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what is the y intercept of the line whose slope is 3 passing points (1,-2)

i need help finding the y intercept of the line whose slope is 3 passing through points (1,-2)

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QUIZ 1: LINEAR EQUATIONS Flashcards | Quizlet

QUIZ 1: LINEAR EQUATIONS. ... The slope of the line passing through the points (-1, 5) and (3, 5) is. 0. ... The y-intercept of the line whose equation is 2x + 5y = 8 ...

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 2x + 5y = 8 is. 8/5. ... Find the slope of the line passing through the points ...

Slope-intercept equation from two points (video ... - Khan ...

Learn how to find the equation of the line that goes through the points (-1, 6) ... Slope-intercept equation from slope & point. Slope-intercept from two points. Up Next.

The x-intercept of the line whose equation is x - y = 3 is -3 0 3

The x-intercept of the line whose equation is x - y = 3 ... 2/3 Weegy: The y-intercept of 3x-4y=-2 is B. 1/2 ... line thru the points. slope, m = diffy/diffx m = (5-1 ...

Complete the equation of the line whose slope is -2, and y ...

Complete the equation of the line whose slope is -2, ... Complete the equation of the line whose slope is -2, and y-intercept is (0,3) 1. ... 5 points 1 hour ago

Section 3.3 The Slope - ALGEBRA ONE - algebrafree.com

... Find the slope of the line passing through points (5, –4) and ... slope. is 2 and the . y-intercept is –3, ... that contains each set of points shown. 1 3 ...

What is the equation of the line whose y-intercept is 3 and ...

What is the equation of the line whose y-intercept is 3 and slope is 1? ... The line passes through these two points, (-3, 0) and (0, -2). Then the slope will be: ...

Slopes and Equations of Lines - University of Missouri–St ...

Slopes and Equations of Lines Some ... Example 1 What is the slope of a line through the points (2,3) and ... we can identify the slope m=2/3 and the y-intercept b=-5/3.

what is the y intercept of the line whose slope is 3 passing points (1,-2)

Problem: what is the y intercept of the line whose slope is 3 passing points (1,-2) i need help finding the y intercept of the line whose slope is 3 passing through points (1,-2) The general equation is y = mx + b. Re-work that equation to find b. b = y - mx b = -2 - 3(1) b = -2 - 3 b = -5 Answer: the y-intercept is -5.

Passing through (-4,4) and parallel to the line whose equation is y=3x+2

strate line thru point=(-4,4) with slope=3 y=3x+konst projekt line from x=-4 tu x=0...deltax=+4 deltay=slope*deltax=3*4=12 y=4+12+16 y=3x+16

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^2=4; 5y^2+4y-3=0, y=(-4±sqrt(16+60)/10=0.47178 ((-2+sqrt(19))/5) and -1.27178 ((-2-sqrt(19))/5). These values give x=1.94356 ((1+2sqrt(19))/5) and -1.54356 ((1-2sqrt(19))/5 respectively, by using the equation x=1+2y. So the points of intersection are (1.94356,0.47178) and (-1.54356,-1.27178). From these we can find the diameter of the cylinder: sqrt(76/25+304/25)=sqrt(380)/5=2sqrt(95)/5=3.9 approx. The radius is sqrt(95)/5. The sides of the cylinder are perpendicular to the diameter.  The slope of the diameter is the same as the slope of the plane x-2y=1, which is 1/2, so the slope of the perpendicular is -2. The equations of the sides of the cylinder where z=1 in the x-y plane are y=-2x+c where c is found by plugging in the intersection points: (-2+sqrt(19))/5=-2(1+2sqrt(19))/5+c; -2+sqrt(19)=-2-4sqrt(19)+5c. So c=sqrt(19) and y=sqrt(19)-2x as one side. Similarly, -2-sqrt(19)=-2+4sqrt(19)+5c, c=-sqrt(19) and y=-(sqrt(19)+2x) for the other side. Consider the view looking along the z axis. The circular cross-section of the cylinder will appear edge on, while the sides will have the slope -2 and will be spaced apart according to the value of z. When z=1±radius of cylinder=1±sqrt(95), the sides will appear to be as one line passing through the x-y plane's origin, the equation of the line being y=-2x when z=1±sqrt(95). The picture shows the view from z=1 looking at the x-y plane. The circle is the cross-section of the sphere and the line passing through (1,0) and (0,-1/2) is the edge of the plane x-2y=1. The diameter of the cylinder is constant and is shown by the two lines perpendicular to each end of the chord where the plane cuts the sphere. These lines represent the sides of the cylinder as they would appear at z=1. Parallel to them and passing through (0,0) is the single line that appears when z is at the extreme limits of the diameter of the cylinder. This line is also the central axis of the cylinder. (The vertical line at y=-2 is just a marker to show the the leftmost limit of the diameter of the sphere.) The general equation of a cylinder is the same as the 2-dimensional equation of a circle: x^2+y^2=a^2. This cylinder, radius a, has the z axis its central axis. x^2+z^2=a^2 is a cylinder with the y axis as its central axis. a=sqrt(95)/5 so a^2=95/25=19/5, making the equation of the cylinder 5x^2+5z^2=19. This is the same size as the cylinder in the problem, but with its central axis as the y axis. 5x^2+5(z-1)^2=19 is the equation of the cylinder with central axis passing through the centre of the sphere. If the cylinder is tilted so that its base coincides with the circle produced by the plane cutting through the original sphere, more calculations need to be made  to transform the coordinates. The picture shows the axial tilt of the cylinder. The central axis of the cylinder, y=-2x, bisects the chord on the line x-2y=1 when x-2(-2x)=1; 5x=1, x=1/5. So y=(x-1)/2=-2/5. If x-2y=1 represents the horizontal axis (we'll call X) and y=-2x represents the vertical axis (Y) we can see that, relative to these X-Y coordinates, the cylinder is upright. The  z value is unaffected by rotation. Take a point P(x,y) in the x-y plane. What are its coordinates in the X-Y plane? To find out we use geometry and trigonometry. The axial tilt of the X-Y axes is angle ø where tanø=1/2, so sinø=1/sqrt(5) and cosø=2/sqrt(5).  X=SP=QPcosø=2(x+y/2)/sqrt(5), Y=PRcosø=2(y-(x-1)/2)/sqrt(5). Z=z In the X-Z plane, 5X^2+5Z^2=19. This transforms to 4(x+y/2)^2+5z^2=19=(2x+y)^2+5z^2. In the X-Y plane the sides of the cylinder are the lines X=-a and X=a, where a^2=19/5. So X^2=19/5, which transforms to 4(x+y/2)^2 or (2x+y)^2=19/5; 2x+y=±sqrt(3.8), corresponding to the equation of two parallel, sloping line forming the sides of the cylinder.

