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Flashcards: Properties of operations with real numbers

Solve the following problem:

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

$-7\tfrac{3}{4}$ $\displaystyle -7\tfrac{3}{4}$

$28\tfrac{1}{4}$ $\displaystyle 28\tfrac{1}{4}$

$28\tfrac{3}{4}$ $\displaystyle 28\tfrac{3}{4}$

$-28\tfrac{1}{4}$ $\displaystyle -28\tfrac{1}{4}$

$7\tfrac{3}{4}$ $\displaystyle 7\tfrac{3}{4}$

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Flashcards: Properties of operations with real numbers

Solve the following problem:

$-7\tfrac{3}{4}$ $\displaystyle -7\tfrac{3}{4}$

$28\tfrac{1}{4}$ $\displaystyle 28\tfrac{1}{4}$

$28\tfrac{3}{4}$ $\displaystyle 28\tfrac{3}{4}$

$-28\tfrac{1}{4}$ $\displaystyle -28\tfrac{1}{4}$

$7\tfrac{3}{4}$ $\displaystyle 7\tfrac{3}{4}$

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Flashcards: Properties of operations with real numbers

Solve the following problem:

Your answer:

$28\tfrac{1}{4}$ $\displaystyle 28\tfrac{1}{4}$

Your answer:

$28\tfrac{3}{4}$ $\displaystyle 28\tfrac{3}{4}$

Your answer:

$-28\tfrac{1}{4}$ $\displaystyle -28\tfrac{1}{4}$

Your answer:

$7\tfrac{3}{4}$ $\displaystyle 7\tfrac{3}{4}$

Correct answer:

$-7\tfrac{3}{4}$ $\displaystyle -7\tfrac{3}{4}$

Explanation:

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.

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

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

Simplify.

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

$=10\tfrac{1}{4}-18$ $\displaystyle =10\tfrac{1}{4}-18$

$=-7\tfrac{3}{4}$ $\displaystyle =-7\tfrac{3}{4}$

Flashcards: Properties of operations with real numbers

Simplify the expression

$\frac{6}{\sqrt{12}}$ $\displaystyle \frac{6}{\sqrt{12}}$

$\displaystyle 3$

$\sqrt{3}$ $\displaystyle \sqrt{3}$

$\displaystyle 2$

$\sqrt{2}$ $\displaystyle \sqrt{2}$

$\sqrt{6}$ $\displaystyle \sqrt{6}$

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Flashcards: Properties of operations with real numbers

Simplify the expression

$\frac{6}{\sqrt{12}}$ $\displaystyle \frac{6}{\sqrt{12}}$

Your answer:

$\displaystyle 3$

Your answer:

$\displaystyle 2$

Your answer:

$\sqrt{2}$ $\displaystyle \sqrt{2}$

Your answer:

$\sqrt{6}$ $\displaystyle \sqrt{6}$

Correct answer:

$\sqrt{3}$ $\displaystyle \sqrt{3}$

Explanation:

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, $\sqrt{12}$ $\displaystyle \sqrt{12}$, as follows:

$\frac{6}{\sqrt{12}} = \frac{6 \times \sqrt{12}}{\sqrt{12} \times \sqrt{12}} = \frac{6 \times \sqrt{12}}{12}$ $\displaystyle \frac{6}{\sqrt{12}} = \frac{6 \times \sqrt{12}}{\sqrt{12} \times \sqrt{12}} = \frac{6 \times \sqrt{12}}{12}$

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

$12 = 2\times 2 \times 3$ $\displaystyle 12 = 2\times 2 \times 3$, so

$\sqrt{12} = \sqrt{2 \times 2 \times 3 } = \sqrt{2 \times 2 } \times \sqrt{ 3 } = 2\sqrt{ 3 }$ $\displaystyle \sqrt{12} = \sqrt{2 \times 2 \times 3 } = \sqrt{2 \times 2 } \times \sqrt{ 3 } = 2\sqrt{ 3 }$

Returning to the original expression and substituting:

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

the correct response.

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