Question

For the following problem, y varies directly as the square of x.
If y = 8 when x = 2, find y when x is 1.

Answer 1

so,
\(\Large y \alpha x^2\)

let the constant of proportionality be 'k'

so, \(\Large y= k x^2\)
when y=8, x= 2
can you find 'k' first ?

Answer 2

k=4

Answer 3

no 2

Answer 4

are you sure ?
\(8 = k\times 2^2 \\ 8 = 4k \\\)

Answer 5

yes! k =2 is correct :)

Answer 6

I forgot about the exponent

Answer 7

now we have
\(\large y = 2x^2\)
just plug in x=1 !

Answer 8

y=2?

Answer 9

thats correct :)

Answer 10

Thank you! :)

Answer 11

welcome ^_^