Katy invests a total of $19,000 in two accounts paying 7% and 5% annual interest, respectively. how much was invested in each account if, after one year, the total interest was $1,210.00

$13,000 at 7%$6,000 at 5%Explanation:

Let x represent the dollar amount invested at 7% (the higher of the two rates). Then (19000-x) is the amount invested at 5%. The total interest earned in one year is the sum of the products of amount and rate. That sum is ...

... 1210 = (x)(.07) + (19000-x)(.05)

... 1210 = 0.02x + 950 . . . . . . collect terms

... 260 = 0.02x . . . . . . . . . . . . subtract 950

... 13000 = x . . . . . . . . . . . . . . divide by 0.02; amount at 7%

Then the amount invested at 5% is

... 19000-x = 19000-13000 = 6000

Comment on solution

Such a problem can be written with two variables, one for each amount. Then the two equations are equivalent to

x + y = 19000 . . . . . . total amount invested.07x +.05y = 1210 . . . total interest earned

Choosing one variable to represent the amount at the higher rate does two things:

It reduces the problem to one equation (equivalent to substituting y=19000-x in the above)It ensures the numbers you're working with are positive numbers.

If you choose x = investment at 5%, then the final arithmetic in the above solution becomes ...

... -120 = -0.02x ⇒ x = 6000 . . . . . same answer, but negative numbers to work with

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.

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