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Y varies directly with the square of x and y is 16 when x is 8.
Please help!

Y varies directly with the square of x and y is 16 when x is 8. Please help!-example-1
User Ali Tor
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Answer:

If y varies directly with the square of x, we can express this relationship using the equation:

y = kx^2

where k is the constant of proportionality. To find the value of k, we can use the given information that y is 16 when x is 8:

16 = k * 8^2

Simplifying the equation, we get:

16 = 64k

Dividing both sides by 64, we get:

k = 16/64 = 1/4

So the equation that relates y and x is:

y = (1/4)x^2

We can use this equation to find the value of y for any given value of x. For example, if x is 10, we have:

y = (1/4) * 10^2 = 25

Therefore, when x is 10, y is 25.

User Cdalto
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