Algebra and Trigonometry 10th Edition

$y \leq x; x \geq 0$ and $x^2+y^2 \leq 16$
Using the intercepts $(0,0), (\sqrt 8, \sqrt 8)$ , we compute the slope $m$. $m=\dfrac{\sqrt 8-0}{\sqrt 8-0}=1$ The slope-intercept form is: $y-0=x-0$ or, $y=x$ Thus, $y \leq x$ Now, we will write the equation of the circle with center $(0,0)$ and radius $4$ as: $x^2+y^2=16$ and $(1)^2+(2)^2 \lt 16$ So, our correct inequalities describing the region are: $y \leq x; x \geq 0$ and $x^2+y^2 \leq 16$