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# Help

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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

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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 ,

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

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May 28, 2019

#1
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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
+3

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