The annual interest on an $20,000 investment exceeds the interest earned on a $4000 investment by $1320. the $20,000 is invested at a 0.6% higher rate of interest than the $4000. what is the interest rate of each investment? $20,000 is invested at what %$4,000 is invested at what %

"8.8%"

Step-by-step explanation:

Let x be the interest rate for the 20,000 and y for the 4,000

The problem says:

x20,000 = y4,000+1320 (1)

and

x = y+0.006 Substitute this value for x on equation (1)

(y+0.006)20,000 = y4,000+1320

20,000y+120 = y4,000+1320

16,000y = 1320

y = 0.0825 So the 4,000 are invested at 8.25%

And since x = y+0.006:

x = .0825+0.006 = 0.0885 or 8.8%. The 20,000 are invested at 8.8%

center= (-2,0)

radius= sqrt2/2

step-by-step explanation: