Ms. Buxx invested a total of $2000
in two savings accounts. The first
account pays 3% interest per year.
The second account pays 5% interest
per year. If the interest from both
accounts totals $84 per year, how
much is invested in each account?

Step-by-step explanation:

Let x represent the amount invested in the account earning the higher interest rate. Then the total interest earned is ...

  0.05x +0.03(2000-x) = 84

  0.02x +60 = 84 . . . . . . . . . . . eliminate parentheses, collect terms

  0.02x = 24 . . . . . . . . . . . . . . . subtract 60

  x = 1200 . . . . . . . . . . . . . . . . . divide by 0.02

Ms Buxx invested $1200 at 5% and $800 at 3%.

$800 was invested at 3%

$1200 was invested at 5%

Step-by-step explanation:

Let x represent the amount that Ms. Buxx invested in the 3% account.

Let y represent the amount that Ms. Buxx invested in the 5% account.

Ms. Buxx invested a total of $2000

in two savings accounts. The first

account pays 3% interest per year.

The second account pays 5% interest

per year. This means that

x + y = 2000

Considering the amount invested in the 3% interest account, the interest after a year would be

(x × 3 × 1)/100 = 0.03x

Considering the amount invested in the 5% interest account, the interest after a year would be

(y × 5 × 1)/100 = 0.05y

If the interest from both

accounts totals $84 per year, it means that

0.03x + 0.05y = 84 - - - - - - - - - - - 1

Substituting x = 2000 - y into equation 1, it becomes

0.03(2000 - y) + 0.05y = 84

60 - 0.03y + 0.05y = 84

- 0.03y + 0.05y = 84 - 60

0.02y = 24

y = 24/0.02 = 1200

x = 2000 - y = 2000 - 1200

x = 800

the red lines denote the population of the bacteria after 35 hours. that would be 2,409.24. the y-intercept is the blue line which is 800 because that is the starting point. the equation used to make the graph is 800(1.032)^h, where h is the number of hours.

step-by-step explanation:

24

step-by-step explanation: