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想問3條中3數學題
發問:
1.If ths sum of three consecutive numbers is less than 49,what is the largest possible value of the smallest number? (Let X be the smallest number)2.Ben has 20 coins which consist only of $2-coins and $5-coins.If the total amount of these coins is not less than $85,how many $2-coins can he have at most?3.The... 顯示更多 1.If ths sum of three consecutive numbers is less than 49,what is the largest possible value of the smallest number? (Let X be the smallest number) 2.Ben has 20 coins which consist only of $2-coins and $5-coins.If the total amount of these coins is not less than $85,how many $2-coins can he have at most? 3.The length of a rectangle is equal to 3 times the width of the rectangle minus 8 cm.If the perimeter of the rectangle is not less than 24 cm,find the minimum possible area of the rectangle.
1. If the sum of threeconsecutive numbers is less than 49,what is the largest possible value of the smallestnumber? The three numbers are integers. Let x be the smallest number. Then, the other two numbers are (x + 1) and (x + 2). x + (x + 1) + (x + 2) < 49 3x + 3 < 49 3x < 46 x < 15 + (1/3) Hence, the largest possible value of the smallestnumber 15. ===== 2. Ben has 20 coins whichconsist only of $2-coins and $5-coins.If the total amount of these coins is notless than $85,how many $2-coins can he have at most? Let n be the number of $2-coins. Then, the number of $5-coins = 20 - n The amount: 2n + 5(20 - n) ≥ 85 2n + 100 - 5n ≥ 85 -3n ≥ -15 3n ≤ 15 n ≤ 5 Hence, he can have at most 5 $2-coins. ===== 3. The length of a rectangleis equal to 3 times the width of the rectangle minus 8 cm. If the perimeter of therectangle is not less than 24 cm, find the minimum possiblearea of the rectangle. Let y cm be the width of the rectangle. Then, the length of the rectangle = (3y - 8) cm perimeter: 2y + 2(3y - 8) ≥ 24 2y + 6y - 16 ≥ 24 8y ≥ 40 y ≥ 5 3y - 8 ≥ 7 Area of the rectangle = y(3y - 8) cm2 ≥ 5 * 7 cm2 = 35 cm2 Hence, the minimum possible area of the rectangle is 35cm2.
其他解答:
想問3條中3數學題
發問:
1.If ths sum of three consecutive numbers is less than 49,what is the largest possible value of the smallest number? (Let X be the smallest number)2.Ben has 20 coins which consist only of $2-coins and $5-coins.If the total amount of these coins is not less than $85,how many $2-coins can he have at most?3.The... 顯示更多 1.If ths sum of three consecutive numbers is less than 49,what is the largest possible value of the smallest number? (Let X be the smallest number) 2.Ben has 20 coins which consist only of $2-coins and $5-coins.If the total amount of these coins is not less than $85,how many $2-coins can he have at most? 3.The length of a rectangle is equal to 3 times the width of the rectangle minus 8 cm.If the perimeter of the rectangle is not less than 24 cm,find the minimum possible area of the rectangle.
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最佳解答:1. If the sum of threeconsecutive numbers is less than 49,what is the largest possible value of the smallestnumber? The three numbers are integers. Let x be the smallest number. Then, the other two numbers are (x + 1) and (x + 2). x + (x + 1) + (x + 2) < 49 3x + 3 < 49 3x < 46 x < 15 + (1/3) Hence, the largest possible value of the smallestnumber 15. ===== 2. Ben has 20 coins whichconsist only of $2-coins and $5-coins.If the total amount of these coins is notless than $85,how many $2-coins can he have at most? Let n be the number of $2-coins. Then, the number of $5-coins = 20 - n The amount: 2n + 5(20 - n) ≥ 85 2n + 100 - 5n ≥ 85 -3n ≥ -15 3n ≤ 15 n ≤ 5 Hence, he can have at most 5 $2-coins. ===== 3. The length of a rectangleis equal to 3 times the width of the rectangle minus 8 cm. If the perimeter of therectangle is not less than 24 cm, find the minimum possiblearea of the rectangle. Let y cm be the width of the rectangle. Then, the length of the rectangle = (3y - 8) cm perimeter: 2y + 2(3y - 8) ≥ 24 2y + 6y - 16 ≥ 24 8y ≥ 40 y ≥ 5 3y - 8 ≥ 7 Area of the rectangle = y(3y - 8) cm2 ≥ 5 * 7 cm2 = 35 cm2 Hence, the minimum possible area of the rectangle is 35cm2.
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