Explanation Method used: Graphing a Linear Inequality in Two Variables: 1. Replace the inequality symbol with an equal sign and graph the corresponding linear equation. Draw a solid line if the original inequality contains a ≥ or ≤ symbol. Draw a dashed line if the original inequality contains a > or < symbol. 2. Choose a test point from one of the half–planes. (Do not choose a point on the line.) Substitute the coordinates of the test point into inequality. 3. If a true statement results, shade the half-plane containing this test point. If a false statement results, shade the half-plane not containing this test point. Calculation: The given inequality is − 2 ≤ x < 5 . The above inequality can be written as in system of inequality as { x ≥ − 2 x < 5 . To sketch the graph of the solution set of the above system, first sketch the graph of the each inequality. Use the above method and obtain the graph of the each inequality is as follows. The graph of the inequality x ≥ − 2 and x < 5 is as follows. Step 1: Replace the inequality symbol by = and graph the linear equation: That is, replace the symbol < by = , and graph the linear equation x = − 2 and x = 5 . Use the intercepts concept to graph the line x = − 2 , Since x = − 2 is a vertical line with x -intercept − 2 , the line passes through the point ( − 2 , 0 ) . Similarly, the line x = 5 is a vertical line with x -intercept 5 , the line passes through the point ( 5 , 0 ) . Use above intercepts and draws the graph of the line x = − 2 as a solid line and x = 5 as a dashed line as shown below in Figure 1. From Figure 1, it is observed that the both lines are parallel to each other. Step 2: Choose a test point from one of the half–planes. (Do not choose a point on the line.) Substitute the coordinates of the test point into inequality. The line x = − 2 divides the x y -plane into three parts, namely the line itself and the two half-planes. The test point in one half-plane satisfy x ≤ − 2 and the test point in other half-plane satisfy x ≥ − 2 . To find half-plane to which the solution of x ≥ − 2 belongs, choose a test point ( 0 , 0 ) from either of the half-planes. Substitute 0 for x and 0 for y into the original inequality. x ≥ − 2 0 ≥ ? − 2 0 ≥ ? − 2 True statement Since, 0 greater than − 2 is true statement, thus the half-plane containing the test point ( 0 , 0 ) is part of the solution set of the inequality x ≥ − 2

College Algebra Essentials (5th Edition)

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ISBN: 9780134469294

Author: Blitzer, Robert F.

Publisher: PEARSON

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Chapter 5.5, Problem 37E

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To sketch: The graph of the solution set of the inequality −2≤x<5 or indicate that the system has no solution.

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Chapter 5.5, Problem 36E

Chapter 5.5, Problem 38E

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Assume a bivariate patch p(u, v) over the unit square [0, 1]² that is given as a tensor product patch where u-sections (u fixed to some constant û; v varying across [0, 1]) are quadratic polynomials Pu:û(v) = p(û, v) while v-sections are lines pv:ô (u) = p(u, v). The boundary lines pv:o(u) and pv:1 (u) are specified by their end points p(0,0) 0.8 and p(1,0) 0.2 as well as p(0, 1) 0.3 and p(1, 1) = 0.8. The boundary quadratics pu:o(v) and pu:1 (v) interpolate p(0,0.5) = 0.1 and p(1, 0.5) = 0.9 in addition to the above given four corner-values. = = = Use Pu:û(v) = (1, v, v² ) Mq (Pu:û(0), Pu:û (0.5), Pu:û(1)) with Ma = 1 0 0 -3 4-1 2 4 2 (Pv:ô as well as pu: (u) = (1, u) M₁ (pv:v (0), P: (1)) with M₁ = = (19) 0 to formulate p(u, v) using the "geometric input" G with G = = (P(0,0%) p(0,0) p(0,0.5) p(0,1) ) = ( 0.39 0.8 0.1 0.3 0.2 0.9 0.8 p(1,0) p(1, 0.5) p(1, 1) See the figure below for (left) a selection of iso-lines of p(u, v) and (right) a 3D rendering of p(u, v) as a height surface…

O Functions Composition of two functions: Domain and... Two functions ƒ and g are defined in the figure below. 76 2 8 5 7 8 19 8 9 Domain of f Range of f Domain of g Range of g 3/5 Anthony Find the domain and range of the composition g.f. Write your answers in set notation. (a) Domain of gof: ☐ (b) Range of gof: ☐ Х Explanation Check 0,0,... Español لكا ©2025 McGraw Hill LLC. All Rights Reserved Torms of lico Privacy Contor Accessibility.

Chapter 5 Solutions

College Algebra Essentials (5th Edition)

Ch. 5.1 - Consider the system: {2x3y=42x+y=4. Determine if...Ch. 5.1 - Prob. 2CP Ch. 5.1 - Prob. 3CP Ch. 5.1 - Prob. 4CP Ch. 5.1 - Solve the system: {5x2y=410x+4y=7.Ch. 5.1 - Prob. 6CP Ch. 5.1 - Prob. 7CP Ch. 5.1 - Prob. 8CP Ch. 5.1 - Prob. 1CVC Ch. 5.1 - Prob. 2CVC

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The graph of the solution set of the inequality − 2 ≤ x < 5 or indicate that the system has no solution.

Solution Summary: The author explains how to graph a linear inequality in two variables.

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To sketch: The graph of the solution set of the inequality −2≤x<5 or indicate that the system has no solution.

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Chapter 5 Solutions

The graph of the solution set of the inequality − 2 ≤ x &lt; 5 or indicate that the system has no solution.

Solution Summary: The author explains how to graph a linear inequality in two variables.

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To sketch: The graph of the solution set of the inequality −2≤x<5 or indicate that the system has no solution.

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Chapter 5 Solutions

The graph of the solution set of the inequality − 2 ≤ x < 5 or indicate that the system has no solution.