-6x^3-y^2-3 ×=2 y=4 what is value

asked Sep 11, 2019 56.6k views

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$x=2, y=4\\-6x^3-y^2-3 = -6\cdot 2^3-4^2-3 = -6\cdot8-16-3=21$

Nikksan answered Sep 14, 2019

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- 67

substitute the given values for x and y into the expression and evaluate

= - 6(2)³ - (4)² - 3 = ( - 6 × 8 ) - 16 - 3 = - 48 - 16 - 3 = - 67

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