When subtracting numbers, accounting for significant digits, the difference must be rounded to the same place as the least precise of the numbers. 10.00 and 1.7889 have their final digits in the hundredths and ten-thousandths place, respectively; this makes 10.00 the less precise measurement, and the difference must be rounded to the nearest hundredth.

Now, subtract the numbers outright:

\(\displaystyle 10.00 - 1.7889= 8.2\textbf{1{\color{Red} 1}}1\)

To the nearest hundredth, this is rounded down to 8.21, the correct response.

When adding numbers, accounting for significant digits, the sum must be rounded to the same place as the least precise of the numbers. 2.3004, 0.64, and 0.307 have their final digits in the ten-thousandths, hundredths, and thousandths place, respectively; this makes 0.64 the least precise measurement, and the sum must be rounded to the nearest hundredth.

Now, add the numbers outright:

\(\displaystyle 2.3004 + 0.64 + 0.307 = 3.2\textbf{4{\color{Red} 7}} 4\)

To the nearest hundredth, this is rounded up to 3.25, the correct response.

Notice that \(\displaystyle \$450.00\) is the price after the \(\displaystyle 25\%\) discount. Therefore, the sale price of \(\displaystyle \$450.00\) is \(\displaystyle 75\%\) of the regular price (or \(\displaystyle 0.75\) times the regular price). The regular price can be represented as \(\displaystyle 100\%\), or \(\displaystyle 1\) times the regular price. 

First, set up a proportion describing the problem, using \(\displaystyle x\) to represent the unknown quantity, the regular price:

\(\displaystyle \frac{\$450.00}{0.75} = \frac{x}{1}\)

Next, cross multiply:

\(\displaystyle \$450.00\cdot 1= 0.75\cdot x\)

Finally, divide both sides by \(\displaystyle 0.75\) to solve for \(\displaystyle x\):

\(\displaystyle x=\frac{\$450.00}{0.75}\)

\(\displaystyle x=\$600\)

Therefore, the regular price is \(\displaystyle \$600.00\).

Set up a proportional relationship, using \(\displaystyle x\) to represent the unknown quantity (the weight of 7 sticks of butter):

\(\displaystyle \frac{12\ ounces}{3\ sticks} = \frac{x\ ounces}{7\ sticks}\)

 Cross multiply:

\(\displaystyle 12\cdot 7 = 3\cdot x\)

\(\displaystyle 84 = 3\cdot x\)

Finally, divide both sides to solve for \(\displaystyle x\):

\(\displaystyle \frac{84}{3} = x\)

\(\displaystyle x = 28\ ounces\)

To say that \(\displaystyle x\) is inversely related to \(\displaystyle y\) is to say \(\displaystyle x=\frac{1}{y}\).

To say that \(\displaystyle x\) is directly related to \(\displaystyle y\) is to say \(\displaystyle x=y\).

Thus, to say \(\displaystyle A\) is inversely related to \(\displaystyle B\) and \(\displaystyle C\) and \(\displaystyle A\) is directly related to \(\displaystyle Z\) is to say

\(\displaystyle A=\frac{Z}{BC}\)

In order to solve this problem we need to follow the correct order of operations. Order of operations are outlined by the acronym PEMDAS: parenthesis, exponents, multiplication, division, addition, and subtraction. These operations should be performed in this order from the left to the right. 

\(\displaystyle \left ( \frac{(3+2-1\times4)}{4} \right )+\left ( (2+3)\times(4-2) \right )-((8\times 3)-2^3+(1+1))\)

Let's start by solving the innermost parenthesis remember to work on exponents first.

\(\displaystyle =\frac{(3+2-4)}{4} +\left ( 5\times2 \right )-(24-8+2)\)

Simplify.

\(\displaystyle =\frac{1}{4} +10-18\)

\(\displaystyle =10\tfrac{1}{4}-18\)

\(\displaystyle =-7\tfrac{3}{4}\)

An expression with a radical expression in the denominator is not simplified, so to simplify, it is necessary to rationalize the denominator. This is accomplished by multiplying both numerator and denominator by the given square root, \(\displaystyle \sqrt{12}\), as follows:

\(\displaystyle \frac{6}{\sqrt{12}} = \frac{6 \times \sqrt{12}}{\sqrt{12} \times \sqrt{12}} = \frac{6 \times \sqrt{12}}{12}\)

\(\displaystyle \sqrt{12}\) can be simplified by taking the prime factorization of 12, and taking advantage of the Product of Radicals Property.

\(\displaystyle 12 = 2\times 2 \times 3\), so

\(\displaystyle \sqrt{12} = \sqrt{2 \times 2 \times 3 } = \sqrt{2 \times 2 } \times \sqrt{ 3 } = 2\sqrt{ 3 }\)

Returning to the original expression and substituting:

\(\displaystyle \frac{6 \times \sqrt{12}}{12} = \frac{6 \times 2 \sqrt{3}}{12} = \frac{12 \sqrt{3}}{12} =\sqrt{3}\),

the correct response.

When simplifying an fraction with a denominator which is the product of an integer and a square root expression, it is necessary to first rationalize the denominator. This is accomplished by multiplying both halves of the fraction by the square root expression. The correct response is therefore \(\displaystyle \sqrt{5}\).

The two expressions comprise the sum and the difference of the same two expressions, and can be multiplied using the difference of squares formula:

\(\displaystyle (7+2 \sqrt{7})(7-2 \sqrt{7})\)

\(\displaystyle = 7^{2} -(2 \sqrt{7})^{2}\)

The second expression can be rewritten by the Power of a Product Property:

\(\displaystyle = 7^{2} -2^{2}( \sqrt{7})^{2}\)

The square of the square root of an expression is the expression itself:

\(\displaystyle =49 -4 (7)\)

By order of operations, multiply, then subtract:

\(\displaystyle =49 -28\)

\(\displaystyle = 21\)

The two expressions comprise the sum and the difference of the same two expressions, and can be multiplied using the difference of squares formula:

\(\displaystyle \left ( \sqrt{7x}-\sqrt{3x} \right )\left ( \sqrt{7x}+\sqrt{3x} \right )\)

\(\displaystyle = ( \sqrt{7x})^{2}-(\sqrt{3x} )^{2 }\)

The square of the square root of an expression is the expression itself:

\(\displaystyle = 7x - 3x\)

By distribution:

\(\displaystyle = (7-3)x\)

\(\displaystyle = 4x\)