课程： 3.8 利用多种标准等量解决现实世界问题 | The Way To Learn

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选择节利用多种标准等量解决现实世界问题

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利用多种标准等量解决现实世界问题
Real-World Application of Multi-Step Equations
::多级等数的实际世界应用

U se strategies for solving to solve real world equations.
::利用解决战略解决现实世界的方程式。

Real-World Application: Profit
::真实世界应用程序:利润

A growers’ cooperative has a farmer’s market in the town center every Saturday. They sell what they have grown and split the money into several categories. 8.5% of all the money taken in is set aside for sales tax. $150 goes to pay the rent on the space they occupy. What remains is split evenly between the seven growers. How much total money is taken in if each grower receives a $175 share?
::种植者合作社每星期六在市中心有一个农民市场,他们出售他们种植的作物,把钱分成几类。所收全部钱的8.5%被留作销售税。 150美元用来支付他们所占用空间的租金。剩下的钱在七个种植者之间平均分配。如果每个种植者得到175美元的股份,他们要花多少钱?

Let’s translate the text above into an equation . The unknown is going to be the total money taken in dollars. We’ll call this $\begin{align*}x\end{align*}$ .
::让我们将上面的文字转换成一个方程式。未知的将是用美元提取的总金额。我们将称之为 x。

“8.5% of all the money taken in is set aside for sales tax." This means that 91.5% of the money remains. This is $\begin{align*}0.915x\end{align*}$ .
::“所收钱的8.5%被留作销售税。”这意味着所收钱的91.5%还剩下。这是0.915x。

“$150 goes to pay the rent on the space they occupy.” This means that what’s left is $\begin{align*}0.915x - 150\end{align*}$ .
::“150美元将支付他们所占用空间的租金。” 这意味着剩下的是0.915x-150。

“What remains is split evenly between the 7 growers.” That means each grower gets $\begin{align*}\frac{0.915x - 150}{7}.\end{align*}$
::“留下的东西在7个种植者之间平均分配。” 这意味着每个种植者得到0.915x-1507。

If each grower’s share is $175, then our equation to find $\begin{align*}x\end{align*}$ is $\begin{align*}\frac{0.915x - 150}{7} = 175.\end{align*}$
::如果每个种植者的份额是175美元,那么我们找到x的方程式是0.915x-1507=175。

First we multiply both sides by 7 to get $\begin{align*}0.915x - 150 = 1225.\end{align*}$
::我们首先将两边乘以7 以获得0.915x-150=1225。

Then add 150 to both sides to get $\begin{align*}0.915x = 1375.\end{align*}$
::然后在双方增加150美元,以获得0.915x=1375。

Finally divide by 0.915 to get $\begin{align*}x \approx 1502.7322\end{align*}$ . Since we want our answer in dollars and cents, we round to two decimal places, or $1502.73.
::最后除以 0. 915 以获得 x1502.7322。既然我们想要用美元和美分回答, 我们四舍五入到小数点后两个位数, 即 1502. 73 。

The workers take in a total of $1502.73.
::工人共领取1502.73美元。

Real-World Application: Ohm’s Law
::真实世界应用程序: Ohm 法

The electrical current, $\begin{align*}I\end{align*}$ (amps), passing through an electronic component varies directly with the applied voltage, $\begin{align*}V\end{align*}$ (volts), according to the relationship $\begin{align*}V = I \cdot R\end{align*}$ where $\begin{align*}R\end{align*}$ is the resistance measured in Ohms $\begin{align*}(\Omega)\end{align*}$ .
::电流I(路标)通过电子部件,与应用电压V(电压)直接不同,根据V=IR的关系,R是Ohms(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

A scientist is trying to deduce the resistance of an unknown component. He labels the resistance of the unknown component $\begin{align*}x \ \Omega\end{align*}$ . The resistance of a circuit containing a number of these components is $\begin{align*}(5x + 20) \Omega\end{align*}$ . If a 120 volt potential difference across the circuit produces a current of 2.5 amps, calculate the resistance of the unknown component.
::科学家正在试图推断未知元件的抗力。他标注未知元件的抗力 x 。含有多个这些元件的电路的抗力是 (5x+20 ) 。如果整个电路120伏的潜在差异产生2.5安培的电流, 计算未知元件的抗力。

