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Discussion in 'Pseudoscience' started by theorist-constant12345, Feb 1, 2015.

Not open for further replies.
1. ### theorist-constant12345BannedBanned

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2,660
v and the 2 ,2's.

2 velocities squared?

I get distance / distance over time = v

should it not be tan 0 for the head over heels?

sin=1 then sin=0 cos-1

Last edited: Feb 6, 2015

3. ### theorist-constant12345BannedBanned

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2,660
Hmmm.....I get Kinetic energy from the bike is transferred into the rider when the bike is on its front wheel propelling the rider forward while the bike flips.

Biker travelling a Tan line while the bike continues in a cos-direction.

5. ### DywyddyrPenguinaciously duckalicious.Valued Senior Member

Messages:
18,742
You don't know what the word "fulcrum" means, do you?

7. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

Messages:
10,450
OK, TC looks at this $\frac{d}{dt}$ and says "I get distance / distance over time = v"

That is simply wonderful. Thank god I wasn't drinking anything it would have been right out my nose.

8. ### Dr_ToadIt's green!Valued Senior Member

Messages:
2,527
I wasn't, and it almost did anyway.

Why do we

continue?

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2,660
motor bike

File size:
10.9 KB
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3
10. ### theorist-constant12345BannedBanned

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2,660
+v=-g=mE+=(Fn=0+)=KE=S=c=+g

Plus velocity = less gravity = more mass energy = force normal energy increase that is equal to kinetic energy that is equal to entropy change that is equal to change at the speed of light.

The faster you go , the faster the rate of exchange with gravity the more magnitude/g-force.

Last edited: Feb 7, 2015

Messages:
18,742
No.

And
No.

12. ### theorist-constant12345BannedBanned

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2,660
You always say no and never why no.

13. ### DywyddyrPenguinaciously duckalicious.Valued Senior Member

Messages:
18,742
I say no because you're wrong.
I never say why because what you post is meaningless nonsense.

You post utter unsupported (and unscientific) drivel and expect some sort of explanation as to WHY you're wrong?
That's the equivalent of asking someone to explain why the statement "Gravity is made of concentrated Mondays and smells of blue mixed with the sound of cat's bark" is wrong - the initial claim is so ridiculous that explanations fail.

14. ### theorist-constant12345BannedBanned

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2,660
Ok, I gave you maths showing that you have light faster than time, how do you explain this if my stuff is meaningless drivel?

15. ### theorist-constant12345BannedBanned

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2,660
I gave you time and space the proper version how do you explain this?

I gave you the nature of light?

I give you what 0 is?

16. ### DywyddyrPenguinaciously duckalicious.Valued Senior Member

Messages:
18,742
No you didn't.
You just posted utterly meaningless sh*t.

See previous comment.

Nope.
You didn't explain (and what you DID post was sheer nonsense).

17. ### theorist-constant12345BannedBanned

Messages:
2,660
I even had to explain what gravity is for you all...

18. ### theorist-constant12345BannedBanned

Messages:
2,660
Only an idiot would believe it is meaningless. I unified everything.

19. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

Messages:
10,450
You say:
Negative gravity equals positive gravity?
Kinetic Energy equals the speed of light?

Who is the idiot again?

20. ### theorist-constant12345BannedBanned

Messages:
2,660
How fast does energy travel again?

starts off -fn, reaches above fn, more energy quicker extraction to remain almost at an equilibrium.

forget gravity being like weight. The bike reduces ''weight'' but the extraction flow increases to the ''ground'' that is why the bike does not fall over. AC and DC

1. ''In alternating current (AC), the flow of electric charge periodically reverses direction. In direct current (DC, also dc), the flow of electric charge is only in one direction. The abbreviations AC and DC are often used to mean simply alternating and direct, as when they modify current or voltage.''
''The power losses (

) in a conductor are a product of the square of the current (I) and the resistance(R) of the conductor, described by the formula''

Last edited: Feb 7, 2015
21. ### theorist-constant12345BannedBanned

Messages:
2,660
Where v is velocity and x is decrease and y is increase in energy E and distance being d and r being rate of energy flow.

+v=+y=+r=+d/t=-n

-v=+x=-r=-d/t=+n

Last edited: Feb 7, 2015
22. ### DywyddyrPenguinaciously duckalicious.Valued Senior Member

Messages:
18,742
What?

Um, no one thinks that "gravity is like weight".

No.

F*ck all to do with anything you've posted so far.

