Answer:
Part 1) The quadratic equation has zero real solutions
Part 2) The solutions are
 and
   and  
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form  is equal to
 is equal to
 
in this problem we have
 
  
so
 
The discriminant is equal to

If D=0 -----> the quadratic equation has only one real solution
If D>0 -----> the quadratic equation has two real solutions
If D<0 -----> the quadratic equation has two complex solutions
<em>Find the value of D</em>
 -----> the quadratic equation has two complex solutions
 -----> the quadratic equation has two complex solutions
<em>Find out the solutions</em>
substitute the values of a,b and c in the formula
 
 
Remember that

 
 
 
 
        
             
        
        
        
<span>1. (f+g)(x) = f(x) +g(x)
.. = (</span>x^2-36) +(<span>x^3+2x^2-10)
.. = x^3 +3x^2 -46
2. </span>(f•g)(x) = f(x)•g(x)
.. = (x^4-9)•(x^3+9)
.. = x^7 +9x^4 -9x^3 -81
<span>3. (f-g)(x) = f(x) -g(x)
.. = (x^3-2x^2+12x-6) -(4x^2-6x+4)
.. = x^3 -6x^2 +18x -10</span>
        
             
        
        
        
Guessing geometric
so
hmm
5 times what=3
divide both sides by 5
what=3/5
what=0.6
each term is multiplied by 0.6
so
1.08 times 0.6=0.648
0.648*0.6=0.3888
the next 2 terms are 0.648 and 0.3888
        
             
        
        
        
Answer:
Step-by-step explanation:
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