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The speed of sound in air is about 332 meters per second ((m)/(s)) at 0 degrees Celsius (^(@)C). If the speed increases by 0.6(m)/(s) for every increase in temperature of 1^(@)C, which inequality best represents the temperatures, T, in degrees Celsius, for which the speed of sound in air exceeds 350(m)/(s) ?

The speed of sound in air is about 332332 meters per second (ms) \left(\frac{\mathrm{m}}{\mathrm{s}}\right) at 00 degrees Celsius (∘C) \left({ }^{\circ} \mathrm{C}\right) . If the speed increases by 0.6ms 0.6 \frac{\mathrm{m}}{\mathrm{s}} for every increase in temperature of 1∘C 1^{\circ} \mathrm{C} , which inequality best represents the temperatures, T T , in degrees Celsius, for which the speed of sound in air exceeds 350ms 350 \frac{\mathrm{m}}{\mathrm{s}} ?

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Q. The speed of sound in air is about 332332 meters per second (ms) \left(\frac{\mathrm{m}}{\mathrm{s}}\right) at 00 degrees Celsius (∘C) \left({ }^{\circ} \mathrm{C}\right) . If the speed increases by 0.6ms 0.6 \frac{\mathrm{m}}{\mathrm{s}} for every increase in temperature of 1∘C 1^{\circ} \mathrm{C} , which inequality best represents the temperatures, T T , in degrees Celsius, for which the speed of sound in air exceeds 350ms 350 \frac{\mathrm{m}}{\mathrm{s}} ?
  1. Given Information: We are given that the speed of sound in air at 00 degrees Celsius is 332332 meters per second and that it increases by 0.60.6 meters per second for every 11 degree Celsius increase in temperature. We need to find the temperature at which the speed of sound exceeds 350350 meters per second. Let's denote the temperature in degrees Celsius as TT and the speed of sound in meters per second as S(T)S(T). The relationship between the speed of sound and the temperature can be expressed as:\newlineS(T)=332+0.6TS(T) = 332 + 0.6T
  2. Speed of Sound Inequality: To find the temperature at which the speed of sound exceeds 350350 meters per second, we need to solve the inequality:\newlineS(T) > 350\newlineSubstituting the expression for S(T)S(T) from the previous step, we get:\newline332 + 0.6T > 350
  3. Solving the Inequality: Now, we solve the inequality for TT:0.6T > 350 - 3320.6T > 18
  4. Isolating T: Divide both sides of the inequality by 0.60.6 to isolate TT: \newlineT > \frac{18}{0.6}\newlineT > 30
  5. Final Temperature Range: The inequality T > 30 represents the temperatures in degrees Celsius for which the speed of sound in air exceeds 350350 meters per second.

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