Note that the 2nd equation can be re-written as y=8x-10.
According to the second equation, y=x^2+12x+30.
Equate these two equations to eliminate y:
8x-10 = x^2+12x+30
Group all terms together on the right side. To do this, add -8x+10 to both sides. Then 0 = x^2 +4x +40. You must now solve this quadratic equation for x, if possible. I found that this equation has NO REAL SOLUTIONS, so we must conclude that the given system of equations has NO REAL SOLUTIONS.
If you have a graphing calculator, please graph 8x-10 and x^2+12x+30 on the same screen. You will see two separate graphs that do NOT intersect. This is another way in which to see / conclude that there is NO REAL SOLUTION to this system of equations.
Yes your correct but thats not a question
8% of 100 is 8, so add that to your initial deposit and you will have $108 after one year.
Answer:
can you plz send the full image
A) No, the root 3 has multiplicity of 3.
B) No, there are two imaginary solutions.
C) No, the root −2 has multiplicity of 3.
D) Yes, the Fundamental Theorem of Algebra does not apply to this equation.
