Precalculus with Limits: A Graphing Approach
7th Edition
ISBN: 9781305071711
Author: Ron Larson
Publisher: Brooks/Cole
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Question
Chapter 5.2, Problem 91E
To determine
To explain: Why the given equation is not an identity and find one value of the variable for which the equation is not true.
Expert Solution & Answer
Explanation of Solution
Given information:
The following equation:
The given equation is not an identity because it is not true for all values of
Let
Substituting value of
Substituting value of
The value on the left hand side is not equal to the value on the right hand side.
Hence for
Also, the given equation is not an identity because the expressions on the left hand side and the right hand side when plotted on a graph, do not coincide.
Graph:
Sketch the graph using graphing utility.
Step 1: Press WINDOW button to access the Window editor.
Step 2: Press
Step 3: Enter the expressions
Step 4: Press GRAPH button to graph the function.
The graph is obtained as:
Interpretation:
The graphs of the left hand side expression and the right hand side expression do no coincide.
Chapter 5 Solutions
Precalculus with Limits: A Graphing Approach
Ch. 5.1 - Prob. 1ECh. 5.1 - Prob. 2ECh. 5.1 - Prob. 3ECh. 5.1 - Prob. 4ECh. 5.1 - Prob. 5ECh. 5.1 - Prob. 6ECh. 5.1 - Prob. 7ECh. 5.1 - Prob. 8ECh. 5.1 - Prob. 9ECh. 5.1 - Prob. 10E
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- 8–23. Sketching vector fields Sketch the following vector fieldsarrow_forward25-30. Normal and tangential components For the vector field F and curve C, complete the following: a. Determine the points (if any) along the curve C at which the vector field F is tangent to C. b. Determine the points (if any) along the curve C at which the vector field F is normal to C. c. Sketch C and a few representative vectors of F on C. 25. F = (2½³, 0); c = {(x, y); y − x² = 1} 26. F = x (23 - 212) ; C = {(x, y); y = x² = 1}) , 2 27. F(x, y); C = {(x, y): x² + y² = 4} 28. F = (y, x); C = {(x, y): x² + y² = 1} 29. F = (x, y); C = 30. F = (y, x); C = {(x, y): x = 1} {(x, y): x² + y² = 1}arrow_forward٣/١ B msl kd 180 Ka, Sin (1) I sin () sin(30) Sin (30) اذا ميريد شرح الكتب بس 0 بالفراغ 3) Cos (30) 0.866 4) Rotating 5) Synchronous speed, 120 x 50 G 5005 1000 s = 1000-950 Copper bosses 5kW Rotor input 5 0.05 : loo kw 6) 1 /0001 ined sove in peaper I need a detailed solution on paper please وه اذا ميريد شرح الكتب فقط ١٥٠ DC 7) rotor a ' (y+xlny + xe*)dx + (xsiny + xlnx + dy = 0. Q1// Find the solution of: ( 357arrow_forward
- ۳/۱ R₂ = X2 2) slots per pole per phase 3/31 B. 180 msl Kas Sin (I) 1sin() sin(30) Sin (30) اذا ميريد شرح الكتب بس 0 بالفراغ 3) Cos (30): 0.866 4) Rotating 5) Synchronous speeds 120×50 looo G 1000-950 1000 Copper losses 5kw Rotor input 5 loo kw 0.05 6) 1 اذا ميريد شرح الكتب فقط look 7) rotor DC ined sove in peaper I need a detailed solution on paper please 0 64 Find the general solution of the following equations: QI//y(4)-16y= 0. Find the general solution of the following equations: Q2ll yll-4y/ +13y=esinx.arrow_forwardR₂ = X2 2) slots per pole per phase = 3/31 B-180 60 msl kd Kas Sin () 2 I sin (6) sin(30) Sin (30) اذا مريد شرح الكتب بس 0 بالفراغ 3 Cos (30) 0.866 4) Rotating ined sove in peaper 5) Synchronous speed s 120×50 6 s = 1000-950 1000 Copper losses 5kw Rotor input 5 0.05 6) 1 loo kw اذا ميريد شرح الكتب فقط Look 7) rotov DC I need a detailed solution on paper please 0 64 Solve the following equations: 0 Q1// Find the solution of: ( y • with y(0) = 1. dx x²+y²arrow_forwardR₂ = X2 2) slots per pole per phase = 3/3 1 B-180-60 msl Ka Sin (1) Isin () sin(30) Sin (30) اذا ميريد شرح الكتب بس 0 بالفراغ 3) Cos (30) 0.866 4) Rotating 5) Synchronous speed, 120 x 50 s = 1000-950 1000 Copper losses 5kw Rotor input 5 6) 1 0.05 G 50105 loo kw اذا ميريد شرح الكتب فقط look 7) rotov DC ined sove in peaper I need a detailed solution on paper please 064 2- A hot ball (D=15 cm ) is cooled by forced air T.-30°C, the rate of heat transfer from the ball is 460.86 W. Take for the air -0.025 Wim °C and Nu=144.89, find the ball surface temperature a) 300 °C 16 b) 327 °C c) 376 °C d) None か = 750 01arrow_forward
- Don't do 14. Please solve 19arrow_forwardPlease solve 14 and 15arrow_forward1. Consider the following system of equations: x13x2 + 4x3 - 5x4 = 7 -2x13x2 + x3 - 6x4 = 7 x16x213x3 - 21x4 = 28 a) Solve the system. Write your solution in parametric and vector form. b) What is a geometric description of the solution. 7 c) Is v = 7 in the span of the set S= [28. 1 HE 3 -5 3 ·6 ? If it is, write v 6 as a linear combination of the vectors in S. Justify. d) How many solutions are there to the associated homogeneous system for the system above? Justify. e) Let A be the coefficient matrix from the system above. Find the set of all solutions to Ax = 0. f) Is there a solution to Ax=b for all b in R³? Justify.arrow_forward
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