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!
+6248 \(\dfrac{y_2}{y_1}=\dfrac{1}{\sqrt{\dfrac{x_2}{x_1}}}\\y_2 = \dfrac{1}{\sqrt{\dfrac{x_2}{x_1}}}y_1 \\ 1 = \dfrac{1}{\sqrt{\dfrac x 2}} 4 \\ 1 = 4 \sqrt{\dfrac 2 x}\\ 1 = 16 \dfrac 2 x \\ x = 32 \)
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+128475 Find the proportionality constant, k
y = k / √x
4 = k / √2
4√2 = k = √32
So
1 = √32 / √x
√x = √32
x = 32
Thanks!! But I am still confuse what “inverse” really is and does in this problem...
+118609 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)
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?
+118609 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)
