205
How could you use Descartes' rule and the Fundamental Theorem of Algebra to predict the number of complex roots to a polynomial as well as find the number of possible positive and negative real roots to a polynomial? |

Descartes rule is really helpful because it eliminates the long list of possible rational roots and you can tell how many positives or negatives roots you will have. Fundamental Theorem of Algebra finds the maximum number of zeros which includes real and complex numbers.

For example:

f(x)= x^3 – 6x^2 + 13x – 20

Yes Yes Yes ( the sign changes from one to the other)
so now we can have 3 or 1 positives.

now to find the negative

f(-x)= -x^3 - 6x^2 – 13x -20

No No No ( the signs did not change at all)
therefore, we have 0 negatives.

Then whenever you make your list, you can eliminate all of the negatives since it would be included in your answer. Since the degree is three, we are going to have three zeros. Positive | 3 | 1 |

Negative | 0 | 0 |
Imaginary | 0 | 2 |
In this example, your answer could have all three positives number or one positive number and two imaginary.

Now let's try another!

f(x)= -x^4 – 2x^3 – 8x + 2

No No Yes (one positive)

f(-x)= -x^4 + 2x^3 + 8x +2

Yes No No (one negative)
Positive | 1 |
Negative | 1 |
Imaginary | 2 |
In this example,the only roots you can have is one positive and one negative the other two are imaginary since the degree of this polynomial is four.

...BIRONDO EDUCATION SUPPORT TUTORIAL CENTER
MATH IV- CHAPTER 2-1 to 2-3
I. Decide whether the relation is a function. If it is a function, state the domain and range.
1. {(-5, -2), (-1, 1), (3, -6), (8, 1)} 2. {(2, -9), (2, -2), (6, 8), (8, 1), (11, -7)}
3. a x 4.
b y
c z
5. 6.
7. 8. 9.
II. FUNCTION NOTATION. Solve the following:
1. f(x) = x3 – x 2– 6x 2. g (x) =
a. f(-2) – f(9) a. g(5) + g(10)
b. -6 [(f(0) + f(9)] b. 8 [g(1) - g(2)]
III. LINE OF BEST FIT. Identify the following:
Name
No. Of Pencils
No. Of Rulers
Abigail
4
1
Bob
16
4
Frank
7
8
Peter
4
5
James
7
2
Alex
27
14
Jack
4
2
Paul
6
3
Lucie
4
2
Jenna
0
1
Claire
2
1
Kym
4
0
a. Sketch a scatter plot for the data.
b. Find the best fit line.
c. Sketch the best fit line on the scatter plot at the left.
d. Use the best fit line to predict the number of rulers does Paul if he has 30 pencils.
e. Find r.
f. Write a sentence which explains the direction and strength of the correlation.
2. Sand lance fish (found in the Northwest Atlantic) were collected and there age and length were recorded.
Age (years) 2 3 4 5 6 7 8
Mean Length (mm) 176 194 212 226 236 244 254...

...Your Name: Jennifer Green MAT 205 Final Examination
Your Score: of 250 points
NOTE: You must show your work on each problem to receive full credit points allocated
for each problem (excluding T/F questions)
Write a matrix to display the information.
1) At a store, Sam bought 3 batteries, 15 60-watt light bulbs, 46 100-watt light bulbs, 8 picture-hanging kits, and a
hammer. Jennifer bought 12 batteries, 3 100-watt light bulbs, and a package of tacks. Write the information as a
2 x 6 matrix.
| |BATTERIES |60watt light bulbs |100 watt light bulbs |Picture hanging kits |Hammer |Package of tacks |
|Sam |3 |15 |46 |8 |1 |0 |
|Jennifer |12 |0 |3 |0 |0 |1 |
In a certain distribution, the mean is 50 with a standard deviation of 6. Use Chebyshev's theorem to tell the probability that a number lies in the following interval. Round your results to the nearest whole percent.
2) Between 35 and 65
Ans: we have z = (35-50)/6 = -2.5 and z = (65-50)/6 = 2.5
So 35 and 65 lies within 2.5 standard deviation of mean
By Chebyshev’s theorem we have the probability as 1 -1/k2 , where k is the standard deviation within which the numbers
In this case k = 2.5...

