CALCULUS+ITS...,EXP.(LL)-W/CODE NVCC
19th Edition
ISBN: 9780136572671
Author: BITTINGER
Publisher: PEARSON C
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Chapter PSDT, Problem 9BP
Find the domain of
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3.12 (B). A horizontal beam AB is 4 m
long and of constant flexural rigidity. It is
rigidly built-in at the left-hand end A and simply supported on a non-yielding support
at the right-hand end B. The beam carries Uniformly distributed vertical loading of
18 kN/m over its whole length, together with a vertical downward load of 10KN at
2.5 m from the end A. Sketch the S.F. and B.M. diagrams for the beam, indicating
all main values. Cl. Struct. E.] CS.F. 45,10,376 KN, B.M. 186, +36.15 kNm.7
Qize
f(x)
=
x + 2x2 - 2
x² + 4x²² -
Solve the equation using Newton
Raphson
-b±√√b2-4ac
2a
@4x²-12x+9=0
27 de febrero de 2025
-b±√√b2-4ac
2a
⑥2x²-4x-1=0
a = 4 b=-12
c=9
a = 2
b = 9
c = \
x=-42±√(2-4 (4) (9)
2(4))
X =
(12) ±√44)-(360)
2(108)
x = ±√
X = =±√√²-4(2) (1)
2()
X = ±√
+
X =
X =
+
X₁ =
=
X₁ =
X₁ =
+
X₁ =
=
=
Chapter PSDT Solutions
CALCULUS+ITS...,EXP.(LL)-W/CODE NVCC
Ch. PSDT - Write an equivalent expression for each of the...Ch. PSDT - Write an equivalent expression for each of the...Ch. PSDT - Write an equivalent expression for each of the...Ch. PSDT - Write an equivalent expression for each of the...Ch. PSDT - Write an equivalent expression for each of the...Ch. PSDT - Write an equivalent expression for each of the...Ch. PSDT - Write an equivalent expression for each of the...Ch. PSDT - Prob. 8APCh. PSDT - Multiply. 9. x5x6Ch. PSDT - Multiply. 10. x2x9
Ch. PSDT - Multiply. 11. 2x35x44x10Ch. PSDT - Divide. 12. a2a2Ch. PSDT - Divide. 13. b3b5Ch. PSDT - Simplify. Express each answer without a negative...Ch. PSDT - Simplify. Express each answer without a negative...Ch. PSDT - Multiply. 16. 3x5Ch. PSDT - Multiply. 17. x5x+3Ch. PSDT - Prob. 18APCh. PSDT - Prob. 19APCh. PSDT - Prob. 20APCh. PSDT - Prob. 21APCh. PSDT - Prob. 22APCh. PSDT - Factor. 23. x25x14Ch. PSDT - Factor. 24. 6x2+7x5Ch. PSDT - Prob. 25APCh. PSDT - Prob. 26APCh. PSDT - Solve. 27. 3x(x2)(5x+4)=0Ch. PSDT - Solve. 28. 4x3=xCh. PSDT - Solve. 29. 2xx36x=18x23xCh. PSDT - Solve 30. 178x5x4Ch. PSDT - Prob. 31APCh. PSDT - Raggs, Ltd., a clothing firm, determines that its...Ch. PSDT - Prob. 1BPCh. PSDT - Prob. 2BPCh. PSDT - Graph. 3. y=x21Ch. PSDT - Prob. 4BPCh. PSDT - Prob. 5BPCh. PSDT - Prob. 6BPCh. PSDT - Graph the function f:...Ch. PSDT - Write interval notation for x4x5.Ch. PSDT - Find the domain of f:f(x)=32x5.Ch. PSDT - Find the slope and y-intercept of the graph of...Ch. PSDT - Prob. 11BPCh. PSDT - Find the slope of the line containing the points...Ch. PSDT - Prob. 13BPCh. PSDT - Prob. 14BPCh. PSDT - Prob. 15BPCh. PSDT - Graph. 16. g(x)=x3Ch. PSDT - Prob. 17BPCh. PSDT - Prob. 18BPCh. PSDT - Graph. 19. f(x)=xCh. PSDT - Suppose that $1000 is earning 5 interest,...
