12.9 承认和应用引物权
章节大纲
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Professor Smith works in a laboratory and is training a new intern, Rakesh, in a specific task. Professor Smith tells Rakesh that when she herself did the task, she started with a very small sample of cobalt. She had 10 grams of it and took one-third of one-third of one-third of one-third of it. How can Rakesh figure out how many grams of the sample Professor Smith actually ended up using?
::Smith教授在实验室工作,正在培训一名新的实习生,Rakesh,从事一项具体的任务。Smith教授告诉Rakesh,当她本人完成这项任务时,她从一个很小的钴样本开始,她有10克的钴,占三分之一的三分之一。Rakesh如何知道Smith教授到底用了多少克?In this concept, you will learn to recognize and apply the power of a quotient property .
::在此概念中,您将学会承认和运用商数财产的力量。Power of a Quotient Property
::引物权Exponents can be applied to both fractions and quotients. For example, evaluate this multiplication , find the product of the numerators and the product of the denominators.
. To
::指数可以适用于分数和商数。 例如, (12) (12) (12) (12) (12) = (12) 4。 要评估此乘法, 请找到分子的产物和分母的产物 。The Power of a Quotient Property says that for any nonzero numbers integer :
and and any
::引号属性的功率表示,对于任何非零数a和b以及任何整数n:
:ab)n=anbn
Let’s look at an example.
::让我们举个例子。Simplify:
.
::简化53)4。
First, apply the product of a quotient property .
::首先,应用商数属性的产物。Next, expand to simplify.
::下一步,扩展到简化。The answer is
.
::答案是62581。Let’s look at another example.
::让我们再看看另一个例子。Simplify:
.
::简化( 3k2j) 4 。First, apply the product of a quotient property.
::首先,应用商数属性的产物。
:3k2j)4=(3k)4(2j)4
Next, expand to simplify.
::下一步,扩展到简化。
:3k)4(2j)4=34k422j4=3}3k422j4=3}333333kkk}k}k}k}k}k}2}2222}j}j}j}j}j}j}=81k416j4}
The answer is
.
::答案是81k416j4。Examples
::实例Example 1
::例1Earlier, you were given a problem about Professor Smith and the original 10 grams of cobalt.
::早些时候,你被问及史密斯教授 和最初的10克钴的问题。Rakesh needs to figure out how much of the cobalt Professor Smith actually ended up using after she took one-third of one-third of one-third of one-third of 10 grams.
::拉凯什需要弄清楚史密斯教授在拿了10克中三分之一的三分之一的三分之一后 究竟用了多少钴教授To figure out the number of grams in the sample, he must use monomials and powers.
::为了找出样本中的克数,他必须使用单项和单项权力。First, set up the problem to model the information given in the story.
::首先,设置问题来模拟故事中所提供的信息。Next, apply the product of a quotient property.
::其次,应用商数属性的产物。Then, expand to simplify.
::然后,扩展到简化。The answer is
.
::答案是1081。Professor Smith’s sample size was grams.
::Smith教授的样本规模为1081克。Example 2
::例2Simplify the following quotient.
::简化下列商数。
:-4x3y)3
First, apply the product of a quotient property.
::首先,应用商数属性的产物。
:-4x3y)3=(-4x3y)3(-3y)3
Next, expand to simplify.
::下一步,扩展到简化。The answer is
.
::答案是 - 64x327y3。Example 3
::例3Simplify the quotient:
.
::简化商数453)
First, apply the product of a quotient property.
::首先,应用商数属性的产物。Next, expand to simplify.
::下一步,扩展到简化。The answer is
.
::答案是64125。Example 4
::例4Simplify the quotient:
.
::简化商数 : (2a3b) 2。First, apply the product of a quotient property.
::首先,应用商数属性的产物。
:2a3b)2=22a232b2
Next, expand to simplify.
::下一步,扩展到简化。The answer is
.
::答案是4a29b2。Example 5
::例5Simplify the quotient:
.
::简化商数a5b) 3。
First, apply the product of a quotient property.
::首先,应用商数属性的产物。
:a5b)3=a353b3
Next, expand to simplify.
::下一步,扩展到简化。
::a353b3=aaaa55_5_5}bb=a3125b3=a3125b3The answer is
.
::答案是a3125b3。Review
::回顾Simplify.
::简化。-
:7k-2m)3
-
:3x-2y)3
-
:4x-3y)4
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:5y-2z)5
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:-2y4z)4
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:4xy-2z5)5
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:12x2y4-6z3)2
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:7x2y-2z3)3
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:2x3y2-2z3)3
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:x119)5
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:-5x33h2j8)5
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 。