Sue has an annual interest income of $3390 from two investments. she has $10,000 more invested at 8% than she has at 6%. find the amount invested at each rate.

  $18500 at 6% and $28500 at 8%

Step-by-step explanation:

Let x represent the amount invested at 6%. Then x+10000 is the amount invested at 8%. The total income from the two investments is the sum of the products of the amount invested and the interest rate:

  x·6% +(x+10000)·8% = 3390

  0.14x + 800 = 3390 . . . . . . . . collect terms

  0.14x = 2590 . . . . . . . . . . . . . .subtract 800

  2590/0.14 = x = 18500 . . . . . divide by the coefficient of x

Sue has invested $18500 at 6% and $28500 at 8%.

16.9 times 2.85

step-by-step explanation:

its 60

4 x 12 = 48

3 x 4 = 12

48 + 12 = 60

yeeeet