课程： 4.9 直接变化线形模型图 | The Way To Learn

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直接变化的线性模型图
Graphs of Linear Models of Direct Variation
::直接变化的线性模型图

Suppose you see someone buy five pounds of strawberries at the grocery store. The clerk weighs the strawberries and charges $12.50 for them. Now suppose you wanted two pounds of strawberries for yourself. How much would you expect to pay for them?
::如果你看到有人在杂货店买5磅草莓,店员会称草莓重量,并收取12.50美元。现在假设你想自己买两磅草莓。你希望花多少钱买这些草莓?

Identify Direct Variation
::识别直接变化

The preceding problem is an example of a . We would expect that the strawberries are priced on a “per pound” basis, and that if you buy two-fifths the amount of strawberries, you would pay two-fifths of $12.50 for your strawberries, or $5.00.
::前述问题就是一个例子。我们期望草莓的价格按“每磅”计算,而且如果你购买五分之二的草莓,你就会为草莓支付五分之二的12.50美元,即5.00美元。

Similarly, if you bought 10 pounds of strawberries (twice the amount) you would pay twice $12.50, and if you did not buy any strawberries you would pay nothing.
::同样,如果你买了10磅的草莓(两倍),你会付两倍的12.50美元,如果你不买任何草莓,你就什么都不付。

If variable $\begin{aligned} y \end{aligned}$ varies directly with variable $\begin{aligned} x \end{aligned}$ , then we write the relationship as $\begin{aligned} y = k \cdot x \end{aligned}$ . $\begin{aligned} k \end{aligned}$ is called the constant of proportionality .
::如果变量 y 与变量 x 直接不同,则将关系写成 y=kx. k 称为相称性常数。

If we were to graph this function , it would pass through the origin, because $\begin{aligned} y = 0 \end{aligned}$ when $\begin{aligned} x = 0 \end{aligned}$ , regardless of the value of $\begin{aligned} k \end{aligned}$ . A direct variation, when graphed, has a single intercept at (0, 0).
::如果我们绘制此函数, 它会通过源代码, 因为 y=0 当 x=0 时 y=0 , 不论 k 值如何。图形显示时, 直接变量在 0 0 时有一次拦截。

Determining the Constant of Proportionality
::确定相称性的常数

If $\begin{aligned} y \end{aligned}$ varies directly with $\begin{aligned} x \end{aligned}$ according to the relationship $\begin{aligned} y = k \cdot x \end{aligned}$ , and $\begin{aligned} y = 7.5 \end{aligned}$ when $\begin{aligned} x = 2.5 \end{aligned}$ , determine the constant of proportionality, $\begin{aligned} k \end{aligned}$ .
::如果 y 根据 y= kx 关系与 x 直接变化, y= 7. 5 当 x= 2. 5 时, y= 7. 5 确定相称性的常数, k 。

We can solve for the constant of proportionality using substitution. Substitute $\begin{aligned} x = 2.5 \end{aligned}$ and $\begin{aligned} y = 7.5 \end{aligned}$ into the equation $\begin{aligned} y = k \cdot x \end{aligned}$ to get $\begin{aligned} 7.5 = k (2.5) \end{aligned}$ . Then divide both sides by 2.5 to get $\begin{aligned} k = \frac{7.5}{2.5} = 3 \end{aligned}$ . The constant of proportionality, $\begin{aligned} k \end{aligned}$ , is 3.
::我们可以用替代来解析相称性的常数。等式的替代值 x=2.5 和y= 7. 5, y=kx 获得 7. 5=k( 2. 5) 。然后将两边除以 2.5 , 以获得 k= 7. 52.5 = 3 。相称性的常数 k, k, 是 3 。

We can graph the relationship quickly, using the intercept (0, 0) and the point (2.5, 7.5). The graph is shown below. It is a straight line with 3.
::我们可以使用截取的(0,0)和点(2.5,7.5)快速绘制关系图。图如下。它是一条直线,直线为3。

The graph of a direct variation always passes through the origin, and always has a slope that is equal to the constant of proportionality, $\begin{aligned} k \end{aligned}$ .
::直接变化图总是通过来源,并且总是有一个相当于相称性不变的斜坡,k。

