Find the rate law that corresponds to the data shown for the following reaction ?
 

Expt	(A)	(B)	Rate
1	0.012	0.035	0.1
2	0.024	0.070	0.8
3	0.024	0.035	0.1
4	0.1	0.070	0.070

$\frac{r_4}{r_1}=\frac{{\left[A_4\right]}^x}{{\left[A_1\right]}^x}\frac{{\left[B_4\right]}^y}{{\left[B_1\right]}^y}$
$\frac{0.8}{0.1}={\left(\frac{0.012}{0.012}\right)}^x{\left(\frac{0.070}{0.035}\right)}^y\Rightarrow 2^3={\left(2\right)}^y$
$\Rightarrow y=3$
$\frac{r_3}{r_1}={\left[\frac{A_3}{A_1}\right]}^x{\left[\frac{B_3}{B_1}\right]}^y$
$\frac{0.1}{0.1}={\left(\frac{0.024}{0.012}\right)}^x{\left(\frac{0.035}{0.035}\right)}^y$
$1={\left(2\right)}^x{\left(1\right)}^3$
$2°=2^x\Rightarrow x=0$
$\therefore rate=K{\left[A\right]}^x{\left[B\right]}^y$
$=K{\left[A\right]}^{°}{\left[B\right]}^3$