...
ANALYSIS
Physics has a lot of topics to cover. In the previous experiments, we discussed Forces, Kinematics, and Motions. In this experiment, the focus is all about Friction. Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There are several types of friction like fluid friction which describes the friction between layers of a viscous fluid that are moving relative to each other; dry friction which resists relative lateral motion of two solid surfaces in contact and is subdivided into static friction between non-moving surfaces, and kinetic friction between moving surfaces; lubricated friction which is a case of fluid friction where a fluid separates two solid surfaces; skin friction which is a component of drag, the force resisting the motion of a fluid across the surface of a body; internal friction is the force resisting motion between the elements making up a solid material while it undergoes deformation and sliding friction.
When surfaces in contact move relative to each other, the friction between the two surfaces converts kinetic energy into heat. This property can have dramatic consequences, as illustrated by the use of friction created by rubbing pieces of wood together to start a fire. Kinetic energy is converted to heat whenever motion with friction occurs, for example when a viscous fluid is stirred. Another important consequence of many types of friction can be wear,...

...
The case between Beauty and Stylish involves concept of a valid contract, pre-contractual statements, express term and misrepresentation.
A valid contract is established between Beauty and Stylish when an offer is accepted and there is intention for both parties to create legal relations. An offer refers to the expression of willingness of the offerer to be contractually bound by an agreement if his or her offer is properly accepted. It has to be clear and certain in terms. It must also be communicated to the offeree before it is being accepted. In addition, the acceptance has to be unqualified, unconditional and made by a positive act. In the case of Beauty and Stylish, a positive act refers to the signing of the contract. All terms of the offer must be accepted without any changes and cannot be subjected to any condition, taking effect only upon fulfillment of that condition. When Beauty and Stylish enter into the agreement, they must intend to bind and bound legally to each other by their agreement. This is the intention to create legal relations between two parties. In the meanwhile, this contract must possess consideration. A contract must therefore be a two-sided affair, with each side providing or promising to provide something of value in exchange for what the other is to provide.
Every contract, whether oral or written, contain terms. The terms of a contract set out the rights and duties of the parties. Terms are the promises and undertakings given by each...

...MATH 122 SYLLABUS
Faculty Information: Name: E-mail: Office: Office Hours:
Section 08 B-1-122 MAK MW 6:00 – 7:15 PM Fall 2013
Corrina Campau campauc@gvsu.edu A-1-132 MAK Phone: (616)331-2052 Tuesday, Thursday 3:45 – 4:30 PM Monday, Wednesday 3:30 - 4:30 PM Monday, Wednesday 7:15 – 8:00 PM by appointment only
Prerequisite:
MTH 110 (a grade of C or better is recommended) or assignment through GVSU Math Placement. You may wish to take the MTH 122 proficiency test which would allow you to waiver 122 and is offered during the first week of class and other times during the semester. For more information visit gvsu.edu/testserv and click on Math placement. College Algebra MTH 122 special edition for Grand Valley State University by John Coburn Students will be required to possess and make use of a TI-83 or TI-84 graphics calculator during the course. You are expected to have and use your calculator every class period. Students will not be allowed to share calculators on tests. Symbolic manipulating calculators (such as the TI-89) and calculators on cell phones, PDA’s, etc. will not be allowed on tests. Math 122 is part of the Mathematical Sciences General Education Foundation Category. Courses in the Foundations Categories introduce students to the major areas of human thought and endeavor. These courses present the academic disciplines as different ways of looking at the world. They introduce students to the...

