If as a straight line equation: 5x+2y = 20 then y = -2.5x+10 whereas -2.5 is the slope and 10 is the y intercept
Do you mean y = -8x+20 Then if so the perpendicular slope is positive 1/8.
Points: (0, -2) and (20, 42) Slope: 11/5 Equation: 5y = 11x-10 or as y = 2.2x-2.
5x - 10 = -20This equation can be restated as 5x = -10 : x = -2This is the equation of a straight line perpendicular to the x axis and passing through the point x = -2. There is no y intercept and the slope is indeterminate.
The average slope in that interval is('y'-value at 30 s) - ('y'-value at 10 s)/20 s
I believe the slope would be: -4 / 3
If as a straight line equation: 5x+2y = 20 then y = -2.5x+10 whereas -2.5 is the slope and 10 is the y intercept
Points: (14, 5) and (20, 4) Slope: -1/6
Parallel lines have the same slope. So if you have y=x+20 for example, the slope is 1 and any parallel line has slope 1 also. I think your equation is x=y+20 but since the+ and - don't show up i am not sure If it is we can rewrite it as -y=-x+20 or y=x-20 and slope is still 1 so any parallel line has slope 1.
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Do you mean y = -8x+20 Then if so the perpendicular slope is positive 1/8.
Points: (0, -2) and (20, 42) Slope: 11/5 Equation: 5y = 11x-10 or as y = 2.2x-2.
5x - 10 = -20This equation can be restated as 5x = -10 : x = -2This is the equation of a straight line perpendicular to the x axis and passing through the point x = -2. There is no y intercept and the slope is indeterminate.
If you mean: 4x+2y = 20 then y = -2x+10 and so the slope is -2
It is 2√5 ≈ 4.47 units.To solve this:Find the equation of the line perpendicular to y = 2x + 10 that passes through the point (2, 4);Find the point where this line meets y = 2x + 10Find the distance from this point to (2, 4) using PythagorasThe slope of the perpendicular line (m') to a line with slope m is such that mm' = -1, ie m' = -1/mFor y = 2x + 10, the perpendicular line has slope -1/2, and so the line that passes through (2, 4) with this slope is given by:y - 4 = -½(x - 2)→ 2y - 8 = -x + 2→ 2y + x = 10To find where this meets the line y = 2x + 10, substitute for y in the equation of the perpendicular line and solve for x:y = 2x + 102y + x = 10→ 2(2x + 10) + x = 10→ 4x + 20 + x = 10→ 5x = -10→ x = -2Now use one of the equations to solve for y:y = 2x + 10→ y = 2(-2) + 10→ y = -4 + 10 = 6This the perpendicular line from (2, 4) meets the line y = 2x + 10 at the point (-2, 6)The distance between these two points is given by:distance = √((-2 - 2)² + (6 - 4)²) = √(16 + 4) = √20 = 2√5 ≈ 4.47 units
The average slope in that interval is('y'-value at 30 s) - ('y'-value at 10 s)/20 s
Any two points on a vertical line. For example, the line x=7 is a vertical line with undefined slope. Every point (7,y) where y is any real number, lies on the same line.