given that x = 7 y = -4 z = -2
evaluate 3x2−2y3z3\frac{3x^2-2y}{3z^3}3z33x2−2y
asked by guest on Nov 18, 2024 at 6:23 pm
MathBot Answer:
3x2−2y3z3=−15524\frac{3 {x}^{2} - 2 y}{3 {z}^{3}} = - \frac{155}{24}3z33x2−2y=−24155
3x2−2y3z3=(3⋅72−2(−4))3(−2)3=(−2(−4)+3⋅49)3(−2)3=(−2(−4)+147)3(−2)3=−(−2(−4)+147)8⋅3=−8+1478⋅3=−15524\begin{aligned}\frac{3 x^{2} - 2 y}{3 z^{3}}&=\frac{\left(3 \cdot 7^{2} - 2 \left(-4\right)\right)}{3 \left(-2\right)^{3}}\\&=\frac{\left(- 2 \left(-4\right) + 3 \cdot 49\right)}{3 \left(-2\right)^{3}}\\&=\frac{\left(- 2 \left(-4\right) + 147\right)}{3 \left(-2\right)^{3}}\\&=-\frac{\left(- 2 \left(-4\right) + 147\right)}{8 \cdot 3}\\&=-\frac{8 + 147}{8 \cdot 3}\\&=- \frac{155}{24}\end{aligned}3z33x2−2y=3(−2)3(3⋅72−2(−4))=3(−2)3(−2(−4)+3⋅49)=3(−2)3(−2(−4)+147)=−8⋅3(−2(−4)+147)=−8⋅38+147=−24155
