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(2 × 3) – (2 × 5x)=15
6-10x=15
6-15=10x
-9=10x
x= - 9/10
x= - 0.9
2. x^{2} + x -20 = 0
x^{2} + 5x -4x -20 = 0
x(x + 5) – 4(x + 5) =0
(x – 4) (x + 5) =0
x – 4 = 0 or x + 5 =0
x = 4 or x = -5
1. y=2x; y= -2x+1
Algebraic method:
y=2x and y= -2x+1
2x = -2x+1
4x=1
x= 1/4 = 0.25 and y=2x
y= 2 × 0.25
y=0.50
Graphical method:
(1) Y= 2x
(2) Y= -2x+1
Algebraic method:
y = 5x + 1 and y =−5x + 1
-5x+1=5x+1
1-1=5x+5x
10x=0
x= 0/10 =0 and y=5x+1
y=5 × 0 + 1
y=0+1=1
y=1 and x=0
Graphical method:
(1) y= 5x+1
(2) y= -5x+1
Algebraic method:
3x+6y=4 and x=2y-5
3(2y-5)+6y=4
6y-15+6y=4
12y=4+15
12y=19
y= 12/19 = 1.58and x=2y-5
x=2 × 19/12 – 5
x= 19/6 -5
Graphical method:
(1) -6y = 3x - 4
(2) 2y = x + 5
3 Find the volume of the following shapes to three significant figures by showing your work step by step
Volume of a cube = a^{3}
= (27)^{3} = 27 × 27 × 27 = 19683 m^{3}
Hence volume of a cube to three significant figures is 197m^{3}.
Pythagoras theorem states that Square of hypotenuse=Square of 1^{st} side+ Square of 2^{nd} side for a Right Angled Triangle
h^{2}= a^{2}+ b^{2}
15 × 15 = 9 × 9 + 12 × 12
225 = 81 + 144
225 = 225
Hence it is proofed that triangle ABC is a right angle triangle.
A Calculate vector c = a + b.
C= (13,20) + (5,-6)
C=(13+5,20-6)
C=(18,14)
B Calculate magnitude of vector c.
Magnitude of vector c = <x, y>
c=<18,14>
C Write a Pseudocode for calculating magnitude of vector c.
Algorithm: magnitudeOfVector(x, y)
Purpose: magnitude of a vector whose 2 components are given
Pre: Given both components of vector
Post: None
Return: Magnitude
{
magnitude ← sqrt(x*x + y*y)
Return magnitude
}
Get assignment help from full time dedicated experts of Locus assignments.
Call us: +44 – 7497 786 317A={ Martin, Marriott, Boast, Preston, Kans}
B= {24k, 25k, 26k, 27k, 30k}
C= {Production, Sales, Finance}
Mr Martin and Mrs Marriott are working at production department, Mrs Boast and Mrs Preston working at sales department and Mr Kans works at Finance department.
Find the Cartesian product of set A and set B. (R=A×B)
Cartesian of a product (R= A × B)
={ Martin, Marriott, Boast, Preston, Kans} × {24k, 25k, 26k, 27k, 30k}, ={Martin , 24k},{Martin , 25k},{Martin , 26k},{Martin , 27k},{Martin , 30k}, {Marriott , 24k},{Marriott , 25k},{Marriott , 26k},{Marriott , 27k},{Marriott , 30k}, {Boast , 24k}, {Boast , 25k}, {Boast , 26k}, {Boast , 27k}, {Boast , 30k}, {Preston , 24k}, {Preston , 25k}, {Preston , 26k}, {Preston , 27k}, {Preston , 30k}, {Kans , 24k}, {Kans , 25k}, {Kans , 26k},{Kans , 27k},{Kans , 30k}
Find the Natural join of R and C.
= (Production, Martin, 24K),(Production, Marriott, 25K), (Sales, Boast, 26K),(Sales, Preston, 27K), (Finance, Kans, 30K)
Fill in the below table by using provided information:
Employee Name | Salary | Department |
Martin | 24,000 | Production |
Marriott | 25,000 | Production |
Boast | 26,000 | Sales |
Preston | 27,000 | Sales |
Kans | 30,000 | Finance |
1. Redbridge,
2. Enfield,
3. Barnet.
Five technicians with following details are working at this company:
Ali (Location: Barnet, age: 25, salary: £21,000),
Steve (Location: Redbridge, age: 45, salary: 23,000),
Mike (Location: Enfield, age: 50, salary: 19,000),
Linda (Location: Barnet, age: 55, salary: 24,000 ),
Carol (Location: Redbridge, age: 43, salary: 27,000)
