Since all of the answer choices look like \(\displaystyle 3^n\), let's find \(\displaystyle n\) in  \(\displaystyle 3^n=81\).

\(\displaystyle 81 = 9 \times 9=3\times3\times3\times3=3^4\)

Then,

\(\displaystyle 81^6=(3^4)^6\)

When you have an exponent being raised to an exponent, multiply the exponents together.

\(\displaystyle (3^4)^6=3^{24}\)

When terms with the same base are multiplied, multiply the coefficients together then add up all the exponents.

For the coefficients: \(\displaystyle 5\times2\times1=10\)

For the exponents: \(\displaystyle 6+2+8=16\)

Thus, the answer is \(\displaystyle 10x^{16}\)

Start by simplifying the numerator. 

When an exponent is raised to another exponent, multiply the two exponents together.

\(\displaystyle (p^2)^4=p^{2\times4}=p^8\)

Now, tackle the denominator. When two numbers of the same base are multiplied, you want to add the exponents.

\(\displaystyle (p^3)(p^6)=p^{3+6}=p^9\)

Put the numerator and denominator together:

\(\displaystyle \frac{p^8}{p^9}\)

When you have a fraction with terms that have the same base, you want to subtract the exponent in the denominator from the exponent in the numerator.

\(\displaystyle \frac{p^8}{p^9}=p^{8-9}=p^{-1}\)

To make a term with a negative exponent in the numerator positive, put it in the denominator.

\(\displaystyle p^{-1}=\frac{1}{p}\)

To factor an expression in the form \(\displaystyle x^2+bx+c\), we need to find factors of \(\displaystyle c\) that add up to \(\displaystyle b\).

In this case, \(\displaystyle c=24\) and \(\displaystyle b=10\).

Start by listing factors of 24 and adding them up. You want the one that adds up to 10.

\(\displaystyle 1+24=25\)

\(\displaystyle 2+12=14\)

\(\displaystyle 3+8=11\)

\(\displaystyle 4+6=10\)

Because 4 and 6 are the factors that we need, you can then write

\(\displaystyle x^2+10x+24=(x+4)(x+6)\)

To check if you factored correctly, you can multiply the two factors together. If you end up with the original expression, then you are correct.

First, we need to factor the numerator and the denominator separately.

To factor an expression with the form \(\displaystyle x^2+bx+c\), we will need to find factors of \(\displaystyle c\) that add up to be.

For the numerator, \(\displaystyle x^2-3x-54\), \(\displaystyle c=-54\) and \(\displaystyle b=-3\).

Write down the factors of \(\displaystyle -54\) and add them up.

\(\displaystyle 1+(-54)=-53\)

\(\displaystyle 2+(-27)=-25\)

\(\displaystyle 3+(-18)=-15\)

\(\displaystyle 6+(-9)=-3\)

Since \(\displaystyle 6\) and \(\displaystyle -9\) add up to \(\displaystyle -3\), \(\displaystyle x^2-3x-54=(x-9)(x+6)\)

Now, do the same thing with the denominator, \(\displaystyle x^2+4x-12\).

\(\displaystyle -1+12=11\)

\(\displaystyle -2 + 6 = 4\)

Since \(\displaystyle -2\) and \(\displaystyle 6\) add up to \(\displaystyle 4\), \(\displaystyle x^2+4x-12=(x+6)(x-2)\).

Now, stack these factors up as fractions:

\(\displaystyle \frac{x^2-3x-54}{x^2+4x-12}=\frac{(x-9)(x+6)}{(x+6)(x-2)}\)

Since both the numerator and denominator have the factors (x+6), they cancel each other out because they divide to 1.

Then,

\(\displaystyle \frac{(x-9)(x+6)}{(x+6)(x-2)}=\frac{x-9}{x-2}\)

When you factor an expression, you are separating it into its basic parts. When you multiply those parts back together, you should obtain the original expression.

The first step when factoring an expression is to see if all of the terms have something in common. In this case, \(\displaystyle \small x^3\), \(\displaystyle \small 14x^2\), and \(\displaystyle \small 48x\) all have an \(\displaystyle \small x\) which can be taken out:

\(\displaystyle \small x^3+14x^2+48x=x(x^2+14x+48)\)

The next step is to focus on what's in the parentheses. To factor an expression of form \(\displaystyle \small x^2+bx+c\), we want to try to find factors \(\displaystyle \small (x+m)(x+n)\), where \(\displaystyle \small m+n=b\) and \(\displaystyle \small m\times n=c\). We therefore need to look at the factors of \(\displaystyle \small 48\) to see if we can find two that add to \(\displaystyle \small 14\):

\(\displaystyle \small 1+48=49\)

\(\displaystyle \small 2+24=26\)

\(\displaystyle \small 3+16=19\)

\(\displaystyle 4+12=16\)

\(\displaystyle 6+8=14\)

We've found our factors! We can therefore factor what's inside the parentheses, \(\displaystyle (x^2+14x+48)\), as \(\displaystyle (x+6)(x+8)\). If we remember the \(\displaystyle x\) we factored out to begin with, our final completely factored answer is:

\(\displaystyle \small x(x+6)(x+8)\)

This question requires you to understand order of operations, which is represented by the acronym "PEMDAS": parentheses, exponents, multiplication and division, addition and subtraction.

Solve the expression within the parenthesis first, beginning with multiplication:

\(\displaystyle 7+66\div(12+21)\)

\(\displaystyle 7+66\div(33)\)

The next operation in the order of operations is division.

\(\displaystyle 7+2\)

Finally, use addition to solve the equation:

\(\displaystyle 9\)

To factor an equation in the form \(\displaystyle ax^2+bx+c\), where \(\displaystyle b=12\) and \(\displaystyle c=36\), you must find factors of \(\displaystyle c\) that add up to \(\displaystyle b\).

List the factors of 36 and add them together:

\(\displaystyle 1+36=37\)

\(\displaystyle 9+4=13\)

\(\displaystyle 12+3=15\)

\(\displaystyle 6+6=12\)

Since \(\displaystyle 6+6=12\), \(\displaystyle 6\) is the factor we need. Plug this factor in to get the final answer. 

\(\displaystyle x^2+12x+36\)\(\displaystyle = (x+6)(x+6)= (x+6)^2\)

First, we need to find how long it took Jimmy to finish the test.

Subtract the hours, then subtract the minutes.

\(\displaystyle 12-11=1 \text{ hour}\)

\(\displaystyle 29-3=26 \text{ minutes}\)

Now, the question asks how many MINUTES it takes Jimmy to finish the test. Convert the hour into minutes.

\(\displaystyle 1 \text{ hour}=60 \text{ minutes}\)

\(\displaystyle 60+26=86\)

To add time, you add the hours together, then you add the minutes together.

\(\displaystyle \text{Hours}=8+1=9\)

\(\displaystyle \text{Minutes}=45+24=69\)

Because there are only 60 minutes in an hour, you cannot have a time whose minute value is greater than 60. In this case, subtract 60 minutes and add 1 more to the hour.

\(\displaystyle \text{Hours} = 9+1=10\)

\(\displaystyle \text{Minutes}: 69-60 = 9\)