You invested $21000 in two accounts paying 5 % and 9 % annual interest, respectively. If the total interest earned for the year was $ 1610 comma how much was invested at each rate?

$7,000 is invested at 5% interest

$14,000 is invested at 9% interest

Explanation:

Let A represent the amount of interest that was invested at 5%

= 5/100

= 0.05

Then 21,000-A will represent the amount that was invested at 9%

= 9/100

= 0.09

The amount invested at 5% can be calculated as follows

0.05A + 0.09(21,000-A)= 1,610

0.05A + 1,890 -0.09A= 1,610

Collect the like terms

0.05A - 0.09A = 1,610-1,890

-0.04A= -280

Divide both sides by the coefficient of A which is -0.04

-0.04A/-0.04= -280/-0.04

A= $7,000

Recall that 21,000-A represents the amount invested at 9%, since we have gotten the value of A then, It can be inputed into the expression

= $21,000-$7,000

= $14,000

Hence $7,000 was invested at 5% interest and $14,000 was invested at 9% interest.

if kj of heat is transferred o 1200 g liquid water at 36'c, what would the new temperature of the water be? (he specific heat capacity of liquid water is 4.186 j/g''c)

a. 15'c

b.100'c

c.51'c

d.36'c

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