Question

Can someone teach me to do these?
y varies directly as the square of x. When x = 4, y = 8. Find x when y = 32.

x = ±4

x = ±6

x = ±8

x = ±12

Answer 1

X=kY
find K by first 2 values and then put it in the above eq for finding the remaining values of Y

Answer 2

what is k ?

Answer 3

the proportionality constant K which u put after removing the sign of proportionality

Answer 4

y varies directly as the `square` of x.\[\Large y=k x^2\]

I'm confused by your equation hamza :o

Answer 5

y=kx^2
when x=4, y=8
8=k*4^2=16k
k=8/16=1/2

Answer 6

o sorry i had missed the square

Answer 7

when y=32
32=1/2 x^2
x^2=64
\[x=\pm 8\]

Answer 8

when something varies directly to something else, means

if a = 5, b = 10
if a changes to 10, b changes to 20
if a changes to 20, b changes to 40
and so on

"varies directly" means, is being affected directly by the change on some other fellow

as "a" changes, the changes ripple to "b"

but if "a" change say 10 units, it doesn't have to ripple or reflect on "b" to another extra 10 units, it changes with a multiplier value, let's call it "k"

b = (k) a
if a =5 then
b = (k) (5)

then b changes to 10, so what's the multiplier "k"
10 = (k) (5)

#solving for "k"
10/5 = k
2 = k


if a = 10, then b is
b = (2)10
b = 20

and so on

Answer 9

The k can be a little confusing if you don't understand the concept.

Like if I told you that `money` is directly related to `pizza`, I would represent it this way:\[\large M=kP\]Where \(\large k\) is maybe the cost of each slice of pizza.

Oh man.. that was the worst example ever lol.. ignore this XD

Answer 10

\[\large y = kx^2\]\[\large 8= (4)^2k\]\[\large 8 = 16k\]\[\large \frac{1}{2} = k\]\[\large y= \frac{x^2}{2}\]\[\large 32 = \frac{x^2}{2}\]\[\large 32 \cdot 2 =x^2\]\[\large \sqrt{64} = x\]\[\large x =\ ? \ \]

Answer 11

@Jhannybean can u tell me how to type like this as u did?

Answer 12

Equation editor. Follow @jdoe0001 's explanation, then look back at mine. It'll make moresense.

Answer 13

@hamza10 http://www.onemathematicalcat.org/MathJaxDocumentation/TeXSyntax.htm

you can try and test at => http://mathb.in/

Answer 14

Woah thats alot to understand lol. @Jhannybean will do .

Answer 15

so x = 8?

Answer 16

yes, \(\pm8\)
:)

Answer 17

thank you :)