Fall 016 P Review nswer Section MULTIPLE HOIE 1. NS: The relationship is proportional, because the data points lie on a line, and the line goes through the origin. To find the constant of proportionality, divide the number of bags by the number of hours. ags Hours = 5 1 = 10 = 0 4 = 5 5 = 5 Thus, the constant of proportionality is 5. To find the constant of proportionality, divide the number of bags by the number of hours. To find the constant of proportionality, divide the number of bags by the number of hours. The data lie on a line, and the line goes through the origin, so the relationship is proportional. NT: NT.SS.MTH.10.7.7.RP..b NT.SS.MTH.10.7.7.RP..a ST: TEKS.7.4. KEY: proportional relationship constant of proportionality unit rate OK: OK 1 1
. NS: Use the given points to find the slope of the line. change in y 11 m = = ( 4 ) = 7 change in x 0 Using the points (x, y) and (0, 4), you can calculate the slope again and set it equal to 7 to get y ( 4) x 0 = 7 as an equation of the line. The change in y should involve the y-coordinate of the point (0, 4), and the change in x should involve the x-coordinate of the point (0, 4). The change in y should involve the y-coordinate of the point (0, 4), and the change in x should involve the x-coordinate of the point (0, 4). lso, the slope of the line is not 7. The slope of the line is not 7. NT: NT.SS.MTH.10.8.8.EE.6 ST: TEKS.7.7 KEY: deriving an equation of a line slope y-intercept OK: OK 1 3. NS: ST: TEKS.7.7 4. NS: NT: NT.SS.MTH.10.8.8.F.3 ST: TEKS.7.7 OK: OK 1 5. NS: ST: TEKS.7.7 6. NS: NT: NT.SS.MTH.10.8.8.NS. ST: TEKS.8.. OK: OK 1 7. NS: NT: NT.SS.MTH.10.8.8.NS. ST: TEKS.8.. OK: OK 1 8. NS: REF: 910fe1e4-6ab-11e0-9c90-001185f0dea NT: NT.SS.MTH.10.8.8.NS. ST: TEKS.8.. TOP: Identifying and Graphing Irrational Numbers KEY: square root approximate root value OK: OK
9. NS: 3 1.7 and 6.4 = 1., so 3 > 6. 3 < 6, but 3 is not less than half of 6. 6 is not the same as 6, so 6 Use decimal approximations of 3 and 3. 6 to determine the relationship. NT: NT.SS.MTH.10.8.8.NS. ST: TEKS.8.. KEY: comparing irrational numbers OK: OK 1 10. NS: 3.6 = 1.96 and 3.7 = 13.69, so 3.6 < 13 < 3.7 and 3.6 < 13 < 3.7. 3.3 = 10.89 and 13 > 10.89, so 13 > 3.3. 3.5 = 1.5 and 13 > 1.5, so 13 > 3.5. 3.8 = 14.44 and 13 < 14.44, so 13 < 3.8. NT: NT.SS.MTH.10.8.8.NS. ST: TEKS.8.. KEY: locating an irrational number on a number line OK: OK 1 11. NS: 5 3 = 15 11. 5 3 3 is not the same as 5 1.7. 5 3 is not the same as 5 3 = 15 3.9. 5 3 is not the same as 3 5 = 43 15.6. NT: NT.SS.MTH.10.8.8.NS. ST: TEKS.8.. KEY: approximating an irrational number OK: OK 1 1. NS: NT: NT.SS.MTH.10.8.8.EE.5 ST: TEKS.8.4. KEY: proportional relationships OK: OK 1 3
13. NS: The line for rand goes through the points Ê Ë Á 0, 0 ˆ and Ê ËÁ 4, 9 ˆ. ivide the difference of the vertical coordinates by the difference of the horizontal coordinates to find the unit rate. 9 0 4 0 = 9 4 =.5 Since the vertical coordinates represent cost in dollars and the horizontal coordinates represent length in meters, the unit rate for rand is $.5 per meter. The line for rand goes through the points Ê Ë Á 0, 0 ˆ and Ê ËÁ, 3 ˆ. ivide the difference of the vertical coordinates by the difference of the horizontal coordinates to find the unit rate. 3 0 0 = 3 = 1.5 Since the vertical coordinates represent cost in dollars and the horizontal coordinates represent length in meters, the unit rate for rand is $1.50 per meter. rand is more expensive than rand because the unit price for rand is greater. rand does not cost $1.75 per meter, and rand does not cost $.5 per meter. rand does not cost $.00 per meter, and rand does not cost $1.50 per meter. The unit rates in this case are dollars per meter, not meters per dollar. NT: NT.SS.MTH.10.8.8.EE.5 ST: TEKS.8.4. KEY: proportional relationship unit rate slope interpreting a graph comparing proportional relationships OK: OK 14. NS: NT: NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.4. OK: OK 1 15. NS: NT: NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.4. OK: OK 1 16. NS: NT: NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.4. OK: OK 17. NS: NT: NT.SS.MTH.10.8.8.EE.5 ST: TEKS.8.5. KEY: proportional relationships OK: OK 1 18. NS: NT: NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.5. OK: OK 1 19. NS: NT: NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.5. OK: OK 0. NS: NT: NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.5. OK: OK 1 1. NS: NT: NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.5. OK: OK. NS: NT: NT.SS.MTH.10.8.8.F.1 NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.5. OK: OK 3. NS: NT: NT.SS.MTH.10.8.8.F.1 NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.5. OK: OK 4
4. NS: Use the first two ordered pairs in the table to calculate the rate of change: 8 9 4 = 1 rate of change is 0.5 block per minute. = 0.5. So, the is the number of minutes per block. The rate of change is not positive. is the number of minutes per block. lso, the rate of change is not positive. NT: NT.SS.MTH.10.8.8.F.4 NT.SS.MTH.10.K-1.MP.4 ST: TEKS.8.5. KEY: linear functions rate of change analyzing tables OK: OK 1 5. NS: NT: NT.SS.MTH.10.8.8.SP. ST: TEKS.8.5. OK: OK 6. NS: NT: NT.SS.MTH.10.8.8.SP. ST: TEKS.8.5. OK: OK 7. NS: NT: NT.SS.MTH.10.8.8.F.1 ST: TEKS.8.5.G OK: OK 8. NS: NT: NT.SS.MTH.10.8.8.F.1 ST: TEKS.8.5.G OK: OK 9. NS: Each x-coordinate of the graph is assigned exactly one y-coordinate. So, the graph represents a function. function assigns to each input exactly one output. It does not matter whether one or more inputs have the same output. graph that has the same y-coordinates for two different x-coordinates can still represent y as a function of x. function assigns to each input exactly one output. NT: NT.SS.MTH.10.8.8.F.1 ST: TEKS.8.5.G KEY: definition of a function analyzing graphs OK: OK 1 5
30. NS: For x > 0, each x-coordinate of the graph is assigned two y-coordinates. So, the graph does not represent a function. For y to be a function of x, the graph must show exactly one y-coordinate assigned to each x-coordinate. For y to be a function of x, the graph must show exactly one y-coordinate assigned to each x-coordinate. For y to be a function of x, the graph must show exactly one y-coordinate assigned to each x-coordinate. NT: NT.SS.MTH.10.8.8.F.1 ST: TEKS.8.5.G KEY: definition of a function analyzing graphs OK: OK 1 31. NS: function assigns exactly one output to each input. ifferent inputs can have the same output, but no input can have different outputs. Review the definition of a function. Review the definition of a function. Review the definition of a function. NT: NT.SS.MTH.10.8.8.F.1 ST: TEKS.8.5.G KEY: functions OK: OK 1 3. NS: NT: NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.5.G OK: OK 1 33. NS: NT: NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.5.G OK: OK 1 34. NS: The rate of change is the average number of gallons of water draining from the bathtub per unit time, which is 3 gallons per minute. This number is negative because the amount of water is decreasing. The initial value is the number of gallons of water in the bathtub before it drains, which is 40 gallons. 40 is the number of gallons initially in the bathtub, and 3 is the average number of gallons the bathtub is draining per minute. positive rate of change means that water is going into the bathtub. 40 is the initial number of gallons in the bathtub, and 3 is the average number of gallons the bathtub is draining per minute. NT: NT.SS.MTH.10.8.8.F.4 NT.SS.MTH.10.K-1.MP.4 ST: TEKS.8.5.G KEY: linear functions rate of change initial value OK: OK 1 6
