Question

найдите область определения функции :
$y = \sqrt{9 - 36x}$

Answer 1

$y= \sqrt{9-36x} \\ 9-36x \geq 0 \\ 9 \geq 36x \\ \frac{9}{36} \geq x \\ \frac{1}{4} \geq x \\ x \in (-\infty; \frac{1}{4}]$