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line d y=2x+4 ,e y=2x+c find c and the lines are parallel and have 6 units between them

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Graphs - NIU

GRAPHS Section 2.3 Lines ... 4) and parallel to the line with equation 3x 5y +2 = 0? (a) ... y = 2x+4 (d) y = 2x (b) y = 2x 4 (e) None of these (c) ...
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College Algebra CLEP Study Flashcards | Quizlet

College Algebra CLEP Study. ... c. y=1/2x-4 Add x to both sides ... Find the equation needed for a parallel line from y=3x+2 that runs through (-3,6). Answer: ...
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SATII Math Flashcards | Quizlet

... [y = 2x + 1], which of the following could be a portion of the graph ... a line B: a point C: two points D: a circle E: ... D when two pairs of parallel lines ...
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Equation and graph of a straight line - A complete course in ...

THE EQUATION AND GRAPH OF A STRAIGHT LINE. ... Vertical and horizontal lines. y = 2x + 6. ... when there are two unknowns, x and y, but only one equation that relates ...
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Coordinate Geometry - GMAT Math Study Guide

Coordinate Geometry - GMAT Math ... Writing the Equation of a Line; Axis Intercepts; Parallel Lines; ... set y = 0 and solve for x. 0 = 2x - 4 x = 2 To find the y ...
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SOLUTION KEYS FOR MATH 105 HW ... = x2+2x−8, f(x)/g ... so the two vectors are indeed parallel. Additional Problem: Find the cosine of the angle between →a ...
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Wolfram|Alpha Examples: Coordinate Geometry

For Coordinate Geometry compute: lines, planes, ... Units & Measures ... compute the area between y=|x| and y=x^2-6.
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Final Exam Study Aid - Department of Mathematics

Math 112 Final Exam Study Aid 4 of 39 ... 2x for x>1 shown below. 6 5 4 3 2 1 1 2 3 4 5 6 6 5 4 3 2 1 1 2 3 4 5 6 x y ... twenty units is: (A) y= 5 3 p x+ 20 (B) y= q 7 2
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OliverKnill Math 21a,Fall 2011

OliverKnill Math 21a,Fall 2011 ... 3 Find the distance d(P,Q) between the points P = (1,2,5) ... 9 Find the center and radius of the sphere x2 +2x+y2−16y +z2 +10z ...
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line d y=2x+4 ,e y=2x+c find c and the lines are parallel and have 6 units between them

line d y=2x+4 ,e y=2x+c find c and the lines are parallel and have 6 units between them There are two lines that satisfy the given parameters, one above and to the left of line d, and one below and to the right of line d. To find a point 6 units away from line d, we need a line that is perpendicular to d. We get the equation of that line by using the negative reciprocal of the m component, 2: that gives us -1/2. The equation of this line is y = -1/2 x + 4, using the same y intercept. A segment of that line that is 6 units represents the hypotenuse of a right triangle. If we are looking at the segment that extends to the left of the given y intercept, we consider the distance to be negative (-6). If we are looking at the segment that extends to the right of the given y intercept, we consider the distance to be positive (+6). To find the x and y co-ordinates, we use the sine and the cosine of the angle (we'll call it a) formed by this new line and the x axis. The slope m of a line is the y distance divided by the x distance. That is also the definition of the tangent of the angle. Using the inverse tangent, we determine that the angle is tan^-1 (-1/2) =  -26.565 degrees. We need to keep all the signs straight in order to get the correct values. The sine of -26.565 degrees is -0.4472. The cosine of -26.565 degrees is 0.8944. y1 / -6 = sin a y1 = -6 sin a = -6 * -0.4472 = 2.6832 This is measured from a horizontal line through the y intercept, because we are constructing a right-triangle with one corner at the y intercept, so the point's y co-ordinate is actually y = 2.6832 + 4 = 6.6832 x / -6 = cos a x = -6 * 0.8944 = -5.3664 Making a quick check using x^2 + y^2 = r^2: (-5.3664)^2 + 2.6832^2 = 28.79825 + 7.19956 = 35.99781   -6^2 = 36 Close enough considering the rounding in the calculations. We have co-ordinate (-5.3664, 6.6832) lying 6 units from line d. Substituting those values into y = 2x + c we have 6.6832 = 2 * (-5.3664) + c 6.6832 = -10.7328 + c 6.6832 + 10.7328 = c = 17.416 The equation of a line 6 units away from d and parallel to it, located above it, is y = 2x + 17.416    <<<<<<<< Working in the other direction, on the line below and to the right of line d, y2 / 6 = sin a y2 = 6 * sin a = 6 * -0.4472 = -2.6832 (that is, 2.6832 below d's y intercept) y = 4 - 2.6832 = 1.3168 x / 6 = cos a x = 6 * cos a = 6 * 0.8944 = 5.3664 The co-ordinates for this point, (5.3664, 1.3168), when substituted into the equation y = 2x + f (f is a new y intercept) gives 1.3168 = 2 * 5.3664 + f 1.3168 - 10.7328 = f = -9.416 The equation of a line 6 units away from d and parallel to it, located below it, is y = 2x - 9.416    <<<<<<<<
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Find the equation of a straight line

