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Mathematics 15 Online
OpenStudy (anonymous):

The value of y varies directly with x2, and y = 36 when x = 3. What is the value of y when x = 5?

Nnesha (nnesha):

what is the equation for direct variation ?

OpenStudy (radar):

A direct variation means (when using y and x for the variable labels) When x increases so does y and vice versa, when x decreases so does y. For the given problem it would appears as:\[y= kx ^{2}. \]First step is to solve for k from the info given.

OpenStudy (anonymous):

so the equation would be like 36=k3^2? or 36=k5^2

OpenStudy (radar):

Your are on the right track, from what is given 36 = k x^2, to solve for k substitute the value given for x. The value was (according to the problem) a 3 Sooo k 3^2 = 36 or 9k = 36 or k=36/9 k=??

OpenStudy (radar):

That value of k will remain the same for all values of x, so it is important to get the correct value. k is called the "constant of proportionally"

OpenStudy (anonymous):

so if you do k=36/9 k would equal 4?

OpenStudy (radar):

Yes k=4 so now you can proceed with the new value of x which is given as 5, use k and plug in the value of 5 for x, and don't forget it is x^2 which is used to solve for y.

OpenStudy (anonymous):

what would the equation look like again?

OpenStudy (radar):

y=kx^2

OpenStudy (anonymous):

so y=4 5^2 or i keep the 36 in the y place

OpenStudy (radar):

No, ditch the 36 that was when x = 3. This is a new situation (except for k) when x = 5.

OpenStudy (anonymous):

i got 100 as my answer cause 5^2 is 25 and 4 * 25 is 100

OpenStudy (radar):

Great, sorry with the slow response, I had visitors. You are correct.

OpenStudy (radar):

As always, good luck with your studies.

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