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0
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avatar+349 

The last one was quite easy, but for the benefit of the people who didn't get it, here's the solution:

 

\(a+b=10, a^2+b^2=44,a^3+b^3=?\)

\((a+b)^2=10^2\)

\(a^2+b^2+2ab=100\)

\(44+2ab=100\)

\(2ab=56,ab=28\)

\(a^3+b^3=(a+b)(a^2-ab+b^2)\)

\(a^3+b^3=(10)(44-28)\)

\(a^3+b^3=10*16\)

\(a^3+b^3=160\)

 

For those who solved it, good job! Now for the next one:

 

\(wxy=10\)

\(wyz=5\)

\(wxz=45\)

\(xyz=12\)

Find \(w+x+y+z\)

 

GOOD LUCK!

Mathhemathh  Aug 15, 2017
 #1
avatar+12 
+1

\(12 {1\over2}\)

DarDragon  Aug 15, 2017
 #2
avatar
0

Divide and substitute and you should get the following:

w = 2 1/2,  x = 6,  y = 2/3,  z = 3, so have:

w+x+y+z = 2 1/2 + 6 + 2/3 + 3 =12 1/6

Guest Aug 15, 2017
 #3
avatar+20153 
0

Now for the next one:

\(\begin{array}{rcr} wxy&=&10 \\ wyz&=&5 \\ wxz&=&45 \\ xyz&=&12 \\ \end{array} \)
Find \( w+x+y+z\)

 

1.

\(\begin{array}{|rcll|} \hline xyz &=& 12 \\ xyz \cdot \frac{wyz}{wxy} \cdot \frac{wxz}{wxy} &=& 12 \cdot \frac{5}{10} \cdot \frac{45}{10} \\ xyz \cdot \frac{z}{x} \cdot \frac{z}{y} &=& 12 \cdot \frac{5}{10} \cdot \frac{45}{10} \\ z^3 &=& 12 \cdot \frac{1}{2} \cdot \frac{9}{2} \\ z^3 &=& 3\cdot 9 \\ z^3 &=& 3^3 \\ \mathbf{z} & \mathbf{=} & \mathbf{3} \\ \hline \end{array} \)

 

2.

\(\begin{array}{|rcll|} \hline && w+x+y+z \\ &=& \left( \frac{wxy}{xyz} + \frac{wxy}{wyz} + \frac{wxy}{wxz} \right) \cdot z + z \\ &=& \left( \frac{10}{12} + \frac{10}{5} + \frac{10}{45} \right)\cdot z + z \\ &=& \left( \frac{5}{6} + 2 + \frac{2}{9} \right)\cdot z + z \quad & | \quad z= 3 \\ &=& \left( \frac{5}{6} + 2 + \frac{2}{9} \right)\cdot 3 + 3 \\ &=& \frac{5}{2} + 6 + \frac{2}{3} + 3 \\ &=& 9 + \frac{5}{2} + \frac{2}{3} \\ &=& 9 + \frac{15+4}{6} \\ &=& 9 + \frac{19}{6} \\ &=& 12 + \frac{1}{6} \\ \hline \end{array}\)

 

laugh

heureka  Aug 16, 2017
edited by heureka  Aug 16, 2017

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