No.

23. ### theorist-constant12345BannedBanned

Messages:
2,660
I am trying get you to see it without much prevail. Spot the forest and not just the trees.

format: plain


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# Bifurcation of convection of fluid in container, when adding temperature

I once read a paper, in which:

• a fluid in a container was heated from below,
• after reaching temperature $T_1$, a circular motion (convection) was clearly distinguishable, in form of cylinder,
• after reaching temperature $T_2$, the circular motion splitted into two circular convections, side by side two cylinders,
• after reaching temperatures $T_3, T_4, \ldots$, etc. waves appeared on cyliders' flat-sides (where the height is measured),
• the frequency of the waves doubled at temperatures $T_4, T_5, \ldots$.

Ratio of every temperature pair $T_1/T_2, T_2/T_3, \ldots$ was the same as in logistic map ratios of parameter $r$, and in every other map - the Feigenbaum number, the period-doubling route to chaos.

I cannot find the paper again.. Does anyone remember such paper? Or maybe other?


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# Vulkan clear color is wrong

I am following https://vulkan-tutorial.com/Introduction to get myself familiar with the Vulkan API. I have gotten to the point where my application is able to draw a triangle. But for some reason the clear color is way brighter than the RGB value that i specified in my code. I am setting the color like this:

    VkClearValue clearColor = {0.2f, 0.2f, 0.2f, 1.0f};

VkRenderPassBeginInfo renderPassBeginInfo = {
.sType = VK_STRUCTURE_TYPE_RENDER_PASS_BEGIN_INFO,
.renderPass = renderPass,
.framebuffer = swapChainFramebuffers[i],
.renderArea = {
.offset = {
.x = 0,
.y = 0
},
.extent = swapChainExtent
},
.clearValueCount = 1,
.pClearValues = &clearColor
};

An RGB value of [0.2, 0.2, 0.2, 1.0] or [50, 50, 50, 255] should be a very dark gray, almost black. Instead i am getting is this:

This looks more like an RGB value of [150,150,150,255]. To get a color close to what i want i have to instead go with something like this:

VkClearValue clearColor = {0.02f, 0.02f, 0.02f, 1.0f};

So it almost seems like everything is shifted by a decimal. Any idea what might cause this issue?


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## Precalculus (6th Edition) Blitzer

We have the expression on the left side $\left( \sec \theta -1 \right)\left( \sec \theta +1 \right)={{\sec }^{2}}\theta -1$ and using the identity ${{\sec }^{2}}\theta ={{\tan }^{2}}\theta +1$ we simplify as: \begin{align} & \left( \sec \theta -1 \right)\left( \sec \theta +1 \right)={{\sec }^{2}}\theta -1 \\ & =1+{{\tan }^{2}}\theta -1 \\ & ={{\tan }^{2}}\theta \end{align} Hence, the expression on the left side is equal to the expression on the right side $\left( \sec \theta -1 \right)\left( \sec \theta +1 \right)={{\tan }^{2}}\theta$.


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# Factorizing a polynomial

1. Feb 10, 2009

### Niles

1. The problem statement, all variables and given/known data
Hi all.

I have to solve $0=-x^4+x^2+\alpha$, where alpha is a constant. I thought of factorizing it, but this wont work.You guys have any tip for solving this?

Sincerely,
Niles.

2. Feb 10, 2009

### Dick

If you substitute u=x^2, it turns into a quadratic equation.

3. Feb 10, 2009

### Niles

Ahh, so simple. I hadn't thought of that.

Thanks.


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Home > Matrix & Vector calculators > Singular Value Decomposition (SVD) of a Matrix calculator

Method and examples
Matrix operations
Method

Singular Value Decomposition (SVD) of a Matrix calculator
Matrix A :

 1 2 3 0 1 0 2 3 1

1. [[4,0],[3,-5]]
2. [[1,0,1,0],[0,1,0,1]]
3. [[1,-2,3],[5,8,-1],[2,1,1],[-1,4,-3]]

 Mode = Decimal Fraction Decimal Place = 0 1 2 3 4 5 6 7 8
SolutionHelp

Dear user,


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# 1st PUC Maths Question Bank Chapter 2 Relations and Functions

Students can Download Maths Chapter 2 Relations and Functions Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka 1st PUC Maths Question Bank Chapter 2 Relations and Functions

Question 1.
Define ordered pair.
Two numbers a and b listed in specific order and enclosed in parentheses form is called an ordered pair (a, b).
Keen Eye: Equality of two ordered pairs:
We have {a, b)-(c,d)⇔a-c and b – d.

