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VARVARA [1.3K]
3 years ago
9

​Joe's annual income has been increasing each year by the same dollar amount. The first year his income was ​$17 comma 90017,900

​, and the 44th year his income was ​$20 comma 30020,300. In which year was his income $ 30 comma 700 question mark
Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
5 0

Answer:

In 17th year, his income was $30,700.

Step-by-step explanation:

It is given that the income has been increasing each year by the same dollar amount. It means it is linear function.

Income in first year = $17,900

Income in 4th year = $20,300

Let y be the income at x year.

It means the line passes through the point (1,17900) and (4,20300).

If a line passes through two points (x_1,y_1) and (x_2,y_2), then the equation of line is

y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

The equation of line is

y-17900=\frac{20300-17900}{4-1}(x-1)

y-17900=\frac{2400}{3}(x-1)

y-17900=800(x-1)

y-17900=800x-800

Add 17900 on both sides.

y=800x-800+17900

y=800x+17100

The income equation is y=800x+17100.

Substitute y=30,700 in the above equation.

30700=800x+17100

Subtract 17100 from both sides.

30700-17100=800x

13600=800x

Divide both sides by 800.

\frac{13600}{800}=x

17=x

Therefore, in 17th year his income was $30,700.

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Factor a perfect square on the left side:

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Calculate the square root of the right side: 6.184658438

Break this problem into two subproblems by setting  

(x + -6.5) equal to 6.184658438 and -6.184658438.


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