To solve this, we need to start with the equation $\begin{align*}V = I \cdot R\end{align*}$ and substitute in $\begin{align*}V = 120, I = 2.5,\end{align*}$ and $\begin{align*}R = 5x + 20\end{align*}$ . That gives us $\begin{align*}120 = 2.5(5x + 20)\end{align*}$ .
::要解决这个问题,我们需要从公式V=IR开始,替换为V=120、I=2.5和R=5x+20。这给我们120=2.5(5x+20)。

Distribute the 2.5 to get $\begin{align*}120 = 12.5x + 50\end{align*}$ .
::分配 2.5 得到 120= 12.5x+50。

Subtract 50 from both sides to get $\begin{align*}70 = 12.5x\end{align*}$ .
::从两边减50 得到70=12.5x。

Finally, divide by 12.5 to get $\begin{align*}5.6 = x\end{align*}$ .
::最后,除以12.5 以获得5.6=x。

The unknown components have a resistance of $\begin{align*}5.6 \ \Omega\end{align*}$ .
::未知部件的抗药性为5.6 。

Real-World Application: Distance, Speed and Time
::真实世界应用:距离、速度和时间

The speed of a body is the distance it travels per unit of time. That means that we can also find out how far an object moves in a certain amount of time if we know its speed: we use the equation “ $\begin{align*}\text{distance} = \text{speed} \times \text{time}\end{align*}$ .”
::身体的速度是它每单位时间行走的距离。这意味着如果我们知道一个物体的速度,我们也可以在一定的时间内知道它移动了多远:我们使用方程式“距离=速度x时间 ” 。

Shanice’s car is traveling 10 miles per hour slower than twice the speed of Brandon’s car. She covers 93 miles in 1 hour 30 minutes. How fast is Brandon driving?
::莎妮丝的车每小时行驶10英里,比布兰登车速的两倍慢。她的车在1小时30分钟内行驶93英里。布兰登的车速有多快?

Here, we don’t know either Brandon’s speed or Shanice’s, but since the question asks for Brandon’s speed, that’s what we’ll use as our variable $\begin{align*}x\end{align*}$ .
::我们不知道布兰登的速度, 或Shanice的速度, 但既然问题要求布兰登的速度,

The distance Shanice covers in miles is 93, and the time in hours is 1.5. Her speed is 10 less than twice Brandon’s speed, or $\begin{align*}2x - 10\end{align*}$ miles per hour. Putting those numbers into the equation gives us $\begin{align*}93 = 1.5(2x - 10)\end{align*}$ .
::仙妮斯的覆盖距离是93英里,时间是1.5小时。她的速度是布兰登速度的10倍以上,或每小时2x-10英里。将这些数字放入方程让我们得出93=1.5(2x-10) 。

First we distribute, to get $\begin{align*}93 = 3x - 15\end{align*}$ .
::首先我们分发93=3x-15。

Then we add 15 to both sides to get $\begin{align*}108 = 3x\end{align*}$ .
::然后我们向双方增加15分以得到108=3x。

Finally we divide by 3 to get $\begin{align*}36 = x\end{align*}$ .
::最后我们除以3 获得36=x。

Brandon is driving at 36 miles per hour.
::布兰登每小时驾驶36英里

We can check this answer by considering the situation another way: we can solve for Shanice’s speed instead of Brandon’s and then check that against Brandon’s speed. We’ll use $\begin{align*}y\end{align*}$ for Shanice’s speed since we already used $\begin{align*}x\end{align*}$ for Brandon’s.
::我们可以通过考虑另一种方式来检查这个答案:我们可以解决Shanice的速度而不是Brandon的速度,然后对照Brandon的速度来检查它。我们将使用 y 来计算Shanice 的速度,因为我们已经用 x 来计算Brandon 的速度。

The equation for Shanice’s speed is simply $\begin{align*}93 = 1.5y\end{align*}$ . We can divide both sides by 1.5 to get $\begin{align*}62 = y\end{align*}$ , so Shanice is traveling at 62 miles per hour.
::仙妮丝速度的方程式是93=1.5y。我们可以将两边除以1.5来获得62=y,所以仙妮丝的飞行速度是每小时62英里。