Question 2.
Define a Cartesian product of two sets.
Let A and B two non-empty sets. Then, the Cartesian product of A and B is the set denoted by Ax B, consisting of all ordered pairs (a, b) such that a e A and be B.
A x B= {(a, b): a ∈ A, b∈B}
Keen Eye:

• If n(A) = p and n(B) = q, then n (A x B) = pq and n (B x A) = pq
• If at least one of A and B is infinite then AxB is infinite and B x A is infinite,
• In general, A x B ≠ B x A
• A x A x A = {(a, b, c) : a, b, c ∈A}. Here (a, b, c) is called an ordered triplet.

Question 3.
If (x + 1, y – 2) = (3,1), find the values of x
Given (x + 1, y – 2) = (3,1)
⇒ x+1=3  ∴x=2
y-2=1  ∴ y=3

Question 4.
If $$\left(\frac{x}{3}+1, y-\frac{2}{3}\right)=\left(\frac{5}{3}, \frac{1}{3}\right)$$
Given $$\left(\frac{x}{3}+1, y-\frac{2}{3}\right)=\left(\frac{5}{3}, \frac{1}{3}\right)$$

Question 5.
If P={a,b,c}and Q={r},from P×Q and Q x P.Are these two products equal?
PxQ = {(a,r),(b,r)(c,r)}
QxP = {(r, a), (r, b), (r, c)}
Clearly PxQ≠QxP

Question 6.
If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in
Given n(A) = 3; n(B) = 3.
∴ n(AxB) = 3×3 = 9

Question 7.
If G=(7, 8} and H={5, 4, 2), find G x II and
GxH = {(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}
HxG = {(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}

Question 8.
State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If P={m, n} and Q = {n, m}, then P x Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then A x B is a non-empty set of ordered pairs (at, y) such that x∈B and y∈A
(iii) If A = {1,2}, B = {3,4}, then A x {B∩φ ) = φ
(i) Given statement is false:
Correct statement:
PxQ={(m, n), (m, m), (n, n), (n, m)}.

(ii) Given statement is false:
Correct statement:
AxB = {(x, y) :x∈A, y ∈B}.                                 ‘

(iii) True statement,

Question 9.
If A x B = {(a, x), (a, y), (b, x), (b, y)}. Find A and
A = {a, b} and B – {x, y}

Question 10.
If A x B = {(p, q), (p, r), (m, q), (m, r)}, find A and
A = set of first elements = {p, m}
B = set of second elements = {q, r}

Question 11.
Let A = (1, 2}, B = [1, 2, 3, 4}, C = { 5, 6} and D = (5,6,7,8}. Verify that
(i) A x (B∩C) = (A x B)∩(A x C).
(ii) A x C is a subset
(i) B∩C = { }
∴ Ax(B∩C)=φ ………….. (1)
A x B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2,4)}
A x C = {(1, 5), (1,6), (2, 5) (2,6)}
∴ (A x B) ∩ (A x C) = φ ………………. (2)
From (1) and (2), we get
A x (B∩C) = (A x B)∩(A x C)

(ii) A x C = {(1, 5), (1,6), (2, 5), (2, 6)}
B x D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}.
Clearly every elements of A x C is an element of B x D.
A x C ⊂B x D.

Question 12.
Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find
(i) A x (B ∩ C)
(ii) (A x B) ∩ (A x C)
(iii) A x (B∪C)
(iv) (A x B)∪(A x C)
(i) B∩C={4}
A x (B∩C) = (1,4), (2, 4), (3,4)}

(ii) A x B = {(1, 3), (1,4), (2, 3), (2, 4), (3, 3), (3, 4)}
A x C = {(1, 4), (1, 5) (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A x B)∩(A x C)= {(1, 4), (2, 4), (3, 4)}

(iii) B ∪ C={3,4, 5, 6}
∴ Ax(B∪C)  = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}

(iv) A x B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
A x C = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A x B)∪(A x C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6) (3, 3), (3, 4), (3, 5), (3,6)}.

Question 13.
Let A = {1, 2} and B = {3, 4}. Write A x B. How many subsets will A x B have? List them.
Given A = {1, 2} and B = {3,4}
A x B= {(1, 3), (1,4), (2, 3), (2, 4)}
∴n (A x B) = 4
Number of subsets of A x B = 24=16
Subsets of A x B are: A x B, φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2,4)}, {(1,4), (2, 3)}, {(1, 4), (2, 4)} {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1,4), (2, 3), (2, 4)}, {(2, 3), (2,4), (1, 3)}.

