idk how to do this.

i tried dividng by 4 but i am pretty sure that is wrong.

also, i am still a bit confuse on what to do if the problem include the word “inverse”... if you can explain this too that will be great!!

thanks!

Guest Oct 10, 2018

#1**+1 **

\(\dfrac{y_2}{y_1}=\dfrac{1}{\sqrt{\dfrac{x_2}{x_1}}}\\y_2 = \dfrac{1}{\sqrt{\dfrac{x_2}{x_1}}}y_1 \\ 1 = \dfrac{1}{\sqrt{\dfrac x 2}} 4 \\ 1 = 4 \sqrt{\dfrac 2 x}\\ 1 = 16 \dfrac 2 x \\ x = 32 \)

Rom
Oct 10, 2018

#2**+2 **

Find the proportionality constant, k

y = k / √x

4 = k / √2

4√2 = k = √32

So

1 = √32 / √x

√x = √32

x = 32

CPhill
Oct 10, 2018

#3**+1 **

Thanks!! But I am still confuse what “inverse” really is and does in this problem...

Guest Oct 10, 2018

#4**+1 **

Inversely proportional simply means that as one gets bigger the other gets smaller.

so if y is inversely proportional to x then

y= k/x (k is just a constant)

here y is inverserely proportional to sqrt(x) so

y = k/ sqrt(x)

Melody
Oct 11, 2018

#5**+1 **

Thanks.

How did it get smaller? How does this get smaller: y = k/sqrt(x). So this is how I am suppose to solve inverse problems? Doing/using this formula/work?

Guest Oct 12, 2018

#6**+1 **

Yes you use this fact

Let k be a constant that you will often need to find using information from the question.

If y is **directly proportional** to x then

\(y=kx\)

If you graph x agains y you will get a straight line.

The y interept is 0

aand the gradient is k

OR

If y is **inversely proportional** to x then

\(y=\frac{k}{x}\)

This time if you graph it it will be a hyperbola.

If k is positive it will be in the first and 3rd quad

and if k is neg it will be in the 2nd and 4th quads.

Now consider the case there k is +1

If x=2 y=1/2

If x=4 y=1/4

See, as x gets bigger, y gets smaller. (that is when k is positive)

Melody
Oct 12, 2018