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.

it's d.

step-by-step explanation:

because: -1(4x + 1y) +2 (1x-3y) distribute each number into the parenthesis.

then you would get: -4x - 1y + 2x -6y combine like terms

and get: 2x -7y

hope my answer has you if not then i'm sorry.

2? if not the correct answer you might need to provide more information

step-by-step explanation: