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MCCC MAT 141 - MAT141 Review Chapters 2 & 3

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MAT141 Review Chapters 2 & 3 Hawkes1. Find the slope-intercept form of the equation of the line through the point– , ,3 4 b gperpendicular to the graph ofy x 9102.[A] y x 910713[B] x y 910713[C] x y 119713[D] y x 1197132. A sidewalk was built around a rectangular garden. If the area of the sidewalk is3602m , find x.(Note: Figure not drawn to scale.)3 6x 3xx  6x[A] 2 m [B] 3 m [C] 9 m [D] 6 m3. The motion of the leg of a robot can be modeled by tl232 where l is the length of the the robot’s leg in feet and t is the number of seconds required for one complete swing. What is the length of the robot’s leg if the time required for one swing is 0.7 second?[A] 0.29 ft [B] 0.40 ft [C] 0.66 ft [D] 0.52 ft4. Solve. x x  81 9 =[A] –17, 19 [B] 19 [C] –17 [D] 05. Find the equation of the vertical line passing through (6, -2).1MAT141 Review Chapters 2 & 3 Hawkes[A] x =6 [B] x = -2 [C] y = 6 [D] y = -26. Find the slope-intercept form of the equation of the line through the point5 3, , b gparallel to theline 2 6 4x y  –.[A] y x  3314[B] y x  13314[C] y x 13143[D] y x  131437. Determine whether the linesL L1 2 and passing through the pair of points are parallel, perpendicular, or neither.L15 3 7 4: , , ,b gb gL28 8 7 10: , , ,b gb g[A] Parallel [B] Perpendicular [C] Neither8. Identify the x- and y-intercepts of the graph of the equation yx335.[A] x-intercept: 518, 0FHGIKJy-intercept: 016, FHGIKJ[B] x-intercept: FHGIKJ185, 0y-intercept: 0 6, b g[C] x-intercept: FHGIKJ518, 0y-intercept: 016, FHGIKJ[D] x-intercept: 185, 0FHGIKJy-intercept: 0 6, b g9. Solve x  32b g = 16.2MAT141 Review Chapters 2 & 3 Hawkes[A]  7, 1[B] 7,  1[C]  7,  1[D] 7, 110. Solve 6 2x 5 3x .[A] 1 [B] 0 [C] 8 [D] No solution11. Dominic bowled 2 games, scoring 188 and 175. What must he bowl in the third game in order to have a 189 average for these 3 games?[A] 184 [B] 200[C] 204 [D] It is impossible to average 189 for these 3 games.12. Use the Quadratic Formula to solve 2 6 2 02x x  .[A] – 3 52 2[B] 3 52 2[C] 3 52[D] – 3 5213. After being shot into the air, the height of a rocket is h t t 96 162 where h is the height above the ground, in feet, after t seconds. After how many seconds will the rocket be at a height of 128 feet?[A] t t 2 4 and [B] t 6[C] t t 4 1 and [D] t 214. Find the slope-intercept form of the equation of the line that passes through the given point and has the indicated slope: 0 9 7, – , – b gm [A] y x 7 9[B] x y 7 9[C] y x 7 9[D] y x17+ 915. Solve 347 4 9x   [A]  8316, [B] 80316, [C] 16 9, [D]  32803, 16. Solve 1 22x x+ = –33MAT141 Review Chapters 2 & 3 Hawkes[A] 0[B] 231, [C] 1[D] 130, 17. Solve the inequality –10  2x + 8  10.[A] –2  x  18 [B] 1  x  –9[C] 18  x  –2 [D] –9  x  118. Solve 2 3 9 02x x  [A] 332, [B] 332, [C]  332, [D]  332, 19. Solve the inequality 1 4 7 2  x.[A] 132  x[B] x x  321, [C] – 252 x[D] x x 52, –220. Express the equation in slope-intercept form and find the slope and the y-intercept.  9 3 2x y[A] y x m   FHGIKJ929029; ; intercept: , –9[B] y x m   FHGIKJ39213029; ; intercept: , [C] y x m   FHGIKJ3233 023; ; intercept: , [D] Given in slope-intercept form; m 3 0 2; intercept:


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