Asked by khalisat Adebayo
The quality y is partly constant and partly veries inversely as the square of x.When y=23/4, x=4 and y=8, x=2.Find the value of x when y=17.
Answers
Answered by
Damon
y = a + b/x^2
23/4 = a + b / 16
8 = a + b/4
so a = 8 - b/4
and then
23/4 = 8 - b/4 + b/16
23 = 32 -b + b/4
3 b/4 = 9
b = 12
a = 8 - 3 = 5
then
17 = 5 + 12/x^2
12 = 12/x^2
x = 1
23/4 = a + b / 16
8 = a + b/4
so a = 8 - b/4
and then
23/4 = 8 - b/4 + b/16
23 = 32 -b + b/4
3 b/4 = 9
b = 12
a = 8 - 3 = 5
then
17 = 5 + 12/x^2
12 = 12/x^2
x = 1
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