Question

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

Answer 1

Answer:-

x varies inversely as y²

$= > x∝ \frac{1}{ {y}^{2} }$

$= > x = \frac{k}{ {y}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: ----(1) \: \\ \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: for \:some \: constant \: k$

$x = 3 \\ y = 4$

According to equation (1)

$= > 3 = \frac{k}{ {4}^{2} }$

$= > k = 3 \times {4}^{2}$

$= > k = 3 \times 16$

$= > k = 48$

Then the equation formed be

$x = \frac{48}{ {y}^{2} } \\ where \: y = 2$

$= > x = \frac{48}{ {2}^{2} }$

$= > x = \frac{48}{4}$

$= > x = 12$

@BengaliBeauty

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