In the year 2005, a company made $6.6 million in profit. for each consecutive year after that, their profit increased by 9%. how much would the company's profit be in the year 2009, to the nearest tenth of a million dollars?

  9.3 million

Step-by-step explanation:

The growth rate of 9% per year corresponds to a growth factor of 1.09, which goes into the exponential equation ...

  profit(t) = 6.6×1.09^t . . . . . for t in years after 2005

2009 is 4 years after 2005, so the profit that year is modeled as ...

  profit(4) = 6.6×1.09^4 ≈ 9.3 . . . . million dollars

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step-by-step explanation:

umm. i dont know

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .