Answer:
2:3
Step-by-step explanation:
Answer:
$9000 at 4$
and
$10000 at 8%
Step-by-step explanation:
Let's assume that "x" is the amount deposited in the 4% account and "y" is the amount deposited in the 8% account.
Recall the formula for interest as : 
where I is the interest, R is the annual rate of interest and t is the number of years.
Since there are two investments, we need to add both interests at the end of the one year: I1 = x (0.04) (1) = 0.04 x and I2 = y (0.08) (1) = 0.08 y
Total Interest = Interest (from the 4% account) + Interest (from the 8% account)
Total Interest = $1160 = 0.04 x + 0.08 y
we also know that the total invested (x + y) adds to $19,000, that is:
$19,000 = x + y
Then we can solve these system of two equations by substitution, for example solving for y in the second equation and using the y substitution in the first equation;
y = 19000 - x
1160 = 0.04 x + 0.08 (19000 - x)
1160 = 0.04 x + 1520 - 0.08 x
0.08 x - 0.04 x = 1520 - 1160
0.04 x = 360
x = 360/0.04 = $9000
Then the other investment was : y = $19000 - $9000 = $10000
A= P(1 + r/n) ^nt
21000=7000(1+0.055/2)^1t
t = 20.52 years
2x+y = -5. Solve this for y. We get y = -2x - 5. Find y^2: 4x^2 + 20x + 25. Substitute 4x^2 + 20x + 25 for y^2 in the first equation:
x^2 + 4x^2 + 20x + 25 = 25
Then 5x^2 + 20x = 0, so that x = 0. Subst. 0 for x in the 2nd eqn and find the value of y. Write your solutions as shown above: ( , ) and ( , ).