bobbymitchel295
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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.

Sagot :

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

±[tex]\sqrt{4}[/tex]=x

2=x and -2=x

Hope that helps :)

Janessa07

Answer:

x = 2

Step-by-step explanation:

y ∝ [tex]\frac{1}{x^{2} }[/tex] ⇒ y = [tex]\frac{k}{x^{2} }[/tex] (where k is the constant of proportionality)

k = x²y

When x = 4 and y = [tex]\frac{1}{4}[/tex],

k = (4)² × [tex]\frac{1}{4}[/tex]

  = [tex]16 * \frac{1}{4}[/tex]

  = 4

So, y = [tex]\frac{4}{x^{2} }[/tex]

When y = 1,

[tex]1 = \frac{4}{x^{2} }[/tex]

x² = 4

√x² = ±√4

∴ x = 2 or -2

Hope this helps

Pls mark as brainliest

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