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  • 单级方位数的应用

    Applications of One-Step Equations 
    ::单级方位数的应用

    Let's use the skills we learned in the last two concepts to solve applications of one-step problems.
    ::让我们利用我们在最后两个概念中学到的技能 解决一步骤问题的应用。

    Real-World Application: Birth Date
    ::真实世界应用:出生日期

    In the year 2017, Anne will be 45 years old. In what year was Anne born?
    ::2017年,安妮将45岁。

    The unknown here is the year Anne was born, so that’s our variable x . Here’s our equation :
    ::这里未知的是Anne出生的年份, 这就是我们的变数x。

    x + 45 = 2017 x + 45 − 45 = 2017 − 45 x = 1972

    ::x+45=2017x+45-45=2017-45x=45x=1972

    Anne was born in 1972.
    ::安妮生于1972年

    Real-World Application: DVD Players 
    ::Real- World 应用程序: DVD 播放器

    A mail order electronics company stocks a new mini DVD player and is using a balance to determine the shipping weight. Using only one-pound weights, the shipping department found that the following arrangement balances:
    ::邮购电子公司储存了一个新的小型DVD播放机,并正在使用余额来确定航运重量。

    How much does each DVD player weigh?
    ::每个DVD播放器的重量是多少?

    Since the system balances, the total weight on each side must be equal. To write our equation, we’ll use x for the weight of one DVD player, which is unknown. There are two DVD players, weighing a total of 2 x pounds, on the left side of the balance, and on the right side are 5 1-pound weights, weighing a total of 5 pounds. So our equation is 2 x = 5 . Dividing both sides by 2 gives us x = 2.5 .
    ::由于系统平衡, 每边的总重量必须是相等的。 为了写入我们的方程式, 我们将使用 x 来计算一个 DVD 播放器的重量, 这一点未知 。 在余额的左侧有两个DVD 播放器, 共重2x磅, 在余额的左侧, 右侧是 5 1磅的重量, 共重5磅。 因此我们的方程式是 2x= 5 。 将两边分成 2 给我们 x= 2.5 。

    Each DVD player weighs 2.5 pounds.
    ::每个DVD播放机重2.5磅。

    Real-World Application: Hot Dogs 
    ::真实世界应用程序:热狗

    In 2004, Takeru Kobayashi of Nagano, Japan, ate 53.5 hot dogs in 12 minutes. This was 3 more hot dogs than his own previous world record, set in 2002. Calculate how many minutes it took him to eat one hot dog.
    ::2004年,日本长野竹鲁小林(Takeru Kobayashi)在12分钟内吃了53.5只热狗。 这比2002年他以前的世界纪录多出3只热狗。 计算一下他吃了1只热狗花了多少分钟的时间。

    We know that the total time for 53.5 hot dogs is 12 minutes. We want to know the time for one hot dog, so that’s x . Our equation is 53.5 x = 12 . Then we divide both sides by 53.5 to get x = 12 53.5 , or x = 0.224   m i n u t e s .
    ::我们知道,53.5热狗的总时间是12分钟。我们想知道一个热狗的时间是多少,这就是x。我们的方程式是53.5x=12。然后我们把两边除以53.5以获得x=1253.5或x=0.224分钟。

    We can also multiply by 60 to get the time in seconds; 0.224 minutes is about 13.5 seconds. So that’s how long it took Takeru to eat one hot dog.
    ::我们也可以乘以60来获得时间;0.224分钟大约为13.5秒。 这就是竹ru吃一只热狗花了多长时间。

    Examples 
    ::实例

    Calculate the following, using the problem from the hot dog example:
    ::使用热狗示例中的问题计算如下 :

    Example 1
    ::例1

    How many hot dogs he ate per minute.
    ::他每分钟吃多少热狗

    For this questions we’re looking for hot dogs per minute instead of minutes per hot dog. We’ll use the variable y instead of x this time so we don’t get the two confused. 12 minutes, times the number of hot dogs per minute, equals the total number of hot dogs, so 12 y = 53.5 . Dividing both sides by 12 gives us y = 53.5 12 , or y = 4.458 hot dogs per minute.
    ::对于这个问题,我们每分钟寻找热狗,而不是每个热狗的分钟。我们将使用变数y而不是x,这样我们就不会混淆这两个问题。 12分钟,每分钟热狗的数量乘以热狗的数量,等于热狗的总数,所以12y=53.5。 将两边分成12个,等于53.512个,或y=4.458个热狗每分钟。

    Example 2
    ::例2

    What his old record was.
    ::他的旧记录是什么?

    We know that his new record is 53.5, and we know that’s three more than his old record. If we call his old record z , we can write the following equation: z + 3 = 53.5 . Subtracting 3 from both sides gives us z = 50.5 . So Takeru’s old record was 50.5 hot dogs in 12 minutes.
    ::我们知道他的新记录是53.5, 我们知道这比他的旧记录多3个。 如果我们叫他的旧记录z,我们可以写下以下等式:z+3=53.5。 从两边减去3后我们就会发现z=50.5。 因此Takeru的旧记录是12分钟内50.5只热狗。

    Review 
    ::回顾

    Peter is collecting tokens on breakfast cereal packets in order to get a model boat. In eight weeks he has collected 10 tokens. He needs 25 tokens for the boat. Write an equation and determine the following information.
    ::Peter正在收集早餐麦片包装上的标牌, 以获得一艘模范船。 八周后, 他收集了10个标牌。 他需要25个标牌。 写一个方程式并确定以下信息 。

    1. How many more tokens he needs to collect, n .
      ::他还要收集多少个纪念品?
    2. How many tokens he collects per week, w .
      ::他每周收集多少象征性物品?
    3. How many more weeks remain until he can send off for his boat, r .
      ::还要等多少个星期 他才能出发上船

    Juan has baked a cake and wants to sell it in his bakery. He is going to cut it into 12 slices and sell them individually. He wants to sell it for three times the cost of making it. The ingredients cost him $8.50, and he allowed $1.25 to cover the cost of electricity to bake it. Write equations that describe the following statements.
    ::胡安已经烤了一个蛋糕,想在面包店里卖掉它,他要切成12片,单独卖掉,要卖出三倍于制作成本。原料花了他8.50美元,他允许1.25美元支付烘烤电费。写出描述以下声明的方程式。

    1. The amount of money that he sells the cake for ( u ) .
      ::他为(u)卖蛋糕的金额。
    2. The amount of money he charges for each slice ( c ) .
      ::每切(c)项他收取的金额。
    3. The total profit he makes on the cake ( w ) .
      ::他在蛋糕上赚的利润总额(w)

    Jane is baking cookies for a large party. She has a recipe that will make one batch of two dozen cookies, and she decides to make five batches. To make five batches, she finds that she will need 12.5 cups of flour and 15 eggs.
    ::Jane正在为一个大型聚会烤饼干。她有一套配方,可以做一批20多个饼干,她决定做五批。为了做五批,她发现她需要12.5杯面粉和15个鸡蛋。

    1. How many cookies will she make in all?
      ::她能做多少饼干?
    2. How many cups of flour go into one batch?
      ::有多少杯面粉进一批?
    3. How many eggs go into one batch?
      ::有多少鸡蛋进一批?
    4. If Jane only has a dozen eggs on hand, how many more does she need to make five batches?
      ::如果Jane手上只有一打鸡蛋 她还要做五批吗?
    5. If she doesn’t go out to get more eggs, how many batches can she make? How many cookies will that be?
      ::如果她不外出买更多鸡蛋,她能做多少批?有多少饼干?

    Review (Answers)
    ::回顾(答复)

    Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
    ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。