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Yuliya22 [10]
3 years ago
6

Joe’s annual income has been increasing each year by the same dollar amount the first year his income was 19,500. 8 years later

his income is 27,900. In what year was his income 30,300
Mathematics
1 answer:
Anna007 [38]3 years ago
6 0

Answer:

The question is either poorly constructed or it is a trick question. We do not know "In what year was his income 30,300" because the years correlating to his income are not disclosed.  Even assuming the question was meant to be posed as "how many years later was his income 30,300" it is still flawed.

Step-by-step explanation:

To arrive at Joe's annual income increase, first subtract 19,500 from 27,900 to get 8400, which is his overall increase over the 8 years. Now divide 8400 by 8. You will get 1050.  Beginning with his income of 27,900, add 1,050 to each successive income amount and you will get the following progression:

27,900, 28,950, 30,000, 31,050, 32,100, 33,150, 34,200, 35,250, 36,300.

Therefore, two years after his income was 27,900 it would be 30,000 and 6 years after that it would be 36,300.  It would  never be 30,300.

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Answer:

a. x=-1,3,\:or\:6

Step-by-step explanation:

The given equation is;

x^3-8x^2+9x+18=0

To solve by the x-intercept method we need to graph the corresponding function using a graphing calculator or software.

The corresponding function is

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We can see from the graph  that; the x-intercepts are;

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