If y varies inversely as the square of x and y=5 when x=2.5, what is the value of Y when X=9?

Question

Abigail Allen · Accepted Answer

undefined To find the value of y when x=9, we can use the inverse variation formula.

The formula for inverse variation is y = k/x^2, where k is the constant of variation.

To find the value of k, we can substitute the given values of y and x into the formula.

5 = k/2.5^2

Simplifying, we get:

5 = k/6.25

To solve for k, we can multiply both sides of the equation by 6.25:

31.25 = k

Now that we have the value of k, we can substitute it back into the inverse variation formula to find the value of y when x=9.

y = 31.25/9^2

Simplifying, we get:

y = 31.25/81

Therefore, when x=9, y is approximately 0.385.

Avery Alexander · Answer

#y=31.25/81# As #yprop1/x^2#, we have #y=k/x^2# or #x^2y=k#. As when #x=2.5#, #y=5#, #2.5^2xx5=k# or #k=6.25xx5=31.25# Therefore #x^2y=31.25# Hence when #x=9#, #9^2xxy=31.25# and #y=31.25/81#

Hazel Anglin · Answer

undefined To find the value of y when x = 9, you can use the inverse variation formula. First, find the constant of variation (k) using the initial values given:

y = k / x^2

Given that y = 5 when x = 2.5:

5 = k / (2.5)^2

Solve for k:

k = 5 * (2.5)^2

Now that you have k, substitute it into the formula and solve for y when x = 9:

y = k / (9)^2

y = (5 * (2.5)^2) / (9)^2

Now calculate the value of y.