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What is the slope of the line given by the equation yx 26?

Anonymous

∙ 12y ago

Updated: 10/17/2024

Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals", "squared", "cubed" etc.

I suspect the answer is 1, but cannot be certain for the reasons given above.

Wiki User

∙ 12y ago

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Q: What is the slope of the line given by the equation yx 26?

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Related Questions

Which equation describes the line with slope 4 that contains the point 7 2?

y = 4x - 26

What is the equation of a line through the point 9 1 that has the slope of three?

3x - y - 26 = 0

What are the equations for the line (-26) (43)?

If you mean points of (-2, 6) and (4, 3) Then its slope is -1/2 and its equation is 2y = -x+10

What is the slope of the line 2y plus 8x equal 26?

-4

What is the slope for points (14) (26) and (38)?

If you mean points of (1,4) (2, 6) and (3, 8) then the slope is 2 and the equation is y = 2x+2

What is the equation of a line with coordinates of 3 -4 and is perpendicular to the line of 5x-2y equals 3?

Equation of given line: 5x - 2y = 3In slope intercept form, that is: 2y = 5x - 3or y = 5/2*x - 3/2So gradient of given line = 5/2Therefore, gradient of perpendicular line = -1/(5/2) = -2/5Also, the point (3, -4) is on this lineSo equation is: y - (-4) = -2/5*(x - 3)y + 4 = -2/5*(x - 3)5y - 20 = -2x + 62x + 5y = 26Equation of given line: 5x - 2y = 3In slope intercept form, that is: 2y = 5x - 3or y = 5/2*x - 3/2So gradient of given line = 5/2Therefore, gradient of perpendicular line = -1/(5/2) = -2/5Also, the point (3, -4) is on this lineSo equation is: y - (-4) = -2/5*(x - 3)y + 4 = -2/5*(x - 3)5y - 20 = -2x + 62x + 5y = 26Equation of given line: 5x - 2y = 3In slope intercept form, that is: 2y = 5x - 3or y = 5/2*x - 3/2So gradient of given line = 5/2Therefore, gradient of perpendicular line = -1/(5/2) = -2/5Also, the point (3, -4) is on this lineSo equation is: y - (-4) = -2/5*(x - 3)y + 4 = -2/5*(x - 3)5y - 20 = -2x + 62x + 5y = 26Equation of given line: 5x - 2y = 3In slope intercept form, that is: 2y = 5x - 3or y = 5/2*x - 3/2So gradient of given line = 5/2Therefore, gradient of perpendicular line = -1/(5/2) = -2/5Also, the point (3, -4) is on this lineSo equation is: y - (-4) = -2/5*(x - 3)y + 4 = -2/5*(x - 3)5y - 20 = -2x + 62x + 5y = 26

What is the slope of line passing through the points (35) and (-26)?

If you mean points of (-2, -1) and (3, 5) then the slope is 6/5

What is the equation of the tangent line that touches the circle x squared plus y squared -6x plus 4y equals 0 at the point of 6 -4 on the Cartesian plane?

Equation: x² + y² -6x +4y = 0 Completing the squares: (x-3)² + (y+2)² = 13 Centre of circle: (3, -2) Contact point: (6, -4) Slope of radius: -2/3 Slope of tangent: 3/2 Tangent equation: y - -4 = 3/2(x-6) => 2y - -8 = 3x-18 => 2y = 3x-26 Tangent line equation in its general form: 3x-2y-26 = 0

What is the maximum number of drainage fixtures units allowed on a 2'' diameter drain line?

The maximum number of drainage fixture unit on a 2" drain line depends on the slope of the drain line. 21 units are allowed if the slope is 1/4' per foot, and 26 units are allowed if the slope of the line is 1/2" per foot.

What is the equation of the tangent line that touches the circle x squared plus 10x plus y squared -2y -39 equals 0 at the point of 3 2 on the Cartesian plane showing work?

First find the slope of the circle's radius as follows:- Equation of circle: x^2 +10x +y^2 -2y -39 = 0 Completing the squares: (x+5)^2 + (y-1)^2 -25 -1 -39 = 0 So: (x+5)^2 +(y-1)^2 = 65 Centre of circle: (-5, 1) and point of contact (3, 2) Slope of radius: (1-2)/(-5-3) = 1/8 which is perpendicular to the tangent line Slope of tangent line: -8 Tangent equation: y-2 = -8(x-3) => y = -8x+26 Tangent equation in its general form: 8x+y-26 = 0

What is the tangent equation line of the circle x2 plus 10x plus y2 -2y -39 equals 0 at the point of contact 3 2 on the Cartesian plane?

Equation of circle: x^2 +10x +y^2 -2y -39 = 0 Completing the squares: (x+5)^2 +(y-1)^2 = 65 Center of circle: (-5, 1) Slope of radius: 1/8 Slope of tangent line: -8 Point of contact: (3, 2) Equation of tangent line: y-2 = -8(x-3) => y = -8x+26 Note that the tangent line meets the radius of the circle at right angles.

What slope options correctly describes a line passing through the points 10 5 and 100 57?

26/45

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