find and graph and equation parallel to 3x-2y=6 and passing through (-2,-1)

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find and graph and equation parallel to 3x-2y=6 and passing through (-2,-1)

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Research, Knowledge and Information :


Find the equation of the line passing through $(-2,6)$ and ...


The question is: Find the equation of the line passing through the point $ ... the equation of the line parallel to $2y - 3x = 8$ that passes through the point $ ...
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Find the Equation of a Line Parallel or Perpendicular to ...


Find the Equation of a Line Parallel or Perpendicular to ... Find the equation of a line passing through the point (–6, 5) parallel to the line 3x – 5y = 9. Step ...
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Find The Equation Of A Line Parallel To 3x - chegg.com


Show transcribed image text Find the equation of a line parallel to 3x ... to 3x - 2y = 8 and passing through (-1, -2). ... Graph the equation. Find the equation ...
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1.Perpendicular to 7x-2y=-2 and passing through the point (7 ...


1.Perpendicular to 7x-2y=-2 and passing through the point (7,1); ... Above equation is slope intercept form of the equation of the required line 2.Parallel to 3y-4x=3 ...
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How would i find an equation of a line passing through point ...


How would i find an equation of a line passing through point B(-2,4) ... is 2 point lower on the graph in ... equation of a line parallel to 3x-2y=6 and ...
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Find the Equation of a Line Parallel or Perpendicular to ...


Find the Equation of a Line Parallel or Perpendicular to Another ... Find the equation of a line passing through ... passing through the point (–7, 2) parallel to ...
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Algebra 2-1 Unit 3 Semester Review Flashcards | Quizlet


Algebra 2-1 Unit 3 Semester Review. ... The y-intercept of the line whose equation is 3x - 2y = 6 is-3. ... Find the slope of the line passing through the points ...
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Standard Form, Parallel/Perpendicular, and X/Y Intercepts ...


Start studying Standard Form, Parallel/Perpendicular, ... 6y = -12 and passes through (-6, 2) ... 3x - 2y = 2. Write in Standard Form Passes
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Suggested Questions And Answer :


write an equation of the line that passes through the given point and is parallel to the give line (6,8), y=-5/2x+10

Since we know the lines are parallel they must have the same slope. That is the first place to start. We know that the slope intercept form of a line is y=mx+b where m is the slope and b is the y itnercept Since we know the slope or m value we can start by writing this as y=-5/2x+b now w
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Find the equation of the cylinder having its base the circle

When z=1, x^2+y^2=4, which is a circle, centre at the origin and radius 2. x-2y=1 is a plane intercepting the x-y plane with y intercept=-1/2 and x intercept=1. The equation x^2+y^2+(z-1)^2=4 is a sphere, centre (0,0,1) and radius 2. The plane cuts through the sphere when (1+2y)^2+y^2=4; 1+4y+4y^2+y
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find the equation of a parabola with axis parallel to y passing through (-1,-3), (1,-2), (2,1)

Problem: find the equation of a parabola with axis parallel to y passing through (-1,-3), (1,-2), (2,1) Pre-calculus, getting the general equation of a parabola We are looking for an equation in the form of y = ax^2 + bx + c. Given three points, we can replace x and y with real values to creat
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find the equation of a line parallel to 6x-7y+14 passing through (-4, 8)

Finding such an equation is very easy, so study the method I use and learn it. Don't just take the answer. The best way to start is rewriting the given function in slope-intercept form. 6x-7y+14 = 0 (I'm assuming it's set equal to 0; you didn't specify!) --> 6x - 7y = -14 &nb
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I need help on this Calculus problem. Thanks.

If f(x)=-4x^2/9+2x/3 as it was in your previous question, f'(x)=-8x/9+2/3, and is the slope of the tangent for all x values. This slope is parallel to y=-2x+4 when -8x/9+2/3=-2, because the line and the tangent line must have the same slope of -2 to be parallel. a) The equation that has to be s
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how do you find the slope of a line sagment?

how do you find the slope of a line sagment? you are given two points which define a line segment. Find the slope of the line sagment, and then use this slope to write the equation of a parallel line [and a] perpendicular line which pass through the new y-intercept 1. line segment (1,8) and (7,-4)
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equation of the line Through (4, 8); parallel to the line passing through (5, 6) and (1, 2)

Find an equation of the line that satisfies the given conditions. Through (4, 8);  parallel to the line passing through  (5, 6) and (1, 2)    
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find and graph and equation parallel to 3x-2y=6 and passing through (-2,-1)

1. 3x-2y=6...-2y=6-3x, 2y=3x-6, y=(3/2)x-3...this giv slope=3/2 2. thru (-2,-1) tu get y-intersept: deltax=+2 deltay=slope*deltax=2*3/2=3 yintersept=-1+3=2 line: y=(3/2)x+2
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what line passes through 1,2 and is parallel to the graph of the line y=3x+8

what line passes through 1,2 and is parallel to the graph of the line y=3x+8 y = 3x + 8     the slope is 3 y = mx + b b = y - mx b = 2 - 3(1) b = 2 - 3 b = -1 The equation is y = 3x - 1
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Find the pts on the graph of z=xy^3+8y^-1 where the tangent plane is parallel to 2x+7y+2z=0.

Question: Find the pts on the graph of z=xy^3+8y^-1 where the tangent plane is parallel to 2x+7y+2z=0. The tangent plane to the surface z=f(x,y)=xy^3+8/y is given by 2x+7y+2z=D Where D is some constant. The equation of the tangent plane to the surface f(x,y), at the point (x_0 y_0 ) ca
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