Conversion Factors
::换算系数

The graph shown below is a conversion chart used to convert U.S. dollars (US$) to British pounds (GB£) in a bank on a particular day. Use the chart to determine:
::下图是用于在某一天将某家银行的美元(美元)转换成英镑(英镑)的换算图。

a) the number of pounds you could buy for $600
:a) 600美元你能买到的磅数

b) the number of dollars it would cost to buy £200
:b) 购买200英镑需要花费的美元数

c) the exchange rate in pounds per dollar
:c) 每美元英镑的汇率

We can read the answers to a) and b) right off the graph. It looks as if at $\begin{aligned} x = 600 \end{aligned}$ the graph is about one fifth of the way between £350 and £400. So $600 would buy £360.
::我们可以从图中读出a)和b)的答案。看似在x=600时,图大约是350英镑到400英镑的五分之一。那么600美元可以买到360英镑。

Similarly, the line $\begin{aligned} y = 200 \end{aligned}$ appears to intersect the graph about a third of the way between $300 and $400. We can round this to $330, so it would cost approximately $330 to buy £200.
::同样,y=200这条线似乎将大约三分之一的300美元到400美元之间的路段的图图相交叉。我们可以绕到330美元,因此购买200英镑大约需要330美元。

To solve for the exchange rate, we should note that as this is a direct variation - the graph is a straight line passing through the origin. The slope of the line gives the constant of proportionality (in this case the exchange rate ) and it is equal to the ratio of the $\begin{aligned} y - \end{aligned}$ value to $\begin{aligned} x - \end{aligned}$ value at any point. Looking closely at the graph, we can see that the line passes through one convenient lattice point : (500, 300). This will give us the most accurate value for the slope and so the exchange rate.
::要解决汇率问题,我们应当指出,由于这是直接变化,因此,图表是一条直线,穿过源头。线的斜坡给出了成比例的常数(在此情况下是汇率),它相当于y-value与x-value在任何时间的比。仔细看图,我们可以看到,线穿过一个方便的拉蒂点500,300),这为我们提供了坡度和汇率的最准确值。

$\begin{aligned} y = k \cdot x \Rightarrow= \frac{y}{x}, \end{aligned}$

::y'y's y's y's, y's y's y's, y's y's y's, y's y's, y's y's, y's y's, y's y's, y's, y's y's, y's, y's, y's, y's, y's, y's, y's, y's, y,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

$\begin{aligned} so rate = \frac{300 pounds}{500 dollars} = 0.60 pounds per dollar. \end{aligned}$

::=300磅500美元=0.60磅每美元。

Graphing Direct Variation Equations
::直接变化方平面图

We know that all direct variation graphs pass through the origin, and also that the slope of the line is equal to the constant of proportionality, $\begin{aligned} k \end{aligned}$ . Graphing is a simple matter of using the point-slope or point-point methods discussed earlier in this chapter.Example C
::我们知道,所有直接变化图都通过来源,而且线的斜坡等于相称性的常数,k。绘图是使用本章前面讨论的点倾斜或点角方法的一个简单问题。

Plot the following direct variations on the same graph.
::在同一图中绘制下列直接变数。

a) $\begin{aligned} y = 3 x \end{aligned}$
::a) y=3x

The line passes through (0, 0), as will all these functions. This function has a slope of 3. When we move across by one unit, the function increases by three units.
::线条和所有这些功能一样通过(0,0)。这个函数的斜坡为3。当我们在一个单位之间移动时,函数会增加3个单位。

b) $\begin{aligned} y = - 2 x \end{aligned}$
:b) y2x

The line has a slope of -2. When we move across the graph by one unit, the function falls by two units.
::线的斜坡为-2。当我们用一个单位跨过图时, 函数会由两个单位产生。

c) $\begin{aligned} y = - 0.2 x \end{aligned}$
::c) y=0.2x

The line has a slope of -0.2. As a fraction this is equal to $\begin{aligned} - \frac{1}{5} \end{aligned}$ . When we move across by five units, the function falls by one unit.
::线的斜坡为-0.2。分数等于- 15。当我们以五个单位移动时, 函数会有一个单位。

d) $\begin{aligned} y = \frac{2}{9} x \end{aligned}$
::d) y=29x

The line passes through (0, 0) and has a slope of $\begin{aligned} \frac{2}{9} \end{aligned}$ . When we move across the graph by 9 units, the function increases by two units.
::线通过(0,0),斜坡为29,当我们通过图9个单位时,函数增加两个单位。

