SOLUTION: May I have some help with this following problem?
In the vicinity of a bonfire, the temperature T in °C at a distance of x meters from the center of the fire was given by the fo
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-> SOLUTION: May I have some help with this following problem?
In the vicinity of a bonfire, the temperature T in °C at a distance of x meters from the center of the fire was given by the fo
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Question 121557: May I have some help with this following problem?
In the vicinity of a bonfire, the temperature T in °C at a distance of x meters from the center of the fire was given by the following formula.
T=325000/(x^2+300)
At what range of distances from the fire's center was the temperature less than 250°C? Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website! 250=325000/(x^2+300) now cross multiply.
250(x^2+300)=325000
250x^2+75000-325000=0
250(x^2+300-1300)=0
using the quadratic equation we get:
x=(-300+-sqrt[300^3-4*1*-1300])/2*1
x=(-300+-sqrt[90000+5200])/2
x=(-300+-sqrt95,200)/2
x=(-300+-308.545)/2
x=(-300+308.545)/2
x=8.545/2
x=4.27 meters is the answer.
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