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Unformatted text preview: - View Panel Style - Student Details Numeric 8-2 Problem Set: Module Eight Score: Duration: 50/50.0 Lessons & Assignments Email: Completed: Student ID: To Be Reviewed: 0 Passed: 0 0 Active: 0 Page: 1 of 1 Rows: 1 - 12 of 12 - 0 Question Q1 4/4.0 View Original Response Unfiltered Response Solve the system of equations by any method. −2x + 5y= −42 7x + 2y= 30 Enter the exact answer as an ordered pair, (x, y) . If there is no solution, enter NS. If there is an infinite number of solutions, enter the gener solution as an ordered pair in terms of x . Include a multiplication sign between symbols. For example, a * x . Auto graded Your response Correct response (6, -6) (6,-6) Grade: 1/1.0 Total grade: 1.0×1/1 = 100% Feedback: Adding these equations as presented will not eliminate a variable. However, we see that the f equation has −2x in it and the second equation has 7x. So if we multiply the first equation by and the second equation by 2, the x -terms will add to zero. 7 (−2x + 5y)= 7 (−42) −14x + 35y= −294 2 (7x + 2y)= 2 (30) 14x + 4y= 60 Multiply both sides by 7. Use the distributive property. Multiply both sides by 2. Use the distributive property. Now, let’s add them. Question −14x + 35y= −294 14x + 4y= 60 39y= −234 y= −6 For the last step, we substitute y = −6 into one of the original equations and solve for x . −2x + 5y= −42 −2x + 5 (−6)= −42 −2x − 30= −42 −2x= −12 x= 6 Our solution is the ordered pair (6, −6). Check the solution in the original second equation. 7x + 2y= 30 7 (6) + 2 (−6)= 30 42 − 12= 30 Q2 4/4.0 View Original Response True Unfiltered Response Solve the system of equations by any method. 5x + 9y = 16 x + 2y = 4 Enter the exact answer as an ordered pair, (x, y) . If there is no solution, enter NS. If there is an infinite number of solutions, enter the gener solution as an ordered pair in terms of x . Include a multiplication sign between symbols. For example, a * x . Your response Correct response (-4, 4) (-4,4) Question Auto graded Grade: 1/1.0 Total grade: 1.0×1/1 = 100% Feedback: In this case we use substitution. First, we will solve the second equation for x . x + 2y = 4 x = −2y + 4 Now we can substitute the expression −2y + 4 for x in the first equation. 5x + 9y = 16 5 (−2y + 4) + 9y = 16 −10y + 20 + 9y = 16 −y = −4 y = 4 Now, we substitute y = 4 into the second equation and solve for x . x + 2 (4) = 4 x + 8 = 4 x = −4 Our solution is (−4, 4). Check the solution by substituting (−4, 4) into both equations. 5x + 9y = 16 5 (−4) + 9 (4) = 16 −20 + 36 = 16 True x + 2y = 4 (−4) + 2 (4) = 4 −4 + 8 = 4 True Question Q3 4/4.0 View Original Response Unfiltered Response Solve the system of equations by any method. −x + 2y = −1 5x − 10y= 6 Enter the exact answer as an ordered pair, (x, y) . If there is no solution, enter NS. If there is an infinite number of solutions, enter the gener solution as an ordered pair in terms of x . Include a multiplication sign between symbols. For example, a * x . Auto graded Your response Correct response NS NS Grade: 1/1.0 Total grade: 1.0×1/1 = 100% Feedback: Adding these equations as presented will not eliminate a variable. However, we see that the f equation has −x in it and the second equation has 5x. So if we multiply the first equation by 5, x -terms will add to zero. −x + 2y = −1 5 (−x + 2y)= 5 (−1) −5x + 10y= −5 Multiply both sides by 5. Use the distributive property. Now, let’s add them. −5x + 10y= −5 5x − 10y= 6 0≠ 1 This statement is a contradiction. Therefore, the system has no solution. View Original Response Unfiltered Response Question Q4 4/4.0 Solve the system of equations by any method. −2x + 4y= 6 x − 2y = −3 Enter the exact answer as an ordered pair, (x, y) . If there is no solution, enter NS. If there is an infinite number of solutions, enter the gener solution as an ordered pair in terms of x . Include a multiplication sign between symbols. For example, a * x . Auto graded Your response Correct response (x, 3/2+x/2) (x, 1/2*x+3/2) Grade: 1/1.0 Total grade: 1.0×1/1 = 100% Feedback: With the addition method, we want to eliminate