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| 2017/06/29 03:25:08瀏覽50｜回應0｜推薦0 | |
| 標題: another 2 math question 發問: 1. x + y + z = 0 x - y + z = 0 x + z = 02. Temperature and Heart RateWhen the temperature , T ( in degree Celius) of a cat is reduced, the cat's heart rate r (in beats per minute) decreases. Under laboratory conditions a cat at a temperature of 36 had a heart rate of 206 and at a temperature of... 顯示更多 1. x + y + z = 0 x - y + z = 0 x + z = 0 2. Temperature and Heart Rate When the temperature , T ( in degree Celius) of a cat is reduced, the cat's heart rate r (in beats per minute) decreases. Under laboratory conditions a cat at a temperature of 36 had a heart rate of 206 and at a temperature of 30 the heart rate was 122. If r is linearly related to T where T is between 26 and 38 a) determine an equation for r in terms of T b) determine the cat's heart rate at temperature of 27 degree celius Thanks a lot
此文章來自奇摩知識+如有不便請留言告知 最佳解答:1. Equation 1 plus equation 2 gives the result 2x + 2z = 0, or x + z = 0 which is the same as equation 3. So there is no solution since the equations are not independent to each other. 2.Since r is linearly related to T, we can let the equation be r = mT + C. When T = 36, r = 206. So 206 = 36m + C................(1) When T = 30, r = 122. So 122 = 30m + C................(2) (1) - (2) we get 84 = 6m m = 14. Sub. into (2), we get 122 = 420 + C So C = -298. So r = 14T - 298. Now T = 27, so r = 14(27) - 298 = 378 - 298 = 80 degree. 2008-10-27 08:38:58 補充： For Q1. The only solution is x = y = z = 0. 其他解答: |
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