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Question 825092: Maria could sew on all the sequins in 10 hours less time than Stephanie. to save time, he gave the job to both women and got all the sequins attached in 17 hr. how long would it have taken Stephanie working alone?
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! ---
rates:
x = maria
y = steph
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time = 1/rate
1/x = 1/y - 10
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working together rate:
x + y = 1/17
y = 1/17 - x
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1/x = 1/y - 10
1/x = 1/(1/17 - x) - 10
1/(1/17 - x) - 1/x = 10
x/x(1/17 - x) - (1/17 - x)/x(1/17 - x) = 10
x - (1/17 - x) = 10x(1/17 - x)
x - (1/17) + x = (10/17)x - 10xx
2x - (1/17) = (10/17)x - 10xx
-10xx - 2x + (10/17)x + (1/17) = 0
-10xx - (34/17)x + (10/17)x + (1/17) = 0
-10xx - (24/17)x + (1/17) = 0
-10xx - (1.411764705882353)x + (0.058823529411765) = 0
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the above quadratic equation is in standard form, with a=-10, b=-1.411764705882353, and c=0.058823529411765
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
-10 -1.411764705882353 0.058823529411765
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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this quadratic has two real roots (the x-intercepts), which are:
x = -0.174823795
x = 0.0336473244
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the negative root does not fit the problem statement, so use the positive root
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rates:
x = 0.0336473244
y = 0.058823529411765 - 0.0336473244
y = 0.025176205011765
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answer:
times:
1/x = maria = 1/0.0336473244 = 29.7 hr
1/y = steph = 1/0.0251762050 = 39.7 hr
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Solve quadratic equations, quadratic formula:
https://sooeet.com/math/quadratic-formula-solver.php
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Solve systems of linear equations up to 6-equations 6-variables:
https://sooeet.com/math/system-of-linear-equations-solver.php
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