10.6 解决双方具有变量的极端等式

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  解决双方有变量的激进方程式

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  A business wants to determine how much money it should spend on advertising to generate $40,000 in profit. If the formula is  $\begin{array}{r}P=500\sqrt{a}-a\end{array}$ , where a is the amount spent on advertising and  P is the profit, how much does this business need to spend on advertising?  
  ::一个企业想要决定它应该花在广告上多少钱来产生40,000美元的利润。 如果公式是P=500a-a,那么广告和P是利润,这个企业需要花在广告上多少钱?

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  In this section, we discuss more complicated radical equations like this one. 
  ::我们在本节讨论更复杂的激进方程式, 比如这个方程式。

  

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  Solving Radical Equations
  ::解决激进等号

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  In this concept, we will continue solving radical equations. Here we will address variables and radicals on both sides of the equation .
  ::在这个概念中,我们将继续解决激进方程式问题,在这里我们将处理方程式两侧的变数和激进方程式问题。

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    Solving Equations with More Than One Radical Expression
  ::用多于一个的激进表达式解析等式

  Isolate only one radical expression at a time, and follow the same steps from the previous section.

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  Example 1
  ::例1

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  Solve $\begin{array}{r}\sqrt{4x+1}-x=\text{-}1\end{array}$ .
  ::解决 4x+1 -x=1。

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  Solution: Now we have an $\begin{array}{r}x\end{array}$ that is not under the radical. Our steps do not change—that is, we still isolate the radical 1st.
  ::解决方案:现在我们有一个不处于激进下的x。我们的步骤不会改变,也就是说,我们仍然孤立了激进的1号。

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  $$\begin{array}{rl}\sqrt{4x+1}-x& =\text{-}1\\ \sqrt{4x-1}& =x-1\end{array}$$

  
  ::4x+1-x=-14x-1=x-1

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  Now, we can square both sides. Be careful when squaring $\begin{array}{r}x-1\end{array}$ ; the answer is not $\begin{array}{r}{x}^2-1\end{array}$ .
  ::现在, 我们可以将两边平开。 当对齐 x - 1 时要小心; 答案不是 x2 - 1 。

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  $$\begin{array}{rl}(\sqrt{4x+1}{)}^2& =(x-1{)}^2\\ 4x+1& =x^2-2x+1\end{array}$$

  
  :4x+1)2=(x-1)24x+1=x2-2x+1

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  This problem has a quadratic term . Combine like terms and set one side equal to 0. From there we can solve by factoring.
  ::这个问题有一个二次词。 将类似条件合并, 并设定一面等于 0。 从那里我们可以通过保理解决 。

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  $$\begin{array}{rl}4x+1& =x^2-2x+1\\ 0& =x^2-6x\\ 0& =x(x-6)\\ x& =0\text{or}x=6\end{array}$$

  
  ::4x+1=x2-2x+10=x2-6x0=x(x-6)xx=0或x=6

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  Checking both solutions remains especially important.
  ::检查这两种解决办法仍然特别重要。

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  $$\begin{array}{rlr}\sqrt{4\left(0\right)+1}-1=\sqrt{0+1}-1=1-1=0\ne \text{-}1& & \text{0 is an extraneous solution.}\end{array}$$

  
  ::4(0)+1-1=0+1-1=1-1=1-1=0-10是一种不相干的解决办法。

  $$\begin{array}{r}\sqrt{4\left(6\right)+1}-6=\sqrt{24+1}-6=5-6=\text{-}1\end{array}$$

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  Therefore , 6 is the only solution.
  ::因此,6是唯一的解决办法。

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  Example 2
  ::例2

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  A business wants to determine how much money it should spend on advertising to generate $40,000 in profit. If the formula is $\begin{array}{r}P=500\sqrt{a}-a\end{array}$ , where $\begin{array}{r}a\end{array}$  is the amount spent on advertising, and P is the profit, how much does this business need to spend on advertising?    
  ::一个企业想要决定它应该花在广告上多少钱来产生40,000美元的利润。 如果公式是P=500a-a,那么在广告上花费的数额在哪里,而P是利润,那么这个企业需要花在广告上多少钱?