Solve Real-World Problems Using Direct Variation Models
::使用直接变化模型解决现实世界问题

Direct variations are seen everywhere in everyday life. Any time one quantity increases at the same rate another quantity increases (for example, doubling when it doubles and tripling when it triples), we say that they follow a direct variation.
::在日常生活中,每处都可以看到直接差异。每当一个数量以同样的速度增加,另一个数量增加(例如,翻一番,翻三番,翻三番 ) , 我们说,它们随直接变化而变化。

Newton's Second Law
::牛顿第二法

In 1687 Sir Isaac Newton published the famous Principia Mathematica. It contained, among other things, his second law of motion. This law is often written as $\begin{aligned} F = m \cdot a \end{aligned}$ , where a force of $\begin{aligned} F \end{aligned}$ Newtons applied to a mass of $\begin{aligned} m \end{aligned}$ kilograms results in acceleration of $\begin{aligned} a \end{aligned}$ meters per second $\begin{aligned} ^{2} \end{aligned}$ . Notice that if the mass stays constant , then this formula is basically the same as the direct variation equation, just with different variables—and $\begin{aligned} m \end{aligned}$ is the constant of proportionality.
::1687年,艾萨克·牛顿爵士发表了著名的Principia Mathematica, 其中包括他的第二项行动法, 这部法律通常以F=ma写成,F牛顿的威力适用于质量为m公斤的重量,加速了每秒1米2。请注意,如果质量保持不变,那么这个公式基本上与直接变异方程式相同,只有不同的变量——M是相称性的。

If a 175 Newton force causes a shopping cart to accelerate down the aisle with an acceleration of $\begin{aligned} 2.5 m / s^{2} \end{aligned}$ , calculate:
::如果175牛顿力造成一辆购物车加速下过道,加速速度2.5米/秒2,计算如下:

a) The mass of a shopping cart.
:a) 购物车质量。

We can solve for $\begin{aligned} m \end{aligned}$ (the mass) by plugging in our given values for force and acceleration. $\begin{aligned} F = m \cdot a \end{aligned}$ becomes $\begin{aligned} 175 = m (2.5) \end{aligned}$ , and then we divide both sides by 2.5 to get $\begin{aligned} 70 = m \end{aligned}$ .
::我们可以通过插入给定的功率和加速度值来解决 m( 质量) 。 F=ma 变成 175=m( 2.5), 然后我们将两边除以 2.5 以获得 70 =m 。

So the mass of the shopping cart is 70 kg.
::因此,购物车的质量为70公斤。

b) The force needed to accelerate the same cart at $\begin{aligned} 6 m / s^{2} \end{aligned}$ .
:b) 加速同一车速为每秒6米/秒所需的部队。

Once we have solved for the mass, we simply substitute that value, plus our required acceleration, back into the formula $\begin{aligned} F = m \cdot a \end{aligned}$ and solve for $\begin{aligned} F \end{aligned}$ . We get $\begin{aligned} F = 70 \times 6 = 420 \end{aligned}$ .
::一旦我们解决了质量问题,我们就简单地用这个值,加上我们所需要的加速度, 回到F=ma的公式,然后解决F。我们得到F=70x6=420。

So the force needed to accelerate the cart at $\begin{aligned} 6 m / s^{2} \end{aligned}$ is 420 Newtons.
::加速6m/s2的车速所需的部队是420牛顿。

