SOLUTION: 1. A rectangular garden has dimensions of 18 feet by 13 feet. A gravel path of uniform width is to be built around the garden. How wide can the path be if there is enough gravel fo

Question 171738: 1. A rectangular garden has dimensions of 18 feet by 13 feet. A gravel path of uniform width is to be built around the garden. How wide can the path be if there is enough gravel for 516 square feet?
2. A business invests $10,000 in a savings account for two years. At the beginning of the second year, an additional $3500 is invested. At the end of the second year, the account balance is $15,569.75. What was the annual interest rate?
3. Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1. A rectangular garden has dimensions of 18 feet by 13 feet. A gravel path of uniform width is to be built around the garden. How wide can the path be if there is enough gravel for 516 square feet?
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Draw the picture of the garden with the path around it.
Area including the path = 18*13 + 516 = 750 sq. ft
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Let the uniform width of the path be "x".
Total area of path and picture : (18+2x)(13+2x)
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EQUATION:
(18+2x)(13+2x) = 750
234 + 36x + 26x + 4x^2 = 750
4x^2 + 62x - 516 = 0
2x^2 + 31x - 258 = 0
x = [-31 +- sqrt(31^2-4*2*-258)]/4
x = [-31 +- 55]/4
Positive solution:
x = [-31+55]/4 = 6 ft
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2. A business invests $10,000 in a savings account for two years. At the beginning of the second year, an additional $3500 is invested. At the end of the second year, the account balance is $15,569.75. What was the annual interest rate?
?
--------------------------
3. Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
Before change DATA:
distance = 200 miles ; rate = x mph ; Time = d/r = 200/x hrs.
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After change DATA:
distance = 200 miles ; rate = "x+10" mph ; time = 200/(x+10) hrs.
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EQUATION:
Before time - After time = 1 hr.
200/x - 200/(x+10) = 1
200x + 2000 - 200x = x(x+10)
x^2 + 10x - 2000 = 0
(x+50)(x-40) = 0
Positive solution:
x = 40 mph (speed of his vehicle before the change)
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Cheers,
Stan H.

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