Question 821710
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L(t) = 0.0004t^2 + 0.16t + 20
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answer a:
t = 100 degrees C:
L(1) = 0.0004(100)^2 + 0.16(100) + 20
L(1) = 4 + 16 + 20
L(1) = 40 ml per hour
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answer b:
L(t) = 53 = 0.0004t^2 + 0.16t + 20
0.0004t^2 + 0.16t - 33 = 0
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the above quadratic equation is in standard form, with a=0.0004, b=0.16, and c=-33
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
0.0004 0.16 -33
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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the two real roots (i.e. the two x-intercepts), of the quadratic are:
t = 150
t = -550
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negative operating temperature doesn't make sense for this problem, so use the positive root:
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operating temperature is 150 degrees C
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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