Ohm's Law
::《 Ohm's 律法》

The electrical current, $\begin{aligned} I \end{aligned}$ (amps), passing through an electronic component varies directly with the applied voltage, $\begin{aligned} V \end{aligned}$ (volts), according to the relationship $\begin{aligned} V = I \cdot R \end{aligned}$ , where $\begin{aligned} R \end{aligned}$ is the resistance (measured in Ohms). The resistance is considered to be a constant for all values of $\begin{aligned} V \end{aligned}$ and $\begin{aligned} I \end{aligned}$ , so once again, this formula is a version of the direct variation formula, with $\begin{aligned} R \end{aligned}$ as the constant of proportionality.
::电流( I ( 标志) ) 通过电子组件, 与应用电压( V ( 伏) ) 直接不同, 根据 V=IR 关系, R 是抗力( 以 Ohms 测量 ) 。阻力被认为是 V 和 I 的所有值的常数, 所以再次, 这个公式是直接变异公式的版本, R 是比例的常数。

A certain electronics component was found to pass a current of 1.3 amps at a voltage of 2.6 volts. When the voltage was increased to 12.0 volts the current was found to be 6.0 amps. Does the component obey Ohm's Law ? What would the current be at 6 volts?
::当电压增加到12.0伏时,电流为6.0安培。电流是否遵守了Ohm法则?电流在6伏时是什么?

Ohm’s law is a simple direct proportionality law, with the resistance $\begin{aligned} R \end{aligned}$ as our constant of proportionality. To know if this component obeys Ohm’s law, we need to know if it follows a direct proportionality rule. In other words, is $\begin{aligned} V \end{aligned}$ directly proportional to $\begin{aligned} I \end{aligned}$ ?
::奥姆法是一个简单的直接相称性法律,抵抗R是我们的相称性常数。要知道这一组成部分是否遵守奥姆法,我们需要知道它是否遵循直接相称性规则。换句话说,V是否与我直接相称?

We can determine this in two different ways.
::我们可以以两种不同的方式来决定这一点。

Graph It: If we plot our two points on a graph and join them with a line, does the line pass through (0, 0)?
::图表 : 如果我们在图表上绘制两点, 并用直线连接它们, 线会通过( 0, 0) 吗 ?

Voltage is the independent variable and current is the dependent variable, so normally we would graph $\begin{aligned} V \end{aligned}$ on the horizontal axis and $\begin{aligned} I \end{aligned}$ on the vertical axis . However, if we swap the variables around just this once, we’ll get a graph whose slope conveniently happens to be equal to the resistance, $\begin{aligned} R \end{aligned}$ . So we’ll treat $\begin{aligned} I \end{aligned}$ as the independent variable, and our two points will be (1.3, 2.6) and (6, 12).
::电压是独立的变量,电压是独立的变量,电流是独立的变量,所以我们通常会在水平轴上绘制V图,我在垂直轴上绘制一幅图。但是,如果我们只用这一次换掉变量,我们就会得到一个其斜坡恰好与阻力相等的图,R。因此,我们将视我为独立的变量,我们的两个点将是(1.3,2.6)和(6,12)。

Plotting those points and joining them gives the following graph:
::绘制并加入这些点的图表如下:

The graph does appear to pass through the origin, so yes, the component obeys Ohm’s law.
::图表似乎通过来源,所以是的,组成部分符合Ohm的法律。

Solve for $\begin{aligned} R \end{aligned}$ : If this component does obey Ohm’s law, the constant of proportionality $\begin{aligned} (R) \end{aligned}$ should be the same when we plug in the second set of values as when we plug in the first set. Let’s see if it is. (We can quickly find the value of $\begin{aligned} R \end{aligned}$ in each case; since $\begin{aligned} V = I \cdot R \end{aligned}$ , that means $\begin{aligned} R = \frac{V}{I} \end{aligned}$ .)
::解决 R : 如果此部件确实符合 Ohm 的法律, 当我们插入第二组值时, 相称性( R) 的常数应该和插插插第一组值时相同。让我们看看是否合适。 ( 我们可以在每种情况下快速找到 R 值; 因为 V= IR, 这意味着 R=VI ) 。 )

$\begin{aligned} Case 1: R & = \frac{V}{I} = \frac{2.6}{1.3} = 2 Ohms \\ Case 2: R & = \frac{V}{I} = \frac{12}{6} = 2 Ohms \end{aligned}$

