节: 与表达式合作综合报告 | 2.6 " 与表达方式合作 " 综述-interactive | The Way To Learn

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The Purpose of this Lesson
::本课程的目的

In this lesson, you will solve simple equations of the various types you explored in the first chapter. You'll also use physics equations to practice simplifying. Your work with expressions in this chapter will help you when working with expressions in this lesson.
::在此课中, 您将解析您在第一章中探讨过的各种类型的简单方程式。您还将使用物理方程式来练习简化。您在本章中的表达方式将帮助您在此课中的表达方式中工作。

Activity 1 : Revisiting Density
::活动1:重新检查密度

Example 1-1
::例1-1

The density of an object is its mass divided by its volume :
::对象的密度是其质量除以其体积:

$\begin{aligned} D = \frac{m}{V} \end{aligned}$
$\begin{align*}D=\frac{m}{V}\end{align*}$
::D=mV =mV =mV =mV

Cork has a density of 225 kilograms per cubic meter (225 kilograms is about 500 pounds, and a cubic meter is about the combined volume of a washer and dryer).
::Cork的密度为每立方米225公斤(225公斤约为500磅,1立方米约为洗衣机和烘干机的合计体积)。

Cork has a density of 225 kg per cubic meter

Laird wants to make a surfboard out of cork. The density is adequate for floating. Laird's design will give the board a volume of 0.08 cubic meters. How many kilograms of cork does Laird need for his board?
::莱尔德想用软木做一个冲浪板。密度足以漂浮。莱尔德的设计将使板体积为0.08立方米。莱尔德的板需要多少公斤的软木?

Cork surfboards are uncommon, but do exist
::滑雪滑板并不常见,但确实存在

Solution: You substitute the known density and volume and solve for mass:
::溶液: 您用已知的密度和音量来替代质量 :

$\begin{aligned} \begin{array}{ll} Equation & Explanation \\ 225 = \frac{m}{0.08} & Substituting known values into the equation. \\ \frac{0.08}{1} \cdot 225 = \frac{m}{0.08} \cdot \frac{0.08}{1} & The inverse of division is multiplication. \\ m = 0.08 \cdot 225 & Re-arranging. \\ m = \frac{8}{100} \cdot \frac{225}{1} & 0.08 = \frac{8}{100} . \\ m = 8 \cdot 2.25 & \frac{225}{100} = 2.25 \\ m = 18 kg & Multiplying. \end{array} \end{aligned}$
$\begin{align*}\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline 225=\frac{m}{0.08} & \text{Substituting known values into the equation.}\\ \frac{0.08}{1}\cdot 225 =\frac{m}{0.08} \cdot \frac{0.08}{1} & \text{The inverse of division is multiplication.}\\ m=0.08 \cdot 225 & \text{Re-arranging.}\\ m = \frac{8}{100}\cdot \frac{225}{1} & 0.08=\frac{8}{100}.\\ m=8 \cdot 2.25 & \frac{225}{100}=2.25\\ m=18~\text{kg} & \text{Multiplying.} \end{array}\end{align*}$
::等式Explation225=m0.08 将已知值转换成等式.0.081225=m0.080.081。分法的反向是乘法.m=0.08225 重新排列.m=81002251.08=8100.m=8225225100=2.25m=18kmMultipliaty。

That makes for a longboard at the top end of acceptable mass, much more than usual, and it's difficult to maneuver the board. In fact, many longboards have a mass closer to 10 kilograms. On the other hand, it would be pretty cool to have a surfboard made of cork! They do exist!
::这使得在可接受的质量的顶端有一个长板,远比往常大得多,而且很难操作板。事实上,许多长板的重量接近10公斤。另一方面,用软木制成的冲浪板是很酷的!它们的确存在!

In this problem, you used an inverse operation to solve for the unknown. Multiplication is the inverse of division .
::在此问题上, 您使用反向操作来解决未知问题。乘法是分裂的反向。

PLIX Interactive
::PLIX 交互式互动

Activity 2 : Using Inverse Operations to Solve Equations
::活动2:使用反向操作来解析等量

Example 2-1

::例2-1

A rocket is accelerating at 60 meters per second, per second. When does the rocket pass the 5-kilometer mark?

::一枚火箭每秒60米加速,每秒60米,火箭何时通过5公里标记?

