標題:
發問:
已知y=ax+b(a為整數)通過(98,19), 它的x軸交點為(p,0),y軸交點為(0,q) p是質數,q是正整數 滿足條件的一次函數有幾多個?
最佳解答:
The equation can be written like this: 19 = 98a + b 0 = pa + b q = b 須符合以上三個條件 q = -pa 19 = 98a - pa 19 = a(98-p) a = 19 , 98-p = 1 --> p = 97 (rejected q = -ve) or 98-p = 19 a = 1 ---> p = 79 (rejected q = -ve) a = -19 , 98-p = -1 ---> p= 99 (rejected) a = -1 , 98-p = -19 --> p = 117 (rejected) Therefore 滿足條件的一次函數 有 0 個
其他解答:
19 = 98a + b b = 19 - 98a x - in : -b/a y - in : b -b/a = p and b = q so,a = 1 or a = -1 -b = p or b = p -(19 - 98(1)) = p or 19 - 98(-1) = p p = 79 or p = 117(rejected since 117 is divisible by 3) q = 79 there is only one possible linear equation
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一次函數的問題發問:
已知y=ax+b(a為整數)通過(98,19), 它的x軸交點為(p,0),y軸交點為(0,q) p是質數,q是正整數 滿足條件的一次函數有幾多個?
最佳解答:
The equation can be written like this: 19 = 98a + b 0 = pa + b q = b 須符合以上三個條件 q = -pa 19 = 98a - pa 19 = a(98-p) a = 19 , 98-p = 1 --> p = 97 (rejected q = -ve) or 98-p = 19 a = 1 ---> p = 79 (rejected q = -ve) a = -19 , 98-p = -1 ---> p= 99 (rejected) a = -1 , 98-p = -19 --> p = 117 (rejected) Therefore 滿足條件的一次函數 有 0 個
其他解答:
19 = 98a + b b = 19 - 98a x - in : -b/a y - in : b -b/a = p and b = q so,a = 1 or a = -1 -b = p or b = p -(19 - 98(1)) = p or 19 - 98(-1) = p p = 79 or p = 117(rejected since 117 is divisible by 3) q = 79 there is only one possible linear equation
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