A picture could hold to 12th of a gallon of water if Roger filled up nine pictures how much water would he have
1 answer:
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Start the graph at the <em>y</em>-intercept. At 0 hours, he will make $0, so (0, 0) is the <em>y</em>-intercept. From this point, he makes $8.10 an hour; go up 8.1 for every 1 hour you go over on the graph. The graph is attached.
Answer:
- CDs — $10,000
- bonds — $80,000
- stocks — $45,000
Step-by-step explanation:
Let the variables c, b, s represent the dollar amounts invested in CDs, stocks, and bonds, respectively. Then the problem statement gives us 3 relations between these 3 variables:
c + b + s = 135000 . . . . . . . . . . . . . . . . . total invested
0.0275c +.045b +0.104s = 8555 . . . . . total income earned
-c + b = 70000 . . . . . . . . . . . . . . . . . . . . . 70,000 more was in bonds than CDs
Using the third equation to write an expression for b, we can substitute into the other two equations.
b = 70000 +c . . . . . . . . . . . . . . . . expression we can substitute for b
c + (70000 +c) +s = 135000 . . . . substitute for b in the first equation
2c +s = 65000 . . . . . . . . . . . . . . . . [eq4] simplify
.0275c +.045(70000 +c) +.104s = 8555 . . . . . substitute for b in 2nd eqn
.0725c +.104s = 5405 . . . . . . . . . . [eq5] simplify
Using [eq4], we can write an expression for s that can be substituted into [eq5].
s = 65000 -2c . . . . . . . expression we can substitute for s
0.0725c +0.104(65000 -2c) = 5405
-0.1355c = -1355 . . . . . . . . . . . . . . . . . . . . subtract 6760, simplify
c = 1355/.1355 = 10,000
s = 65000 -2×10000 = 45,000
b = 70000 +10000 = 80,000
The amounts invested in stocks, bonds, and CDs were $45,000, $80,000, and $10,000, respectively.
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Alternatively, you can reduce the augmented matrix for this problem to row-echelon form using any of several calculators or on-line sites. That matrix is ...