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heureka

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Имя пользователяheureka
Гол22537
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Вопросов 11
ответы 4358

 #2
avatar+22537 
+3

Ms. Forsythe gave the same algebra test to her three classes.
The first class averaged 80%, the second class averaged 85%, and the third 89%.
Together, the first two classes averaged 83%, and the second and third classes together averaged 87%.
What was the average for all three classes combined?
Express your answer to the nearest hundredth.

 

\(\text{The students in the first class $ = s_1$ } \\ \text{The students in the second class $ = s_2$ } \\ \text{The students in the third class $ = s_3$ } \\ \text{The sum of the points in the first class $ = p_1$ } \\ \text{The sum of the points in the second class $ = p_2$ } \\ \text{The sum of the points in the third class $ = p_3$ } \\ \text{The maximal points of the test $ = p$ } \)

 

\(\begin{array}{|lrcl|lrcl||lrcl|} \hline & \dfrac{\frac{p_1}{s_1}} {p} &=& 80\% & & \dfrac{\frac{p_2}{s_2}} {p} &=& 85\% & & \dfrac{\frac{p_3}{s_3}} {p} &=& 89\% \\ (1) & p_1 &=& 80\%ps_1 & (2) & p_2 &=& 85\%ps_2 &(3) & p_3 &=& 89\%ps_3 \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \dfrac{\frac{p_1+p_2}{s_1+s_2}} {p} &=& 83\% \quad | \quad p_1 = 80\%ps_1,\ p_2 = 85\%ps_2 \\ \dfrac{\frac{80\%ps_1+85\%ps_2}{s_1+s_2}} {p} &=& 83\% \\ \dfrac{80\%ps_1+85\%ps_2}{(s_1+s_2)p} &=& 83\% \\ \dfrac{80\%s_1+85\%s_2}{(s_1+s_2)} &=& 83\% \\ 80\%s_1+85\%s_2 &=& 83\% (s_1+s_2) \\ 80\%s_1+85\%s_2 &=& 83\%s_1+83\%s_2 \\ 85\%s_2-83\%s_2 &=& 83\%s_1-80\%s_1 \\ 2\%s_2&=& 3\%s_1 \\ 2s_2&=& 3s_1 \\ \mathbf{s_2} &=& \mathbf{1.5s_1} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \dfrac{\frac{p_2+p_3}{s_2+s_3}} {p} &=& 87\% \quad | \quad p_2 = 85\%ps_2,\ p_3 = 89\%ps_3 \\ \dfrac{\frac{85\%ps_2+89\%ps_3}{s_2+s_3}} {p} &=& 87\% \\ \dfrac{85\%ps_2+89\%ps_3}{(s_2+s_3)p} &=& 87\% \\ \dfrac{85\%s_2+89\%s_3}{(s_2+s_3)} &=& 87\% \\ 85\%s_2+89\%s_3 &=& 87\%(s_2+s_3) \\ 85\%s_2+89\%s_3 &=& 87\%s_2+87\%s_3 \\ 89\%s_3-87\%s_3 &=& 87\%s_2-85\%s_2 \\ 2\%s_3 &=& 2\%s_2 \\ \mathbf{s_3} &=& \mathbf{s_2} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \dfrac{\frac{p_1+p_2+p_3}{s_1+s_2+s_3}} {p} &=& x \quad | \quad p_1 = 80\%ps_1,\ p_2 = 85\%ps_2,\ p_3 = 89\%ps_3 \\ \dfrac{\frac{80\%ps_1+85\%ps_2+89\%ps_3}{s_1+s_2+s_3}} {p} &=& x \\ \dfrac{80\%ps_1+85\%ps_2+89\%ps_3}{(s_1+s_2+s_3)p} &=& x \\ \dfrac{80\%s_1+85\%s_2+89\%s_3}{(s_1+s_2+s_3)} &=& x \\ 80\%s_1+85\%s_2+89\%s_3 &=& x(s_1+s_2+s_3) \quad | \quad s_2 =1.5s_1,\ s_3=s_2=1.5s_1 \\ 80\%s_1+85\%1.5s_1+89\%1.5s_1 &=& x(s_1+1.5s_1+1.5s_1) \\ 80\%s_1+85\%1.5s_1+89\%1.5s_1 &=& x(4s_1) \\ 80\% +85\%1.5 +89\%1.5 &=& 4x \\ 80\% +127.5\% +133.5\% &=& 4x \\ 341\% &=& 4x \quad | \quad : 4 \\ 85.25\% &=& x \\ \mathbf{x} &=& \mathbf{85.25\%} \\ \hline \end{array}\)

 

The average for all three classes combined is \(\mathbf{85.25\%}\)

 

laugh

18 июл. 2019 г., 7:21:15
 #1
avatar+22537 
+2

3)

Let a,b,c be positive real numbers such that \(\log_a b + \log_b c + \log_c a = 0\).
Find \((\log_a b)^3 + (\log_b c)^3 + (\log_c a)^3\).

 

\(\text{Let $\log_a b = \mathbf{x} $} \\ \text{Let $\log_b c = \mathbf{y} $} \\ \text{Let $\log_c a = \mathbf{z} $} \)

 

\(\begin{array}{|rcll|} \hline \log_a b + \log_b c + \log_c a &=& 0 \\ \mathbf{x+y+z} &=& \mathbf{0} \qquad (1) \\ \hline \end{array}\)

 

\(\begin{array}{|rclrcl|} \hline \log_a b &=&\dfrac{\log_c b}{\log_c a} \quad&| \quad \log_c b &=&\dfrac{\log_b b}{\log_b c} \\\\ \log_a b &=&\dfrac{\log_b b}{\log_b c\log_c a} \quad&| \quad \log_b b = 1 \\\\ \log_a b &=&\dfrac{1}{\log_b c\log_c a} \\\\ \log_a b\log_b c\log_c a &=& 1 \\\\ \mathbf{xyz} &=& \mathbf{1} \qquad (2) \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline (x+y+z)^3 &=& (x+y+z)^2(x+y+z) \\ &=& \left(x^2+y^2+z^2+2(xy+yz+xz)\right)(x+y+z) \\ &=& (x^2+y^2+z^2)(x+y+z)+ 2(x+y+z)(xy+yz+xz) \\\\ &=& x^3+y^3+z^3 \\ && +x^2y+x^2z+y^2x+y^2z+z^2x+z^2y + 2(x+y+z)(xy+yz+xz) \\\\ &=& x^3+y^3+z^3 \\ && +(x+y+z)(xy+yz+xz) -3xyz + 2(x+y+z)(xy+yz+xz) \\\\ (x+y+z)^3&=& x^3+y^3+z^3 -3xyz + 3(x+y+z)(xy+yz+xz) \\ \hline \mathbf{x^3+y^3+z^3} &=& \mathbf{(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz} \qquad (3) \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{x^3+y^3+z^3} &=& \mathbf{(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz} \\ && \boxed{x+y+z = 0} \\ x^3+y^3+z^3 &=& 0^3-3\cdot 0\cdot (xy+yz+xz)+3xyz \\ x^3+y^3+z^3 &=& 3xyz \\ && \boxed{xyz = 1} \\ x^3+y^3+z^3 &=& 3\cdot 1 \\ x^3+y^3+z^3 &=& 3 \\ \mathbf{(\log_a b)^3 + (\log_b c)^3 + (\log_c a)^3} &=& \mathbf{3} \\ \hline \end{array}\)

 

laugh

18 июл. 2019 г.
 #13
avatar+22537 
+2
17 июл. 2019 г.
 #13
avatar+22537 
+3
16 июл. 2019 г.