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On a 40Fahrenheit40 \, \text{Fahrenheit} day, a 25mile per hour25 \, \text{mile per hour} wind creates a wind chill of 29Fahrenheit29 \, \text{Fahrenheit}. To approximate the wind chill at various temperatures caused by a 25mile per hour25 \, \text{mile per hour} wind, Elon uses this rule: take 2424 less than 43\frac{4}{3} of the air temperature(Fahrenheit\text{Fahrenheit}). If a 25mile per hour25 \, \text{mile per hour} wind creates a chill of 5Fahrenheit-5 \, \text{Fahrenheit}, which of the following best approximates the corresponding air temperature?

Full solution

Q. On a 40Fahrenheit40 \, \text{Fahrenheit} day, a 25mile per hour25 \, \text{mile per hour} wind creates a wind chill of 29Fahrenheit29 \, \text{Fahrenheit}. To approximate the wind chill at various temperatures caused by a 25mile per hour25 \, \text{mile per hour} wind, Elon uses this rule: take 2424 less than 43\frac{4}{3} of the air temperature(Fahrenheit\text{Fahrenheit}). If a 25mile per hour25 \, \text{mile per hour} wind creates a chill of 5Fahrenheit-5 \, \text{Fahrenheit}, which of the following best approximates the corresponding air temperature?
  1. Understand the rule: Understand the rule given by Elon for approximating the wind chill at various temperatures caused by a 2525 mile per hour wind. The rule is to take 2424 less than 43\frac{4}{3} of the air temperature (Fahrenheit).
  2. Set up the equation: Set up the equation using the rule to find the air temperature when the wind chill is 5-5 Fahrenheit.\newlineLet TT be the air temperature in Fahrenheit. According to the rule:\newlineWind chill = (4/3)T24(4/3)T - 24\newlineWe know the wind chill is 5-5, so we can write:\newline5=(4/3)T24-5 = (4/3)T - 24
  3. Solve the equation: Solve the equation for TT. Add 2424 to both sides to isolate the term with TT on one side: 5+24=(43)T24+24-5 + 24 = \left(\frac{4}{3}\right)T - 24 + 24 19=(43)T19 = \left(\frac{4}{3}\right)T
  4. Multiply to find TT: Multiply both sides by the reciprocal of 43\frac{4}{3} to solve for TT.34×19=T\frac{3}{4} \times 19 = TT=574T = \frac{57}{4}T=14.25T = 14.25

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