Answer:
Step-by-step explanation:
-1/5 
-1/7
-1/8
-1/9
-1/10
0
1/6
1/5
1/4
1/3
 
        
             
        
        
        
You could simplify this work by factoring "3" out of all four terms, as follows:
3(x^2 + 2x - 3) =3(0) = 0
Hold the 3 for later re-insertion.  Focus on "completing the square" of x^2 + 2x - 3.
1.  Take the coefficient (2) of x and halve it:  2 divided by 2 is 1
2.   Square this result:  1^2 = 1
3.   Add this result (1) to x^2 + 2x, holding the "-3" for later:
                    x^2 +2x 
4    Subtract (1) from x^2 + 2x + 1:     x^2 + 2x + 1               -3 -1  =    0, 
       or      x^2 + 2x + 1 - 4 = 0
5.   Simplify, remembering that x^2 + 2x + 1 is a perfect square:
                        (x+1)^2 - 4 = 0
We have "completed the square."  We can stop here.  or, we could solve for x:  one way would be to factor the left side:
            [(x+1)-2][(x+1)+2]=0     The solutions would then be:
             x+1-2=0=> x-1=0, or x=1, and
             x+1 +2 = 0 => x+3=0, or x=-3.  (you were not asked to do this).
        
             
        
        
        
Answer: option D. 2x^2 + (3/2)x  - 5
Explanation:
1) polynomials given:
 f(x) = x/2 - 2 and g(x) = 2x^2 + x - 3
2) question: find (f + g) (x) 
That means that f(x) + g(x), so you have to add up the two polynomials given.
3) x/2  - 2 + 2x^2 + x - 3
4) Combine like terms:
a) terms with x^2: you only have 2x^2, so it is not combined with other term.
b) terms with x: x/2 + x
that is a sum of fractions: x/2 + x = [x + 2x] / 2 = 3x / 2 = (3/2)x 
c) constant terms: - 2 + (-3) = - 2 - 3 = - 5
5) Result: 2x^2 + (3/2)x - 5
That is the option d.
        
             
        
        
        
Answer:
it has infinitely many solutions
 
        
             
        
        
        
We have that
<span> |x+6| >= 5
step 1 
resolve for (x+6)>=5------> x>=5-6-------> x>= -1
the solution is the interval </span>(-1, ∞)
<span>
step 2
resolve for -(x+6) >=5------> -x-6 >=5----> -x >= 5+6---> -x>=11----> x<=-11
</span>the solution is the interval (-∞, -11)
<span>
using a graph tool
see the attached figure
the solution is the interval (-</span>
∞, -11) ∩ (-1, ∞)