ysin30=-xcos30+2; y=-xtan60+2cosec30=-xtan60+4=-x√3+4. When x=0, y=4 (y-intercept), y=0, x=4√3/3. (i) The slope of the line is -tan60=-√3 and this is also the slope of the parallel line which has the equation: y-3=-tan60(x-4) in slope-intercept form. Therefore y=3-xtan60+4tan60=3+4√3-x√3 or xcos30+ysin30=3sin30+4cos30. For a general point P(x,y) on the line, OX=YP=x and OY=XP=y. OT=OS+ST=OS+QP. OS=xcos30 and QP=ysin30, so OT=xcos30+ysin30=2. OT is the radius of the circle of which the line xcos30+ysin30=2 is a tangent. (ii) ON is the length of the perpendicular on to the parallel line xcos30+ysin30=4cos30+3sin30. By analogy, ON=4cos30+3sin30=2√3+3/2 just as OT=2 for the other line. (iii) The distance between the lines must therefore be the difference in the two radii, ON-OT=2√3+3/2-2=2√3-1/2.  
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I just need help getting started on this problem. I don't even know how to begin!

The graph is centered over the line x=-3. This line acts like a mirror so that the two halves of the inverted U-shaped graph are reflected in it. The graph has to be shifted 3 units to the right so that the line x=-3 becomes the line x=0, which is the y (f(x)) axis. The equation of the graph changes to f(x)=-0.5x^2+8. When x=0 f(0)=8, and this value on the vertical axis is the highest point of the curve (vertex). To picture f(x+k) you need a picture, a moving picture. Think of the horizontal axis, the x axis, and the line f(x)=8, which is a line parallel to the x axis a distance of 8 units above it. The two lines resemble a track, like a rail track. The curve  is, as we've established, an inverted U shape where the arms of the U are moving further apart the further away the curve is from its vertex. The curve has a constant shape and remains in contact with the line f(x)=8. We start with the curve right in the middle so that the vertical axis bisects the U curve. This is when k=0.  Now, we're going to slide the curve so that its vertex runs along the track f(x)=8 left and right. The curve cuts the axis at two points separated by 8 units, the gap between the zeroes of the function. When the curve is centrally positioned, the points on the x axis are -4 and 4 and k=0. Slide the curve to the left (negative side) and this is equivalent to positive values of k; slide it to the right and we're into negative values of k. If we move the y (f(x)) axis with the curve it becomes the movable axis of symmetry, the mirror I mentioned earlier. The points where the curve cuts the x axis remain 8 units apart. When the curve moves leftward one unit, the value of k increase by 1, and rightward one unit when k decreases by 1. The vertical line x=2 is unmovable, but as the curve slides from left to right this line touches or cuts through the curve. Consider only the part of the curve lying on or above the x axis. When does it touch the vertical line? Move the curve so that the right part of the curve just touches the line. This is the zero with the higher value. So it must be when x=2 is a root, i.e., -0.5(2+k)^2+8=0. 0.5(2+k)^2=8, so (2+k)^2=16, 2+k=+4, and k=4-2 or -4-2, which is 2 or -6. If we slide the curve past the line x=2, till the left part of the curve touches the line on the right side, the point where it touches is the other root, 8 lower than the the right-hand root. So if 2 was the right root then 2-8=-6 is the left root, and if 2 was the left root then 10 is the right root. So we have the functions f(x)=-0.5(x+2)^2+8 and f(x)=-0.5(x-6)^2+8, where the values of k have been substituted. When the curve is on the left of x=2, the roots of f(x+2) are -6 and 2 and when on the right the roots of f(x-6) are 2 and 10. In between these values of k, 2 and -6, the curve touches or is above the x axis, so this the range for k: -6 Read More: ...