A. Draw required number of tables and fit in the above information there.
Employee Name | Age | Location | Salary (pounds) |
Ali | 25 | Barnet | 21,000 |
Steve | 45 | Redbridge | 23,000 |
Mike | 50 | Enfield | 19,000 |
Linda | 55 | Barnet | 24,000 |
Carol | 43 | Redbridge | 27,000 |
B List individuals satisfying the conditions below:
1. (Age<46) AND (Salary> £ 23,000)
{Carol}
2. (Age>26 ) OR (Salary < £24,000)
Steve ,Mike,Linda,Carol,Ali
3. (Age< 53) AND (Salary>29) OR (Location=1)
Steve,Carol
4. (Age> 25) XOR (Salary>30) OR (Location=2)
Steve ,Mike,Linda,Carol
0101 15 14 p
A= 12 06 q r
08 s t u
X 03 y z
Magic numbers 4X4 , the total should be equal to 34 from all the sides. All the number 1 to 16 should be used to solve the matrix using each number only one.As per the grid 01, 15, 14, 12, 06, 08 and 03 are already present. The rest numbers to be used. Left are 2, 4, 5, 7, 9, 10,11,13,16. These numbers when used in combination gives the total 34
01+15+14+p = 34 => p = 34 – 30 => p = 4
01+12+8+x = 34 => x = 34 – 21 => x = 13
15+06+s+3 = 34 +> s = 34 – 24 => s = 10
x + s + q + p = 34 => 13+10+q+4 = 34 => q = 34 – 27 => q = 7
12+06+q+r = 34 => 12+06+7+r = 34 => r = 34 – 25 => r = 9
We have left to find the values for t, u, y, z.
Also from the 16 numbers that we have in our Matrix, we have left 2, 5, 11, 16.
08 + s + t + u = 34 => 08 + 10 + t + u = 34 => 18 + t + u = 34 => t + u = 34 – 18 =>
t + u = 16 => t = 11, u = 05;
x + 03 + y + z = 34 => 13 + 03 + y + z = 34 => 16 +y + z = 34 => y + z = 34 – 16 =>
y + z = 18 => y = 02, z = 16.
0 01 15 14 04
A = 12 06 07 09
08 10 11 05
13 03 02 16
Hence p = 04, q = 07, r = 09, s = 10, t = 11, u = 05, x = 13, y = 02, z = 16
Then P is the inverse of Q. [P2.2]
If P × Q = I and Q × P = I then P = Q^{-1} and Q = P^{-1}(P and Q are inverse of each other)
If P is inverse of Q then below statements should hold true: P×Q=I where I = as below
11 0
0 1
P × Q = 1 2 −2 1
3 4 1.5 −0.5
P × Q = 1×(−2)+2×1.5 1×1+2×(−0.5)
3×(−2)+4×1.5 3×1+4×(−0.5)
−2+3 1+(−1) P × Q = 1 0
−6+6 3+(−2) 0 1
P × Q = I
Q × P = −2 1 1 2
1.5 −0.5 3 4
Q × P = (−2)×1+1×3 (−2)×2+1×4
1.5×1+(−0.5)×3 1.5×2+(−0.5)×4
Q × P = −2+3 −4+4 Q × P = 1 0
1.5+(−1.5) 3+(−2) 0 1
Q × P = I
P × Q = I and Q × P = I then P = Q^{-1} and Q = P^{-1}(P and Q are inverse of each other)
Hence P is inverse of Matrix Q
A = { g, e, r, m, a, n, i }
B = { p, o, l, a, n, d }
Identify the following statements as true or false:
{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27} and consider two sets P and O defined as follows:
P = “all multiples of 3”
O = “the first ten even numbers”
Represent all of the elements in a Venn diagram and identify the elements in P?O, P?O
and PΔO.
P = {3, 6, 9, 12, 15, 18, 21, 24, 27}
O= {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}
Those elements that they make the union of elements in P and O.
where P?O = { 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 27}
Those elements that they are common in P and O.
and P?O = {6, 12, 18}
PΔO= P?O \ P?O
VENN DIAGRAM
S1 = {x: x = 2n, where 1 ≤ n ≤ 6}
Values for n = (1, 2, 3, 4, 5, 6)
Replacing with the n values in x = 2n => x = 2 × (1, 2, 3, 4, 5, 6)
S1 = {2, 4, 6, 8, 10, 12}
S2 = {x: x = 3n^{2}, where 1 ≤ n ≤ 5}
Values for n = (1, 2, 3, 4, 5)
Replacing with the n values in x = 3n^{2}=> x = 3 × [(1^{2}), (2^{2}), (3^{2}), (4^{2}), (5^{2})] =>
x = 3 × (1, 4, 9, 16, 25)
S2 = {3, 12, 27, 48, 75}
S3 = {y: y = 5n^{3}, where 1 ≤ n ≤ 4}
Values for n = (1, 2, 3, 4)
Replacing with the n values in x = 5n^{3} => x = 5 × [(1^{3}), (2^{3}), (3^{3}), (4^{3})] =>
x = 5 × [(1×1×1), (2×2×2), (3×3×3), (4×4×4)] → x = 5 × (1, 8, 27, 64)
S3 = {5, 40, 135, 320}
S4 = {x: x = √n, where 3 < n < 5} → Values for n = 4
Replacing with the values for n in x = √n => x = √4
S4 = {2}
A | B | C | B·C | A·B·C | Not |
0 | 0 | 0 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 0 | 1 |
0 | 1 | 0 | 0 | 0 | 1 |
0 | 1 | 1 | 1 | 0 | 1 |
1 | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 0 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 0 |
Find the number of different ways that the salesman can leave home, visit two different customers and then return home.
