If x varies inversely as y^2 and x = 3 when y = 4, find x when y = 2.
please tell me the answer
please tell me in process
(HINT:ANSWER IS X=12)
jinhe answer nahi aata to maat likho kuch bhi please i am requesting to you
please tell me the answer
Share
Verified answer
Answer:-
x varies inversely as y²
[tex] = > x∝ \frac{1}{ {y}^{2} } [/tex]
[tex] = > x = \frac{k}{ {y}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: ----(1) \: \\ \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: for \:some \: constant \: k[/tex]
[tex]x = 3 \\ y = 4[/tex]
According to equation (1)
[tex] = > 3 = \frac{k}{ {4}^{2} } [/tex]
[tex] = > k = 3 \times {4}^{2} [/tex]
[tex] = > k = 3 \times 16[/tex]
[tex] = > k = 48[/tex]
Then the equation formed be
[tex]x = \frac{48}{ {y}^{2} } \\ where \: y = 2[/tex]
[tex] = > x = \frac{48}{ {2}^{2} } [/tex]
[tex] = > x = \frac{48}{4} [/tex]
[tex] = > x = 12[/tex]
@BengaliBeauty
Feel free to ask your doubts anytime