::案例1:R=VI=2.61.3=2OmsCase 2:R=VI=126=2Oms

The values for $\begin{aligned} R \end{aligned}$ agree! This means that we are indeed looking at a direct variation. The component obeys Ohm’s law.
::R的价值观是一致的! 这意味着我们确实在研究一种直接的变异。组件符合Ohm 的法律。

b) Now to find the current at 6 volts, simply substitute the values for $\begin{aligned} V \end{aligned}$ and $\begin{aligned} R \end{aligned}$ into $\begin{aligned} V = I \cdot R \end{aligned}$ . We found that $\begin{aligned} R = 2 \end{aligned}$ , so we plug in 2 for $\begin{aligned} R \end{aligned}$ and 6 for $\begin{aligned} V \end{aligned}$ to get $\begin{aligned} 6 = I (2) \end{aligned}$ , and divide both sides by 2 to get $\begin{aligned} 3 = I \end{aligned}$ .
:b) 现在发现目前的电流为6伏,只需将V和R的值替换为V=IQR。我们发现R=2, 因此我们插入R2, V6的插件为6=I(2), 将两边除以2, 3=I。

So the current through the component at a voltage of 6 volts is 3 amps.
::6伏电压的电流是3安培。

Example
::示例示例示例示例

Example 1
::例1

The volume of water in a fish-tank, $\begin{aligned} V \end{aligned}$ , varies directly with depth, $\begin{aligned} d \end{aligned}$ . If there are 15 gallons in the tank when the depth is 8 inches, calculate how much water is in the tank when the depth is 20 inches.
::鱼缸V的水量随深度而异, d. 如果罐体深8英寸时有15加仑水,则计算罐体深20英寸时有多少水。

This is a good example of a direct variation, but for this problem we’ll have to determine the equation of the variation ourselves. Since the volume, $\begin{aligned} V \end{aligned}$ , depends on depth, $\begin{aligned} d \end{aligned}$ , we’ll use an equation of the form $\begin{aligned} y = k \cdot x \end{aligned}$ , but in place of $\begin{aligned} y \end{aligned}$ we’ll use $\begin{aligned} V \end{aligned}$ and in place of $\begin{aligned} x \end{aligned}$ we’ll use $\begin{aligned} d \end{aligned}$ :
::这是一个直接变异的好例子,但对于这个问题,我们必须自己决定变异的方程式。由于体积V取决于深度,d,我们将使用y=kx的方程式,但代替y,我们将使用V,取代x,我们将使用d:

$\begin{aligned} V = k \cdot d \end{aligned}$

::V=kd

We know that when the depth is 8 inches the volume is 15 gallons, so to solve for $\begin{aligned} k \end{aligned}$ , we plug in 15 for $\begin{aligned} V \end{aligned}$ and 8 for $\begin{aligned} d \end{aligned}$ to get $\begin{aligned} 15 = k (8) \end{aligned}$ . Dividing by 8 gives us $\begin{aligned} k = \frac{15}{8} = 1.875 \end{aligned}$ .
::我们知道,当深度为8英寸时,体积为15加仑,为了解决k,我们用15加仑为V加插,用8加8加丁为D加插15加仑为15加仑(8加仑)。分差为8等于8等于K=158=1.875。

Now to find the volume of water at the final depth, we use $\begin{aligned} V = k \cdot d \end{aligned}$ again, but this time we can plug in our new $\begin{aligned} d \end{aligned}$ and the value we found for $\begin{aligned} k \end{aligned}$ :
::现在,为了在最后深度找到水量, 我们再次使用 V=kd, 但这次我们可以插入我们的新 d 和我们找到的 k 的值 :

$\begin{aligned} V & = 1.875 \times 20 \\ V & = 37.5 \end{aligned}$

::V=1.875x20V=37.5

At a depth of 20 inches, the volume of water in the tank is 37.5 gallons.
::水箱深20英寸,水量为37.5加仑。

Review
::回顾

For 1-4, plot the following direct variations on the same graph.
::1-4,在同一图中绘制下列直接变数。

$\begin{aligned} y = \frac{4}{3} x \end{aligned}$
::y=43x y=43x

$\begin{aligned} y = - \frac{2}{3} x \end{aligned}$
::y 23x

$\begin{aligned} y = - \frac{1}{6} x \end{aligned}$
::y 16x

$\begin{aligned} y = 1.75 x \end{aligned}$
::y=1.75x y= 1.75x

Dasan’s mom takes him to the video arcade for his birthday.