Question 14.
Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A x B, find A and B, where x, y, z are distinct elements.
A = {x, y, z} and B = {1, 2}.

Question 15.
The Cartesian product A x A has 9 elements along which are found (-1, 0) and (0,1). Find the set A and the remaining elements Ax A.
Given n(A x A) = 9 = 32
⇒n(A) = 3
But (-1, 0) and (0, 1) are in A x A
∴ A= {-1,0,1}.
Remaining elements of A x A: (-1, -1), (-1, 1),
(0,-1), (0,0), (1,-1), (1,0), (1,1).

Question 16.
If P = {1,2}, form the set
P x P x P = {(1, 1, 1), (1, 1, 2), (1, 2, 1),
(1, 2,2), (2, 1,1), (2, 1, 2), (2, 2,1), (2, 2, 2)}

Question 17.
If A = {-1,1}, find A x A x A.
A x A x A = {(-1, -1, -1), (-1, -I, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1,1,-1), (1,1,1)}

Question 18.
If R is the set of all real numbers, what do the Cartesian products R x R and R x R x R represent?
We have R x R = {(x, y) : x, y ∈ R } which represents the coordinates of all the points in two dimensional space and R x R x R = {(x, y, z)  x,y,z ∈ R } which represents the coordinates of all the points in three-dimensional space.

Question 19.
Define a relation.
A relation R from a non-empty set A to non empty set B is a subset of the Cartesian product A x B.

Question 20.
Define domain of a relation.
The set of all first elements of the ordered pairs in a relation R from a set A to a set B is called the domain of the relation R.

Question 21.
Define range of a relation.
The set of all second elements in a relation R from a set A to a set B is called the range of the relation R. The whole set B is called the co-domain of the relation R.

Question 22.
Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by R = {(x,y): y = x + 1}
(i) Depict this relation using an arrow diagram
(ii) Write down the domain, condomain and range of
Given R = {(x, y): y = x + 1}
= {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

Domain = {1, 2, 3,4, 5,}; Co-domain = A
Range = {2, 3,4, 5, 6}.

Question 23.
Let A = {1, 2, 3, ………….14}. Define a relation R from A to A by R = {(x, y): 3x – y = 0, x,
y ∈ A}. Write down its domain, co-domain and range.
Given R = {(x, y): 3x -y = 0, x, y ∈ A}
= {(1, 3), (2, 6), (3, 9), (4,12)}
Domain = {1, 2, 3,4}
Co-domain = A Range = {3, 6, 9,12}

Question 24.
Define a relation R on the set N of natural numbers by R – {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ . N}. Depict this relationship using roster form. Write down the domain and the range.
Given R = {(x, y): x, y ∈ N and y = x + 5, x < 4}
= {(1,6), (2,7), (3, 8)}
Domain = {1, 2, 3}
Range = {6, 7, 8}

Question 25.
A = {1, 2, 3, 5} and B = (4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y∈ B}. Write R in roster form.
Given A = {1, 2, 3, 5} and B = {4, 6, 9}
R = {(x, y): the difference between x and y is odd; x∈ A,y∈B}
= {(1,4), (1, 6), (2, 9), (3, 4), (3, 6), (5,4), (5, 6)}

Question 26.
The figure shows a relationship between the sets P and Write this relation
(i) in set- builder-form
(ii) roster form. What is its domain and range?

Given P={5,6,7} and Q={3,4,5}
(i) Set builder form
R= {(x,y):x-y = 2; x∈P,y ∈Q)

(ii) Roster form
R = {(5, 3), (6,4), (7, 5)}
Domain of R = P
Range of R = Q.

