Given that ( x = sqrt[3]{27} ) and ( y = cos(frac{pi}{3}) ), solve for ( z = 3x^2 - 2y ).

Question

Accepted Answer

Step:1 ：First, evaluate the cube root: ( x = sqrt[3]{27} = 3 ).Step:2 ：Next, evaluate the cosine function: ( y = cos(frac{pi}{3}) = frac{1}{2} ).Step:3 ：Finally, substitute ( x ) and ( y ) into the equation: ( z = 3x^2 - 2y = 3(3)^2 - 2(frac{1}{2}) = 27 - 1 = 26 ).

Answer

Step: 1 ：First, evaluate the cube root: ( x = sqrt[3]{27} = 3 ).Step: 2 ：Next, evaluate the cosine function: ( y = cos(frac{pi}{3}) = frac{1}{2} ).Step: 3 ：Finally, substitute ( x ) and ( y ) into the equation: ( z = 3x^2 - 2y = 3(3)^2 - 2(frac{1}{2}) = 27 - 1 = 26 ).

Given that $x = \sqrt[3]{27}$ and $y = \cos (\frac{π}{3})$ , solve for $z = 3 x^{2} - 2 y$ .

Answer

Popular Problems

Given that x=273 and y=cos⁡(π3), solve for z=3x2−2y.

Answer

Popular Problems

Given that $x = \sqrt[3]{27}$ and $y = \cos (\frac{π}{3})$ , solve for $z = 3 x^{2} - 2 y$ .