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不等式的性质
(8分)某手机专营店代理销售A、B两种型号手机.手机的进价、售价如下表:(1)第一季度:用36000元购进A、B两种型号的手机,全部售完后获利6300元,求第一...
2023-05-24 11:12
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数学
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这道
初中
数学的题目是:
(8分)某手机专营店代理销售A、B两种型号手机.手机的进价、售价如下
表:
(1)第一季度:用36000元购进A、B两种型号的手机,全部售完后获利6300元,求
第一季度购进A、B两种型号手机的数量;
(2)第二季度:计划购进A、B两种型号手机共34部,且不超出第一季度的购机总费用,则A型号手机最多能购多少部?
1条回答
霞映澄塘
2023-05-24 11:24
这道
初中
数学题的正确答案为:
(1)解:设该专营店第一季度购进A、B两种型号手机的数量分别为x部和y部. 1分
答:该专营店本次购进A、B两种型号手机的数分别为15部和18部.······························································ 4分
(2)解:设第二季度购进A型号手机a部.···················································· 5分
由题意可知:1200a+1000(34-a)≤36000,····················································· 6分
解得:a≤10··························································································· 7分
不等式的最大整数解为10
答:第二季度最多能购A型号手机10部.······················································ 8分
解题思路
略
加载中...
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相关问题
答:该专营店本次购进A、B两种型号手机的数分别为15部和18部.······························································ 4分
(2)解:设第二季度购进A型号手机a部.···················································· 5分
由题意可知:1200a+1000(34-a)≤36000,····················································· 6分
解得:a≤10··························································································· 7分
不等式的最大整数解为10
答:第二季度最多能购A型号手机10部.······················································ 8分
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