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.

a line that passes though the points (0,2) and (-3,5) has the equation :

the equation of a line in slope intercept form is y=mx+b m=slope b=y intercept The first thing we can do is calculate the slope which is the change in y over the change in x or "rise over run" 2 moving up to 5 is a difference of 3 0 moving left to -3 is a -3 3/-3 we can simplify this to -1 so we know that y=-1x+b We still need to find the y intercept. In this case one of our points (0,2) is on the y axis. so we know that the y intercept is 2 because that is where the line crosses the y axis. The equation then is y=-1x+2 the coefficent for x isn't really necessary so you could write this as y=-x+2 The graph of this line looks like this:

write an equation in point-slope form for the line that passes through (0,-2), (3,2). Then use the same set of points to write the equation in standard form and again in slope-intercept form.

The equation is of the form y=mx+c where m is the gradient and c the intercept. To find m we use the points: m=(difference of y values)/(difference of x values)=(2-(-2))/(3-0)=4/3. So now we have y=4x/3+c. Take one point and plug in x, y. Use (0,-2): -2=c so c=-2 and y=4x/3-2 is the equation in slope-intercept form. (This can be written: 3y=4x-6.) Note that (3,2) also fits.

what is the slope intercept form that is parallel to the line and that passes through 5=2/3(4)+b

?????????? wi did yu put up that misleeding hedline ??????? werds belo giv us the stuf us need tu figger problem line: 2x+3y=7....y=(7-2x)/3 but want line thru point=(4,5) slope=-2/3 or -0.6666666 projejt line from x=4 tu x=0. delta x=-4 deltay=slope*deltax=(-2/3)*(-4)=8/3=2.6666666 nu y=5+2.666666=7.66666 now point=(0,7.66666666) y=-(2/3)x +7.66666

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 we need to find the y intercept or b value. The easiest way to do this is to graph the line by plotting the point and using the slope to find other points.  Here is a graph of the two lines:   The y intercept of the unknown line, or the point at which it crosses the y axis is 23. The equation then for the unknown line is y=-5/2x+23

the line passes through (7,-2) and is parallel to the line whose equation y=4x+2 is

start: point=(7,-2) & slope=4 need tu get y-intersept...x=0 deltax=7, deltay=slope*deltax=28 y-intersept=-2 -28=-30 line: y=4x-30

Find the slope-intercept passes through the point (-5,4) and is perpendicular to the line y+=5/7x -3

want strate line 90 deg tu line4.....y=(5/7)x+konst refer line hav slope=5/7, so want slope=-(7/5) tu be 90 deg from that thang y=(-7/5)x + konst this line go thru point=(-5,4) projekt line from x=-5 tu x=0...deltax=+5 deltay=slope*deltax=(-7/5)*5=-7 y=4-7=-3 point=(0,-3) y=-(7/5)x-3