suppose X varies inversely as square of Y and when Y=3, then X=4. Find X, when Y=6.
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Answered by
2
equation of x varies inversely as square of y means
x=k/y^2
substituting y=3 and x=4
we get
4=k/3^2
4=k/9
multiplying both sides by 9
4*9=k
k=36
finding x when y=6
so substituting y=6 and k=36 in x=k/y^2
x=36/6^2
x=36/36
x=1
hence the value of x when y=6 is 1
x=k/y^2
substituting y=3 and x=4
we get
4=k/3^2
4=k/9
multiplying both sides by 9
4*9=k
k=36
finding x when y=6
so substituting y=6 and k=36 in x=k/y^2
x=36/6^2
x=36/36
x=1
hence the value of x when y=6 is 1
Answered by
1
X varies inversely as y^2 therefore X=d/y^2.therefore replace X=4 when y=3 therefore d= 36. Now replace k=36 and y= 6 therefore X=6
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