Solution:

$\begin{aligned} \begin{array}{ll} Equation & Explanation \\ y = \frac{1}{2} a x^{2} & Quadratic model, distance as a function of time. \\ 5000 = \frac{1}{2} (60) x^{2} & Substituting known values. \\ 5000 = 30 x^{2} & Simplifying. \\ \frac{500}{3} = x^{2} & Division is the inverse of multiplication. \\ x = \pm \sqrt{\frac{500}{3}} & Square-rooting is the inverse of squaring. \\ x = \sqrt{\frac{500}{3}} & Only the positive result makes sense. \\ x = \frac{\sqrt{500}}{\sqrt{3}} & The root of a quotient is the quotient of the roots. \\ x = \frac{10 \sqrt{5}}{\sqrt{3}} & Simplifying. \\ x = \frac{10 \sqrt{5}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} & Rationalizing the denominator. \\ x = \frac{10 \sqrt{1} 5}{3} & The product of roots is the root of the product. \end{array} \end{aligned}$
$\begin{align*}\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline y=\frac12 ax^2 & \text{Quadratic model, distance as a function of time.}\\ 5000=\frac12 (60) x^2 & \text{Substituting known values.}\\ 5000=30x^2 & \text{Simplifying.}\\ \frac{500}{3}=x^2 & \text{Division is the inverse of multiplication.}\\ x=\pm\sqrt{\frac{500}{3}} & \text{Square-rooting is the inverse of squaring.}\\ x=\sqrt{\frac{500}{3}} & \text{Only the positive result makes sense.}\\ x=\frac{\sqrt{500}}{\sqrt{3}} & \text{The root of a quotient is the quotient of the roots.}\\ x=\frac{10\sqrt5}{\sqrt3} & \text{Simplifying.}\\ x=\frac{10\sqrt5}{\sqrt3}\cdot \frac{\sqrt3}{\sqrt3} & \text{Rationalizing the denominator.}\\ x=\frac{10\sqrt15}{3} &\text{The product of roots is the root of the product.} \end{array}\end{align*}$
::解析度 = 12ax2 二次曲线模型, 距离作为时间的函数 5000= 12( 60x2) , 以时间的函数 5000= 30x2 = 30x2 简化 50.03 =x2 分数是乘法的反值。 x\\ 50003 Square 根部是 quarring.x= 5003 的反值。xx = 5003 纯正结果是有道理的。商数的根是根的商数。 x= 1053.Simplication.x= 1053.3 分母化。x= 10153 根的产物是产品的根。

In the above problem, you used two inverse operations : division and square rooting. The rest of your work was in simplifying and rationalizing the denominator, using the tools you learned in this chapter.
::在上述问题中,你使用了两种反向操作:分裂和平方根。你的其他工作是利用本章中学到的工具简化分母并使之合理化。

Interactive
::交互式互动

In the previous examples, you've used inverse operations to solve for the unknown. You've probably used inverse operations in every equation you've ever solved. In the past two chapters, you've expanded the types of equations and expressions you can work with. Use the interactive below to review a variety of inverse operations you will use to solve equations.
::在前几个例子中, 您用反向操作来解决未知的操作。您很可能在您曾经解答过的方程式中都使用了反向操作。在过去的两章中, 您扩大了您可以使用的方程式和表达式类型。使用下面的交互操作来审查您用来解答方程式的各种反向操作。

Work it Out
::工作出来

Each equation below can be solved with a single inverse operation. State the inverse operation required, and solve the equation.
$\begin{align*}\quad\begin{array}{ll} \text{a.} & 45=30a\\ \text{b.} & \frac{60}{49}=b^2\\ \text{c.} & c^2=\frac{81}{8}\\ \text{d.} & \sqrt{d}=6\\ \text{e.} & 42e=21\\ \text{f.} & 64=f^3\\ \text{g.} & 0.001=g^2\\ \text{h.} & h^7=2^7\\ \end{array}\end{align*}$
::下面的每一个方程式都可用一个反向操作解答。请指出所需的反向操作, 并解析方程式。 a. 45=30ab. 6049=b2c. c2=818d. d=6e.42e=21f. 64=f3g. 0.001=g2h. h7=27

Create 3 equations. Each equation must require the application of 2 inverse operations to solve. Then solve each equation.
::创建 3 个方程式。每个方程式必须应用两个反向操作才能解答。然后解答每个方程式。

Determine if each of the following equations is linear, quadratic, cubic, square root , or rational. Describe the nature and order of the inverse operations required to solve each equation below. Then solve. Simplify answers completely, and rationalize denominators where necessary.
::确定以下方程式中的每一个方程式是线性、二次方程式、立方方方程式、平方根或理性方程式。描述解决下面每个方程式所需的反向操作的性质和顺序。然后解答。完全简化答案, 必要时使分母合理化。

