bobbymitchel295
Answered

Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

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

We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.