Because the number of days goes down as the number of children goes up, this problem type is inverse variation. We can solve this problem by the following steps:

20*6=24*x

120=24x

x=120/24

x=5

In this equation, x represents the total number of days that can be weathered by 24 students. This is down from the 6 days that 20 students could take shelter together. So the difference is 1 day less.

If \(\displaystyle A\) varies inversely as the cube of \(\displaystyle B\), then, if \(\displaystyle A ,B\) are the initial values of the variables and \(\displaystyle A' ,B'\) are the final values,

\(\displaystyle A'B' ^{3} = AB^{3}\)

Substitute \(\displaystyle A = 8, B = 66, B'= 100\) and find \(\displaystyle A'\):

\(\displaystyle A' \cdot 100 ^{3} = 8 \cdot 66^{3}\)

\(\displaystyle A' =\frac{ 8 \cdot 66^{3}}{100^{3}} = \frac{ 8 \cdot 287,496}{1,000,000} =2.299968\)

This rounds to 2.3.

If \(\displaystyle A\) varies inversely as the cube root of \(\displaystyle B\), then, if \(\displaystyle A ,B\) are the initial values of the variables and \(\displaystyle A' ,B'\) are the final values,

\(\displaystyle A'\sqrt[3]{B'} = A\sqrt[3]{B}\)

Substitute \(\displaystyle A = 8, B = 66, B'= 100\) and find \(\displaystyle A'\):

\(\displaystyle A'\sqrt[3]{100} = 8\sqrt[3]{66}\)

\(\displaystyle A'=\frac{ 8\sqrt[3]{66}}{\sqrt[3]{100} } \approx \frac{ 8 \cdot 4.0412}{4.6416} \approx 7.0\)

To find the inverse we first switch the x and y variables

\(\displaystyle x=\frac{1}{2}y-4\)

Now we add 4 to each side

\(\displaystyle x+4=\frac{1}{2}y\)

From here to isolate y we need to multiply each side by 2

\(\displaystyle 2(x+4)=y\)

By distributing the 2 we get our final solution:

\(\displaystyle y=2x+8\)

If \(\displaystyle A\) varies inversely as the square of \(\displaystyle B\), then, if \(\displaystyle A ,B\) are the initial values of the variables and \(\displaystyle A' ,B'\) are the final values,

\(\displaystyle A'B' ^{2} = AB^{2}\).

Substitute \(\displaystyle A = 8, B = 66, B'= 100\) and find \(\displaystyle A'\):

\(\displaystyle A' \cdot 100 ^{2} = 8 \cdot 66^{2}\)

\(\displaystyle A' =\frac{ 8 \cdot 66^{2}}{100^{2}} =\frac{ 8 \cdot 66^{2}}{100^{2}} = \frac{ 8 \cdot 4,356}{10,000} =3.4848\)

This rounds to 3.5.

If \(\displaystyle A\) varies inversely with the square root of \(\displaystyle B\), then, if \(\displaystyle A ,B\) are the initial values of the variables and \(\displaystyle A' ,B'\) are the final values,

\(\displaystyle A'\sqrt{B'} = A\sqrt{B}\).

Substitute \(\displaystyle A = 8, B = 66, B'= 100\) and find \(\displaystyle A'\):

\(\displaystyle A'\sqrt{100} = 8\sqrt{66}\)

\(\displaystyle A'=\frac{ 8\sqrt{66}}{\sqrt{100} } \approx \frac{ 8 \cdot 8.1240}{10} \approx 6.5\)

To find the inverse of a function we need to first switch the \(\displaystyle x\) and \(\displaystyle y\). Therefore, \(\displaystyle y=2x+1\) becomes

\(\displaystyle x=2y+1\)

We now solve for y by subtracting 1 from each side

\(\displaystyle x-1=2y\)

From here we divide both sides by 2 which results in

\(\displaystyle y=\frac{x-1}{2}\)

To find the inverse we first switch the variables then solve for y.

\(\displaystyle 9x=3y+12\)

Then we subtract \(\displaystyle 12\) from each side

\(\displaystyle 9x-12=3y\)

Now we divide by \(\displaystyle 3\) to get our final answer. When we divide \(\displaystyle 9\) by \(\displaystyle 3\) we are left with \(\displaystyle 3\). When we divide \(\displaystyle -12\) by \(\displaystyle 3\) we are left with \(\displaystyle -4\). Thus resulting in:

\(\displaystyle y=3x-4\)

To solve for an inverse, we switch x and y and solve for y. Doing so yields:

\(\displaystyle \frac{3x-20}{2}=y\)

\(\displaystyle 5x-4y=18\)

1. Switch the \(\displaystyle x\) and \(\displaystyle y\) variables in the above equation.

\(\displaystyle 5y-4x=18\)

2. Solve for \(\displaystyle y\):

\(\displaystyle 5y-4x=18\)

\(\displaystyle 5y-4x+4x=18+4x\)

\(\displaystyle 5y=4x+18\)

\(\displaystyle \frac{5y}{5}=\frac{4x+18}{5}\)

\(\displaystyle y=\frac{4x+18}{5}\)