zan81289
x^2(x-2)y"+(5x-7)y'+2(3+5x^2)y=0
I'm getting it all right except for the classification of one of the singular points. Can anyone help me with this. (Answer from book: 0,regular 2,regular)
Thanks Ted, that's what I got as well, I guess its just a mistake in the answer key.
Answer
as typed 2 is regular but 0 is irregular as lim { x---> 0 } [ x(5x-7) / x²] does not exist...that is 0 is a pole of order 2 for p_1 (x) =(5x-7) / (x² [x-2]) and 2 is a pole of order 1
as typed 2 is regular but 0 is irregular as lim { x---> 0 } [ x(5x-7) / x²] does not exist...that is 0 is a pole of order 2 for p_1 (x) =(5x-7) / (x² [x-2]) and 2 is a pole of order 1
How to find perimeter of figure in coordinate plane?
Kayla H
Im homeschooled and all i have is a math book it doesnt really describe how to find the perimeter if its a triangle in a coordinate plane.
X(0,2), Y(4,-1), Z(-2,-1).
Step by step please!!!
Thanks!
Answer
First graph it out, see where each point is. Based on the graph I sketeched out from those three points, zy, yx, and xz are all the line segments of the triangle. Since you know the perimeter is the sum of all the line segments of a shape, just add up the distances of those three line segments to find the perimeter. The distance formula is â(y2-y1)^2+(x2-x1)^2 (the whole thing is squared)
y2 and x2 are a pair of points, x1 and y1 are a pair of points. Now just pick a line segment and plug in the points.
Ill start with zy. (it doesnt matter which pair of points is x1 and y1, just pick one).
â(-1- -1)^2+ (-2-4)^2
simplify : â(0)^2+ (-6)^2
: â36
= 6
the distance for the line segment zy is 6. Now find the distance for the other line segments of the triangle.
Next Ill do yx.
: â(-1-2)^2+ (4-0)^2
simplify : â(-3)^2 + (4)^2
â9+16
â25
=5
the distance for the line segment yx is 5. Now find the last one, xz
: â(2- -1)^2 + (0- -2)^2
simplify : â(3)^2 + (2)^2
simplify again : â9+4
: â15
Now add up all the line segments together to find the perimeter : 6+5+ â15 = 14.87298335
First graph it out, see where each point is. Based on the graph I sketeched out from those three points, zy, yx, and xz are all the line segments of the triangle. Since you know the perimeter is the sum of all the line segments of a shape, just add up the distances of those three line segments to find the perimeter. The distance formula is â(y2-y1)^2+(x2-x1)^2 (the whole thing is squared)
y2 and x2 are a pair of points, x1 and y1 are a pair of points. Now just pick a line segment and plug in the points.
Ill start with zy. (it doesnt matter which pair of points is x1 and y1, just pick one).
â(-1- -1)^2+ (-2-4)^2
simplify : â(0)^2+ (-6)^2
: â36
= 6
the distance for the line segment zy is 6. Now find the distance for the other line segments of the triangle.
Next Ill do yx.
: â(-1-2)^2+ (4-0)^2
simplify : â(-3)^2 + (4)^2
â9+16
â25
=5
the distance for the line segment yx is 5. Now find the last one, xz
: â(2- -1)^2 + (0- -2)^2
simplify : â(3)^2 + (2)^2
simplify again : â9+4
: â15
Now add up all the line segments together to find the perimeter : 6+5+ â15 = 14.87298335
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