# Mathematics-2ct1-Set-theory-option

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Module 1 Critical Thinking Option 1

This paper is a review of some basic algebra properties in which algebraic problems will be solved through simplifying. In algebra, simplifying an equation or problem makes it easier to read and sometimes comprehend which allows the problem to be solves more easily. Through using properties like the distributive property, difference of cubes, addition and multiplication properties, as well as the properties of radicals, algebraic equations can be simple to solve.

Rationalize the Denominator

Radicals, or the square root, should be taken out of the denominator and in doing so, this is called rationalizing the denominator. The first step in rationalizing the denominator is to multiply the expression by 1 to get the radical out of the denominator. When we have an expression in the denominator, we multiply by the conjugate which is a property of the sum of a number plus a radical. In this problem, 1 will equal . The FOIL method will be used to multiply across.

After simplifying the expression, we find that the denominator is equal to a -1. In order to make it a +1 in the denominator, we will again multiply by 1 which in this case will be .

Simplify the Expression in terms of Positive Exponents Only

In order to get rid of negative exponents and solve this problem, we will be utilizing properties of exponents. First, we will work with the denominator. Using a property of exponents, we need to make the -1 and -2 exponents into positives by making each base into a fraction. After this, we can multiply the fractions across and get a more manageable denominator.

Our next step is to get the fraction out of the denominator so we will then multiply by the divisor. After this, we will multiply the two terms. In doing so, we will add the exponents of like bases together as the product rule in the properties of exponents where (Bittinger, Beecher, Ellengoben, & Penna, 2016).

The problem state that we need to have all positive exponents. As property of exponents that we used earlier in the denominator, we again multiply by the divisor to get the exponent of to a positive number and reach our answer.

Multiply and Simplify the Expression by Combining like Terms

Using the distributive property in algebra, we will be able to multiply this binomial by the trinomial In doing so, we will multiply 5 by to get , then 5 by to get , then multiply 5 by -1 to get -5. This completes the first half of the polynomial. Now we need to distribute through the trinomial in this same manner.

=

In order to finish off the problem, we merge like terms and because there are not any like terms in this problem, the last thing we need to do is to write it in the correct order, starting with the highest exponent.

Factor

Using the difference of cubes which is , we are able to solve this problem (Khan, 2011). We can recognize that 8 is equal to while -27 is also equal to . Therefore, we are able to rewrite this as . Now because we know that the difference of cubes is true, we follow the pattern in which our and our After the order of operations is completed, we then come up with our answer for the factoring

Solve the following for x:

When solving for x, we need to isolate x and have it by itself on one side of the equation in order to see what x is equal to. In this problem, we will use the addition/subtraction principle where if , then (Bittinger, et al., 2016). This principle also holds true for subtraction. The first step in solving this problem is to get rid of the -10 on the left side of the equation since it is furthest removed from x in the reverse order of operations. To do so, we will at a positive 10 to each side of the equation.

In the next step, we will utilize the multiplication/division principle where if , (Bittinger, et al., 2016). This principle also holds true for division. The next thing we will do is divide by 5 as well as dividing 10 by 5.

In the final step in solving for x, we need to get rid of the exponent. In order to do so, we will use properties of radicals. Assuming that and n is an even number, then (Bittinger, et al., 2016). To undo , we need to square root both sides of the equation. With a positive number being square rooted, there are two square roots, positive and negative.

Conclusion

Without knowing or utilizing algebraic properties, equations or expressions are difficult or impossible to solve. It is used in our everyday lives without even realizing it. For example, a finding the time it will take to travel from point A to point B while driving x miles per hour will use an algebraic equation. While some problems can be more difficult to solve, it is also important to understand why properties work and to write up the steps in which the answer was solved so that, again, there is an understanding of how the answer was reached. Simple algebraic equations are in our everyday lives.

References

Bittinger, M. L., Beecher, J. A., Ellengoben, D. J., and Penna, J. (2016). Algebra & trigonometry: Graphs and models (6th ed.). [E-reader version]. Upper Saddle River, NJ: Pearson. Retrieved from http://www.pearsonmylabandmastering.com/northamerica/mymathlab/ISBN-13: 9780321199911

Option #1: Set Theory Option 1

1. Use the Venn diagram below to answer the following questions • What regions represent (A B) C?
• What regions represent (A B) C-THE SECOND PART OF THIS ONE WAS INCORRECT THE FIRST TIME

Find the cardinal number for each of the following sets.

1. C = {x I x is letter in the word six}
2. B = {x I x < 4 and x â‰¥ 12}

Determine the number of subsets and proper subsets for each of the following.

1. {2,4,6,8,10,12}
2. {x I x is a day of the week}

Requirements:

• Show all your work so that the instructor clearly sees how you solved the problem. 