You invested $6000 between two accounts paying 3% and 9% annual interest, respectively. If the total interest earned for the year was $480, how much was invested at each rate?

$1,000 and $5,000

Step-by-step explanation:

let the 2 amounts be x and y (which add  to $6,000)

i.e,

x + y = 6000 (eq1)

assume x is invested at 3% and y is invested at 9%

Assuming simple interest, the following formula is applicable.

I = Prt

where

P = principal ( i.e either $x or $y)

r = rate (i.e 3% or 9%)

t = time elapsed (given as 1 year)

since t = 1 year, the equation reduces to

I = Pr

For $x invested at 3% (i.e 0.03) ---> I = 0.03x

For $y invested at 9% (i.e 0.09) ---> I = 0.09y

given that the total interest for the year = $480,

0.03x + 0.09y = 480   (multiplying both sides by 100)

3x + 9y = 48,000 (eq2)

solving the system of equations made up of eq 1 and eq 2 by elimination,

eq 1 x 3

3x + 3y = 18,000 (eq 3)

eq2 - eq3

9y - 3y = 48,000 - 18,000

6y = 30,000

y = 30,00 / 6 = $5,000 (answer)

substitute y = 5,000 into eq 1.

x + 5000 = 6000

x = $1,000 (answer)

d) 30 units^2

area of rectangle = l × b

we know,

length = 10 unit

breadth = 3 unit

applying the formula,

= l × b

= 10 × 3

= 30 unit^2

happy to

pls mark as brainliest

it will cost dollar $540

the living room is 3 which means if we divide 325/3 then we get $108

therefore if 1part of the living room cost $108 and 3 costed $325

then total 5part of the room will cost 5*108 which gives $540