Melody08
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Given that y is inversely proportion to the square of ( x + 1),
(a) Express y in terms of x and k, where k is a constant
(b) Given that x = 4 when y = 2, find an equation connecting y to x.
(c) Hence, find the value of x when y = 4

Please help and thank you:)​

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DᴀʀᴋPᴀʀᴀᴅᴏx

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

See the solution below ~

[tex] \large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}[/tex]

a.) The required expression is ~

  • [tex]y \propto \dfrac{1}{(x + 1) {}^{2} } [/tex]

  • [tex]y = k \times \dfrac{1}{(x + 1) {}^{2} } [/tex]

here, k = proportionality constant ~

  • [tex]y = \dfrac{k}{(x + 1) {}^{2} } [/tex]

b.) Let's find the equation by Plugging the values as " x = 4 and y = 2 ~

  • [tex]2 = \dfrac{ k}{(4 + 1 {}^{} ) {}^{2} } [/tex]

now, the value of k can be determined ~

  • [tex]2 = \dfrac{k}{ {5}^{2} } [/tex]

  • [tex]2 = \dfrac{k}{25} [/tex]

  • [tex]k = 50[/tex]

c.) find the value of x when y = 4 ~

Plugging the value of x as 4 and k as 50, we get ~

  • [tex]y = \dfrac{50}{(4 + 1) {}^{2} } [/tex]

  • [tex]y = \dfrac{50}{ {5}^{2} } [/tex]

  • [tex]y = \dfrac{50}{25} [/tex]

  • [tex]y = 2[/tex]

I hope it helped ~

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