ou invested 7000 between two accounts paying 4% and 9% annual interest, respectively. If the total interest earned for the year
was $430 how much was invested at each rate? $ nothing was invested at and $ nothing was invested at
1 answer:
9514 1404 393
Answer:
Step-by-step explanation:
Let x represent the amount invested at 9%. Then 7000-x was invested at 4% and the interest earned was ...
9%·x +4%(7000-x) = 430
5%·x +280 = 430 . . . . . . . . . simplify
0.05x = 150 . . . . . . . . . . . subtract 280
x = 3000 . . . . . . . . . . divide by 0.05
$3000 was invested at 9%; $4000 was invested at 4%.
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Answer:
15
Step-by-step explanation:
=> (-3)²+(-4+7)(2)
=> 9+(3)(2)
=> 9+6
=> 15
First divide 2,727 by 27, which gives you 101. Then multiply 101 by $4.04.
This gives you D, $408.04
Step-by-step explanation:
- 2x + 4 + 3 = -4
-2x+7= -4
-2x= -4-7
-2x= -11
x=-11/-2
x=11/2
x=5.5 (if u want the answer in decimal form)
Do you need both of the equations or just the entire solution?
Pemdas helps so first you do Parentheses Exponet multiply divide add and subtract
