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avatar+46 

The value of y varies inversely as \(\sqrt{x}\) and when \(x =2\), \(y=4\). What is x when y=1?

 Apr 5, 2022
 #1
avatar+36416 
+1

y = k / (sqrtx)

y sqrt x = k

4 (sqrt2) = k

 

1 = 4 sqrt 2  / x

x = 4 sqrt 2

 Apr 6, 2022
 #2
avatar+46 
0

Unfortunately, that answer is wrong. The correct answer is 32.

SpaceXGeek  Apr 6, 2022
 #4
avatar+36416 
+1

Yeppers.....As Melody stated.....I left out the sqrt on the denominator

 

1 = 4 sqrt 2  / sqrt x

sqrt x = 4 sqrt 2            now, square both sides

x = 32

 

surprise

ElectricPavlov  Apr 6, 2022
 #3
avatar+117105 
+1

You forgot the sqrt on the bottom EP, just a trivial mistake that SpaceX should have picke  up for himeself.

 

\(y=\frac{k}{\sqrt x}\\ (2,4)\\ 4=\frac{k}{\sqrt 2}\\ k=4\sqrt 2\\ y=\frac{4\sqrt 2}{\sqrt x}\\ when\;y=1\\ 1=\frac{4\sqrt 2}{\sqrt x}\\ 1=\frac{32}{x}\\ x=32\)

 Apr 6, 2022

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