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anonymous
 one year ago
Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form?
a. 7x + 2y = –1
b. 7x + 2y = 1
c. 14x + 4y = –1
d. 14x + 4y = 1
anonymous
 one year ago
Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form? a. 7x + 2y = –1 b. 7x + 2y = 1 c. 14x + 4y = –1 d. 14x + 4y = 1

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Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.0use the slope formula to calculate the slope between those two points

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i still can't get it @Vocaloid

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.0slope = m = (y2y1)/(x2x1)

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.0(x1,y1) = (3,11) (x2,y2) = (1,3)

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.0just plug in the values for x1,x2,y1, and y2, then find the slope.

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.0show me how you got that answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm on a ezam i can't show you i just need to know is that right I'm timed @Vocaloid

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.0not allowed to help on exams, sorry :( best of luck~

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0here this helps me...
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