Question 27.
Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b): a, b ∈A, b is exactly divisible by a},
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R
Given A = { 1, 2, 3,4, 6}
R- {(a, b),a,b∈A,bis exactly divisible by a}

(i) Roster form:
R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4,4), (6, 6)}

(ii) Domain of R = {1, 2, 3,4, 6} = A

(iii) Range of R = {1, 2, 3,4, 6} = A

Question 28.
Determine the domain and range of the relation R .defined by R = {(x, x + 5): x e {0,1, 2,3,4,5}}.
Given R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}.
Domain of R = {0, 1, 2, 3,4, 5}
Range of R = {5, 6, 7, 8, 9, 10}

Question 29.
Write the relation R = {(x, x3) : x is a prime number less than 10} in roster form.
Given R = {(x, x3): x g {2, 3, 5, 7}}
= {(2, 8), (3, 27), (5, 125), (7, 343)}

Question 30.
Let A – {x, y, z} and B = (1, 2}. Find the number of relations from A to
Given n(A) = 3 and n(B) = 2
∴ n (A x B) = 3 x 2 = 6
Number of relations from A to B = 2n (A x B) = 26 = 64

Question 31.
Let R be the relation on Z defined by R = {(a, b): a, b ∈ Z, a – b is an integer}. Find the domain and range of
Given: R = {(a, b): a, b ∈Z,a-b is an integer} Domain of R-Z Range of R = Z

Question 32.
Let R be a relation from Q to Q defined by R = {(a, b)\ a,b ∈Q and a – b ∈ Z}. Show that (a, a) g R, for all a ∈Q
(ii) (a, b) ∈ R implies that (b, a) ∈R
(iii) (a, b) ∈ R and (b, c) ∈R implies that (a, c) ∈ R.
Given R = {{a, b): a,b∈Q and a-b ∈ Z)
(i) ∀ a ∈ Q, a – a = 0 ∈ Z ⇒ (a, a)∈R

(ii) Let (a, b) ∈ R⇒ a- b ∈ Z
b – a ∈ Z ⇒ (b, a) ∈ R

(iii) Let (a, b) and (b, c) g R ⇒ a – b ∈Z and
b – c ∈ Z
a- c = (a-b) + (b – c) ∈Z
∴ (a, c)∈ R

Question 33.
Let R be a relation from A to A defined by R = {(a, b): a,b ∈ N and a = b2} Are the following true?
(i) (a, a) ∈ R for all a ∈ A
(ii) (a, b) ∈ R, implies (b, a) ∈ R
(iii) (a, b) ∈ R, (b, c) ∈ R, implies (a, c) ∈ R.
Given R= {(a, b): a,b∈N and a = b2}
= {(1,1), (2,4), (3, 9), (4, 16),…}
(i) (a, a)∈ R for all a ∈ N is not true because (2, 2) ∉ R.
(ii) (a, b) ∈ R implies (b, a) ∈ R is not true, because  (2,4) ∈ R but (4,2) ∉ R
(iii) (a, b) ∈ R, (b, c) ∈ R implies (a,c) ∈ R is not true because (2,4) and (4,16) ∈ R but (2,16) ∉ R.

Question 34.
Define a function.
A relation/from a set A to a set B is said to be a function if every element of set A has one and only one image in set B, and
we write f : A → B

Question 35.
Define
(i) a real valued function
(ii) a real function
A function which has either R or one of its subsets as its range is called a real valued function. A function f: A → B is said to be a real function if both A and B are subsets of R.

Question 36.
Let N be the set of natural numbers and the relation R be defined on N such that
R = {(x,y):y = 2x,x,y ∈ N} What is the domain, co-domain and range of R? Is this relation a function?
Given R = {x, y): y = 2x; x, y ∈ N}
Domain of R = set of natural numbers Co-domain of
R = set of natural number
Range of R = set of even natural numbers
Clearly, every natural number is related to unique image, so this relation is a function.

Question 37.
Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not?
(i) R = {(2,1), (3,1), (4,2)}
(ii) R = {(2,2), (2,4), (3,3), (4,4)}
(iii) R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6,7)}
(i) Given R= {(2,1), (3,1), (4, 2)}
Here every element of domain is related to unique element of co-domain, so it is a function.

(ii) Given R = {(2, 2), (2,4), (3, 3), (4, 4)}
Since the element 2 has two images namely 2 and 4, so it is not a function.

(iii) Given R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)}
Since every element of domain is related to unique

Question 38.
Let N be the set of natural numbers. Define a real valued function f: N → N by
f(x) = 2x + 1. Using this definition, complete the table given below:

 X 1 2 3 4 5 6 7 y F : (1) = F : ( 2) = F : (3) = F : (4) = F : (5) = F : (6) = F : (7) =

Given function is f(x) = 2x + 1.