In the first 10 minutes, he spends $3.50 playing games. If his allowance for the day is $20, how long can he keep playing games before his money is gone?
::在头十分钟里,他花3.50美元玩游戏。如果他一天的津贴是20美元,那么他还能在钱走之前玩游戏多久?

He spends the next 15 minutes playing Alien Invaders. In the first two minutes, he shoots 130 aliens. If he keeps going at this rate, how many aliens will he shoot in fifteen minutes?
::他在接下来的15分钟里扮演外星人入侵者。在头两分钟里,他射杀130名外星人。如果他继续这样下去,他15分钟内会射杀多少外星人?

The high score on this machine is 120000 points. If each alien is worth 100 points, will Dasan beat the high score? What if he keeps playing for five more minutes?
::这台机器的高分是120000分,如果每个外星人都值100分,大山会胜得高分吗?如果他再玩5分钟呢?

::Dasan的妈妈带他去视频街机过生日。在头10分钟里,他花3.50美元玩游戏。如果他一天的津贴是20美元,那么在钱走光之前他还能继续玩游戏多久?他接下来的15分钟玩外星人入侵者游戏。在头两分钟里,他射杀130名外国人。如果他继续这样下去,他将在15分钟内射杀多少外国人?这个机器的高分是120 000分。如果每个外国人都值100分,Dasan会打高分吗?如果他再玩5分钟又打5分钟呢?

The current standard for low-flow showerheads is 2.5 gallons per minute.

How long would it take to fill a 30-gallon bathtub using such a showerhead to supply the water?
::用这种淋浴头顶填满30加仑浴缸需要多长时间才能供水?

If the bathtub drain were not plugged all the way, so that every minute 0.5 gallons ran out as 2.5 gallons ran in, how long would it take to fill the tub?
::如果浴缸排水管没有一直通通通通通通堵塞,那么每分钟,每分钟有0.5加仑的水流流出2.5加仑,填满浴缸需要多长时间?

After the tub was full and the showerhead was turned off, how long would it take the tub to empty through the partly unplugged drain?
::在浴缸满满后,淋浴池被关了, 洗浴缸要多久才能清空, 才能清空部分未插管的排水管?

If the drain were immediately unplugged all the way when the showerhead was turned off, so that it drained at a rate of 1.5 gallons per minute, how long would it take to empty?
::如果在淋浴头关机时,排水管立即被拔掉,以每分钟1.5加仑的速度排水,那么要多久才空出来?

::低流量淋浴头目前的标准是每分钟2.5加仑。使用这样的淋浴头填满30加仑浴缸需要多长时间才能提供水? 如果浴缸排水管不完全堵塞, 因此每分钟有0.5加仑流出2.5加仑, 填满浴缸需要多长时间? 浴缸满了,淋浴头关了关机,浴缸要用多久才能清空? 如果在淋浴头关闭时立即拔出排水管, 排水管以每分钟1.5加仑的速度排水, 要用多久才能清空?

Amin is using a hose to fill his new swimming pool for the first time. He starts the hose at 10 PM and leaves it running all night.

At 6 AM he measures the depth and calculates that the pool is four sevenths full. At what time will his new pool be full?
::早上6点,他测量了水池的深度,计算出水池是七分之四。他的新水池什么时候会满?

At 10 AM he measures again and realizes his earlier calculations were wrong. The pool is still only three quarters full. When will it actually be full?
::早上10点,他再次测量并意识到他早先的计算是错误的。池子仍然只有四分之三满。它什么时候会满?

After filling the pool, he needs to chlorinate it to a level of 2.0 ppm (parts per million). He adds two gallons of chlorine solution and finds that the chlorine level is now 0.7 ppm. How many more gallons does he need to add?
::填满水池后,他需要将其氯化到2.0ppm的水平(百万分之一),他增加了两加仑氯溶液,并发现氯含量现在是0.7 ppm。他还需要加多少加仑?

