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# Rates/ratios problem

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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
+3190
+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
+92759
+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
+94110
+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
+94110
+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

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
#7
+94110
+1

In your question, y varies inversly to sqrtx

$$y=\frac{k}{\sqrt{x}}$$

then sub in (2,4) to determine the value of k

Once you have k then you have a formula that you can use for determining the y value of any x value.

Melody  Oct 12, 2018