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  Solution:  We proceed as we did in the previous example. 
  ::解决办法:我们与前一个例子一样行事。

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  $$\begin{array}{rl}40,000& =500\sqrt{a}-a\\ 40,000+a& =500\sqrt{a}\\ \frac{40,000+a}{500}& =\sqrt{a}\\ (\frac{40,000+a}{500}{)}^2& =(\sqrt{a}{)}^2\\ \frac{(40,000+a{)}^2}{{500}^2}& =a\\ \frac{1,600,000,000+80,000a+a^2}{250,000}& =a\\ 6,400+0.32a+\frac{a^2}{250,000}& =a\\ \frac{a^2}{250,000}-0.68a+6,400& =0\end{array}$$

  
  ::40,000=500a-a40,000+a=500a40,000+a500=a(40,000+a500)2=(a)(a)(40,000+a)25002=a1,600,000,000+80,000+a2250,000=a6,400,000+a(a)+0.32a+a2250,000=aa2250,000-0.68a+6,400=0

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  Let's use the quadratic formula to solve this. 
  ::我们用二次方程式解决这个问题吧

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  $$\begin{array}{rl}a& =\frac{\text{-}(\text{-}0.68)\pm \sqrt{(\text{-}0.68{)}^2-4(\frac{1}{250,000})(6,400)}}{2(\frac{1}{250,000})}\\ & =\frac{0.68\pm \sqrt{0.464-0.1024}}{\frac{1}{125,000}}\\ & =\frac{0.68\pm 0.60}{\frac{1}{125,000}}\\ a=10,000& \text{or}a=160,000\end{array}$$

  
  ::a=-(-0.68)](-0.68)/2-4(1250,000)(6,400)(2,250,000)=0.680.464-0.1024.1125,000=0.68×0.0.0.60115,000a=10,000或a=160,000)

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  Let's hope that, for $40,000 in profit, you need to spend only $10,000.  
  ::让我们希望,对于4万美元的利润, 你只需要花一万块。

  $$\begin{array}{rl}40,000& =500\sqrt{10,000}-10,000\\ 40,000& =500\cdot 100-10,000\\ 40,000& =50,000-10,000\\ 40,000& =40,000\\ \\ 40,000& =500\sqrt{160,000}-160,000\\ 40,000& =500\cdot 400-160,000\\ 40,000& =200,000-160,000\\ 40,000& =40,000\end{array}$$

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  Even though $10,000 and $160,000 both check, it makes more sense to keep costs down and spend only $10,000.  
  ::虽然两张支票都核对了10 000美元和160 000美元,但降低成本和只支出10 000美元更有意义。

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  by cpfaffinator demonstrates how to solve radical . 
  ::由折叠器演示如何解析激进 。

   

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  Example 3
  ::例3

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  Solve $\begin{array}{r}\sqrt{8x-11}-\sqrt{3x+19}=0\end{array}$ .
  ::解决 8x- 11- 3x+19=0 。

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  Solution: In this example, you need to isolate one radical at a time. To do this, add the 2nd radical to both sides. Then square both sides.
  ::解决方案 : 在此示例中, 您需要一次孤立一个激进分子 。 要做到这一点, 请将第二激进分子加入到两边 。 然后将两边平开 。

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  $$\begin{array}{rl}\sqrt{8x-11}-\sqrt{3x+19}& =0\\ (\sqrt{8x-11}{)}^2& =(\sqrt{3x+19}{)}^2\\ 8x-11& =3x+19\\ 5x& =30\\ x& =6\end{array}$$

  
  ::8-11-3x+19=0(8x-11)2=(3x+19)28x-11=3x+19x=3x=195x=30x=6

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  Check: $\begin{array}{r}\sqrt{8\left(6\right)-11}-\sqrt{3\left(6\right)+19}=\sqrt{48-11}-\sqrt{18+19}=\sqrt{37}-\sqrt{37}=0\end{array}$
  ::检查时间: 8(6)-11-3(6)+19=48-11-18+19=37-37=0

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  Example 4
  ::例4

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  Solve  $\begin{array}{r}\sqrt[3]{4x^3-24}=x\end{array}$ .
  ::解决 4x3 -243=x