 X 1 2 3 4 5 6 7 y F:(1)=3 F:(2)=5 F:(3)=7 F:(4)=9 F:(5)=11 F:(6)=13 F:(7)=15

Question 39.
Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14,7)}
(iii) {(1,3), (1,5), (2,5)}.
(i) Clearly, every element of domain is related to unique element of co-domain, so it is a function.
Domain = {2, 5, 8, 11, 14, 17}
Range = {1}

(ii) Clearly, every element of domain is related to unique element of co-domain, so it is a function.
Domain = {2,4, 6, 8,10,12,14}
Range = {1,2, 3,4, 5, 6,7}

(iii) 1 is related to two elements of co-domain, namely 3 and 5, so it is not a function.

Question 40.
Let A={1,2,3,4), B={1,5,9,11,15,16} and f = {(1,5),(2,9), (3,1), (4,5),(2, 11)). Are the
following true?
(i) f is a relation from A to B
(ii) f is a function from A to B. Justify your answer in each case.
(i) Every element off is an element of A x B, so f is a relation.
(ii) ‘f’ is not a function the element 2 has two images.

Question 41.
Let f be the subset of Z x Z defined by f ={(ab,a +b):a,b ∈z) Is f a function from Z to Z? Justify your answer.
Given f={(ab,a+b):a,b∈Z)
If a = 1 and b = 4 = ab = 4 and a+b=5
∴ (4,5) ∈ fIf a=2 and b=2⇒ab=4 and a+b=4
∴ (4,4) ∈ f∴ The element 4 has two images, so f is not a function.

Question 42.
The relation f is defined by
$$f(x)=\left\{\begin{array}{ll}{x^{2},} & {0 \leq x \leq 3} \\ {3 x,} & {3 \leq x \leq 10}\end{array}\right.$$
The relation g is defined by
$$g(x)=\left\{\begin{array}{ll}{x^{2},} & {0 \leq x \leq 2} \\ {3 x,} & {2 \leq x \leq 10}\end{array}\right.$$ .
Show that/is a function and g is not a function.
Since f(x) is unique for 0 ≤ x ≤ 10.
f(x) is a function. g(2) = 22 =4 and g(2) = 3(2) = 6
∴ z has two images under g.
∴ g is not a function.

Question 43.
Let f= {(1,1),(2,3),(0,-1),(-1,-3)} be a linear function from Z into Z. Find f(x) ).
Since/is a linear function.
∴f{x) = ax + b
But (0,-1) ∈ f . f(0) ∴ a(0) + b ⇒ -1 = b
Similarly, (1,1) ∈ f ∴ f(1) = a(1) +a(1)+b
⇒ 1=a+b ∴ a=2 ∴ f(x) = 2x -1

Question 44.
A function f is defined by f(x) = 2x-5. Write down the values of (i) f(0) (ii) f(17) (iii) f(-3).
Given: f(x) = 2x-5

• f(0) = 2(0) – 5 = -5
• f(17) = 2(17)-5 = 29
• f(-3) = 2(-3) – 5 = -11

Question 45.
The function ‘t’ which maps temperature in degree Celsius into temperature in degree. Fahrenheit is defined by $$t(c)=\frac{9 c}{5}+32$$. Find
(i) t (0)
(ii) t (28)
(iii) t (-10)
(iv) The value of c, when t(c) – 212

Question 46.
$$\text { If } f(x)=x^{2}, \text { find } \frac{f(1 \cdot 1)-f(1)}{1 \cdot 1-1}$$

Question 47.
Find the range of each of the following functions:
(i) f{x) = 2-3x, x∈R,x>0
(ii) f(x) = x2 + 2, x is a real number
(iii) f(x) = x, x is a real number.
(i) Given f(x) = 2-3x, x∈R,x>0
For x > 0,f(x) = 2 – 3x < 2
∴ Range of f= (-∞, 2)

(ii) Given f(x) = x2 + 2, x ∈ R
For x ∈ R, f(x) = x2 + 2 ≥ 2
Range of f = [2, ∞)

(iii) Given f(x) = x, x∈ R For r∈ E, f(x) = x∈R
Range of f = R

Question 48.
Let A = (9, 10, 11, 12, 13} and let f:A→N be defined by f(n) = the highest prime factor of n. Find the range of f.
Given f(n) = the highest prime factor of n.
f(9) = the highest prime factor of 9 = 3
f(10) = the highest prime factor of 10 = 5
f(11) = the highest prime factor of 11 = 11
f(12) = the highest prime factor of 12 = 3
f(13) = the highest prime factor of 13 = 13
Range of f ={3,5,11,13}

Question 49.
$$\text { Let } f=\left\{\left(x, \frac{x^{2}}{1+x^{2}}\right): x \in \mathbb{R}\right\}$$ be a function from R to R .Determine the range of f.

