线段的斜率为3/4。该段具有端点D(8,-5)和E(k,2)。 k的价值是多少? [请帮忙!谢谢!!]
K = 52/3>“使用”颜色(蓝色)“梯度公式计算斜率m”•颜色(白色)(x)m =(y_2-y_1)/(x_2-x_1)“let”(x_1,y_1) )=(8,-5)“和”(x_2,y_2)=(k,2)rArrm =(2 - ( - 5))/(k-8)= 7 /(k-8)“我们给出“m = 3/4 rArr7 /(k-8)= 3 / 4larrcolor(蓝色)”交叉乘法“rArr3(k-8)= 28”将两边除以3“rArrk-8 = 28/3”加8对双方来说“rArrk = 28/3 + 24/3 = 52/3
证明: - sin(7 theta)+ sin(5 theta)/ sin(7 theta)-sin(5 theta)=?
(sin7x + sin5x)/(sin7x-sin5x)= tan6x * cotx rarr(sin7x + sin5x)/(sin7x-sin5x)=(2sin((7x + 5x)/ 2)* cos((7x-5x)/ 2) )/(2sin((7x-5x)/ 2)* cos((7x + 5x)/ 2)=(sin6x * cosx)/(sinx * cos6x)=(tan6x)/ tanx = tan6x * cottx
K的价值是多少?
答案是k in(2,4)对于二次方程f(x)> 0,图形必须接触x轴判别式必须是= 0 f(x)= x ^ 2-2(4k-1) x + 15k ^ 2-2k-7 Delta =(2(4k-1))^ 2-4(1)(15k ^ 2-2k-7)= 0 =>,4(16k ^ 2-8k + 1) -60k ^ 2 + 8k + 28 = 0 =>,64k ^ 2-32k + 4-60k ^ 2 + 8k + 28 = 0 =>,4k ^ 2-24k + 32 = 0 =>,4(k ^ 2 -6k + 8)= 0 =>,(k-2)(k-4)= 0 =>,{(k = 2),(k = 4):}当k = 2 =>时,x ^ 2- 14x + 49 = 0图{x ^ 2-14x + 49 [-14.48,21.55,-4.62,13.4]}当k = 4 =>时,x ^ 2-30x + 225 = 0图{x ^ 2-30x + 225 [-14.48,21.55,-4.62,13.4]}当k = 3 =>时,x ^ 2-22x + 122 = 0图{x ^ 2-22x + 122 [-14.48,21.55,-4.62,13.4]}