She needs \(\displaystyle 25+10+10+1+1\) to make \(\displaystyle 47\) cents.

You can only add like terms. Therefore, different variables are treated as different types of terms. Since \(\displaystyle 2x\) and \(\displaystyle 3x\) both end in the variable \(\displaystyle x\), they can be added together. The \(\displaystyle 5b\) cannot be added to these numbers; however, because it has a different variable. The answer is:

\(\displaystyle 5x + 5b\)

The associative property of addition allows us to group the numbers with the same variables together: \(\displaystyle 2x+3x+4y+6xy\)

The like terms in this expression are:

• \(\displaystyle 2x\) and \(\displaystyle 3x\)
• \(\displaystyle 4y\)
• \(\displaystyle 6xy\)

Terms with different variables cannot be grouped together.

As a result, the only way to simplify this expression is to add \(\displaystyle 2x\) and \(\displaystyle 3x\).

When adding variables of the same type, they are simply added together, and the variable remains to the first power. This is known as combining like-terms.

\(\displaystyle 7x+2x+5x=9x+5x=14x\)

Thus, the correct answer is \(\displaystyle 14x\).

In order to solve this addition problem, a common denominator must first be found. 

\(\displaystyle \frac{1}{4}\) should be converted to a fraction in which the denominator is 16.

\(\displaystyle \frac{1}{4}+\frac{3}{16}\)

\(\displaystyle \frac{(1\times4)}{(4\times4)}+\frac{3}{16}\)

\(\displaystyle \frac{4}{16}+\frac{3}{16}\)

Finally, add the fractions.

\(\displaystyle \frac{4+3}{16}\)

\(\displaystyle \frac{7}{16}\)

By the order of operations, the operation within parentheses, which is addition, is carried out first; of the remaining two, exponentiation - squaring here - precedes multiplication.

By the order of operations, any expressions within grouping symbols, such as parentheses and brackets, are carried out from the inside outward. Therefore, the operation in the innermost set of grouping symbols - the subtraction within parentheses - will be carried out first, followed by the remaining operation within the brackets - the multiplication. The remaining operation - the addition - is last.

In base five, each place value is a power of five, starting with 1 at the right, then, going to the left, \(\displaystyle 5 ^{1}=5,5 ^{2}=25,5 ^{3}=125,...\).

\(\displaystyle 2131 _{\textrm{five}}\) can be calculated in base ten as

\(\displaystyle 2 \times 125 + 1 \times 25 + 3 \times 5 + 1 \times 1 = 250 + 25 + 15 + 1 = 291\).

In modulo 15 arithmetic, a number is congruent to the remainder of the divison of that number by 15. Since 

\(\displaystyle 9+ 7+5+ 3 + 1= 25\)

and

\(\displaystyle 25 \div 15 = 1 \textrm{ R }10\),

\(\displaystyle 9+ 7+5+ 3 + 1 \equiv 10 \mod 15\).

This makes 10 the correct choice.

Add the numbers and keep the variable:

\(\displaystyle 7n+3n=10n\)

Answer: \(\displaystyle 10n\)