Question 50.
Find the domain of the function $$f(x)=\frac{x^{2}+3 x+5}{x^{2}-5 x+4}$$
Given $$f(x)=\frac{x^{2}+3 x+5}{x^{2}-5 x+4}$$
f(x) is defined for all real numbers except x2 – 5x + 4 = 0
But x2 -5x + 4 = (x-4) (x-1)
Domain of f= R-{1,4}

Question 51.
Find the domain of the function
$$f(x)=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$$
$$f(x)=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$$
is not defined when
x2 – 8+12 = 0.
∴ (x-6)(x-2) = 0
∴ Domain of f = R-{2,6}

Question 52.
Find the domain and range of the following real functions:
(i) $$f(x)=-|x|$$
(ii) $$f(x)=\sqrt{9-x^{2}}$$
(iii) $$f(x)=\sqrt{x-1}$$
(iv) $$f(x)=|x-1|$$

Remark: Operations of functions:
(i) Addition of two real functions:
Let f: X → R and g : X → R Then.
(f + g):X → R; (f + g)(x) = f(x) + g(x) for all x ∈ X

(ii) Difference of two real functions:
Let f : X → R and g : X → R Then.
(f – g): X → R; (f- g)(x) = f(x) – g(x) for all x ∈ X

(iii) Scalar multiplication of a function:
Let f: X → R and let ‘a’ he a scalar. Then,
(αf):X → R; (fg)(x) = f(x) for all x ∈ X

(iv) Multiplication of two real functions:
Let f: X → R and g : X → R .Then
(fg): X → R; (fg)(x)= f(x)g(x), for all x ∈ X

(v) Quotient of two real functions:
Let f: X → R and  g : X → R for all x for which g(x) ≠ 0. Then
$$\left(\frac{f}{g}\right): X \rightarrow \mathbb{R} ;\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}$$

Question 53.
Let f(x) = x2 and g(x) = 2x + 1 be two real functions. Find (f+g)(x),(f-g)(x)
$$(f g)(x),\left(\frac{f}{g}\right)(x)$$

Question 54.
Let $$f(x)=\sqrt{x} \text { and } g(x)=x$$ be two functions defined over the set of non-negative real numbers. Find (f+ g)(x), (f – g)(x) $$(f g)(x) \text { and }\left(\frac{f}{g}\right)(x)$$

Question 55.
Let f,g: R→R be defined, respectively by f(x) = x + 1,g(x) = 2x-3- Find f + g,f-g and $$\frac{f}{g}$$
Given: f(x) = x + 1,g(x) = 2x-3
(f + g)(x) = f (x) + g(x) = x + 1 + 2x-3 = 3x-2
(f – g)(x) = fix) – g(x) = x + 1-2x + 3 = -x + 4
$$\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x+1}{2 x-3}, x \neq \frac{3}{2}$$

Question 56.
Define an identity function and draw its graph also find its domain and range.
The function f: M → R; f(x) = x for a II x∈R is called an identity function on R.

Question 57.
Define a constant function and draw its graph also find its domain and range.
Let c be a fixed real number. Then, the function f : R→R, f(x) = c for all x∈R is called the constant function.

f(x) is defined for all real number,
∴ Domain = M
Range = { c }

Question 58.
Define a polynomial function.
A function f : M → R is said to be polynomial function if for each x in IR,
y = f (x) = a0 + axx + a2x2 + ……………. + anxn
where n is a non-negative integer and a0,a1,a2………………….an ∈ R .

Question 59.
Draw the graph of the function f (x) = x2 and write its domain and range.
Given function: f(x)= x2

Domain = R
Range = set of non-negative reals.

Question 60.
Draw the graph of the function f: R → R defined by f(x) = x3 Find its domain and range.
Let f : R → R: f(x) = x3, ∀ ∈ R
Then, domain of f = R and range of f = R . we have

Question 61.
Define a rational function
The functions of the type $$\frac{f(x)}{g(x)}$$ where f(x) g(x) and g(x) are polynomial functions of x, defined in a domain, where g(x) ≠ 0

Question 62.
Let f : R – {0} → R defined by $$f(x)=\frac{1}{x}, \forall x \in \mathbb{R}-\{0\}$$ . Find its domain and x range. Also, draw its graph.
Given function is f : R – {0} → R defined by $$f(x)=\frac{1}{x}$$
∴ Domain = R-{0} and range =R-{0}

