MATH 221 ENTIRE COURSE ASSIGNMENTS AND DQS

November 8, 2016   Comments Off on MATH 221 ENTIRE COURSE ASSIGNMENTS AND DQS

MATH 221 ENTIRE COURSE ASSIGNMENTS AND DQS

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MATH 221 ENTIRE COURSE ASSIGNMENTS AND DQS

MAT 221 Week 1 Assignment 1 Simplifying Expressions

This assignment is comprised off of the properties of real numbers. In this assignment we will use these properties to simplify using 2a (a – 5) + 4(a – 5), 2w – 3 + 3(w – 4) – 5(w – 6), & 0.05(0.3m + 35n) – 0.8(-0.09n – 22m).

Simplifying: 2a(a-5)+4(a-5)

2a2-10a+4a-20

2a2-6a-20

In this expression we will use the distribution property which will remove the parentheses.  After this you will combine all of the like termwhich is done by adding the coefficients, upon completion this will finish the simplifying process.

Simplifying 2w-3+3(w-4)-5(w-6)

2w-3+3w-12-5w+30

2w+3w-5w-3-12+30

5w-5w-15+30

15

At this point you will use the distribution property andremove the parenthesis. After removing the parenthesis you will use the commutative property to compact the like terms.  In this expression you will add two variable and the two constant terms. Once this is complete you will have simplified this expression.

1. .05(.3m+35n)-.8(.09n-22m)

.015m+1.75n+.072n+17.6m

.015m+17.6m+1.75n+.072n

17.615m+1.822n

In this expression, much like the first one, thedistribution property will be imperative to remove the parentheses.  Again, you will use the commutative property to match up the like termstogether. From here you will use the coefficients to add the like terms.  At this point you will have simplified the expression.

MAT 221 Week 1 DQ 1

A=12

B=-28

C=88

1.          a3-b3

123-(-28)3

1728+(-21,952)

23,680

1. (a-b)(a2+ab+b2)

[12-(-28)] [122+12(-28)+(-28)2]

[12+28] [144+(-336)+784]

40(144-336+784)

40(592)

23,680

MAT 221 Week 2 Assignment 2 Inequalities

The Body Mass Index (BMI) is an indicator to help people to determine if they might have a longer life span than average, are probably not overweight, are probably overweight, or are obese. The intervals for each are from 17 to 22, 23 to 24.999, 25 to 29.9, and over 30 respectively. Notice that it is between 17 and 22. That is not inclusive but rather a compound inequality statement which is 17 < BMI < 22. Moreover, over 30 is an inequality statement with a positive infinity which is any BMI that is greater than 30, or BMI > 30 which will be written as (30, +∞). Anyway, my BMI will be calculated, and I will explain how I arrived at the results. Sometimes, a person’s BMI can be misleading, so reasons will be provided about why. Finally, there is an evaluation of the regions outside of the “probably not overweight” range by using the set and interval notations along with a simple graph of the regions.

Now, I am five feet and eleven inches tall, and I weigh 180 pounds. Remember that one foot is equivalent to twelve inches. Since I am five feet tall, we will multiply five with twelve to get sixty. Now, I am an additional eleven inches taller than five feet, that is, sixty inches. Hence, we will add eleven inches to sixty inches to make that seventy one inches. The formula is:

BMI = (703W)/(H^2) where BMI is the Body Mass Index, W is the weight in pounds, and H is height in inches. Since I am seventy one inches tall, we will denote that as H = 71. Since I am 180 pounds, we will denote that as W = 180. Hence, plug both of the values into the formula, which is BMI = (703*180)/(71^2). 71^2 is 71 squared which means that 71 times 71 is 5,041. 703*180 is 703 times 180 equals 126,540. Hence, we will have BMI =

MAT 221 Week 2 DQ 1 Formulas

The formula for Cowling’s Rule is d=D(a+1)/24

1. A)    To calculate the child’s dose:

a=6

D=1000

d=child’s dose

d=D(a+1)/24

d=1000(6+1)/24

d=1000(7)/24

d=7000/24

d=291.66

d=292mg of acetaminophen for a 6 year old.

To calculate the child’s dose I substituted a for 6 and D for 1000, I then added 6 and 1 for the answer of 7, multiplied 7 by 1000 and finally divided that by 24 for the answer of 291.66 I then rounded this up to 292 for my final answer.

1. B)    To calculate the child’s age:

a=child’s age

D=600

d=200

MAT 221 Week 3 Assignment 3 Two-Variable Inequality

Ozark Furniture Company

In this paper we will discuss a linear inequality word problem. We will discuss and determine the regions which are safe and unsafe. From there we will know what lies in the safe region and we will produce an understanding of the safe region to solve the linear inequality.

The first step is much like any other step; we must attain a understanding of what we are working with. The text states that “Ozark Furniture Company can obtain at most 3,000 board feet of maple lumber for making its classic and modern maple rocking chairs. A maple classic rocker requires 15 board feet of maple, and a maple modern rocker requires 12 board feet of maple. Write an inequality that limits the possible number of maple rockers of each type that can be made, and graph the inequality in the first quadrant.”

