Crack problems concerning boundaries of convex lens like formsijtsrd
The singular stress problem of aperipheral edge crack around a cavity of spherical portion in an infinite elastic medium whenthe crack is subjected to a known pressure is investigated. The problem is solved byusing integral transforms and is reduced to the solution of a singularintegral equation of the first kind. The solution of this equation is obtainednumerically by the method due to Erdogan, Gupta , and Cook, and thestress intensity factors are displayed graphically.Also investigated in this paper is the penny-shaped crack situated symmetrically on the central plane of a convex lens shaped elastic material. Doo-Sung Lee"Crack problems concerning boundaries of convex lens like forms" Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456-6470, Volume-2 | Issue-3 , April 2018, URL: http://www.ijtsrd.com/papers/ijtsrd11106.pdf http://www.ijtsrd.com/mathemetics/applied-mathamatics/11106/crack-problems-concerning-boundaries-of-convex-lens-like-forms/doo-sung-lee
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Crack problems concerning boundaries of convex lens like formsijtsrd
The singular stress problem of aperipheral edge crack around a cavity of spherical portion in an infinite elastic medium whenthe crack is subjected to a known pressure is investigated. The problem is solved byusing integral transforms and is reduced to the solution of a singularintegral equation of the first kind. The solution of this equation is obtainednumerically by the method due to Erdogan, Gupta , and Cook, and thestress intensity factors are displayed graphically.Also investigated in this paper is the penny-shaped crack situated symmetrically on the central plane of a convex lens shaped elastic material. Doo-Sung Lee"Crack problems concerning boundaries of convex lens like forms" Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456-6470, Volume-2 | Issue-3 , April 2018, URL: http://www.ijtsrd.com/papers/ijtsrd11106.pdf http://www.ijtsrd.com/mathemetics/applied-mathamatics/11106/crack-problems-concerning-boundaries-of-convex-lens-like-forms/doo-sung-lee
EDGE-NEIGHBOR RUPTURE DEGREE ON GRAPH OPERATIONSmathsjournal
Vulnerability and reliability parameters measure the resistance of the network to disruption of operation after the failure of certain stations or communication links in a communication network. An edge subversion strategy of a graph , say , is a set of edge(s) in whose adjacent vertices which is incident with the removal edge(s) are removed from . The survival subgraph is denoted by . In this paper we give some results for the edge-neighbor-rupture degree of the graph operations and Thorny graph types are examined.
Some Properties of M-projective Curvature Tensor on Generalized Sasakian-Spac...inventionjournals
The object of this paper is to study the M -projective curvature tensor on generalized Sasakian-space-forms. We study M -projectively semisymmetric, M -projectively pseudosymmetric and -M - projectively semisymmetric generalized Sasakian-space-form.
Pre-Calculus: Conics - Introduction to Conics and Determining & Graphing Circ...Myrrhtaire Castillo
This PowerPoint contains an introduction to conical sections: the conics formed from double-napped circular cone - the Parabola, Hyperbola, Circle, & Ellipse. It also contains the basic parts of Circle. Identifying the standard form of circle's radius and center. Graphing a circle from its standard form. Transforming General Equation of Circle to Standard Form and some of the special cases.
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In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
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7. RECTANGLE
A parallelogram with four right angles.
Four right angles
Opposite sides are parallel and congruent
Opposite angles are congruent
8. RECTANGLE
A parallelogram with four right angles.
Four right angles
Opposite sides are parallel and congruent
Opposite angles are congruent
Consecutive angles are supplementary
9. RECTANGLE
A parallelogram with four right angles.
Four right angles
Opposite sides are parallel and congruent
Opposite angles are congruent
Consecutive angles are supplementary
Diagonals bisect each other
11. THEOREMS
6.13 - Diagonals of a Rectangle: If a parallelogram is a
rectangle, then its diagonals are congruent
6.14 - Diagonals of a Rectangle Converse:
12. THEOREMS
6.13 - Diagonals of a Rectangle: If a parallelogram is a
rectangle, then its diagonals are congruent
6.14 - Diagonals of a Rectangle Converse: If diagonals of a
parallelogram are congruent, then the parallelogram is a
rectangle
13. EXAMPLE 1
A rectangular garden gate is reinforced with diagonal
braces to prevent it from sagging. If JK = 12 feet and
LN = 6.5 feet, find KM.
