Compare the investment below to an investment of the same principal at the same rate compounded annually principal: $9000 annual interest: 8% interest periods: 6 number of years: 12

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

Considering the investment compounded annually,

From the information given,

P = 9000

r = 8% = 8/100 = 0.08

n = 1 because it was compounded once in a year.

t = 12 years

Therefore,

A = 9000(1 + 0.08/1)^1 × 12

A = 9000(1.08)^12

A = $22664

For the second investment,

n = 6

Therefore,

A = 9000(1 + 0.08/6)^6 × 12

A = 9000(1 + 0.0133)^72

A = 9000(1.0133)^72

A = $23301

The second investment gave higher returns.

(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

y/x of point b= y/x of point a

step-by-step explanation:

(i got the same question, this was the correct answer.)