Fuseina has ¢40000 to invest. She decides to put part of the money in an account paying 4% interest and the reminder into stocks paying 6% interest . Her financial advisor tells her that the total annual income from these investments should be¢2040. How much should she invest at each rate?

The amounts invested in the account = ¢ 18000

and the amount invested in the stock = ¢40000-¢18000= ¢22000.

Step-by-step explanation:

Given that the account paying 4% interest and the stocks paying 6% interest.

Assuming she put ¢ x in the account, so

Income (interest) from the account = 4% of x = ¢ 0.04x...(i)

She has a total of ¢ 40000 to invest, so the remaining amount = ¢(40000-x).

As she put the remaining amount into stock, so

Income (interest) from the stock=6% of (40000-x)=¢(2400-0.06x)...(ii)

As total income from the investment is ¢ 2040, so on adding both the incomes from equations (i) and (ii), we have

So, the amount invested in the account = ¢ 18000

and the amount invested in the stock = ¢40000-¢18000= ¢22000.

the answer is 48m^2!

step-by-step explanation:

c: y > - 1/6x + 1

step-by-step explanation:

dashed line - meaning this < or > also not equal

solid line - meaning this < or > also equal

your answer is c because y > means up & it's shaded up, but y < means down

hope this ! : )