Five times the first of three consecutive even integers is 2 less than twice the sum of the second and third. find the 3 integers

As we are being asked to solve for the middle of three consecutive even integers, let this be x.  Then, the first even integer is x - 2, and the third even integer is x + 2.

Then, we are given that 5 times the second of three consecutive even integers is six more than twice the sum of the first and third integers.

Translating this into an equation, we have

5 * x = 6 + 2 * (x-2 + x+2)

5x = 6 + 2 * 2x

5x = 6 + 4x

5x - 4x = 6

1x = 6

x = 6

Check the answer; the middle even integer is 6, so the first even integer is 4 and the third is 8

Then, 5 times the second even integer is 5 * 6 = 30

6 + 2 * the sum of the first and third integers is

6 + 2 *(4 + 8) = 6 + 2 * 12 = 30

30 = 30, so it checks

the reason why your question keeps getting deleted because this is not a school related question : )

i will answer your question though.

i never tried church's

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.