Astate employees pension fund invested a total of one million dollars into two accounts that earned 3.5 and 4.5 annual simple interest. at the end of the year, the total interest earned was 42,000. how much was invested at each rate

Let the investment in first account be x and this account is earning 3.5% interest

So Interest on this account is 3.5% of x = 0.035x

So the investment in second account would be (1,000,000-x) and this account is earning 4.5% interest

So the interest on this account is 4.5% of (1,000,000-x) = 0.045*(1,000,000-x)

Given that total interest for the year is $42,000

So we have

0.035x + 0.045(1,000,000-x) = 42,000

0.035x - 0.045x + 0.045*1,000,000 = 42,000

-0.01x + 45,000 = 42,000

-0.01x = 42,000-45,000

-0.01x = - 3000

0.01x = 3000

x = 3000/0.01

x= 3,00,000

Therefore Investment in one account s 300,000 and the investment in other account is 1,000,000-300,000 = 700,000

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

answer is : muchas/tipicas

have a nice day!

~recklesschicken