Solve These ACT Math Questions Flashcards

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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?
The answer is 408.04
If you divide 2,727 miles by 27 miles per gallon you will get the number of gallons: The answer is 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 is the correct answer. 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 ?
The corect answer is 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.
What is the greatest common factor of 42, 126, and 210 ?
42 is the correct answer since it is the largest number that is a factor of all three numbers given. You can find the greatest common factor by writing out the prime factorization of all three numbers, and then taking each of the common prime factors to the lowest power that appears for that factor:
42 = 2 × 3 × 7; 126 = 2 × 32 × 7; and 210 = 2 × 3 × 5 × 7. So the greatest common factor is 2 × 3 × 7 = 42.
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 is the correct answer. 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:
The correct answer is that it is 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.
The correct answer is that it is equal. If you draw line segments DF and FG, you can show The correct answer is that it is 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 The correct answer is that it is 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. The correct answer is that it is 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, The correct answer is that it is 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 The correct answer is that it is 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. The correct answer is that it is 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.