Can someone please help me with high school math - Direct Variation question...?

Question

I'm a 3rd year mathematics uni student (albeit not a very good one or I wouldn't be asking)

My cousin just asked me a few question for her end of year exams and there was a few I haven't had any experience in for about 15 years and I've totally forgotten...

The question is

If y varies directly as the square of x, y=100 when x=5, find the value of y when x=3

:) I know it has something to do with the greek alpha symbol, well at least that's what I remember? Any help will be great!!

Answer 1

Here you go:

y varies directly as x means: y = kx^2, where k is a constant.

Substitute in your values: 100 = 25k, and solve: k = 4.

So your variation equation is: y = 4x^2.

Find the value of y when x=3: y = 4(9) = 36

Answer 2

IF y varies directly as the square of x then

y = k times X squared, where K is a constant. call this equation 1

If y = 100 when X = 5 then substituting the figures into equation 1, you get 100=25 K. This means k=100/25 = 4

So equation 1 becomes y=4X. SO when x=3 then y=4times 3 squared.

This means y = 4 times 9 which is 36. so when x=3, y=36

Answer 3

y varies directly or is directly proportional to x squared
y=kxx Solve for k when x=5, y=100: 100=k5*5 => k=100/25=4
So y=4xx. Then for x=3 y=4*3*3=36

Answer 4

I'm not sure what alpha symbol you referred to, but this is the method I would normally use.

The general formula for direct variation is y = kx, where k is a constant.

Since y varies directly as x^2, the formula becomes y = kx^2.

Now, to find k, substitute with the known values of y and x.

y = kx^2
100 = k * 5^2
100 = 25k
k = 4

To find y when x = 3, once again substitute the known values into the equation;

y = kx^2
y = 4 * 3^2
y = 4 * 9
y = 36

Therefore, when x = 3, y = 36.

Answer 5

y=k*x^2 and 100= k*25 so k=4
and
y=4*x^2 so for x=3 y = 36

Answer 6

Argh im doing this for maths too