If the chlorine level in the pool decreases by 0.05 ppm per day, how much solution will he need to add each week?
::如果池内的氯含量每天减少0.05 ppm,他每周需要增加多少解决办法?

::Amin第一次用水管来填充新的游泳池。他于下午10时开始水管, 并离开水管, 整个晚上运行。他于上午6时测量深度, 计算游泳池是满七分之四。他的新游泳池何时会满? 什么时候他的新游泳池会满? 在上午10时他再次测量, 并意识到他早先的计算是错误的。游泳池仍然满满四分之三。当水池填满后, 他需要将其氯化到2.0 ppm的水平( 每百万分之一 ) 。他增加了两加仑氯溶液, 并发现氯含量现在是0.7 ppm ppm 。他还需要增加多少加仑? 如果池中的氯含量每天减少0.05 ppm ppm, 那么他每周需要增加多少溶液 ?

Land in Wisconsin is for sale to property investors. A 232-acre lot is listed for sale for $200,500.

Assuming the same price per acre, how much would a 60-acre lot sell for?
::假设每英亩价格相同, 60英亩的地要卖多少钱?

Again assuming the same price, what size lot could you purchase for $100,000?
::再假设同样的价钱, 你买多少? 10万美元?

::威斯康辛州的地盘将出售给房地产投资者。 232英亩的地块以200,500美元的价格出售。假设每英亩价格相同,60英亩的地块将卖多少钱? 再假设同样的价格,10万美元能买到多少大小的地块?

The force $\begin{aligned} (F) \end{aligned}$ needed to stretch a spring by a distance $\begin{aligned} x \end{aligned}$ is given by the equation $\begin{aligned} F = k \cdot x \end{aligned}$ , where $\begin{aligned} k \end{aligned}$ is the spring constant (measured in Newtons per centimeter, or N/cm). If a 12 Newton force stretches a certain spring by 10 cm, calculate:

The spring constant, $\begin{aligned} k \end{aligned}$
::春季常数 k

The force needed to stretch the spring by 7 cm.
::这支部队需要把春天拉长7厘米

The distance the spring would stretch with a 23 Newton force.
::弹簧的距离将延伸与23牛顿的力量。

::F=kx 方程式给出了将弹簧伸展到距离x 所需的力 (F) 。方程式将弹簧常数( 以牛顿每厘米计, 或以牛顿每厘米计, 或以牛顿每厘米计) 。如果12 牛顿力将某一弹簧拉展10厘米, 计算: 弹簧常数, k 将弹簧伸展到7厘米。弹簧的距离将用23牛顿力延伸。

Angela’s cell phone is completely out of power when she puts it on the charger at 3 PM. An hour later, it is 30% charged. When will it be completely charged?
::安吉拉的手机在下午3点投入充电器时完全停电。一小时后,手机将充电30%。什么时候完全充电?

It costs $100 to rent a recreation hall for three hours and $150 to rent it for five hours.

Is this a direct variation?
::这是直接的变异吗?

Based on the cost to rent the hall for three hours, what would it cost to rent it for six hours, assuming it is a direct variation?
::根据租用大厅三个小时的费用,假定是直接变动,租用6小时的费用是多少?

Based on the cost to rent the hall for five hours, what would it cost to rent it for six hours, assuming it is a direct variation?
::根据租用大厅5小时的费用计算,假定有直接变化,租用6小时的费用是多少?

Plot the costs given for three and five hours and graph the line through those points. Based on that graph, what would you expect the cost to be for a six-hour rental?
::将给出的3小时和5小时的费用计为3小时和5小时的费用,然后通过这些点绘制线条图。根据这个图,你对6小时租房的费用有什么期望?

::租用娱乐厅3小时的费用为100美元,租用娱乐厅5小时的费用为150美元。这是直接的变数吗?根据租用3小时的费用,如果是直接变数,租6小时的费用为多少?根据租用5小时的费用,租6小时的费用为多少?假定是直接变数,租6小时的费用为多少?计算3小时和5小时的费用并通过这些点绘制线路图。根据这个图,你预计6小时租金的费用为多少?

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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