Question

Find the value of p(2/3) for p(y) = 2y³ - y² - 13y - 6

Answer 1

Answer:

The value of $p(\frac{2}{3})\: is\: \frac{-392}{27}$

Step-by-step explanation:

$p(y)=2y^3-y^2-13y-6\\\\\\\\Now\\\\p(\frac{2}{3})\\\\=2(\frac{2}{3})^3-(\frac{2}{3})^2-13(\frac{2}{3})-6\\\\=2(\frac{8}{27})-(\frac{4}{9})-\frac{26}{3}-6\\\\=\frac{16}{27}-(\frac{4}{9})-\frac{26}{3}-6\\\\=\frac{16-12-234-162}{27}\\\\=\frac{16-408}{27}\\\\=\frac{-392}{27}$

Answer 2

Answer:

-392/27

Step-by-step explanation:

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