How can you use a function to represent a linear relationship?
1 answer:
You might be interested in
Answer:
3.51 (round to the nearest hundreds)
Step-by-step explanation:
PART A
slope of the line passing through AC : (6-1)/(1-2) = 5/-1 = -5
equation of the line passing through B and perpendicular to AC:
y-3 = 1/5(x-3)
y-3 = 1/5x -3/5
y = 1/5 x - 3/5 + 3
y = 1/5 x + (-3+15)/5
y = 1/5 x + 12/5
PART B
line that passes through the points A and C
y-1 = -5(x-2)
y = -5x+10 + 1
y = -5x + 11
Interception of the lines
y = 1/5x + 12/5
y = -5x + 11
-5x + 11 = 1/5x + 12/5
-25x +55 = x + 12
26x = 43
x = 43/26
y = -5(43/26) + 11
y= -215/26 + 11
y = (-215+286)/26 = 71/26
D (1.65, 2.73)
PART C
AC = = 5,09902
BD = = 1,376735
PART D
AREA = (5,09902 * 1.376735)/2 = 3,509999