We know that \(\displaystyle x^2\) is always positive for all values of \(\displaystyle x\). Therefore \(\displaystyle (-x^2)\) would be negative for all values of \(\displaystyle x\). From this conclusion, we know:

\(\displaystyle -x^2< x^2\)

So we have:

\(\displaystyle x^2+2>-x^2+2\)

\(\displaystyle (b)>(a)\)

\(\displaystyle (b)\) is the greater quantity.

A positive number raised to the third power will be positive, while a negative number raised to the third power will remain negative.

If \(\displaystyle x>0\), then \(\displaystyle x^3>0\) and \(\displaystyle -x^3< 0\).

If \(\displaystyle x< 0\), then \(\displaystyle x^3< 0\) and \(\displaystyle -x^3>0\).

Since we do not know if \(\displaystyle x\) is positive or negative, we cannot draw a conclusion about which option is greater.

If \(\displaystyle x>0\), then \(\displaystyle (b)\) is greater.

If \(\displaystyle x< 0\), then \(\displaystyle (a)\) is greater.

When \(\displaystyle x>0\) we can write:

\(\displaystyle x^3>0\ \text{and}\ -x^3< 0\)

We know that \(\displaystyle x^3>-x^3\) and \(\displaystyle x>-x\). Based on this, we can compare the two given quantities.

\(\displaystyle x^3+x+1>-x^3-x+1\)

\(\displaystyle (b)>(a)\)

\(\displaystyle (b)\) is the greater quantity.

We know that \(\displaystyle x\) is greater than \(\displaystyle 3\). We can easily test a few values for \(\displaystyle x\) to determine if the values are increasing or decreasing.

If \(\displaystyle {x=3}\):

\(\displaystyle (a)=3^2-6(3)+10=9-18+10=1\)

If \(\displaystyle x=4\):

\(\displaystyle (a)=4^2-6(4)+10=16-24+10=2\)

If \(\displaystyle x=5\):

\(\displaystyle (a)=5^2-6(5)+10=25-30+10=5\)

The value of \(\displaystyle (a)\) is increasing, with the smallest possible value being \(\displaystyle 1\). From this, we know that \(\displaystyle (a)>0\), so \(\displaystyle (a)>(b)\).

Using the distributive property:

\(\displaystyle 3t + 5t = (3 + 5)t = 8t\)

and

\(\displaystyle 80t - 10t = (80-10)t = 70t\)

Using the associative property of multiplication:

\(\displaystyle 4 \cdot 4t =\left ( 4 \cdot 4 \right ) t= 16t\)

We can rewrite \(\displaystyle 16t \div 8\) as \(\displaystyle 16t \cdot \frac{1}{8}\); using the commutative and associative properties of multiplication:

\(\displaystyle 16t \cdot \frac{1}{8} = \left ( 16 \cdot \frac{1}{8} \right ) \cdot t = 2t\)

\(\displaystyle 6 + 2t\) is the sum of unlike terms and cannot be simplified.

\(\displaystyle 3t + 5t\) is the correct choice.

Depending on the value of \(\displaystyle t\), it is possible for either expression to be greater or for both to be equal.

Case 1: \(\displaystyle t = 1\)

\(\displaystyle 5t + 8t + 7 = 5 \cdot 1 + 8 \cdot 1 + 7 = 5 + 8 + 7 = 20\)

and 

\(\displaystyle 5+ 8t + 7t = 5+ 8 \cdot 1 + 7 \cdot 1 = 5 + 8 + 7 = 20\)

So the two are equal.

Case 2: \(\displaystyle t = 2\)

\(\displaystyle 5t + 8t + 7 = 5 \cdot 2 + 8 \cdot 2 + 7 = 10 + 16 + 7 = 33\)

and 

\(\displaystyle 5+ 8t + 7t = 5+ 8 \cdot 2 + 7 \cdot 2 = 5 + 16 + 14 = 35\)

So (B) is greater. 

The correct response is that it cannot be determined which is greater.

\(\displaystyle 6t+ 5t + 4t + 3t + 2 t + t = (6+5+4+3+2+1)t = 21t\)

\(\displaystyle 7t+5t+3t+t = (7 + 5 + 3 + 1)t = 16t\)

Since \(\displaystyle 21 > 16\), and \(\displaystyle t\) is positive,

then by the multiplication property of inequality,

\(\displaystyle 21 t > 16t\)

making (A) greater regardless of the value of \(\displaystyle t\).

\(\displaystyle 7t + 11 + 4t = 7t + 4t + 11= (7+4)t + 11 = 11t + 11\)

\(\displaystyle 7 + 11t + 4 = 11t + 7 + 4 = 11t + 11\)

Regardless of the value of \(\displaystyle t\), the expressions are equal.

The expression is the sum of two unlike terms, and therefore cannot be further simplified. None of these responses is correct.

\(\displaystyle 8t + 7 + 5t + 9 = \left ( 8 + 5 \right ) t + 9+ 7 = 13t + 16\)

\(\displaystyle 10 + 4t + 9t +5 = (4+9)t + 10 + 5 = 13t + 15\)

Since \(\displaystyle 16 > 15\), \(\displaystyle 13t+ 16 > 13t+ 15\), so (A) is greater regardless of the value of \(\displaystyle t\).