A) 0.5
B) 0.25
C) 4
D) 2
Correct Answer: A
Solution :
\[16qxy={{(2x+y)}^{2}}-{{(2x-y)}^{2}}\] \[\Rightarrow \]\[16qxy=4\times 2x\times y\] \[[\because \,{{(a+b)}^{2}}-{{(a-b)}^{2}}=4ab]\] \[\Rightarrow \]\[2q=1\] \[\Rightarrow \]\[q=\frac{1}{2}=0.5\]
You need to login to perform this action.
You will be redirected in
3 sec
