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kuta software infinite algebra 1 graphing lines
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Video About Algebra: graphing lines 1  Encyclopedia.
Graphing Linear Equations; Part 1 · (Algebra) Number Lines Part 1
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Question:Hey, I can't seem to get graphing perpendicular lines on the Cartesian Graph. I understand that the slopeintercept formula for the second line must be the reciprocal and the opposite positive or negative sign than the original line, but when I get past this point, I am not able to lengthen the line to pass through the correct Yaxis intercept... How can I get it to pass through the right Yaxis intercept?? Hope i explained it enough, if not, just ask, thanks! :) Catbek: Yes, I understand that completly, but even when I have that part 100% correct I STILL don't get it to pass through the correct Yintercept... :/ Ummm... my formula is: y=mx /+ b
Answers:So you have the gradient and two points I am assuming. Use y  y1 = m(x  x1) to find the equation of the line. Then substitute in 0 for x into the equation to find the y intercept as when the line passes through the y intercept, x = 0.
Answers:So you have the gradient and two points I am assuming. Use y  y1 = m(x  x1) to find the equation of the line. Then substitute in 0 for x into the equation to find the y intercept as when the line passes through the y intercept, x = 0.
Question:I have to graph two equations, and write the solution. for coincident lines, my teacher said you write either "all points fall on x = #" or "all points equal x = #" but I forget which one it is. which one is it, or are they both right?
Answers:I don't understand "all points fall on x = #" or "all points equal x = #" but for coincident lines, the equations are equivalent  can be shown to be the same equation. The solution points for both equations fall on the same line.
Answers:I don't understand "all points fall on x = #" or "all points equal x = #" but for coincident lines, the equations are equivalent  can be shown to be the same equation. The solution points for both equations fall on the same line.
Question:Line 1: x 7y +9 = 0
Line 2: 4x+3y+14 =0
Line 1, Line 2, both lines or neither?
2,1
5,2
7,5 I have to make a choice, if Line one is at 2,1, 5,2, 7,5, same with Line 2, or both lines at the same place or neither lines at none of these choices
Answers:Yes, both are lines. .
Answers:Yes, both are lines. .
Question:I don't know how to do this and th etecher didn't even teach us. It's not in our books either:// Please dont just give me an answer, I do my homework. Bu tplease tell me why or how you get an answer. THANKS
problem:
graph by plotting:
y=x^2
y=x1
Can you describe how to graph/solve.
Answers:1) y=x^2 The easiest way to do these is to make a table of the values, then graph the points. For this one since x is squared, you would consider both the positive and negative of the same number  i.e. you would get the same value for y. For example, x=3 and x= 3, both give y = 9 x  y  points to graph 0  0  (0,0) + 1/2  1/4  (1/2,1/4) (1/2, 1/4) + 1  1  (1,1 ) (1,1) + 2  4  (2,4) (2,4) + 3  9  (3,9) (3,9) You can do a few more x values, but this gives you an idea.  2) y=x1 Please read over the reference as it is one of the better ones I've found to explain y=mx+b, where m is the slope (rise over run) and b is the yintercept. The yintercept is just the point on the y axis when x is 0. So the yintercept is 1 or the point (0, 1). The slope is 1. 1 times x is just x. You can think of it a 1/1, and go up one and over one  to (1,0). Or you can think of it as 1 / 1 and go down one and back one  to (1, 2). Again you could make a table with x and y values and get these points x  y  point 2  3  (2, 3) 1  2  (1, 2) 0  1  (0, 1) 1  0  (1, 0) 2  1  (2, 1)
Answers:1) y=x^2 The easiest way to do these is to make a table of the values, then graph the points. For this one since x is squared, you would consider both the positive and negative of the same number  i.e. you would get the same value for y. For example, x=3 and x= 3, both give y = 9 x  y  points to graph 0  0  (0,0) + 1/2  1/4  (1/2,1/4) (1/2, 1/4) + 1  1  (1,1 ) (1,1) + 2  4  (2,4) (2,4) + 3  9  (3,9) (3,9) You can do a few more x values, but this gives you an idea.  2) y=x1 Please read over the reference as it is one of the better ones I've found to explain y=mx+b, where m is the slope (rise over run) and b is the yintercept. The yintercept is just the point on the y axis when x is 0. So the yintercept is 1 or the point (0, 1). The slope is 1. 1 times x is just x. You can think of it a 1/1, and go up one and over one  to (1,0). Or you can think of it as 1 / 1 and go down one and back one  to (1, 2). Again you could make a table with x and y values and get these points x  y  point 2  3  (2, 3) 1  2  (1, 2) 0  1  (0, 1) 1  0  (1, 0) 2  1  (2, 1)
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Algebra: graphing lines 1 :Graphing linear equations
Algebra graphing lines 1.flv :Learn math (algebre) so easy. :)