Provide answer. THE ANSWER IS NOT 5 OR 16. If y varies inversely as x^2, and x = 4 while y = 1/4, find x when y = 1.

Step-by-step explanation:

if y varies inversely as x^2 then 1/y=x^2

note: there needs to be a constant

so let's just say that the constant is k

so we multiply k by x^2 to equal 1/y

1/y=(x^2)k

put in the values for x and y

x=4 and y=1/4

1/(1/4)=(4^2)k

4/16=k

1/4=k

now we insert k back into the first equation that we made

1/y=(x^2)(1/4)

put in the value for y

y=1

x=?

1/1=(x^2)(1/4)

4=x^2

±=x

2=x and -2=x

Hope that helps :)

x = 2

Step-by-step explanation:

y ∝ ⇒ y = (where k is the constant of proportionality)

k = x²y

When x = 4 and y = ,

k = (4)² ×

  =

  = 4

So, y =

When y = 1,

x² = 4

√x² = ±√4

∴ x = 2 or -2

Hope this helps

Pls mark as brainliest

answer: idk

step-by-step explanation: