解答一 举报 在点(3,4/81)求y=4/x^4 对x的偏导d(4x^(-4))/dx=-16x^(-5)把x=3代入,球的直线斜率为-16/243所以equation:y-4/81=-16/243(x-3)同理d(3x^(-4))/dx=-12x^(-5)equation:y-3/625=-12/(5^5) (x-5) 解析看不懂?免费查看同类题视频解析查看解答 ...
8. (a) Find an equation of the line join singA(7,4) and B(2, 0), giving your answer in the form ax+by+c=0, where a, b and c are integers.(3)(b) Find the length of AB, leaving your answer in surd form.(2)The point C has coordinates (2, t), where t 0, and ...
In Problems, find an equation of the line L.L is perpendicularto y=2x 相关知识点: 试题来源: 解析 x+2y=5 or y=-12x+52 反馈 收藏
Find an equation of the line that passes through {eq}\, (5, -1)\, {/eq} and is perpendicular to the line {eq}\, -7x + 8y = -43.\, {/eq} Perpendicular Line: A line is perpendicular to another when the slope of both lines are related th...
The equation of a line that passes through two points has the form: {eq}y= mx+ n {/eq}. The first step to know the form of the equation of the line is to find the slope using the coordinates of these two points: {eq}A(x_1,y_1) {/eq} and {eq}B(x_2,...
Find an equation of the line that contains the given point and is perpendicular to the given line (2,-6), 3x-5y=2Follow • 1 Add comment 1 Expert Answer Best Newest Oldest Raymond B. answered • 02/25/21 Tutor 5 (2) Math, microeconomics or criminal justice ...
相关知识点: 试题来源: 解析 y= 12x 结果一 题目 Find an equation of the line: parallel to the line x-2y-1=0 passing through (0,0) 答案相关推荐 1Find an equation of the line: parallel to the line x-2y-1=0 passing through (0,0) 反馈 收藏 ...
解析 y= 12x 结果一 题目 Find an equation of the line: parallel to the line x-2y-1=0 passing through (0,0) 答案 1 y= 2 C相关推荐 1Find an equation of the line: parallel to the line x-2y-1=0 passing through (0,0) 反馈 收藏 ...
Find an equation of the line containing the points {eq}P = (0, -3, 1) {/eq} and {eq}Q = (1, 1, 0) {/eq}.Line Between Two Points:Suppose that {eq}P_1(x_1,y_1,z_1) {/eq} and {eq}P_2(x_2,y_2,z_2) {/eq} are two points in 3-space. The...
An Equation of the Line Tangent to the Curve:We can find the equation of the line tangent to the curve by using the point-slope form of the equation. We differentiate the curve at the point to get the slope. We insert all the values in the point-slope form ...