Let
x-------> the first number
y-------> the second number
we know that

equation 1

equation 2
substitute the equation 1 in equation 2
![[-5-y]*y=-90 \\ -5y- y^{2} =-90 \\ y^{2} +5y-90=0](https://tex.z-dn.net/?f=%5B-5-y%5D%2Ay%3D-90%20%5C%5C%20-5y-%20y%5E%7B2%7D%20%3D-90%20%5C%5C%20y%5E%7B2%7D%20%2B5y-90%3D0)
using a graphical tool to resolve the second order equation
see the attached figure
the numbers are-12.3117.311
28,000 x 25% = 7,000
28,000 + 7,000 = 35,000
Sec(theta) = 1 / cos (theta) = hypotenuse / x -coordinate
hypotenuse = 1 (because it is the radius of the unit circle)
sec (theta) = 1 / (-3/5) = - 5/3
cot (theta) = 1 / tan(theta) = x-coordinate / y - coordinate
cot (theta) = -3/5 / y
y^2 + (-3/5)^2 = 1 => y^2 = 1 - 9/25 = 16/25 = y = +/- 4/5
Third quadrant => y = -4/5
=> cot (theta) = (-3/5) / (-4/5) = 3/4
Answer:
not possible
Step-by-step explanation:
f(2)=2^2-4=0
g(f(2))=5/0 it's not possible