Find an equation for the line through point (3, -2, 1) parallel to the line x=1+2t, y=2-t, z=3t

Given two parametric lines defined by p = p1 + t(A) q = q1 + s(B) these lines will be parallel if the unit direction vectors A and B are the same. Let us define p as p = p1 + t(p2 - p1) where p1 = (1, 2, 0) and p2 - p1 = (2, -1, 3) taken from x = 1 + 2t, y = 2 - t, z = 3t Let A = (p2-p1)/norm(p2-p1) be the normalized direction vector of p2-p1 where norm(p2-p1) = sqrt(4 + 1 + 9) = sqrt(14) Then A = (2/sqrt(14), -1/sqrt(14), 3/sqrt(14)) which is the normalized direction vector for line p so that p = p1 + t(A) If we set B = A, then the parametric line q passing through the point q1 = (3, -2, 1) is given by q = q1 + s(B) Because B = A, line q is parallel to the line p given above. Note that when the direction vector is normalized, the parameters "s" and "t" for lines q and p represent the Euclidian distance from the points q1 and p1 respectively.
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Find the angle and arc measurements. a = _____ b = _____ c = _____ d = _____ e = _____ f = _____ g = _____ h = _____ j = _____ k = _____ m = _____ n = _____

Angles: h=a=90 (tangent/radius); c=d=f=2*28.65=57.3 (external angle of isosceles triangle and angle, same side angles on parallel lines l1 and l2); e=180-2*57.3=65.4; g=180-57.3=122.7 (d+e=g, opposite angles on intersecting lines). Arcs, assuming r=1 (unit circle) (57.3º=1 radian approx): b=j=k=c=d=f in radians=1 approx; m=g=pi-1=2.142 approx; n=e=pi-2=1.142 approx. If r is not equal to 1, multiply all arc lengths by r.
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(-4,-2) (0,0) and (2,7) (k,5) what is k

(-4,-2) (0,0) and (2,7) (k,5) what is k i have to find the value of k, so that the two lines will be parallel. The first observation to make is that parallel lines have the same slope. We can determine the slope of the first line and use that to find the unknown x value for the second line. Let's give identities to the points so we can refer to them by name. P1 is (x1, y1)   (-4,-2) P2 is (x2, y2)   (0, 0) P3 is (x3, y3)   (2, 7) P4 is (x4, y4)   (k, 5) The x and y designations are so you can follow along with the general equation to calculate the slope, m. 1) m = (y2 - y1) / (x2 - y2) 2) m = (0 - (-2)) / (0 - (-4)) 3) m = (0 + 2) / (0 + 4) 4) m = 2 / 4 = 1/2 Now that we know the slope of the two lines, we can use the second set of points to find the value of k. 5) m = (y4 - y3) / (x4 - x3)   Of couse, x4 is k. 6) m = (5 - 7) / (k - 2) We know that m is 1/2, so we will now substitute that into the equation. 1/2 = (5 - 7) / (k - 2) 1/2 = -2 / (k - 2) We multiply both sides by (k - 2) 1/2 * (k - 2) = (-2 / (k - 2)) * (k - 2) 1/2 * (k - 2) = -2 Multiply both sides by 2 1/2 * (k - 2) * 2 = -2 * 2 (k - 2) = -4 Add 2 to both sides (k - 2) + 2 = -4 + 2 k = -2 Substitute that value into equation 6, for the slope. m = (5 - 7) / (k - 2) m = (5 - 7) / (-2 - 2) m = -2 / -4 m = 1/2 When k is -2, the second line segment has the same slope as the first line segment, therefore, the two lines are parallel.
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How do I graph the equation x=2?

How do I graph the equation x=2? If X = 2 how do I figure out what Y is? Find the point two units to the right of the origin on the x-axis. Draw a vertical line through that point. That line is parallel to the y-axis. You can pick any point along that line, above or below the x-axis, and that point will have its own y value.
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what is enough information to find out if two lines in a triangle are parallel?

Triangles can't have parallel lines, at least not in the type of math you're dealing with. You could argue that a triangle on a sphere can have lines that are parallel at certain points, but these lines won't be parallel throughout the triangle.  Consider a triangle (northern hemisphere) formed by the Earth's equator, the 0 degree longitude line, and the 90 degree longitude line.  Assume for the moment that the Earth is a perfect sphere (it isn't in real life).  The 0 degree line and 90 degree line are parallel to each other at the points where they hit the equator, but they are perpendicular to each other where they hit the north pole.
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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)  y-intercept of new lines: (0,-7) The slope is (y2 - y1) / (x2 - x1) m = (-4 - 8) / (7 - 1) = -12/6 = -2 This line segment is y = -2x + b. We don't know the y-intercept, but we are not concerned with that. A parallel line has the same slope but a different y-intercept. We are given the y-intercept, so the equation is y = -2x - 7. A line perpendicular to the given line segment has a slope that is the negative reciprocal: 1/2. The equation of that line is y = (1/2)x - 7.
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Write the equation of the line parallel to the line 10x - 5y = 8 and passing through point (2,4).

Rewrite the equation: 5y=10x-8 and divide by 5: y=2x-8/5. The slope of the line is 2, and the slope of the line we need to find is also 2, because parallel lines have the same slope. We know that the slope of the line we have to find is 2. We can write y=2x+c where we have to find c. Substitute the coords of the given point: x=2, y=4: 4=4+c, so c=0 and the equation of the line is simply y=2x.
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