The first number is multiplied by three 

\(\displaystyle \left ( 2*3=6 \right )\).  

Then it is divide by two 

\(\displaystyle \left (6/2=3 \right )\).  

The following is multiplied by three 

\(\displaystyle \left ( 3*3=9\right )\) 

then divided by two 

\(\displaystyle \left ( 9/2=4.5\right )\).  

That makes the next step to multiply by three which gives us 

\(\displaystyle 4.5*3=13.5\).

Rewrite the first term in fraction form: \(\displaystyle 0.3 = \frac{3}{10}\).

The sequence now begins 

\(\displaystyle \frac{3}{10}, \frac{1}{3}\),...

Rewrite the terms with their least common denominator, which is \(\displaystyle LCM (3, 10) = 30\):

\(\displaystyle \frac{3}{10} = \frac{9}{30}\)

\(\displaystyle \frac{1}{3} = \frac{10}{30}\)

The common difference \(\displaystyle d\) of the sequence is the difference of the second and first terms, which is

\(\displaystyle d = \frac{10}{30} - \frac{9}{30} = \frac{1}{30}\).

The rule for term \(\displaystyle a_{n}\) of an arithmetic sequence, given first term \(\displaystyle a _{1}\) and common difference \(\displaystyle d\), is 

\(\displaystyle a_{n} = a_{1} + (n-1)d\);

Setting \(\displaystyle a_{1}= \frac{3}{10}\), \(\displaystyle n = 10\), and  \(\displaystyle d = \frac{1}{30}\), we can find the tenth term \(\displaystyle a_{10}\) by evaluating the expression:

\(\displaystyle a_{10} = \frac{3}{10} + (10-1) \frac{1}{30 }\)

\(\displaystyle = \frac{3}{10} + 9 \cdot \frac{1}{30 }\)

\(\displaystyle = \frac{3}{10} + \frac{9}{30 }\)

\(\displaystyle = \frac{9}{30} + \frac{9}{30 }\)

\(\displaystyle = \frac{18}{30 }\)

\(\displaystyle = \frac{3}{5}\),

the correct response.

Using the Product of Radicals principle, we can simplify the first two terms of the sequence as follows:

\(\displaystyle a_{1}=\sqrt{20} = \sqrt{4} \cdot \sqrt{5} = 2 \sqrt{5}\)

\(\displaystyle a_{2}= \sqrt{180} = \sqrt{36} \cdot \sqrt{5} = 6 \sqrt{5}\)

The common ratio \(\displaystyle r\) of a geometric sequence is the quotient of the second term and the first:

\(\displaystyle r = \frac{a_{2}}{a_{1}}= \frac{6 \sqrt{5}}{2\sqrt{5} } = 3\)

Multiply the second term by the common ratio to obtain the third term:

\(\displaystyle a_{3} = r a_{2} = 3 \cdot 6 \sqrt{5} = 18 \sqrt{5}\)

The common ratio \(\displaystyle r\) of a geometric sequence is the quotient of the third term and the second:

\(\displaystyle r =\frac{a_{3} }{a_{2}} = \frac{-6}{3i} = \frac{-2}{i}\)

Multiplying numerator and denominator by \(\displaystyle -i\), this becomes

\(\displaystyle r = \frac{-2}{i} = \frac{-2(-i)}{i (-i)} = \frac{-2i}{ -i^{2}} = \frac{-2i}{ -(-1 )} = \frac{-2i}{1} = -2i\)

The second term of the sequence is equal to the first term multiplied by the common ratio:

\(\displaystyle r \cdot a_{1} = a_{2}\).

so equivalently:

\(\displaystyle a_{1} =\frac{ a_{2}}{r}\)

Substituting:

\(\displaystyle a_{1} =\frac{ 3i}{-2i} = -\frac{3}{2}\),

the correct response.

The common ratio \(\displaystyle r\) of a geometric sequence is the quotient of the second term and the first:

\(\displaystyle r = \frac{a_{2}}{a_{1}}= \frac{1}{4i}\)

Simplify this common ratio by multiplying both numerator and denominator by \(\displaystyle i\):

\(\displaystyle r = \frac{1}{4i} = \frac{1 \cdot i}{4i \cdot i } = \frac{i}{4 i^{2}} = \frac{i}{4 (-1)} = \frac{i}{-4 } = -\frac{1}{4} i\)

Multiply the second term by the common ratio to obtain the third term:

\(\displaystyle a_{3} = r a_{2} = -\frac{1}{4} i \cdot 1 = -\frac{1}{4} i\)

Let \(\displaystyle r\) be the common ratio of the geometric sequence. Then 

\(\displaystyle a_{2} = r a_{1}\)

and 

\(\displaystyle a_{3} = r a_{2}= r \cdot r a_{1} = r ^{2}a_{1}\)

Therefore, 

\(\displaystyle r ^{2} = \frac{a_{3}}{a_{1}}\),

and

\(\displaystyle r = \pm \sqrt{ \frac{a_{3}}{a_{1}}}\)

Setting \(\displaystyle a_{1}= -12, a_{3} = 24\):

\(\displaystyle r = \pm \sqrt{ \frac{24}{-12} } = \pm\sqrt{-2} = \pm \sqrt{-1} \cdot \sqrt{2} = \pm i \sqrt{2}\).

Substituting for \(\displaystyle r\) and \(\displaystyle a_{1}\), either

\(\displaystyle a_{2} = r a_{1} = \pm i \sqrt{2} \cdot (-12) = \mp 12 i \sqrt{2}\).

The second term can be either \(\displaystyle -12 i \sqrt{2}\) or \(\displaystyle 12 i \sqrt{2}\), the former of which is a choice.

Let \(\displaystyle r\) be the common ratio of the geometric sequence. Then 

\(\displaystyle a_{2} = r a_{1}\)

and 

\(\displaystyle a_{3} = r a_{2}= r \cdot r a_{1} = r ^{2}a_{1}\)

Therefore, 

\(\displaystyle r ^{2} = \frac{a_{3}}{a_{1}}\),

Setting \(\displaystyle a_{1}= \sqrt{20}, a_{3} = \sqrt{180}\), and applying the Quotient of Radicals Rule:

\(\displaystyle r ^{2} = \frac{a_{3}}{a_{1}} = \frac{\sqrt{180}}{\sqrt{20}} = \sqrt{9} = 3\)

Taking the square root of both sides:

\(\displaystyle r = \pm \sqrt{3}\)

Substituting, and applying the Product of Radicals Rule:

\(\displaystyle a_{2} = r a_{1}\)

\(\displaystyle a_{2} = \pm \sqrt{3} \cdot \sqrt{20 } =\pm \sqrt{60 } = \pm \sqrt{4} \cdot \sqrt{15}= \pm 2 \sqrt{15}\)

Since all elements of the sequence are positive, \(\displaystyle a_{2} = 2 \sqrt{15}\).

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

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

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

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