Question

X varies inversely with the square of y. when y is 2, x is 48. find x when y is 4.

Answer 1

X is inversly proportional to y^2
X1=48 when Y1=2 and X2=? when Y2=4.
X1/X2=Y1^2/Y2^2.
48/x=4^2/2^2.
X=12.

Answer 2

HEY BUDDY...!!!

HERE IS UR ANSWER.

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▶️ According to statement "x varies inversely with the square of y. " , i.e


$x \: \infty \: 1 \div y ^{2} \\ \\ x = k \times \: 1 \div y ^{2}$

▶️ Where k is a constant,

⏺️ Now given , when y is 2 x is 48 , Putting values in relation , so we could find value of ( k )

=> x = k × 1 / y^2

=> x . y^2 = k

=> 48 × ( 2 )^2 = k

=> 48 × 4 = k

=> [ k = 192 ]✔️

▶️ Now we have to find x , when y = 4 , and he had find value of k , putting values in relation ,

=> x = k × 1 / y^2

=> x = 192 × 1 / ( 4 )^2

=> x = 192 / 16

=> [ x = 12 ]✔️


HOPE HELPED.


JAI HIND!!


:-)