14. EXAMPLE 1
A rectangular garden gate is reinforced with diagonal
braces to prevent it from sagging. If JK = 12 feet and
LN = 6.5 feet, find KM.
Since we have a rectangle, the
diagonals are congruent.
15. EXAMPLE 1
A rectangular garden gate is reinforced with diagonal
braces to prevent it from sagging. If JK = 12 feet and
LN = 6.5 feet, find KM.
Since we have a rectangle, the
diagonals are congruent.
The diagonals also bisect each other,
so JN = LN and KN = MN.
16. EXAMPLE 1
A rectangular garden gate is reinforced with diagonal
braces to prevent it from sagging. If JK = 12 feet and
LN = 6.5 feet, find KM.
Since we have a rectangle, the
diagonals are congruent.
The diagonals also bisect each other,
so JN = LN and KN = MN.
So JN = LN = KN = MN = 6.5 feet and KM = KN + MN.
17. EXAMPLE 1
A rectangular garden gate is reinforced with diagonal
braces to prevent it from sagging. If JK = 12 feet and
LN = 6.5 feet, find KM.
Since we have a rectangle, the
diagonals are congruent.
The diagonals also bisect each other,
so JN = LN and KN = MN.
So JN = LN = KN = MN = 6.5 feet and KM = KN + MN.
KM = 13 feet
19. EXAMPLE 2
Quadrilateral RSTU is a rectangle. If m∠RTU = (8x + 4)°
and m∠SUR = (3x − 2)°, find x.
m∠RTU + m∠SUR = 90
20. EXAMPLE 2
Quadrilateral RSTU is a rectangle. If m∠RTU = (8x + 4)°
and m∠SUR = (3x − 2)°, find x.
m∠RTU + m∠SUR = 90
8x + 4 + 3x − 2 = 90
21. EXAMPLE 2
Quadrilateral RSTU is a rectangle. If m∠RTU = (8x + 4)°
and m∠SUR = (3x − 2)°, find x.
m∠RTU + m∠SUR = 90
8x + 4 + 3x − 2 = 90
11x + 2 = 90
22. EXAMPLE 2
Quadrilateral RSTU is a rectangle. If m∠RTU = (8x + 4)°
and m∠SUR = (3x − 2)°, find x.
m∠RTU + m∠SUR = 90
8x + 4 + 3x − 2 = 90
11x + 2 = 90
−2 −2
23. EXAMPLE 2
Quadrilateral RSTU is a rectangle. If m∠RTU = (8x + 4)°
and m∠SUR = (3x − 2)°, find x.
m∠RTU + m∠SUR = 90
8x + 4 + 3x − 2 = 90
11x + 2 = 90
−2 −2
11x = 88
24. EXAMPLE 2
Quadrilateral RSTU is a rectangle. If m∠RTU = (8x + 4)°
and m∠SUR = (3x − 2)°, find x.
m∠RTU + m∠SUR = 90
8x + 4 + 3x − 2 = 90
11x + 2 = 90
−2 −2
11x = 88
11 11
25. EXAMPLE 2
Quadrilateral RSTU is a rectangle. If m∠RTU = (8x + 4)°
and m∠SUR = (3x − 2)°, find x.
m∠RTU + m∠SUR = 90
8x + 4 + 3x − 2 = 90
11x + 2 = 90
−2 −2
11x = 88
11 11
x = 8
26. EXAMPLE 3
Some artists stretch their own canvas over wooden
frames. This allows them to customize the size of a
canvas. In order to ensure that the frame is rectangular
before stretching the canvas, an artist measures the sides
of the diagonals of the frame. If AB = 12 inches, BC = 35
inches, CD = 12 inches, and DA = 35 inches, how long do
the lengths of the diagonals need to be?
27. EXAMPLE 3
Some artists stretch their own canvas over wooden
frames. This allows them to customize the size of a
canvas. In order to ensure that the frame is rectangular
before stretching the canvas, an artist measures the sides
of the diagonals of the frame. If AB = 12 inches, BC = 35
inches, CD = 12 inches, and DA = 35 inches, how long do
the lengths of the diagonals need to be?
The diagonal forms a right triangle
with legs of 12 and 35. We need to find
the hypotenuse.