35. NS: First, find the function s rate of change using the points (, 75) and (4, 115). 115 75 = 40 4 = 0 Then, find the initial value using the equation y = mx + b, the rate of change, and the point (, 75). 75 = 0 ( ) + b 75 = 40 + b 35 = b So, Vincent initially has $35. The initial value is not necessarily 0. Find the function y = mx + b with slope m and y-intercept b. 0 is the function s rate of change, not its initial value. The initial value is the amount at week 0, not week. NT: NT.SS.MTH.10.8.8.F.4 NT.SS.MTH.10.K-1.MP.4 ST: TEKS.8.5.G KEY: linear functions initial value analyzing tables OK: OK 36. NS: First, find the function s rate of change using the first two ordered pairs in the table. 5 19 = 6 6 3 3 = Then, find the function s initial value using the equation s = mt + b, the slope, and one ordered pair from the table. 19 = ( 3) + b 19 = 6 + b 13 = b So, the function represented by the table is s = t + 13. The rate of change is not negative. The initial value is not. The initial value is not 0. NT: NT.SS.MTH.10.8.8.F.4 NT.SS.MTH.10.K-1.MP.4 ST: TEKS.8.5.G KEY: constructing linear functions analyzing tables OK: OK 1 7
37. NS: NT: NT.SS.MTH.10.8.8.F.3 ST: TEKS.8.5.I OK: OK 1 38. NS: The equations y = 4 x + 1 and y = 3 are both in the form of y = mx + b, and the graphs of these 5 equations are straight lines. So, these equations represent linear functions. The equation x = 5, although its graph is linear, does not represent a function (because the x-value 3 is paired with infinitely many y-values), and its equation cannot be put in the form y = mx + b. So, the equation for the shorter leg does not represent a linear function. The equation for the hypotenuse does represent a linear function. The equation for the longer leg does represent a linear function. One of the equations does not represent a linear function. NT: NT.SS.MTH.10.8.8.F.3 ST: TEKS.8.5.I KEY: linear functions OK: OK 39. NS: NT: NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.5.I OK: OK 1 40. NS: NT: NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.5.I OK: OK 41. NS: NT: NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.5.I OK: OK 4. NS: NT: NT.SS.MTH.10.8.8.F.1 NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.5.I OK: OK 43. NS: NT: NT.SS.MTH.10.8.8.F.4 ST: TEKS.8.5.I OK: OK 3 44. NS: The line includes the points (0, 0) and (, 50), so the slope of the line is m = 50 0 0 = 5. So, the rate of change is 5 miles per gallon of gasoline. To find the rate of change, determine the slope of the line. To find the rate of change, determine the slope of the line. To find the rate of change, determine the slope of the line. NT: NT.SS.MTH.10.8.8.F.4 NT.SS.MTH.10.K-1.MP.4 ST: TEKS.8.5.I KEY: linear functions rate of change analyzing graphs OK: OK 1 45. NS: NT: NT.SS.MTH.10.8.8.EE.8 ST: TEKS.8.9 OK: OK 3 8
46. NS: For an ordered pair to be a solution of a system of two equations, it must satisfy both equations. For an ordered pair to be a solution of a system of two equations, it must satisfy both equations. For an ordered pair to be a solution of a system of two equations, it must satisfy both equations. NT: NT.SS.MTH.10.8.8.EE.8.a ST: TEKS.8.9 KEY: system of equations solution of a system OK: OK 1 47. NS: The ordered pair ( 5, 3) does not satisfy the system of equations because the solution satisfies only the equation y = 3 x. The ordered pair has to satisfy both equations to be a solution of the system. 5 t least one of the equations is not satisfied by the ordered pair ( 5, 3). The ordered pair ( 5, 3) does not satisfy the equation y = x 1. t least one of the equations is satisfied by the ordered pair ( 5, 3). NT: NT.SS.MTH.10.8.8.EE.8.a ST: TEKS.8.9 KEY: system of equations solution of a system OK: OK 1 9
48. NS: The equation of the line that passes through the points ( 5, 6) and ( 3, ) is y = 4x + 14. Solve the Ï y = 4x + 14 system of equations Ì Ô by substituting x 4 for y in the equation y = 4x + 14. ÓÔ y = x 4 x 4 = 4x + 14 x x 4 = 4x x + 14 4 14 = 3x 18 3 = 3x 3 6 = x Substitute 6 for x in the equation y = x 4. y = 6 4 = 10 The lines intersect at the point ( 6, 10). The line y = x + 4 does not pass through the point ( 5, 6). The line that passes through the points ( 5, 6) and ( 3, ) does not pass through the point (, ). The line y = x + 4 does not pass through the point ( 3, ). NT: NT.SS.MTH.10.8.8.EE.8.c ST: TEKS.8.9 KEY: system of equations intersection of lines OK: OK 1 10