^{2} - 13y - 6 \\ = 2( { \frac{2}{3}) }^{3} - ( { \frac{2}{3}) }^{2} - 13( \frac{2}{3} ) - 6 \\ = 2( \frac{8}{27} ) - \frac{4}{9} - \frac{26}{3} - 6 \\ = \frac{16}{27} - \frac{4}{9} - \frac{26}{3} - \frac{6}{1} \\ = \frac{16 - 12 - 234 - 162}{27} \\ \\ = \frac{ - 392}{27} \\ = - 14 \frac{14}{27}$

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