6m + m + m + 4y + n + y + 12 + 2n Group like terms: 6m + m + m + 2n + n + 4y + y + 12 Add: 8m + 3n + 5y + 12
standard form (slope-intercept) is y = mx + b where m = slope and b = y intercept 6x + y = 12 subtract 6x from both sides y = -6x + 12 m = -6; b = 12
The equation is in standard form y = mx + b where m is the slope and b is the y-intercept. For y = 12x + 1, m = 12 so the slope is 12.
.888kg/m
35
Surely that's months in a year
A
16
The slope m' of a line perpendicular to a line with slope m is such that m'm = -1 → m' = -1/m A line with slope m' through point (X, Y) has equation: y - Y = m'(x - X) → The line perpendicular to 7x - 8y = 12 through p (-3, 1) is found: 7x - 8y = 12 → 8y = 7x - 12 → y = (7/8)x - 3/2 → The slope of 7x - 8y = 12 is m = 7/8 → The slope of the perpendicular line is m' = -1/m = -1/(7/8) = -8/7 → The line through P with this slope m' is: y - 1 = -8/7(x - -3) → y = (-8/7)x - 24/7 + 1 → y = (-8/7)x - 17/7 This can be rearranged into: y = (-8/7)x - 17/7 → 7y = -8x - 17 → 7y + 8x + 17 = 0
90 = degrees in a right angle
5:3:1:1
It is an area not a linear distance. It could be 4 m by 12 m or perhaps 6m by 8m or any value x m by y m wherethis constraint holds true.... x*y = 48.