MAT 221 Week 3 DQ 1 Parallel and Perpendicular

y-y1 = m(x-x1) is the slope formula to be used in both problems.

For the first problem I will show the math to find the a line parallel to y= -1/3x – 4 , that passes through the ordered pair (-6, -3). The slope for the given line and the line parallel to is -1/3, to be parallel the lines must follow the same slope so that they always run side by side with an equal space between them.

• y-y1 = m(x-x1)
• y-(-3) = -1/3(x-(-6)
• y + 3 = -1/3x – (-1/3)(-6)
• y + 3 = -1/3x – 2
• y + 3 – 3 = -1/3x – 2 – 3
• y = -1/3x – 5

MAT 221 Week 4 Assignment 4 Financial Polynomials

Compounded Semiannual Interest

In this paper we are given three problems to figure out. Two of these problems entail the use of compound interest, with the other problem dividing two polynomials. Through this paper we will discuss the steps needed in which to solve these problems. The following formula will be used P(1 + r/2)^2. With the fact that this has an exponent, we must remove this exponent to get an accurate answer. We will now set the formula up to (1 + r/2) * (1 + r/2). Using the foilmethod we will use multiplication. We will use (A + B)*(C + D) = (1 + r/2)*(1 + r/2) where A = 1, B = r/2, C = 1, and D = r/2. When A is multiplied by C, it basically boils down to 1 multiplied by 1 which results in 1. From there we will multiply A by D which is 1 times r/2, and that equals r/2. Last we will multiply B by D which would result in r/2 multiplied by r/2. The result of this is two fractions is the result of the product of the terms of the numerator is R and R which comes to r squared because there are two of R’s. The product of the two denominators is 2 times 2 which come to 2 squared. The squared becomes 4. So when everything is put where it needs to be it should create the fraction r squared over 4, or r^2/4. When one uses the foilmethod, it creates AC + AD + BC + BD which is the same as 1 + r/2 + r/2 + r^2/4. If one noticed, r/2 and r/2 are the same, or like terms which causes them to be added together.

MAT 221 Week 4 DQ 1 Initial Investment

My desired item is a RV

In 12 years it will cost about \$7000

My interest rate is 5%

P= Principal amount needed

A= Amount desired item will cost in 12 years

r= Interest rate

n= Number of years

P=A(1+r)-n

P=7000(1+.05)-12

P=7000(1.05)-12

P=7000

1.79585

P=3897.88

MAT 221 Week 5 Assignment 5 Pythagorean Quadratic

Buried Treasure

For this week’s Assignment we are given a word problem involving buried treasure and the use of thePythagorean Theorem. We will use many different ways to attempt to factor down the three quadratic expressions which is in this problem. The problem is as, ““Ahmed has half of a treasure map, which indicates that the treasure is buried in the desert 2x + 6 paces from Castle Rock. Vanessa has the other half of the map. Her half indicates that to find the treasure, one must get to Castle Rock, walk x paces to the north, and then walk 2+ 4 paces to the east. If they share their information, then they can find x and save a lot of digging. What is x?”

First we will look at the equation so we know how far Ahmed will have to walk which is 2x + 6 paces from Castle Rock.  If Ahmed used string and tied it to the Castle Rock point and labeled it as point “A” on paper it would be basically 2x + 6 paces.  With this being the radius you will know that it is the same anyway you fit this from the circle. So to find the Treasure we will use “C”. Vanessa will use her compass to find north from Castle Rock. From there she will walk “X” paces in a straight line northward. At the end of her distance she will call this point “B”. Vanessa will now turn 90 degrees to the right and will walk 2x+4 paces east until she is at point “C”. We have now acquired a line segment which we will call AB which is basically the line from A to B, the line segment from B to C is considered BC. However, the lines AB and BC intersect to form a perpendicular angel, and we will use line AC as Ahmed’s route. The end state of the line segments if one was to draw them out would equal a triangle. With the face that Vanessa had turned in a 90 degree angle that makes this triangle a right angle ABC. Lines AB and BC are to be considered as the legs and we will think of AC as the hypotenuse.

MAT 221 Week 5 DQ 1 Factoring

In this week’s discussion, I will be factoring polynomials using the appropriate strategy associated with the assigned problem. Since the third letter of my name is “n,” on pages 345-6 I will be working on problems 80 and 106 and on page 353 I will be working problem 68.

Pages 345-6:

1. w+ 30w + 81                             The given problem.  This trinomial is also in the form of a perfect square. Also each term has a GCF of 3.

81 = 27?3                                    81 is the product of 27?3.

30 = 27 + 3                                 30 is the sum of 27 + 3.

(w + 27)(w + 3)                           This is polynomial is now completely factored. Since all the terms are positive, we know that both of the factors will be positive.

1. ac + xc + aw2+ xw2                      This is also a perfect square polynomial and we will factor this by grouping.

C(a + x) + w2(a + x)                      The first thing we did is factor out the GCF from both the first and second pair of terms.

(c + w2)(a + x)                             This polynomial is now completely factored. No further work is needed since these terms are prime factors.

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