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  Solution: The radical is isolated. Cube both sides to eliminate the cube root .
  ::解答:激进分子是孤立的。立方体两边消灭立方根。

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  $$\begin{array}{rl}(\sqrt[3]{4x^3-24}{)}^3& =x^3\\ 4x^3-24& =x^3\\ \text{-}24& =\text{-}3x^3\\ 8& =x^3\\ 2& =x\end{array}$$

  
  :4x3-2433)3=x34x3-24=x3-24=-3x38=x32=x

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  Check: $\begin{array}{r}\sqrt[3]{4{\left(2\right)}^3-24}=\sqrt[3]{32-24}=\sqrt[3]{8}=2\end{array}$
  ::查询:4(2)3-243=32-243=83=2

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  Example 5
  ::例5

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  Solve $\begin{array}{r}(2x^2-1{)}^{\frac{1}{4}}=x\end{array}$ .
  ::溶解 (2x2- 1) 14=x.

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  Solution: The radical is isolated. To eliminate it, we must raise both sides to the 4th power.
  ::解决方案:激进分子是孤立的。为了消灭它,我们必须将双方提升到第四势力。

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  $$\begin{array}{rl}((2x^2-1{)}^{\frac{1}{4}}{)}^4& =x^4\\ 2x^2-1& =x^4\\ 0& =x^4-2x^2+1\\ 0& =(x^2-1)(x^2-1)\\ 0& =(x-1)(x+1)(x-1)(x+1)\\ x& =1\text{or}\text{-}1\end{array}$$

  
  :( 2x2 - 1) 14) 4=x42x2 - 1=x40=x40=x4=- 2x2+10=( x2 - 1) (x2 - 1) (x2 - 1) 0=( x-1) (x1) (x1) (x1) (x+1) (x+1) (x+1)x=1 或 -1

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  Check:

  $$\begin{array}{r}\sqrt[4]{2(1{)}^2-1}=\sqrt[4]{2-1}=\sqrt[4]{1}=1\\ \sqrt[4]{2(-1{)}^2-1}=\sqrt[4]{2-1}=\sqrt[4]{1}=1\end{array}$$

  
  ::检查: 2(1)2 - 14=2 - 14=14=14=12( - 1) - 2 - 14=2 - 14=14=1

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  Example 6
  ::例6

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  Solve $\begin{array}{r}\text{-}2(x-5{)}^{\frac{3}{4}}+48=\text{-}202\end{array}$ .
  ::解决 - (2x-5) 34+48=-202。

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  Solution: Isolate $\begin{array}{r}(x-5{)}^{\frac{3}{4}}\end{array}$ by subtracting 48 and dividing by $\begin{array}{r}\text{-}2\end{array}$ .
  ::解决办法:孤立(x-5-5)34,减去48,除以-2。

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  $$\begin{array}{rl}\text{-}2(x-5{)}^{\frac{3}{4}}+48& =\text{-}202\\ \text{-}2(x-5{)}^{\frac{3}{4}}& =\text{-}250\\ (x-5{)}^{\frac{3}{4}}& =125\end{array}$$

  
  :2x-5534+48=-202-2(x-55)34=-250(x-55)34=125

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  To undo the 3/4 power, raise everything to the 4/3 power.
  ::推翻3/4权力 将一切提升到4/3权力

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  $$\begin{array}{rl}{\left[{\left(x-5\right)}^{\frac{3}{4}}\right]}^{\frac{4}{3}}& ={125}^{\frac{4}{3}}\\ x-5& =625\\ x& =630\end{array}$$

  
  ::[(x--534)]43=12543x-5=625x=630

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  Check: $\begin{array}{r}\text{-}2(630-5{)}^{\frac{3}{4}}+48=\text{-}2\cdot {625}^{\frac{3}{4}}+48=\text{-}2\cdot 125+48=\text{-}250+48=\text{-}202\end{array}$
  ::检查: -2( 630- 55) 3434+48=-262534+48=-2125+48=- 250+48=- 202

