Expert answer:Consumer Mathematics Option 2
Solved by verified expert:Question #1 – Simple InterestSarah borrowed $700 at a simple interest rate of 8%. At the end of the loan term she owed $812. What was the term of her loan?Question #2 – Effective Annual YieldLisa has the opportunity to invest some of her money. She is trying to decide which type of investment would give her the best annual yield. Calculate the effective annual yield for both investments and decide which one is the best.8% compounded monthly8.25% compounded monthlyQuestion #3 – Annuity EarningsAubrey is saving money in an annuity and is earning 4% annual interest compounded semi-annually. If she deposits $600 in the account every six months for five years, what will the future value of her account equal? How much interest will she have earned?Requirements: Show all your work so that the instructor clearly sees how you solved the problem.Make sure your final answer is clear and visible.use attached document to submit answer
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Running head: MODULE 1 CRITICAL THINKING OPTION 1
Module 1 Critical Thinking Option 1
MTH122: College Algebra
Colorado State University-Global Campus
LeslieAnn Humphreys
October 20, 2017
1
MODULE 1 CRITICAL THINKING OPTION 1
2
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
3
2 + √5
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
2−√5
sum of a number plus a radical. In this problem, 1 will equal 2−√5. The FOIL method will be
used to multiply across.
3
2 + √5
∙
2 − √5
2 − √5
=
6 − 3√5
4 + −2√5 + 2√5 + −5
=
6 − 3√5
4−5
=
6 − 3√5
−1
MODULE 1 CRITICAL THINKING OPTION 1
3
After simplifying the expression, we find that the denominator is equal to a -1. In order
−1
to make it a +1 in the denominator, we will again multiply by 1 which in this case will be −1.
6 − 3√5 −1 −6 + 3√5
∙
=
−1
−1
1
−6 + 3√5 3√5 − 6
Simplify the Expression in terms of Positive Exponents Only
2 2 −3
(4 )−1 −2
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.
2 2 −3
2 2 −3
=
1 1
1
(4 )( 2 ) (
)
4 2
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).
=
2 2 −3 4 2
∙
= (2 2 −3 )(4 2 )
1
1
(
)
4 2
MODULE 1 CRITICAL THINKING OPTION 1
4
= 8 4 −2
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.
8 4 −2 = 8 4 (
=
1
)
2
8 4
2
Multiply and Simplify the Expression by Combining like Terms
(5 + 3 )( 4 − 3 2 − 1)
Using the distributive property in algebra, we will be able to multiply this binomial
(5 + 3 ) by the trinomial ( 4 − 3 2 − 1). In doing so, we will multiply 5 by 4 to get 5 4 ,
then 5 by −3 2 to get −15 2 , then multiply 5 by -1 to get -5. This completes the first half of
the polynomial. Now we need to distribute 3 through the trinomial in this same manner.
(5 + 3 )( 4 − 3 2 − 1) = 5 4 + −15 2 + −5 + 3 5 + −9 3 + −3
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.
3 5 + 5 4 − 9 3 − 15 2 − 3 − 5
Factor
MODULE 1 CRITICAL THINKING OPTION 1
5
8 3 − 27
Using the difference of cubes which is 3 − 3 = ( − )( 2 + + 2 ), we are able to
solve this problem (Khan, 2011). We can recognize that 8 is equal to 23 while -27 is also equal
to 33 . Therefore, we are able to rewrite this as (2 )3 − 33 . Now because we know that the
difference of cubes is true, we follow the pattern in which our = 2 and our = −3. After the
order of operations is completed, we then come up with our answer for the factoring 8 3 − 27.
(2 )3 − 33 = (2 − 3)((2 )2 + (2 ∙ 3) + (3)2 )
= (2 − 3)(4 2 + 6 + 9)
Solve the following for x:
5 2 − 10 = 0
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.
5 2 − 10 = 0
5 2 = 10
MODULE 1 CRITICAL THINKING OPTION 1
6
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 5 2 by 5 as well as dividing 10 by 5.
5 2 10
=
5
5
2 = 2
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 ≠ 1 and n is an even number, then √ = | |
(Bittinger, et al., 2016). To undo 2 , 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.
√ 2 = √2
= ±√2
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.
MODULE 1 CRITICAL THINKING OPTION 1
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
Khan, S. [Khan Academy]. (2011, February 7). Factoring the difference of cubes [Video file].
Retrieved from https://www.youtube.com/watch?v=APJx1y78ovA
7
…
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