$\begin{align*}\quad\begin{array}{l} \quad \text{a.} & 5=6x+\frac57\\ \quad \text{b.} & 5\sqrt2=6x+3\sqrt2\\ \quad \text{c.} & 2\pi-x=3\pi\\ \quad \text{d.} & 5=4+\sqrt{x}\\ \quad \text{e.} & \sqrt7=2\sqrt{x}\\ \quad \text{f.} & \frac12 x^2=\frac23\\ \quad \text{g.} & \frac{5}{x}=-3\\ \quad \text{h.} & \frac{5}{11}=\frac{3}{x}\\ \quad \text{i.} & -4\sqrt{x}=-14\\ \quad \text{j.} & 5x^3=625\\ \quad \text{k.} & \frac{x}{3}-\frac12=4\\ \quad \text{l} & 8x^2=27\\ \quad \text{m.} & 0.09x^2=98\\ \quad \text{n.} & 5(x+2)=30\\ \quad \text{o.} & \sqrt{x}=3\\ \quad \text{p.} & \sqrt{x+2}=3\\ \end{array}\end{align*}$
::a.5=6x+57b.52=6x+32c.2**x=3d.5=4+xe.7=2x12x2=23g.5x=3h.511=3xxx.-4x14j.5x3=625k.x3-12=4l8x2=27m.0.09x2=98n.5(x+2)=30o.x=3p.x+2=3xxxxxxxx3xxxxxx3xxxxx3xx4x4x4x625k.x3=4l8x2=27m.0.09x2=98n.5(x+2)=30o.x=3p.x+2=3

Activity 3 : Solve Physics and Geometry Equations
::活动3:解决物理和几何等等同

Work it Out
::工作出来

There are many powerful equations from the disciplines of physics and geometry that describe relationships in our world. Substitute the given values into the given equations, then solve and simplify to find the result.
::物理学和几何学的学科中有许多强有力的方程式来描述我们世界的关系。将给定的值替换为给定方程式,然后解决和简化以寻找结果。

The circumference of a circle is $\begin{align*}\pi\end{align*}$ times the diameter: $\begin{align*}C=\pi d.\end{align*}$ If the circumference of a circle is $\begin{align*}2\pi\end{align*}$ , what's the diameter?
::圆的环绕值是直径的乘以 : Cd。如果圆的环绕值是 2, 直径是什么 ?

Two sides of a right triangle measure 1 and 2. One is the hypotenuse . Find the remaining side.
::右三角形措施1和2的两边,一个是下限,找到另一边。

Two sides of a right triangle measure $\begin{align*}\frac{\sqrt2}{2}.\end{align*}$ Find the remaining side.
::右三角措施22的两边找到另一边

The force being applied to an object is its mass times its acceleration. If a 100-kilogram object is accelerating at 10 meters per second, per second, find the force. Write your result in scientific notation . The units for force are $\begin{align*}\frac{\text{kg}\cdot\text{m}}{\text{s}^2}\end{align*}$ otherwise known as Newtons (N).
::用于对象的强制力是其质量乘以其加速度。如果一个100公斤的物体加速速度为每秒10米, 每秒10米, 则会找到它。写上你的科学标记。强制力单位是kgms2, 也称为牛顿( N ) 。

The mass of a comet is $\begin{align*}5.76\cdot10^{12}\end{align*}$ kilograms. It's accelerating towards a stationary object at $\begin{align*}7.324\cdot10^5\end{align*}$ meters per second, per second. Determine the force of the comet's impact. Write your result in scientific notation.
::彗星的质量为5. 761012公斤。它正在加速向固定物体前进, 每秒7. 324. 105米。确定彗星撞击的威力。将结果写在科学符号上。

Given two objects, Newton's law of universal gravitation tells us the relationship between their masses, their centers of gravity, a nd the force of their attraction. The equation is $\begin{align*}F=\frac{G\cdot m_1\cdot m_2}{r^2}.\end{align*}$ $\begin{align*}G\end{align*}$ is a constant : $\begin{align*}6.67\cdot10^{\text{-}11} \tfrac{\text{N}\cdot m^2}{kg^2}.\end{align*}$ The other variables are the mass of the two objects, $\begin{align*}m_1 \text{ and }m_2,\end{align*}$ in kg, the distance between them, $\begin{align*}r,\end{align*}$ in meters, and the force of their attraction, $\begin{align*}F,\end{align*}$ in Newtons. Jim has a mass of 85 kilograms. His mass applies a force of 834 Newtons to a scale. He's standing at sea level. The mass of the earth is $\begin{align*}5.98\cdot10^{24}\end{align*}$ kilograms. What's the radius of the earth?
::牛顿的宇宙引力定律告诉我们它们的质量、重力中心及其吸引力之间的关系。方程式是 F = Gcm1m2r2. G 是一个常数 : 6. 6710- 11Nm2kg2. 其他变量是两个物体的质量, 以公斤计, m1 和 m2, 它们之间的距离, 以米计, r, 和它们的吸引力, F 。吉姆的重量为 85 公斤。他的重量将834 牛顿的威力应用到一个比例。他站在海平面上。地球的重量是 5. 981024公斤。地球的半径是什么?

Summary
::摘要

Solving equations often requires the use of inverse operations.
As you learn about new types of functions and equations, you expand your inverse operation toolbox accordingly.
::解析方程式通常需要使用反向操作。随着您了解新的函数和方程式类型,您会相应扩展反向操作工具箱。

Equations from geometry and physics provide many opportunities to practice simplifying with fractions, exponents, and more.
::几何和物理学的等同为采用简化方法提供了许多机会,包括分数、指数等等。

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