Answer:
x = 6
y = -4
Step-by-step explanation:
Plug y into the first equation.
5x - 7(-x+2) = 58
Then, solve for x.
Plug in x back to any one of the equations to find y.
Using the Pythagorean theorem you would find X by taking the square root of the base squared (Y^2) subtracted from the hypotenuse squared ( 10^2 = 100)
The equation would be x = √100-y^2
The second answer is the correct one.
There are 150 teachers because 20 * 150 is 3000.
If you ever have a polynomial with a solution of bi, then it will also have a solution of -bi. Imaginary solutions always come in pairs.
So, yes, you could have a polynomial with solutions 2i and 5i, as long as -2i and -5i are also solutions.
(x-2i)(x+2i)(x-5i)(x+5i) would be the most basic polynomial you could form.
(x-2i)(x+2i)(x-5i)(x+5i) = (x^2+4)(x^2+25)
= x^4 + 29x^2 + 100
So the equation would be x^4 + 29x^2 + 100 = 0.
Now, if the question was "only the solutions of 2i and 5i and no others," then the answer is no, for the previously stated reason.