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  by HCCMathHelp demonstrates how to solve and rational exponents. 
  ::HCCMathHelp展示了如何解决和理性的推手。

   

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  Summary
  ::摘要

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  • To solve rational equations with variables on both sides, isolate one radical expression at a time and follow the steps as in the previous section.
    ::解决双方有变量的合理方程式,一次孤立一个激进表达式,并遵循前一节中的步骤。

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  Review
  ::回顾

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  Solve the radical equations below. Be sure to check for extraneous solutions.
  ::解决下面的激进方程式。 请务必检查不相干的解决办法 。

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  1.  $\begin{array}{r}\sqrt{x-3}=x-5\end{array}$
  ::1. x-3=x-5

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  2.  $\begin{array}{r}\sqrt{x+3}+15=x-12\end{array}$
  ::2. x+3+15=x-12

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  3.  $\begin{array}{r}\sqrt[4]{3x^2+54}=x\end{array}$
  ::3. 3x2+544=x

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  4.  $\begin{array}{r}\sqrt{x^2+60}=4\sqrt{x}\end{array}$
  ::4. x2+60=4x

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  5.  $\begin{array}{r}\sqrt{x^4+5x^3}=2\sqrt{2x+10}\end{array}$
  ::5. x4+5x3=22x+10

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  6.  $\begin{array}{r}x=\sqrt{5x-6}\end{array}$
  ::6. x=5x-6

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  7.  $\begin{array}{r}\sqrt{3x+4}=x-2\end{array}$
  ::7. 3x+4=x-2

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  8.  $\begin{array}{r}\sqrt{x^3+8x}-\sqrt{9x^2-60}=0\end{array}$
  ::8. x3+8x-9x2-60=0

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  9.  $\begin{array}{r}x=\sqrt[3]{4x+4-x^2}\end{array}$
  ::9. x=4x+4-x23

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  10.  $\begin{array}{r}\sqrt[4]{x^3+3}=2\sqrt[4]{x+3}\end{array}$
  ::10. 3+34=2x+34

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  11.  $\begin{array}{r}{x}^2-\sqrt{42x^2+343}=0\end{array}$
  ::11. x2 - 42x2+343=0

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  12.  $\begin{array}{r}x\sqrt{x^2-21}=2\sqrt{x^3-25x+25}\end{array}$
  ::12.xx2-21=2x3-25x+25

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  13.  $\begin{array}{r}3x^{\frac{1}{3}}+5=17\end{array}$
  ::13. 3x13+5=17

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  14.  $\begin{array}{r}(7x-3{)}^{\frac{2}{5}}=4\end{array}$
  ::14. (7x-3-3)25=4

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  15.  $\begin{array}{r}(4x+5{)}^{\frac{1}{2}}=x-4\end{array}$
  ::15. (4x+5)12=x-4

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  16.  $\begin{array}{r}(7x-8{)}^{\frac{2}{3}}=4(x-5{)}^{\frac{2}{3}}\end{array}$
  ::16. (7x-8)23=4(x-5)23

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  17.  $\begin{array}{r}4997=5x^{\frac{3}{2}}-3\end{array}$
  ::17.4997=5x32-3

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  For questions 18-20, you will need to use the method illustrated in the example below.
  ::对于问题18-20,你将需要使用下面示例中说明的方法。

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  $$\begin{array}{rl}\sqrt{x-15}& =\sqrt{x}-3\\ {\left(\sqrt{x-15}\right)}^2& ={\left(\sqrt{x}-3\right)}^2\\ x-15& =x-6\sqrt{x}+9\\ \text{-}24& =\text{-}6\sqrt{x}\\ (4{)}^2& ={\left(\sqrt{x}\right)}^2\\ 16& =x\end{array}$$

  
  ::x- 15=x-3(x- 15) 2=(x-3) 2x- 15=x-6x+9-24=6-24=6x(4)2=(x) 216=x

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  1. Square both sides
     ::两边广场
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  2. Combine like terms to isolate the remaining radical.
     ::结合一些术语来隔离剩下的激进分子
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  3. Square both sides again to solve.
     ::双方重开广场解决

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  Check: Do not forget to check your answers for extraneous solutions.
  ::选中 : 不要忘记检查您对不相干解决方案的答案 。

  $$\begin{array}{rl}\sqrt{16-15}& =\sqrt{16}-3\\ \sqrt{1}& =4-3\\ 1& =1\end{array}$$

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  by CK-12 demonstrates how to solve special cases of radical equations.
  ::CK-12展示了如何解决激进方程式的特殊情况。

   

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  18.  $\begin{array}{r}\sqrt{x+11}-2=\sqrt{x-21}\end{array}$
  ::18. x+11-2=x-21

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  19.  $\begin{array}{r}\sqrt{x-6}=\sqrt{7x}-22\end{array}$
  ::19. x-6=7x-22

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  20.  $\begin{array}{r}2+\sqrt{x+5}=\sqrt{4x-7}\end{array}$
  ::20. 2+x+5=4x-7

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  Explore More
  ::探索更多

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  1.  The legs of a right triangle measure 12 and $\begin{array}{r}\sqrt{x+1}\end{array}$ . The hypotenuse measures $\begin{array}{r}\sqrt{7x+1}\end{array}$ . What are the lengths of the sides with the unknown values?
  ::1. 右三角的双腿测量12和x+1。 下限测量7x+1。 具有未知值的两边长度是多少?

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  2.  The period (in seconds) of a pendulum with a length of L (in meters) is given by the formula ¹ $\begin{array}{r}P=2\pi {(\frac{L}{9.8})}^{\frac{1}{2}}\end{array}$ . If the period of a pendulum is   $\begin{array}{r}10\pi ,\end{array}$  what is  the length of the pendulum?
  ::2. 长度为L(米)的钟摆的时段(秒)由公式1 P=2(L9.8)12给出。如果钟摆的时段为10,钟摆的时段长度是多少?

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  3.  The hull speed, $\begin{array}{r}s\end{array}$ , in nautical miles per hour of a sailboat can be modeled by the formula  $\begin{array}{r}s=1.34\sqrt{l}\end{array}$ , where $\begin{array}{r}l\end{array}$  is the length in feet of the sailboat's waterline ² . Find the speed of a boat whose hull length is 10 feet. Round your answer to the nearest tenth of a nautical mile per hour.
  ::3. 帆船每小时的船体速度,S,每海里的船体速度,可按公式S=1.34l模拟,其中我为帆船水系足长2。 找到船体长度为10英尺的船只的速度。 请将回答回至每小时10海里的最接近10海里的速度。

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  4.  The number of bacteria in a swimming pool after disinfection can be modeled by the equation $\begin{array}{r}y=5,550\sqrt{0.025x+0.1}\end{array}$ , where $\begin{array}{r}x\end{array}$  is the number of minutes past. How many minutes later are there 3,000 bacteria left in the swimming pool?
  ::4. 消毒后游泳池中的细菌数量可以通过y=5 5500.025x+0.1等式进行模拟,后者是过去分钟的x。

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  5.  The average amount of sewage discharged by a manufacturing plant (in tons per month) from 2001 to 2010 can be modeled by the equation $\begin{array}{r}y=235\sqrt{2x+0.75}\end{array}$ , where $\begin{array}{r}x\end{array}$  is the number of years since 2000. In what year was 610 tons of sewage discharged per month?
  ::5. 以y=2352x+0.75的公式为模型模拟了2001年至2010年制造厂平均排污量(每月吨),2000年以来的年数是xx,2000年是610吨每月排污量?

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  Answers for Review and Explore More Problems
  ::回顾和探讨更多问题的答复

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  Please see the Appendix. 
  ::请参看附录。

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  PLIX
  ::PLIX

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  Try this interactive that reinforces the concepts explored in this section:
  ::尝试这一互动,强化本节所探讨的概念:

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  References
  ::参考参考资料

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  1. "Pendulum," last edited May 30, 2017,
  ::1. 2017年5月30日 上一次编辑的“Pendulum”,

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  2. "Crunching Numbers: Hull Speed & Boat Length," by Charles Doane, posted March 26, 2010,
  ::2. Charles Doane于2010年3月26日上传“清理数字:超速和船长”,