Write a pseudocode for calculating the answer for the previous section.
Algorithm: PermutationOfTwo(x,y)
Purpose: No of ways of y out of x with repetition
Pre: Given both x and y
Post: None
Return: permutation
{
b What is the Mode value of number of children’s per household?
Mode value = None
So 5^{th} no will be median
=140
X = Sales | mean | X - mean | (X-mean)^{2} | |X – mean | |
150 | 150 | 0 | 0 | 0 |
130 | 150 | -20 | 400 | 20 |
140 | 150 | -10 | 100 | 10 |
150 | 150 | 0 | 0 | 0 |
140 | 150 | -10 | 100 | 10 |
300 | 150 | 150 | 22500 | 150 |
110 | 150 | -40 | 1600 | 40 |
120 | 150 | -30 | 900 | 30 |
140 | 150 | -10 | 100 | 10 |
120 | 150 | -30 | 900 | 30 |
Propositional logic: That branch of mathematics which is mainly concerned with the binary terms ‘true’ and ‘false’ is popularly known as Propositional Logic. But its scope is not just limited to mathematics, and extends to the field of English and Computer Science also. We will be discussing here the uses of Propositional Logic in the area of computers. For a sound understanding of logic in day to day life, English language can be used to well organize it. Logic can be divided as: and (conjunction), or (disjunction), not (negation), and if (material conditional). Propositional calculus is the basis for categorizing in the field of computer science.
Computers and Propositional Logic: With the introduction of automation of computer systems , logical mathematics has strongly impacted the concept of Artificial Intelligence. With the realisation of the fact that in solving conclusive operations on the computer, logical inferences can be highly effective, this process was initiated.
When FOL (first order logic) was accepted as a fine way of evaluating and analyzing information, a same kind of representation was expressed to the pinnacle in actual. Research projects like Knowledge Based Software Assistant also uses it. Propositional Logic can help in transforming computer specifications into a diversified range of codes. The use of Propositional Logic extends in the field of High Frame Languages where complex semantics of Frame languages can be simplified, ‘KL One’ for instance in this regard. A calculus form had been derived which helps in analyzing the sets, subsets etc. which is a characteristics of this. Propositional Logic is linked with the field of computers in many ways, where the most effective in all are as presented here.
Attributes of a Propositional Logic: The attributes of Propositional Logic in the field of computer science are as follows:
The science of atomic propositions is the appropriate definition of Propositional Logic in the field of computer science. Some complex issues termed as connectives that are associated with Propositional Logic, are discussed below for a better understanding.
And: It can be termed as a conjunctive form of Propositional Logic. For example, if A and B are two propositions, then they can be expressed as:
A∧B
For the whole system to be true, both A and B need to be true.
The concept of Boolean algebra is used to present it like the binary system.
Or: It is termed as a disjunctive form of Propositional Logic and the relation between A and B here can be expressed as:
A∨B
For the overall logical system to be true, either one of them should may be true.
If and Then: Propositional Logic having this combination is termed as conditional logic. If A and B are two arbitrary elements, then they can be expressed as:
A ⇒B
This is true only if either A is false or B is true, that is, a conditional relationship between A and B is depicted here.
Not: Above two considerations are binary while this is exclusively Unary and it is a negation statement, that is, totally opposite for an element.
When we write, ¬B, it means that if A is false, then it can be true.
Applications and Analysis:To calculate logical consequences, this concept is effectively applicable where tools like Proof Assistants follow logic that is based on propositions. An error-free proof script van be developed effectively because the validity of proofs can be checked with the development of a proof-checker.
Industry applications can take the benefit of languages like HML, Scheme etc. that are logic based. The designing of Java Card specifications uses an effective method of Jakarta Toolset. The ad-hoc model transformations that are highly structure based can be easily passed into virtual machine operations with the help of this Jakarta Toolset. Thus proof obligations can be developed effectively.
The Pythagorean Theorem[Online] Available at [HYPERLINK http://www.mathsisfun.com/pythagoras.html] [Accessed on 15/11/2014]
Graphical drawing [online] Available at [HYPERLINK
http://www.math-aids.com/Graph_Paper/Coordinate_Plane_Graph_Paper.html
[Accessed on 08/11/2014]
Mean, Median, Mode, and Range [online] Available at [HYPERLINK http://www.cimt.plymouth.ac.uk/projects/mepres/book8/bk8i5/bk8_5i2.htm] [Accessed on
15/11/2014]
Standard Deviation and Variance [online] Available at [HYPERLINK http://www.quickmba.com/stats/standard-deviation] [Accessed on 08/11/2014]
Calculating Magnitude with Vectors [online] Available at [HYPERLINK http://hotmath.com/hotmath_help/topics/magnitude-and-direction-of-vectors.html] [Accessed on
15/11/2014]
Brown, F. (2003), Boolean Reasoning: The Logic of Boolean Equations, 1st edition, Kluwer Academic Publishers, Norwell, MA. 2nd edition, Dover Publications, Mineola, NY.
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