To find the domain you need to set the denominator not equal to zero
4+I-15I÷3
4+15÷3
4+5
9
The answer is 9
Check the picture below.
so let's find the lengths of those two sides in red, since are the length and width of the rectangle.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-3}~,~\stackrel{y_2}{6})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d = \sqrt{[-3-(-6)]^2+[6-3]^2}\implies d=\sqrt{(-3+6)^2+(6-3)^2} \\\\\\ d=\sqrt{9+9}\implies \boxed{d=\sqrt{18}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-6%7D~%2C~%5Cstackrel%7By_1%7D%7B3%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-3%7D~%2C~%5Cstackrel%7By_2%7D%7B6%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%20%3D%20%5Csqrt%7B%5B-3-%28-6%29%5D%5E2%2B%5B6-3%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-3%2B6%29%5E2%2B%286-3%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B9%2B9%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B18%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-1})~\hfill d=\sqrt{[-2-(-6)]^2+[-1-3]^2} \\\\\\ d=\sqrt{(-2+6)^2+(-1-3)^2}\implies d=\sqrt{16+16}\implies \boxed{d=\sqrt{32}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the rectangle}}{(\sqrt{18})(\sqrt{32})}\implies \sqrt{18\cdot 32}\implies \sqrt{576}\implies 24](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-6%7D~%2C~%5Cstackrel%7By_1%7D%7B3%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-2%7D~%2C~%5Cstackrel%7By_2%7D%7B-1%7D%29~%5Chfill%20d%3D%5Csqrt%7B%5B-2-%28-6%29%5D%5E2%2B%5B-1-3%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B%28-2%2B6%29%5E2%2B%28-1-3%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B16%2B16%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B32%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20rectangle%7D%7D%7B%28%5Csqrt%7B18%7D%29%28%5Csqrt%7B32%7D%29%7D%5Cimplies%20%5Csqrt%7B18%5Ccdot%2032%7D%5Cimplies%20%5Csqrt%7B576%7D%5Cimplies%2024)
The answer is -6/5
First convert the equation to y=mx+b form
you’ll get y=5/6x+27
then to find the line perpendicular to this equation, you have the flip the slope.
ending with -6/5
The abscissa of the ordered pair, that is the x-coordinate, is equal to 1 and the ordinate, the y-coordinate, is equal to -1. In the cartesian plane, this point lies in the fourth (IV) quadrant. The standard position of the angle is that which has one of its side is in the x-axis.
Solve for the hypotenuse of the right triangle formed.
h = sqrt((-1)² + (1)²) = √2
Below items show the calculation for each of the trigonometric functions.
sin θ = opposite/hypotenuse = y/h = (-1)/(√2) = -√2/2
cos θ = adjacent/hypotenuse = x/h = (1)/√2 = √2/2
tan θ = opposite/adjacent = y/x = -1/1 = -1
