Ask a New Question
Menu

How do I find JK for J(3, 8)and K(-1, 4)? Please help!

1 answer

J(3,8),K(-1,4).

(JK)^2 = (x2-x1)^2 + (y2-y1)^2.
(JK)^2 = (-1-3)^2 + 4-8)^2.
(JK)^2 = 16 + 16 = 32.
JK = sqrt32 = 5.66.
Similar Questions
  1. Given the function: f(x) = x^2 + 1 / x^2 - 9a)find y and x intercepts b) find the first derivative c) find any critical values
    1. answers icon 1 answer
  2. Given the function: f(x) = x^2 + 1 / x^2 - 9a)find y and x intercepts b) find the first derivative c) find any critical values
    1. answers icon 0 answers
  3. Let equation of an hyperbola be y^2-4x^2+4y+24x-41=0a. Find the standard form b. Find the center c. Find the vertices d. Find
    1. answers icon 0 answers
  4. For the following graph:a. Find the domain of f. b. Find the range of f. c. Find the x-intercepts. d. Find the y-intercept. e.
    1. answers icon 1 answer
more similar questions