The problem tells us that Shanice is traveling 10 mph slower than twice Brandon’s speed; that would mean that 62 is equal to 2 times 36 minus 10. Is that true? Well, 2 times 36 is 72, minus 10 is 62. The answer checks out.
::问题告诉我们,Shanice的行驶速度比Brandon速度的两倍慢了10米;这意味着62等于2乘以36减10,这是真的吗?2乘以36等于72,减以10等于62。答案已经查出来了。

In algebra, there’s almost always more than one method of solving a problem. If time allows, it’s always a good idea to try to solve the problem using two different methods just to confirm that you’ve got the answer right.
::在代数中,解决问题的方法几乎总是不止一种。如果时间允许,用两种不同的方法试图解决问题总是一个好主意,只要确认你得到了正确的答案。

Real-World Application: Speed of Sound
::真实世界应用:声音速度

The speed of sound in dry air, $\begin{align*}v\end{align*}$ , is given by the equation $\begin{align*}v = 331 + 0.6T\end{align*}$ , where $\begin{align*}T\end{align*}$ is the temperature in Celsius and $\begin{align*}v\end{align*}$ is the speed of sound in meters per second.
::等式 v=331+0.6T给出了干空中声音的速度。 T是摄氏度的温度, v是每秒米声音的速度。

Tashi hits a drainpipe with a hammer and 250 meters away Minh hears the sound and hits his own drainpipe. Unfortunately, there is a one second delay between him hearing the sound and hitting his own pipe. Tashi accurately measures the time between her hitting the pipe and hearing Mihn’s pipe at 2.46 seconds. What is the temperature of the air?
::塔什用锤子敲下排水管,距离明亮250米就听到声音并击中自己的排水管。不幸的是,他听到声音和打自己的管子之间还有一秒钟的延迟。塔什准确地测量了她敲下管子和听Mihn的管子之间在2.46秒的时间。空气的温度是多少?

This is a complex problem and we need to be careful in writing our equations. First of all, the distance the sound travels is equal to the speed of sound multiplied by the time, and the speed is given by the equation above. So the distance equals $\begin{align*}(331 + 0.6T) \times \text{time}\end{align*}$ , and the time is $\begin{align*}2.46 - 1\end{align*}$ (because for 1 second out of the 2.46 seconds measured, there was no sound actually traveling). We also know that the distance is $\begin{align*}250 \times 2\end{align*}$ (because the sound traveled from Tashi to Minh and back again), so our equation is $\begin{align*}250 \times 2 = (331 + 0.6T)(2.46 - 1)\end{align*}$ , which simplifies to $\begin{align*}500 = 1.46(331 + 0.6T)\end{align*}$ .
::这是一个复杂的问题, 我们需要在写我们的方程时小心。首先, 音速旅行的距离等于音速乘以时间, 速度由以上方程式给出。因此, 距离等于( 331+0. 6T) xx 时间, 时间是2.46- 1( 因为测量的2.46 秒中有一秒没有声音实际移动 ) 。我们还知道距离是 250x2 ( 因为声音从塔希传到明明, 然后再传回) , 因此我们的方程是 250x2 =( 331+0. 6T)( 2. 46- 1) , 简化为 500= 1.46 ( 331+0. 6T) 。

Distributing gives us $\begin{align*}500 = 483.26 + 0.876T\end{align*}$ , and subtracting 483.26 from both sides gives us $\begin{align*}16.74 = 0.876T\end{align*}$ . Then we divide by 0.876 to get $\begin{align*}T \approx 19.1\end{align*}$ .
::分配给了我们500=483.26+0.876T,从两边减去483.26 给了我们16.74=0.876T。然后我们除以0.876 以获得T19.1。

The temperature is about 19.1 degrees Celsius.
::温度约为19.1摄氏度。

Example
::示例示例示例示例

Example 1
::例1

A factory manager is packing engine components into wooden crates to be shipped on a small truck. The truck is designed to hold sixteen crates, and will safely carry a 1200 lb cargo. Each crate weighs 12 lbs empty. How much weight should the manager instruct the workers to put in each crate in order to get the shipment weight as close as possible to 1200 lbs?
::一名工厂经理将发动机部件包装在小卡车上装运的木箱中,该卡车设计成持有16箱,并将安全运载1200磅货物,每箱重量为12磅空。经理应指示工人将每箱重量提高到1200磅,以使装运重量尽可能接近1200磅?