one of the variables by adding the equations. In t case, let’s focus on eliminating x . If we multiply both sides of the second equation by 2, then will be able to eliminate the x -variable. x − 2y = −3 2 (x − 2y)= 2 (−3) 2x − 4y= −6 Multiply both sides by 2. Use the distributive property. Now, add the equations. −2x + 4y= 6 2x − 4y= −6 0= 0 We can see that there will be an infinite number of solutions that satisfy both equations. Solve the second equation for y . Question x − 2y = −3 −2y= −x − 3 y= So the general solution is (x, Q5 4/4.0 View Original Response 1 2 x + 3 2 ) 1 2 x + 3 2 . Unfiltered Response A fast-food restaurant has a cost of production C (x) = 14x + 144 and a revenue functio R (x) = 6x . When does the company start to turn a profit? Enter the exact answer. If there is no solution, enter NS. If there is an infinite number of solutions, enter IS. x = Auto graded Your response Correct response NS NS Grade: 1/1.0 Total grade: 1.0×1/1 = 100% Feedback: Write the system of equations using y to replace function notation. y = 14x + 144 y = 6x Substitute the expression 14x + 144 from the first equation into the second equation and solve x. 14x + 144 = 6x 8x = −144 x = −18 Since x is negative, we conclude the company never turns a profit. View Original Response Unfiltered Response Question Q6 4/4.0 Use a system of linear equations with two variables and two equations to solve. A number is 8 more than another number. Twice the sum of the two numbers is 40 . Find the tw numbers. Enter the numbers in increasing order. First Number: Auto graded Your response Correct response 6 6 Your response Correct response 14 14 Grade: 1/1.0 Second Number: Auto graded Grade: 1/1.0 Total grade: 1.0×1/2 + 1.0×1/2 = 50% + 50% Feedback: Use the given information to create two equations with two variables. Let x be one number, and y be the other. y = x + 8 (1) 2 (x + y) = 40 x + y = 20 (2) Solve the system using substitution. x + (x + 8) = 20 2x = 12 x = 6 For the last step, we substitute x = 6 into one of the original equations and solve for y . y = x + 8 y = 6 + 8 y = 14 Question Therefore, the two numbers are 6 and 14 . Q7 4/4.0 View Original Response Unfiltered Response Use a system of linear equations with two variables and two equations to solve. students enrolled in a freshman-level chemistry class. By the end of the semester, 7 time the number of students passed as failed. Find the number of students who passed, and th number of students who failed. 328 Passing Students: Auto graded Your response Correct response 287 287 Grade: 1/1.0 Failing Students: Auto graded Your response Correct response 41 41 Grade: 1/1.0 Total grade: 1.0×1/2 + 1.0×1/2 = 50% + 50% Feedback: Use the given information to create two equations with two variables. Let x be the number of failing students, and y be the number of passing students. y = 7x x + y = 328 (1) (2) Solve the system using substitution. x + (7x) = 328 8x = 328 x = 41 For the last step, we substitute x = 41 into one of the original equations and solve for y . Question y = 7x y = 7 (41) y = 287 Therefore, 287 students passed and 41 students failed. Question Q8 4/4.0 View Original Response Unfiltered Response Determine whether the given ordered triple is a solution to the system of equations. x − y = 0 x − z = 10 and (8, 8, −2) x − y + z = −2 Your response Correct response Yes, it is a solution. Yes, it is a solution. Feedback: Correct. Auto graded Grade: 1/1.0 Total grade: 1.0×1/1 = 100% Feedback: Check each equation by substituting in the values of the ordered triple (8, 8, −2) for x , y , and z. x y = 0 (8) − (8) = 0 True x − z = 10 (8) − (−2) = 10 8 + 2 = 10 True x − y + z = −2 (8) − (8) + (−2) = −2 True The ordered triple (8, 8, −2) satisfies all of the equations, so it is a solution to the system. Q9 5/5.0 View Original Response Unfiltered Response Question Solve the system of equations by any method. 