y^2/3 + 3y^-1 -2 when y =1/8
asked by guest on Dec 07, 2024 at 9:44 pm
You asked:
MathBot Answer:
y23+3y−1−2=4225192\frac{{y}^{2}}{3} + 3 {y}^{-1} - 2 = \frac{4225}{192}3y2+3y−1−2=1924225
y23−2+3y=(18)23+318−2=3⋅8+13⋅82−2=3⋅8+1364−2=3⋅8+1192−2=4225192\begin{aligned}\frac{y^{2}}{3} - 2 + \frac{3}{y}&=\frac{\left(\frac{1}{8}\right)^{2}}{3} + \frac{3}{\frac{1}{8}} - 2\\&=3 \cdot 8 + \frac{1}{3 \cdot 8^{2}} - 2\\&=3 \cdot 8 + \frac{\frac{1}{3}}{64} - 2\\&=3 \cdot 8 + \frac{1}{192} - 2\\&=\frac{4225}{192}\end{aligned}3y2−2+y3=3(81)2+813−2=3⋅8+3⋅821−2=3⋅8+6431−2=3⋅8+1921−2=1924225
