The sum of three consecutive intergers is 57 find the intergers

We know that "consecutive" means next to each other.

Let's make "x" the first integer, x+1 the second, and x+2 the third. For example, we could make x = 1.

Then we can plug this into our equation:

x + (x+1) + (x+2) = 57.

1 + (1+1) + (1+2) = 57

1 + 2 + 3 = 57.

So as you can tell, this does NOT work.

Let's continue the original equation:

x + (x+1) + (x+2) = 57.

Combine like terms:

x + x + 1 + x + 2 = 57.

3x = 57 - 2 - 1.

3x = 54.

x = 18.

x + 1 = 19

x + 2 = 20.

Add these up for proof: 18 + 19 + 20 = 57

:)

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation:

b. f(x) = –√x + 3this is the correct answer