Math, asked by smritakumari1986, 1 month ago

$( \sqrt{32} ) {}^{x} \div 2 {}^{y + 1 } =1 and \: 8 {}^{y} - 16 {}^{4 - x |2 } = 0$
find x and y

Answers

Answered by sadikshaikh226023

0

Answer:

22÷$32÷$35/%35^323=3%%4/%54/÷$^%/53355

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