35. EXAMPLE 3
a2
+ b2
= c2
122
+ 352
= c2
144 +1225 = c2
1369 = c2
1369 = c2
c = 37
The diagonals must both be 37 inches
36. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
37. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
Diagonals must be congruent
38. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
Diagonals must be congruent
39. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
Diagonals must be congruent
40. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25
Diagonals must be congruent
41. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
Diagonals must be congruent
42. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
Diagonals must be congruent
43. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
= 12
+ 72
Diagonals must be congruent
44. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
= 12
+ 72
= 1+ 49
Diagonals must be congruent
45. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
= 12
+ 72
= 1+ 49 = 50
Diagonals must be congruent
46. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
= 12
+ 72
= 1+ 49 = 50
Diagonals must be congruent
Opposite sides parallel, consecutive sides perpendicular
47. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
= 12
+ 72
= 1+ 49 = 50
m( JK) =
4 − 3
1+ 2
Diagonals must be congruent
Opposite sides parallel, consecutive sides perpendicular
48. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
= 12
+ 72
= 1+ 49 = 50
m( JK) =
4 − 3
1+ 2
=
1
3
Diagonals must be congruent
Opposite sides parallel, consecutive sides perpendicular
49. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
= 12
+ 72
= 1+ 49 = 50
m( JK) =
4 − 3
1+ 2
=
1
3
m(LM) =
−3+ 2
0− 3
Diagonals must be congruent
Opposite sides parallel, consecutive sides perpendicular
50. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
= 12
+ 72
= 1+ 49 = 50
m( JK) =
4 − 3
1+ 2
=
1
3
m(LM) =
−3+ 2
0− 3
=
−1
−3
Diagonals must be congruent
Opposite sides parallel, consecutive sides perpendicular
51. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
= 12
+ 72
= 1+ 49 = 50
m( JK) =
4 − 3
1+ 2
=
1
3
m(LM) =
−3+ 2
0− 3
=
−1
−3
=
1
3
Diagonals must be congruent
Opposite sides parallel, consecutive sides perpendicular
52. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
= 12
+ 72
= 1+ 49 = 50
m( JK) =
4 − 3
1+ 2
=
1
3
m(LM) =
−3+ 2
0− 3
=
−1
−3
=
1
3
m(KL) =
−2 − 4
3−1
Diagonals must be congruent
Opposite sides parallel, consecutive sides perpendicular
53. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
= 12
+ 72
= 1+ 49 = 50
m( JK) =
4 − 3
1+ 2
=
1
3
m(LM) =
−3+ 2
0− 3
=
−1
−3
=
1
3
m(KL) =
−2 − 4
3−1
=
−6
2
Diagonals must be congruent
Opposite sides parallel, consecutive sides perpendicular
54. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
= 12
+ 72
= 1+ 49 = 50
m( JK) =
4 − 3
1+ 2
=
1
3
m(LM) =
−3+ 2
0− 3
=
−1
−3
=
1
3
m(KL) =
−2 − 4
3−1
=
−6
2
= −3
Diagonals must be congruent
Opposite sides parallel, consecutive sides perpendicular
55. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
= 12
+ 72
= 1+ 49 = 50
m( JK) =
4 − 3
1+ 2
=
1
3
m(LM) =
−3+ 2
0− 3
=
−1
−3
=
1
3
m(KL) =
−2 − 4
3−1
=
−6
2
m( JM) =
−3− 3
0+ 2
= −3
Diagonals must be congruent
Opposite sides parallel, consecutive sides perpendicular
56. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
= 12
+ 72
= 1+ 49 = 50
m( JK) =
4 − 3
1+ 2
=
1
3
m(LM) =
−3+ 2
0− 3
=
−1
−3
=
1
3
m(KL) =
−2 − 4
3−1
=
−6
2
m( JM) =
−3− 3
0+ 2
=
−6
2
= −3
Diagonals must be congruent
Opposite sides parallel, consecutive sides perpendicular
57. EXAMPLE 4
Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2),
and M(0, −3). Determine whether JKLM is a rectangle by
using the distance formula, then slope.
JL = (−2 − 3)2
+ (3+ 2)2
= (−5)2
+ 52
= 25 + 25 = 50
KM = (1−0)2
+ (4 + 3)2
= 12
+ 72
= 1+ 49 = 50
m( JK) =
4 − 3
1+ 2
=
1
3
m(LM) =
−3+ 2
0− 3
=
−1
−3
=
1
3
m(KL) =
−2 − 4
3−1
=
−6
2
m( JM) =
−3− 3
0+ 2
=
−6
2
= −3
= −3
Diagonals must be congruent
Opposite sides parallel, consecutive sides perpendicular