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You are given x is directly proportional to y^3, and y is inversely proportional to sqrt{z}. If the value of x is 3 when z is 12, what is the value of x when z is equal to 75?

 May 28, 2019

Best Answer 

 #2
avatar+8095 
+2

x  is directly proportional to  y3     so     \(x=ky^3\)     where  k  is a constant.

 

y  is inversely proportional to  √z     so     \(y=\dfrac{s}{\sqrt{z}}\)     where  s  is a constant.

 

\(x=ky^3\\~\\ x=k\Big(\frac{s}{\sqrt{z}}\Big)^3 \\~\\ x=\frac{ks^3}{z\sqrt{z}}\)

 

When   z = 12 ,   x = 3     So...

 

\(3=\frac{ks^3}{12\sqrt{12}}\\~\\ ks^3= 3\cdot12\sqrt{12}\\~\\ks^3=72\sqrt3\)

 

When   z = 75 ,

 

\(x=\frac{ks^3}{z\sqrt{z}}\\~\\ x=\frac{72\sqrt3}{75\sqrt{75}}\\~\\ x=\frac{72\sqrt3}{375\sqrt{3}}\\~\\ x=\frac{24}{125}\)

.
 May 28, 2019
 #1
avatar+18274 
+1

X = k y^3       y = k/sqrtz      combine to

 

x = k /sqrt z ^3

or

xsqrt z^3 = k   always

 

3(sqrt12)^3 = xsqrt( 75) ^3

 

x = 3 (sqrt12)^3 /(sqrt75)^3

x = 3 *8 /125 = 24/125

 May 28, 2019
edited by Guest  May 28, 2019
edited by Guest  May 28, 2019
edited by Guest  May 28, 2019
 #2
avatar+8095 
+2
Best Answer

x  is directly proportional to  y3     so     \(x=ky^3\)     where  k  is a constant.

 

y  is inversely proportional to  √z     so     \(y=\dfrac{s}{\sqrt{z}}\)     where  s  is a constant.

 

\(x=ky^3\\~\\ x=k\Big(\frac{s}{\sqrt{z}}\Big)^3 \\~\\ x=\frac{ks^3}{z\sqrt{z}}\)

 

When   z = 12 ,   x = 3     So...

 

\(3=\frac{ks^3}{12\sqrt{12}}\\~\\ ks^3= 3\cdot12\sqrt{12}\\~\\ks^3=72\sqrt3\)

 

When   z = 75 ,

 

\(x=\frac{ks^3}{z\sqrt{z}}\\~\\ x=\frac{72\sqrt3}{75\sqrt{75}}\\~\\ x=\frac{72\sqrt3}{375\sqrt{3}}\\~\\ x=\frac{24}{125}\)

hectictar May 28, 2019

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