The unknown quantity is the weight to put in each box, so we’ll call that $\begin{align*}x\end{align*}$ .
::未知数量是每个盒子的重量, 所以我们称它为x。

Each crate when full will weigh $\begin{align*}x + 12 \ lbs\end{align*}$ , so all 16 crates together will weigh $\begin{align*}16(x + 12) \ lbs\end{align*}$ .
::每箱装满时重量为 x+12 lbs, 所以所有16 箱一起重量为 16 (x+12) lbs。

We also know that all 16 crates together should weigh 1200 lbs, so we can say that $\begin{align*}16(x + 12) = 1200\end{align*}$ .
::我们还知道,所有16个箱子一起应该重1200磅, 所以我们可以说16(x+12)=1200。

To solve this equation, we can start by dividing both sides by 16: $\begin{align*}x + 12 = \frac{1200}{16} = 75\end{align*}$ .
::为了解决这个方程式,我们可以首先将两边除以 16: x+12=120016=75。

Then subtract 12 from both sides: $\begin{align*}x = 63\end{align*}$ .
::然后从两边减去12:x=63。

The manager should tell the workers to put 63 lbs of components in each crate.
::经理应该让工人把63磅的部件放进每箱里

Review
::回顾

For 1-6, solve for the variable in the equation.
::1 -6, 解答方程中的变量。

$\begin{align*}\frac{s - 4}{11} = \frac{2}{5} \end{align*}$
::s-411=25

$\begin{align*}\frac{2k}{7} = \frac{3}{8} \end{align*}$
::2k7=38

$\begin{align*}\frac{7x + 4}{3} = \frac{9}{2} \end{align*}$
::7x+43=92

$\begin{align*}\frac{9y - 3}{6} = \frac{5}{2} \end{align*}$
::9-36=52

$\begin{align*}\frac{r}{3} + \frac{r}{2} = 7\end{align*}$
::r3+r2=7

$\begin{align*} \frac{p}{16} - \frac{2p}{3} = \frac{1}{9} \end{align*}$
::p16-2p3=19

An engineer is building a suspended platform to raise bags of cement. The platform has a mass of 200 kg, and each bag of cement is 40 kg. He is using two steel cables, each capable of holding 250 kg. Write an equation for the number of bags he can put on the platform at once, and solve it.
::工程师正在建造一个悬浮平台,以抬举水泥袋。平台质量为200公斤, 每袋水泥为40公斤。他使用的是两根钢缆, 每条钢缆都能够持有250公斤。写一个他可以同时放在平台上的袋子数量的方程式, 并解决它。

A scientist is testing a number of identical components of unknown resistance which he labels $\begin{align*}x\Omega\end{align*}$ . He connects a circuit with resistance $\begin{align*}(3x + 4)\Omega\end{align*}$ to a steady 12 volt supply and finds that this produces a current of 1.2 amps. What is the value of the unknown resistance?
::科学家正在测试一些他标明为 x_ 的未知抗力的相同部件。他将电路与抗力( 3x+4) / 连接到稳定的 12伏供应中, 并发现这会产生1.2安普的电流。未知抗力的价值是什么 ?

Lydia inherited a sum of money. She split it into five equal parts. She invested three parts of the money in a high-interest bank account which added 10% to the value. She placed the rest of her inheritance plus $500 in the stock market but lost 20% on that money. If the two accounts end up with exactly the same amount of money in them, how much did she inherit?
::莉蒂亚继承了一笔钱。她把它分成五等分。她把三分钱投资于一个高息银行账户,增加了10%的价值。她把其余的遗产加上500美元放在股票市场,但损失了20%。如果这两个账户最后的金额完全相同,她继承了多少钱?

Pang drove to his mother’s house to drop off her new TV. He drove at 50 miles per hour there and back, and spent 10 minutes dropping off the TV. The entire journey took him 94 minutes. How far away does his mother live?
::Pang开车到他母亲家送她的新电视机。他开着每小时50英里的车,然后又从电视机上下车10分钟。整个旅程花了94分钟。他母亲住多远?

Review (Answers)
::回顾(答复)

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::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。

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