One number is 4 times a first number. A third number is 100 more than the first number. If the sum of the three numbers is 400, find the numbers

Need ! give step by step solutions on how to solve number one [tex]\frac{9-2\sqrt{3} }{12+\sqrt{3} }[/tex] number two [tex]x+4=\sqrt{13x-20}[/tex] number three (domain and range) [tex]f(x)=2\sqrt[3]{x} +1[/tex]

How do you find the area of a triangle

Just tell me how to set up the equation.