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PreCalculusChapter 7 Practice TestName:Note the exponents of 2 in the equations. Any exponent greater than 1 will identify a system asnon‐linear. This system is non‐linear.Answer B.Note: You can download a Microsoft Excel file from the following linkthat will allow you to explore systems of 3 equations. Give it a , you would begin solving a set of three equations by selecting pairs of equations andeliminating the same variable in each pair. I’ll start this one by eliminating the variable .Add 1st and 3rd equationsAdd 2nd and 3rd equations1329342212We got lucky here because we were able to eliminate two variables at the same time by addingthe first and third equations. Normally, this would not happen, and you would have to solve theset of two simultaneous equations which result. Let’s continue.Solve for :Then, solve for :242222 2422Then, solve for :1221122124281134Finally, test your results in one of the original equations, but not the one used to solve for .2Second equation:Solution:Page 1,,439
PreCalculusChapter 7 Practice TestName:For problems 3 to 5, we need only write the form of the decomposition. We do not need to solve thedecomposition for the constants , and .This rational function has no repeated linear factors in the denominator, so the decomposition isstraightforward:2636This rational function has a repeated linear factor7 , so the decomposition must includeeach integral exponent of7 , up to the exponent of the term in the rational function 2 :3517This rational function has a quadratic function in the denominator so the decomposition musttake this into account:22Page 254
PreCalculusChapter 7 Practice TestName:Write the form of the decomposition:Multiply both sides by15 :Simplify:15391539Write the simultaneous equations and solve them:15Solve for :5Then, solve for :15539424156So, the partial fraction decomposition is:151Page 33939515515
PreCalculusChapter 7 Practice TestName:Write the form of the decomposition:3 :Multiply both sides byExpand and simplify:77367363636393639Write the simultaneous equations and solve them:0Solve for :9637Then, solve for :364936Then, solve for :0606 424So, the partial fraction decomposition is:7Page 436337312477
PreCalculusChapter 7 Practice TestName:Write the form of the decomposition:1Multiply both sides by1 :Expand and simplify:1021021021Write the simultaneous equations and solve them:0102form the 2nd equation with some help from the 3rd equation.Now, let’s eliminate102212Then,Solve for :2Then, solve for :12003412So, the partial fraction decomposition is:101Page 5Then, solve for :2102421
PreCalculusChapter 7 Practice TestName:Write the form of the decomposition:4 :Multiply both sides by3Expand and simplify:313114444Write the simultaneous equations and solve them:01Solve for :4343So, the partial fraction decomposition is:14Page 614 1434Then, solve for :34 041114
PreCalculusChapter 7 Practice Test6881671005216602, 52, we get:When26, so4 2, 4 is a solution1 5, 1 is a solution5, we get:When56, so,So, our solutions are:,15,561515155656155607807, 8When7, we get:15When77, 8 is a solution7 8, 7 is a solution8, we get:158So, our solutions are:Page 78 ,,,Name:
PreCalculusChapter 7 Practice Test266Name:41Let’s use the Addition (i.e., Elimination) Method266multiply by 141multiply by2661412555, we get:When541254116 45, 4 and 5, 4 are solutions5, we get:When541254116 5, 4 and45, 4 are solutionsSo, the entire solution set is:,,,,,,,Note: you can graph multiple conic sections (and lines) using the Algebra App, available php. On the opening page, click on the “Conic Sections” button in the “More Algebra” column. On the left hand side of the Conic Sections page, click on the “Graph Multiple Equations”button You may enter up to 4 equations, using either General Form or Standard Form for each.Page 8
PreCalculusChapter 7 Practice Test294Name:17Let’s use the Addition (i.e., Elimination) Method429multiply by 117multiply by29141744212120602, 62, we get:When174421781725 2, 5 and52, 5 are solutions6, we get:When1744 61724177Result: no real solutions whenSo, the entire solution set is:,Page 9,,6
PreCalculusChapter 7 Practice TestLet’s start by solving this for .2632To graph this, do the following: Graph the line:3. The line will be solid because thereis an “equal sign" included in theinequality. Fill in the portion of the graphabove the line because of the“greater than” portion of theinequality.To graph this inequality, do the following: Graph the curve:2 . Some points on the curve:2, 0.25 , 0, 1 , 2, 4 The curve will be solid becausethere is an “equal sign" included inthe inequality. Fill in the portion of the graphbelow the curve because of the“less than” portion of theinequality.Page 10Name:
PreCalculusChapter 7 Practice TestName:is the interior of a circle withcenter,and radius .To graph this inequality, do the following: Graph the circle:49. Some points on the curve:0, 7 , 0, 7 , 7, 0 , 7, 0 The curve will be solid because there is an“equal sign" included in the inequality. Fill in the interior of the circle because ofthe “less than” portion of the inequality.(orange and green areas) Graph the line:4. The line will be dashed because there is no“equal sign" included in the inequality. Fill in the portion of the graph below theline because of the “less than” portion ofthe inequality.(violet and green areas)23. Graph the line: The line will be dashed because there is no“equal sign" included in the inequality. Fill in the portion of the graph above theline because of the “greater than” portionof the inequality.The green area is the area of intersection of the two linear inequalities.Page 11
PreCalculusChapter 7 Practice TestName:(orange and green areas) Graph the parabola:. Some points on the curve:0, 0 , 2, 4 , 2, 4 The curve will be dashed because there isno “equal sign" included in the inequality. Fill in the portion of the graph above thecurve because of the “greater than”portion of the inequality.(violet and green areas) Put this in slope intercept form10660660 105103. Graph the line:10 The line will be solid because there is an“equal sign" included in the inequality. Fill in the portion of the graph below the line because of the “less than” portion of theinequality.The green area (and contiguous magenta line) is the area of intersection of the two linearinequalities.Page 12
PreCalculusChapter 7 Practice TestName:(orange and green areas) Graph the circle:49. Some points on the curve:0, 7 , 0, 7 , 7, 0 , 7, 0 The curve will be solid because there isan “equal sign" included in theinequality. Fill in the interior of the circle because ofthe “less than” portion of the inequality.(violet and green areas) Put this in “” form0 Graph the parabola:. Some points on the curve:0, 0 , 2, 4 , 2, 4 The curve will be dashed because there is no “equal sign" included in the inequality. Fill in the portion of the graph above the curve because of the “greater than” portion ofthe inequality.The green area (and contiguous orange curve) is the area of intersection of the two linearinequalities.Page 13
PreCalculusChapter 7 Practice TestName:Let’s use the Addition (i.e., Elimination) Method520273273Since the ship is in Q1, we must have positivepositive . So,3. Then,5and a20205 3452025 5 in Q17So,,is the exact location144Let’s use the Substitution Method77771441441441609016, 916, we get:When7When169 9, we get:7916 9, 16 is a solutionSo, the two numbers are:Page 1416, 9 is a solutionand
PreCalculusChapter 7 Practice Test37Name:5Let’s use the Substitution Method55102102512563737006106, 16, we get:When6When51 6, 1 is a solution6 1, 6 is a solution1, we get:15So, the two numbers are:2242and90with the dimensions of the rectangle being:Let’s use the Substitution Method21212190902190156006, 15When6, we get:21When615 15, we get:21156 15, 6 is a solutionSo, the dimensions are:Page 156, 15 is a solutionft. byft.by
PreCalculus Chapter 7 Practice Test Name:_ Page 10 Let's start by solving this for U. E
