Find three consecutive even integers such that the sum of the smallest number and twice the middle number is more than the largest number.

Hey there!!

Let's take the first number as 'x'

Then, the second consecutive integer would be 'x+2'

Hence, the third consecutive integer would be 'x+4'

As the question states, the sum of the smallest integer and the second integer is 20 greater than the third number.

Hence, the equation would be:

... x + x + 2 = x + 4 + 20

... 2x + 2 = x + 24

Subtracting x on both sides:

... 2x - x + 2 = 24

... x + 2 = 24

Subtracting 2 on both sides:

... x = 24 - 2

... x = 22

Hence, the smallest number is 22.

The second number is = 22 + 2

... 24

The third number is = 22 + 4

... 26

Hence, the numbers are 22,24 and 26.

Hope my answer helps!

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places

cross multiplying, we get :

⇒ (3) × (5) = (9) × (x)

⇒ 15 = 9x