ACT Math Practice Question Flashcards

Are you preparing for the ACT math exams? The flashcards below have some practice questions that will come in handy during your revision; they not only have some well-designed questions but will offer you the solutions once you are trying the questions. Be sure to take as much time as you need as you work out each. All the best!

5 cards | Total Attempts: 188 | Created by jade26 | Last updated: Jan 28, 2020 | Total Attempts: 188

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A Car Averages 27 Miles Per Gallon. If Gas Costs $4.04 Per Gallon, Which Of The Following Is Closest To How Much The Gas Would Cost For This Car To Travel 2,727 Typical Miles?

Cards In This Set

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A car averages 27 miles per gallon. If gas costs $4.04 per gallon, which of the following is closest to how much the gas would cost for this car to travel 2,727 typical miles?	$408.04 -If you divide 2,727 miles by 27 miles per gallon you will get the number of gallons: $$408.04-If you divide 2,727 miles by 27 miles per gallon you will get the number of gallons:= 101. Then, multiply the number of gallons by the cost per gallon: 101(4.04) = 408.04. This gives the cost of gas for this car to travel 2,727 typical miles.$ = 101. Then, multiply the number of gallons by the cost per gallon: 101(4.04) = 408.04. This gives the cost of gas for this car to travel 2,727 typical miles.
When x = 3 and y = 5, by how much does the value of 3x² – 2y exceed the value of 2x² – 3y ?	14 -When you use x = 3 and y = 5 in the given expressions, 3x² – 2y = 3(3)² – 2(5) = 27 – 10 = 17 and 2x² – 3y = 2(3)² – 3(5) = 18 – 15 = 3. Then subtract 3 from 17 to get 14.
What is the value of x when 2x + 3 = 3x – 4 ?	7 -You can solve this problem by first subtracting 2x from each side of the equation to get 3 = x – 4. Then add 4 to each side, so x = 7.
Sales for a business were 3 million dollars more the second year than the first, and sales for the third year were double the sales for the second year. If sales for the third year were 38 million dollars, what were sales, in millions of dollars, for the first year?	16 -If x = sales for the first year, then x + 3 = sales for the second year. Since sales for the third year were double the sales for the second year, sales for the third year = 2(x + 3). Sales for the third year were 38, so 2(x + 3) = 38. To solve this equation, you could first divide each side by 2 to get x + 3 = 19. Then, by subtracting 3 from both sides, x = 16.
In the figure below, ray $In the figure below, ray was constructed starting from rays and . By using a compass, D and G were marked equidistant from E on rays and . The compass was then used to locate a point F, distinct from E, so that F is equidistant from D and G. For all constructions defined by the above steps, the measures of DEF and GEF:$ was constructed starting from rays $In the figure below, ray was constructed starting from rays and . By using a compass, D and G were marked equidistant from E on rays and . The compass was then used to locate a point F, distinct from E, so that F is equidistant from D and G. For all constructions defined by the above steps, the measures of DEF and GEF:$ and $In the figure below, ray was constructed starting from rays and . By using a compass, D and G were marked equidistant from E on rays and . The compass was then used to locate a point F, distinct from E, so that F is equidistant from D and G. For all constructions defined by the above steps, the measures of DEF and GEF:$ . By using a compass, D and G were marked equidistant from E on rays $In the figure below, ray was constructed starting from rays and . By using a compass, D and G were marked equidistant from E on rays and . The compass was then used to locate a point F, distinct from E, so that F is equidistant from D and G. For all constructions defined by the above steps, the measures of DEF and GEF:$ and $In the figure below, ray was constructed starting from rays and . By using a compass, D and G were marked equidistant from E on rays and . The compass was then used to locate a point F, distinct from E, so that F is equidistant from D and G. For all constructions defined by the above steps, the measures of DEF and GEF:$ . The compass was then used to locate a point F, distinct from E, so that F is equidistant from D and G. For all constructions defined by the above steps, the measures of $In the figure below, ray was constructed starting from rays and . By using a compass, D and G were marked equidistant from E on rays and . The compass was then used to locate a point F, distinct from E, so that F is equidistant from D and G. For all constructions defined by the above steps, the measures of DEF and GEF:$ DEF and $In the figure below, ray was constructed starting from rays and . By using a compass, D and G were marked equidistant from E on rays and . The compass was then used to locate a point F, distinct from E, so that F is equidistant from D and G. For all constructions defined by the above steps, the measures of DEF and GEF:$ GEF: $In the figure below, ray was constructed starting from rays and . By using a compass, D and G were marked equidistant from E on rays and . The compass was then used to locate a point F, distinct from E, so that F is equidistant from D and G. For all constructions defined by the above steps, the measures of DEF and GEF:$	Are equal -If you draw line segments DF and FG, you can show $Are equal-If you draw line segments DF and FG, you can show DEF GEF by SSS (side-side-side congruence). So, DEF GEF because corresponding parts of congruent triangles are congruent.$ DEF $Are equal-If you draw line segments DF and FG, you can show DEF GEF by SSS (side-side-side congruence). So, DEF GEF because corresponding parts of congruent triangles are congruent.$ $Are equal-If you draw line segments DF and FG, you can show DEF GEF by SSS (side-side-side congruence). So, DEF GEF because corresponding parts of congruent triangles are congruent.$ GEF by SSS (side-side-side congruence). So, $Are equal-If you draw line segments DF and FG, you can show DEF GEF by SSS (side-side-side congruence). So, DEF GEF because corresponding parts of congruent triangles are congruent.$ DEF $Are equal-If you draw line segments DF and FG, you can show DEF GEF by SSS (side-side-side congruence). So, DEF GEF because corresponding parts of congruent triangles are congruent.$ $Are equal-If you draw line segments DF and FG, you can show DEF GEF by SSS (side-side-side congruence). So, DEF GEF because corresponding parts of congruent triangles are congruent.$ GEF because corresponding parts of congruent triangles are congruent.