Answer:
Step-by-step explanation:
Each successive year, he
earned a 5% raise. It means that the salary is increasing in geometric progression. The formula for determining the nth term of a geometric progression is expressed as 
Tn = ar^(n - 1)
Where
a represents the first term of the sequence(amount earned in the first year).
r represents the common ratio.
n represents the number of terms(years).
From the information given, 
a = $32,000
r = 1 + 5/100 = 1.05
n = 20 years
The amount earned in his 20th year, T20 is
T20 = 32000 × 1.05^(20 - 1)
T20 = 32000 × 1.05^(19)
T20 = $80862.4
To determine the his total
earnings over the 20-year period, we would apply the formula for determining the sum of n terms, Sn of a geometric sequence which is expressed as
Sn = (ar^n - 1)/(r - 1)
Therefore, the sum of the first 20 terms, S20 is 
S20 = (32000 × 1.05^(20) - 1)/1.05 - 1
S20 = (32000 × 1.653)/0.05
S20 = $1057920
 
        
             
        
        
        
6.667 yards I think I got it of safari
        
                    
             
        
        
        
The complete proof statement and reason for the required proof is as follows:
Statement                                    Reason
m<PNO = 45                               Given
MO                                              Given
<MNP and <PNO are a
linear pair of angles                     Definition of linear pairs of angles
<MNP and <PNO are
supplementary angles                 Linear Pair Postulate
m<MNP + m<PNO = 180°          Definition of supplementary angles
m<MNP + 45° = 180°                 Substitution property of equality
m<MNP = 135°                          Subtraction property of equality
        
                    
             
        
        
        
Answer:
640
Step-by-step explanation:
you say 100% = 800
what about 80%(i get 80% after substracting 100% that's marked price - 20% discount)
then u say 80×800÷100
 
        
             
        
        
        
Answer:
1.  =
 = 
2.  =
 = 
3.  =
 = 
4.  =
 = 
Step-by-step explanation:
1. 
Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)

Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.
2. 
Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)
Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)
Putting factors

Prove closure: The value of x≠-2 and x≠4 because if there values are -2 and 4 then the denominator will be zero.
3. 
Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)
Factors of x^2-5x+6 = x^2-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)
Putting factors

Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)


Prove closure: The value of x≠3 and x≠-3 and x≠2 because if there values are -3,3 and 2 then the denominator will be zero.
4. 
Factors of x^2-5x+6 = x^2-3x-2x+6 = x(x-3)-2(x-3) = (x-2)(x-3)
Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

Converting ÷ sign into multiplication we will take reciprocal of the second term

Prove Closure: The value of x≠2 and x≠4 because if there values are 2 and 4 then the denominator will be zero.