Question

********************HELP**************************HELP****************

y varies directly as the square of x. When x = 1, y = 8. Find x when y = 128.

Answer 1

finally someone to the rescue

Answer 2

\(\bf y = (n)x^2\)
if you were to solve for "n", what would "n" be?

Answer 3

well, hold one

Answer 4

n= y\[n=y\]

Answer 5

n=y srt(x)?

Answer 6

$$\bf
y = (n)x^2\\
\text{when y=1, x=8, thus}\\
1 = (n)8^2
$$
solve for "n" now :)

Answer 7

im still confused ):

Answer 8

1=(n)64
now what ?

Answer 9

well yes, when something "varies directly to something
meaning it moves at some proportion of something
so as "x" changes value, "y" changes by some value

Answer 10

$$ \bf
y = (n)x^2\\
\text{when y=1, x=8, thus}\\
1 = (n)8^2\\
\frac{1}{64} =n\\
\text{now what is "x" when y = 128}\\
128 = (n)x \implies x = \cfrac{128}{n}
$$

Answer 11

hmm one sec, lemme revise that :/

Answer 12

im still lost

Answer 13

ok, what part ... you confuses you?

Answer 14

why is that 128=(n)x

Answer 15

i am so lost

Answer 16

ohh, one sec -> Find x when y = 128. <---

the equation would be \( \bf y = (n)x^2\)
so when y =128
the equation will look like \(\bf 128 = (n)x^2\)

Answer 17

128 = 8*x²

Answer 18

x² = 16
x = 4

Answer 19

got it now

Answer 20

$$\bf
\text{now what is "x" when y = 128}\\
128 = (n)x^2 \implies x = \sqrt{\cfrac{128}{n}} \implies x = \sqrt{\cfrac{128}{\frac{1}{64}}}\\
\implies \sqrt{\frac{128}{1} \times \frac{64}{1}} \implies x = \sqrt{8192}
$$

Answer 21

wait... smokes, I ... .messed up :/

Answer 22

when x=1 y = 8

Answer 23

so
y = nx

8 = n1
n = 8

Answer 24

so much for my simplifying

Answer 25

lol

Answer 26

heheh, but yes, you're correct :)

Answer 27

I misread the terms values my bad