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- 3.9 (A/B). A beam ABCDE, with A on the left, is 7 m long and is simply supported at Band E. The lengths of the various portions are AB 1-5m, BC = 1-5m, CD = 1 m and DE : 3 m. There is a uniformly distributed load of 15kN/m between B and a point 2m to the right of B and concentrated loads of 20 KN act at 4 and 0 with one of 50 KN at C. (a) Draw the S.F. diagrams and hence determine the position from A at which the S.F. is zero. (b) Determine the value of the B.M. at this point. (c) Sketch the B.M. diagram approximately to scale, quoting the principal values. [3.32 m, 69.8 KNm, 0, 30, 69.1, 68.1, 0 kNm.]arrow_forward4. Verify that V X (aẢ) = (Va) XẢ + aV X Ả where Ả = xyz(x + y + 2) A and a = 3xy + 4zx by carrying out the detailed differentiations.arrow_forward3. For each of the arrow or quiver graphs shown below, determine analytically V°C and V X Č. From these analytical solutions, identify the extrema (+/-) and plot these points on the arrow graph. (a) C = −✰CosxSiny + ŷSinxCosy -π<ׂу<π Ty (b) C = −xSin2y + ŷCos2y x, y<π -π< (c) C = −xCosx + ŷSiny -π< x, y < πarrow_forward
- 7.10 (B/C). A circular flat plate of diameter 305 mm and thickness 6.35 mm is clamped at the edges and subjected to a Uniform lateral pressure of 345 kN/m². Evaluate: (a) the central deflection, (b) the position and magnitude of the maximum radial stress. C6.1 x 10 m; 149.2 MN/m².] 100 200arrow_forward3.15 (B). A beam ABCD is simply supported at B and C with ABCD=2m; BC 4 m. It carries a point load of 60 KN at the free end A, a Uniformly distributed load of 60 KN/m between B and C and an anticlockwise moment of 80 KN m in the plane of the beam applied at the free end D. Sketch and dimension the S.F. and B.M. diagrams, and determine the position and magnitude of the maximum bending moment. CEL.E.] CS.F. 60, 170, 70KN, B.M. 120, +120.1, +80 kNm, 120.1 kNm at 2.83 m to right of 8.7arrow_forward7.1 (A/B). A Uniform I-section beam has flanges 150 mm wide by 8 mm thick and a web 180 mm wide and 8 mm thick. At a certain section there is a shearing force of 120 KN. Draw a diagram to illustrate the distribution of shear stress across the section as a result of bending. What is the maximum shear stress? [86.7 MN/m².arrow_forward
- 1. Let Ả = −2x + 3y+42, B = - - 7x +lý +22, and C = −1x + 2y + 42. Find (a) Ả X B (b) ẢX B°C c) →→ Ả B X C d) ẢB°C e) ẢX B XC.arrow_forward3.13 (B). A beam ABC, 6 m long, is simply-supported at the left-hand end A and at B I'm from the right-hand end C. The beam is of weight 100 N/metre run. (a) Determine the reactions at A and B. (b) Construct to scales of 20 mm = 1 m and 20 mm = 100 N, the shearing-force diagram for the beam, indicating thereon the principal values. (c) Determine the magnitude and position of the maximum bending moment. (You may, if you so wish, deduce the answers from the shearing force diagram without constructing a full or partial bending-moment diagram.) [C.G.] C240 N, 360 N, 288 Nm, 2.4 m from A.]arrow_forward5. Using parentheses make sense of the expression V · VXVV · Å where Ả = Ã(x, y, z). Is the result a vector or a scaler?arrow_forward
- 3.10 (A/B). A beam ABCDE is simply supported at A and D. It carries the following loading: a distributed load of 30 kN/m between A and B, a concentrated load of 20 KN at B, a concentrated load of 20 KN at C, a concentrated load of 10 KN at E; a distributed load of 60 kN/m between 0 and E. Span AB = 1.5 BC = CD = DE 1 m. Calculate the value of the reactions at A and D and hence draw the S.F. and B.M. diagrams. What are the magnitude and position of the maximum B.M. on the beam? [41.1, 113.9 KN, 28.15 kNm; 1.37 m from A.J m,arrow_forward3.14 (B). A beam ABCD, 6 m long, is simply-supported at the right-hand end and at a point B Im from the left-hand end A. It carries a vertical load of 10 KN at A, a second concentrated load of 20 KN at C, 3 m from D, and a uniformly distributed load of 10 kN/m between C and D. Determine: (a) the values of the reactions at B and 0, (6) the position and magnitude of the maximum bending moment. [33 KN, 27 KN, 2.7 m from D, 36.45k Nm.]arrow_forward3.17 (B). A simply supported beam has a span of 6 m and carries a distributed load which varies in a linea manner from 30 kN/m at one support to 90 kN/m at the other support. Locate the point of maximum bendin moment and calculate the value of this maximum. Sketch the S.F. and B.M. diagrams. [U.L.] [3.25 m from l.h. end; 272 KN m 30. 90arrow_forward
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