3x − 4y + 2z = −17 (1) 2x + 4y + z = 21 (2) 2x + 3y + 5z = 28 (3) Enter the exact answer as an ordered triple, (x, y, z). Hint: There are multiple ways to solve this system of equations. A strategy is to eliminate one of the variables and end up with 2 equations in 2 variables. One way to do that is to begin with equation 2 to get z = 21 − 2 x − 4 y. Then substitute 21 − 2 x − 4 y for z in equations 1 and 3. You now have 2 equations (the modified equations 1 and 3) in 2 variables (x and y ). Solve this smaller system for x and y and then use z = 21 − 2 x − 4 y to calculate z. Auto graded Your response Correct response (-1, 5, 3) (-1,5,3) Grade: 1/1.0 Total grade: 1.0×1/1 = 100% Feedback: First, we will solve the second equation for z. 2x + 4y + z = 21 z = −2x − 4y + 21 (2) (4) Now we can substitute the expression −2x − 4y + 21 for z into the other equations. Question 3x − 4y + 2z = −17 (1) 3x − 4y + 2 (−2x − 4y + 21) = −17 3x − 4y − 4x − 8y + 42 = −17 −x − 12y + 42 = −17 x = −12y + 59 2x + 3y + 5z = 28 (5) (3) 2x + 3y + 5 (−2x − 4y + 21) = 28 2x + 3y − 10x − 20y + 105 = 28 −8x − 17y + 105 = 28 8x + 17y = 77 (6) Now, we substitute equation (5) into equation (6) and solve for y . 8 (−12y + 59) + 17y = 77 −96y + 472 + 17y = 77 −79y = −395 y = 5 Now, we substitute y = 5 into equation (5) and solve for x . x = −12 (5) + 59 = −60 + 59 = −1 Now, we substitute x = −1, y = 5 into equation (4) and solve for z. z = −2 (−1) − 4 (5) + 21 = 2 − 20 + 21 = 3 Therefore, our solution is (−1, 5, 3) . Check the solution by substituting (−1, 5, 3) into all equations. Question 3x − 4y + 2z = −17 3 (−1) − 4 (5) + 2 (3) = −17 −3 − 20 + 6 = −17 True 2x + 4y + z = 21 2 (−1) + 4 (5) + (3) = 21 −2 + 20 + 3 = 21 True 2x + 3y + 5z = 28 2 (−1) + 3 (5) + 5 (3) = 28 −2 + 15 + 15 = 28 True Question Q10 4/4.0 View Original Response Unfiltered Response Use the matrices below to perform the indicated operation if possible. −9 8 B = [ −4 ,C 0 10 7 1 = [ 3 If the operation is undefined, enter NA. Hint: To create a matrix in Mobius, click on the icon that has three rows of three dots - it next to the infinity symbol. Specify the appropriate number of rows (horizontal) and column (vertical). 2C + B = Your response −9 28 10 5 [ Auto graded Correct response −9 28 10 5 [ Grade: 1/1.0 Total grade: 1.0×1/1 = 100% Feedback: First perform the scalar multiplication for 2C . Multiply each entry in C by 2. 0 10 7 1 2C = 2 [ 0 20 14 2 = [ Now perform the addition for 2C + B . Add the corresponding entries. 0 20 2C + B = [ = [ −9 8 −4 3 + [ 14 2 −9 28 10 5 Question Q11 4/4.0 View Original Response Unfiltered Response Use the matrices below to perform matrix multiplication. 4 6 5 B = [ −8 0 ,C ⎡ 5 = ⎢ −3 12 ⎣ 6 10 ⎤ 7⎥ 8 ⎦ If the operation is undefined, enter NA. Hint: To create a matrix in Mobius, click on the icon that has three rows of three dots - it next to the infinity symbol. Specify the appropriate number of rows (horizontal) and column (vertical). BC = Your response 32 122 32 16 [ Auto graded Correct response 32 122 32 16 [ Grade: 1/1.0 Total grade: 1.0×1/1 = 100% Feedback: The dimensions of B are 2 × 3 and the dimensions of C are 3 × 2 . The inner dimensions match the product is defined and will be a 2 × 2 matrix. 4 6 5 BC = [ ⎡ 5 ⎢ −3 −8 0 12 ⎣ 6 10 ⎤ 7⎥ 8 ⎦ 4 (5) + 6 (−3) + 5 (6) 4 (10) + 6 (7) + 5 (8) −8 (5) + 0 (−3) + 12 (6) −8 (10) + 0 (7) + 12 (8) = [ 32 122 32 16 = [ Question Q12 5/5.0 View Original Response Unfiltered Response Use the matrices below to perform matrix multiplication. 2 7 B = [ −10 0 4 −5 , D = ⎢8 5 ⎡ 3 11 ⎣ 0 7 12 ⎤ 1⎥ −10 ⎦ If the operation is undefined, enter NA. Hint: To create a matrix in Mobius, click on the icon that has three rows of three dots - it next to the infinity symbol. Specify the appropriate number of rows (horizontal) and column (vertical). BD = Your response Correct response 64 46 1 −40 127 −230 [ Auto graded 64 46 1 −40 127 −230 [ Grade: 1/1.0 Total grade: 1.0×1/1 = 100% Feedback: The dimensions of B are 2 × 3 and the dimensions of D are 3 × 3 . The inner dimensions match the product is defined and will be a 2 × 3 matrix. 2 7 3 BD = [ −10 0 11 4 −5 ⎢8 5 ⎡ ⎣ 0 7 12 ⎤ 1⎥ −10 ⎦ 2 (4) + 7 (8) + 3 (0) 2 (−5) + 7 (5) + 3 (7) 2 (12) + 7 (1) + 3 (−1 −10 (4) + 0 (8) + 11 (0) −10 (−5) + 0 (5) + 11 (7) −10 (12) + 0 (1) + 11 (−1 = [ 64 46 1 −40 127 −230 = [ ...
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