In modulo 6 arithmetic, a number is congruent to the reainder of its division by 6.

Therefore, since $\displaystyle 4 \times 5 \times 3 = 60$ and $\displaystyle 60 \div 6 = 10 \textrm{ R }0$,

$\displaystyle 4 \times 5 \times 3 \equiv 0 \mod 6$.

The correct response is 0.

A set is closed under addition if and only if the sum of any two (not necessarily distinct) elements of the set is also an element of the set.

$\displaystyle \left \{ 0\right \}$ is closed under addition, since $\displaystyle 0 + 0 = 0$

The set of all negative integers is closed under addition, since any two negative integers can be added to yield a third negative integer.

The set of all positive even integers is closed under addition, since any two positive even integers can be added to yield a third positive even integer.

The remaining set is the set of all integers between 1 and 10 inclusive. It is not closed under addition, as can be seen by this counterexample:

$\displaystyle 1,10 \in \left \{1,2,3...10 \right \}$

but 

$\displaystyle 1+10 = 11\notin \left \{1,2,3...10 \right \}$

$\displaystyle T$ varies directly as $\displaystyle \sqrt{L}$, which means that for some constant of variation $\displaystyle k$,

$\displaystyle \frac{T}{\sqrt{L}} = k$

We can write this relationship alternatively as

$\displaystyle \frac{T_{1}}{\sqrt{L_{1}}} = \frac{T_{2}}{\sqrt{L_{2}}}$

where the initial conditions can be substituted on the left side and final conditions, on the right. We will be solving for $\displaystyle T_{2}$ in the equation

$\displaystyle \frac{13}{\sqrt{42}} = \frac{T_{2}}{\sqrt{12}}$

$\displaystyle \frac{13}{\sqrt{42}} \cdot \sqrt{12}= \frac{T_{2}}{\sqrt{12}} \cdot \sqrt{12}$

$\displaystyle T_{2} = \frac{13\sqrt{12}}{\sqrt{42}} \approx \frac{13 \cdot 3.4641}{6.4807} \approx 6.9$

$\displaystyle N$ varies inversely as $\displaystyle m^{2}$ and directly as the cube of $\displaystyle b^{3}$. This means that for some constant of variation $\displaystyle k$, 

$\displaystyle \frac{Nm^{2}}{b^{3}} = k$

We can write this relationship alternatively as

$\displaystyle \frac{N_{1}m_{1}^{2}}{b_{1}^{3}} = \frac{N_{2}m_{2}^{2}}{b_{2}^{3}}$

where the initial conditions can be substituted on the left side and final conditions, on the right. We will be solving for $\displaystyle N_{2}$ in the equation

$\displaystyle \frac{9,000 \cdot 6^{2}}{2^{3}} = \frac{N_{2} \cdot 4^{2}}{4^{3}}$

$\displaystyle \frac{9,000 \cdot 36}{8} = \frac{N_{2} \cdot 16}{64}$

$\displaystyle 40,500 = \frac{N_{2} }{4}$

$\displaystyle N_{2} = 40,500 \cdot 4 = 162,000$

$\displaystyle \frac{1}{4}\; in=1\; mile$ is the same as $\displaystyle 1\; in = 4\; miles$

So to find out the distance between the cities

$\displaystyle 12.5\; in \cdot \frac{4\; miles}{1\;in }=50\; miles$

Let $\displaystyle M,L$ be the mass of the weight and the elongation of the spring. Then for some constant of variation $\displaystyle k$, 

$\displaystyle L = kM$

We can find $\displaystyle k$ by setting $\displaystyle M = 20,L=7.2$ from the first situation:

$\displaystyle 7.2 = k \cdot 20$

$\displaystyle k = 7.2 \div 20 = 0.36$

so $\displaystyle L = 0.36 M$

In the second situation, we set $\displaystyle M = 32$ and solve for $\displaystyle L$:

$\displaystyle L = 0.36 \cdot32 =11.52$ 

which rounds to 11.5 centimeters.

First set up the proportion:

$\displaystyle \frac{3\; parts \; yellow}{1 \; part\; red}=\frac{x \; quarts \; yellow}{2 \; quarts \; red}$

x = $\displaystyle 6 \; quarts\;yellow\; paint$

Then convert this to gallons:

$\displaystyle 6 \; quarts\; \cdot \frac{1 \; gallon}{4\; quarts}= 1.50\; gallons$

To find the percentage increase, divide the number of new books by the original amount of books:

$\displaystyle 192-160=32$

She has 32 additional new books; she originally had 160.

$\displaystyle \frac{32}{160} = \frac{16}{80} = 0.20=20\%$

To find x we need to find the direct proportion. In order to do this we need to cross multiply and divide.

$\displaystyle \frac{x}{100}=\frac{1}{4}$

From here we mulitply 100 and 1 together. This gets us 100 and now we divide 100 by 4 which results in 

$\displaystyle x=25$

If the real distance between the two cities is $\displaystyle 2400$ $\displaystyle miles$, and $\displaystyle 250$ $\displaystyle miles$ = $\displaystyle 1$ $\displaystyle inch$, then we can set up the proportional equation:

$\displaystyle \frac{1 in}{250 miles}=\frac{xin}{2400 miles}$

$\displaystyle 250x=2400$ $\displaystyle in$

$\displaystyle x = 9.6$ $\displaystyle in$