...MATHEMATICS
SAMPLE TEST PAPER (SEMSTER II)
CLASS VI
Class:6
Time :2hrs
Max Mks:45
No of pages: 3
General Instructions:
All questions are compulsory.
Questions 1to 4 carry 1mark each.
Questions 5to 7 carry 2mark each.
Questions 8 to 12 carry 3mark each.
Questions 13 to17 carry 4 mark each.
rit
e.
co
m
Ò
Ò
Ò
Ò
Ò
.e
du
1. Rohit has 8 blue balls, 5 red balls and 12 yellow balls. What is the ratio of blue and yellow
balls?
2. Name the side, vertices ad diagonals of given figure.
w
w
w
3. How many right angles do you make if you start facing south and turns clockwise to east.
4. Construct a line segment having length of 2a
5. Write the number of sides and angles of below figures
6. Draw the mirror image of the given letter
w
w
w
.e
du
rit
e.
co
m
7. Find the perimeter of the rectangle with l= 3.m-6m B= 2.6m
8. Draw a line segment XY of any length. Now,without measuring it , draw a copy of XY
9. Given RS = 2.3 draw AB such that the length of XY is thrice of RS verify by measurement.
10. Find the perimeter of a regular octagon with sides 12.8 cm.
11. Find the No of faces, No of corners, No of edges of a cuboid
12. Measure each of the marked angle with the help of protractor
13. Prove that a square is a rhombus with all its angles at right angles.
14. Draw a rough sketch of a pentagon and draw all it diagonals. How many diagonals can you
draw
15. Draw a circle of radius 5.4 cm mark points A at...

...Chapter 11
Four Decades of the Defence of
Australia: Reflections on Australian
Defence Policy over the Past 40 Years
Hugh White
The serious academic study of Australian defence policy can be said to have
begun with the publication of a book by the SDSC’s founder, Tom Millar, in
1965. The dust jacket of that book, Australia’s Defence, posed the following
question: ‘Can Australia Defend Itself?’ Millar thus placed the defence of Australia
at the centre of his (and the SDSC’s) work from the outset. Much of the SDSC’s
effort over the intervening 40 years, and I would venture to say most of what
has been of value in that effort, has been directed toward questions about the
defence of the continent. This has also been the case for most of the work by
Australian defence policymakers over the same period. In this chapter I want
to reflect on that work by exploring how the idea of the ‘defence of Australia’
has evolved over that time, and especially how its role in policy has changed,
from the mid-1960s up to and including the most recent comprehensive statement
of defence policy, Defence 2000: Our Future Defence Force.
This is no dry academic question. The key question for Australian defence
policy today is how we balance priority for the defence of Australia against
priority for the defence of wider strategic interests. The starting point for that
debate is the policies of the 1970s and 1980s, which placed major emphasis on
the defence of the continent....

...Yr 10
Mathematics
Assignment
LCR Maths
By Adonis Chigeza
Understanding and Fluency Tasks
Task A
1. y = 1.2𝑥 + 2.57
2. Interpolation: y = -3.43
Extrapolation: y = -8.23
Task B
a) The equation for the path of the ball is h = -0.1t^2 + 0.9t + 1 (h = height, t = time)
b) The vertical height of the ball after 2. seconds2.664m
c) The maximum height reached by the ball is 3.025m
d) The time of with the ball is at maximum height of 3.025 is 4.5 seconds
e) The total time in which the ball was in the air is 10 seconds
f) The two times in which the ball was 1 metre above ground is 0 and 9
Adonis Chigeza 10C
LCR Mathematics
Problem Solving and Reasoning Task
1.
Equation: y = -1.2𝒙2 + 8.4𝒙
a. The bridge is 7 metres wides so therefore it will successfully span the river with 2
metres to spare.
b. If a yacht has a 15 metre mask it will be unable to pass safely under the bridge
because the bridge only has a vertical height 14.7 metres.
Adonis Chigeza 10C
LCR Mathematics
2. Equation: v= -0.2h2 + 2.4h
a. The horizontal distance covered by the rocket when it reached its maximum
height of 7.2 metres was 6 metres.
b. The maximum height reached by the rocket was 7.2 metres.
c. At the horizontal distance of 9 metres from the launch site, there is a 5.2 metre
wall and at that vertical distance, the rocket has a vertical distance 5.4 metre.
That is not taking to account the dimensions of the rocket, however the rocket
cannot have...

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