Paul has $20000 to invest. He can invest part of his money at 6% and the rest at 4%. How much does paul have to invest at each rate if he wants to get a total of $1100 interest on his $20000

he should invest $15000 at 6% and $5000 at 4%

Step-by-step explanation:

Let x represent the part of his money that he should invest at 6%.

Let y represent the part of his money that he should invest at 4%.

Paul has $20000 to invest. He can invest part of his money at 6% and the rest at 4%. It means that

x + y = 20000

The interest that he would earn on the amount invested at 6% after 1 year is

6/100 × x = 0.06x

The interest that he would earn on the amount invested at 4% after 1 year is

4/100 × y = 0.04y

if he wants to get a total of $1100 interest on his $20000, it means that

0.06x + 0.04y = 1100- - - - - - - - - 1

Substituting x = 20000 - y into equation 1, it becomes

0.06(20000 - y) + 0.04y = 1100

1200 - 0.06y + 0.04y = 1100

- 0.06y + 0.04y = 1100 - 1200

- 0.02y = - 100

y = - 100/-0.02

y = 5000

x = 20000 - y = 20000 - 5000

x = 15000

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

8

step-by-step explanation:

just put √(2) * √(32) into a calculator. that will make the answer be 8.