The sum of the interior angles of a triangle is 180 degrees
2x-9+3x+x+3=180
6x-6=180
6x=186
x=31
A=2(31)-9
A=62-9
A=53
        
             
        
        
        
Answer:
See sample table below.
Step-by-step explanation:
The function is given as :
f(x) = 3ˣ
A table of values can be formed as ;
x                <u>calculations</u>           f(x)
-4               3⁻⁴                          0.0123
-3               3⁻³                          0.0370
-2               3⁻²                          0.1111
-1                3⁻¹                           0.3333
0                 3⁰                            1.000
1                  3¹                             3.000
2                 3²                             9.000
3                 3³                              27.00
4                 3⁴                              81.00
 
        
             
        
        
        
At Gino's you pay $16 plus $8 per pizza.
At Venetian's you pay $24 plus $6 per pizza.
Let the number of pizzas be x.
At Gino's you pay 16 + 8x
At Venetian's you pay 24 + 6x
Set the two costs equal and solve for x to find out the number of pizzas for which both costs are the same.
16 + 8x = 24 + 6x
16 + 2x = 24
2x = 8
x = 4
Each place gives you a free pizza.
4 pizzas plus the free pizza equals 5 pizzas.
If you need 5 pizzas (including the free one), both parlors cost the same.
If you need fewer than 5 pizzas, use Gino's.
If you need more than 5 pizzas, use Venetian's.
        
             
        
        
        
Answer:
D 144 in hope this helps you on your test
 
        
             
        
        
        
Answer:
   x = 2
Step-by-step explanation:
These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.
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<h3>Squaring</h3>
The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.
   
The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.
   x = -1: √(-1+2) +1 = √(3(-1)+3)   ⇒   1+1 = 0 . . . . not true
   x = 2: √(2+2) +1 = √(3(2) +3)   ⇒   2 +1 = 3 . . . . true . . . x = 2 is the solution
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<h3>Substitution</h3>
Another way to solve this is using substitution for one of the radicals. We choose ...
   
Solutions to this equation are ...
   u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution
The value of x is ...
   x = u² -2 = 2² -2
   x = 2 . . . . the solution to the equation
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<em>Additional comment</em>
Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.