 X -4 -2 -1 -0-5 -0-25 0-25 0-5 1 2 f(x)=1/x -0-25 -0-5 -1 -2 -4 4 2 1 0-5

Question 63.
Define a modulus function. Find its domain and range. Also, draw its graph.
Let f: R → R defined by f(x) =1 x I, for each x ∈ R, is called modulus function.
$$\text { i.e., } f(x)=|x|=\left\{\begin{array}{ll}{x,} & {\text { if } x \geq 0} \\ {-x,} & {\text { if } x<0}\end{array}\right.$$

Domain = R
Range = set of non negative real numbers

Question 64.
Define Signum function. Draw its graph and find its domain and range.
The function f: R → R defined by
$$f(x)=\left\{\begin{array}{lll}{1,} & {\text { if }} & {x>0} \\ {0,} & {\text { if }} & {x=0} \\ {-1,} & {\text { if }} & {x<0}\end{array}\right.$$ is called signum function
We have

Domain = R
Range = {-1,0,1}

Question 65.
Define a greatest integer function. Draw its graph and find its domain and range.

The function f : R → R define by f(x) = [x], x∈ R assumes the value of the greatest integer, less than or equal to x. Such a’ function is called the greatest integer function or step function. We have
[x] = -2 for – 2 ≤ x < -1
[x] = -1 for -1≤x<0
[x] = 0 for 0≤x≤1
[x] = 1 for 1 ≤ x < 2
[x] = 2 for 2 ≤ x < 3.
Hence, domain of f = R and range = Z.

Question 66.
Define a linear function.
The function f : R → R defined by f(x) = mx + c, x ∈ R is called linear function, where m and c are constant.

Question 67.
Let R be the set of real numbers. Define the real function f : R →R by f(x) = x + 10 and sketch the graph of this function.
$$f(x)=\left\{\begin{array}{cl}{1-x,} & {x<0} \\ {1,} & {x=0} \\ {x+1,} & {x>0}\end{array}\right.$$.


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# What Happens on North Pole official

By khelpedia
The gravity of Earth, denoted g, refers to the acceleration that the Earth imparts to objects on or near its surface. In SI units this acceleration is measured in metres per second per second (in symbols, m/s2 or m·s-2) or in newtons per kilogram (N/kg or N·kg-1). It has an approximate value of 9.81 m/s2, which means that, ignoring air resistance, the speed of an object falling freely near the Earth's surface increases by about 9.81 metres per second every second. This quantity is informally known as little g (contrasted with G, the gravitational constant, known as big G). There is a direct relationship between gravitational acceleration and the downwards weight force experienced by objects on Earth (see Conversion between weight and mass). However, other factors such as the rotation of the Earth also contribute to the net acceleration. The precise strength of the Earth's gravity varies depending on location. The nominal "average" value at the Earth's surface, known as standard gravity is, by definition, 9.80665 m/s2 (32.1740 ft/s2). This quantity is denoted variously as gn, ge (though this sometimes means the normal equatorial value on Earth, 9.78033 m/s2), g0, gee, or simply g (which is also used for the variable local value). The symbol g should not be confused with g, the abbreviation for gram (which is not italicized)
09/14/2011


text:
## Polynomial interpolation and a priori bootstrap for computer-assisted proofs in nonlinear ODEs.(English)Zbl 1398.65092

Summary: In this work, we introduce a method based on piecewise polynomial interpolation to enclose rigorously solutions of nonlinear ODEs. Using a technique which we call a priori bootstrap, we transform the problem of solving the ODE into one of looking for a fixed point of a high order smoothing Picard-like operator. We then develop a rigorous computational method based on a Newton-Kantorovich type argument (the radii polynomial approach) to prove existence of a fixed point of the Picard-like operator. We present all necessary estimates in full generality and for any nonlinearities. With our approach, we study two systems of nonlinear equations: the Lorenz system and the ABC flow. For the Lorenz system, we solve Cauchy problems and prove existence of periodic and connecting orbits at the classical parameters, and for ABC flows, we prove existence of ballistic spiral orbits.

### MSC:

 65G20 Algorithms with automatic result verification 65P99 Numerical problems in dynamical systems 65D30 Numerical integration 37M99 Approximation methods and numerical treatment of dynamical systems 37C27 Periodic orbits of vector fields and flows

### Software:

CAPD; PolyPaver; CkAnalytic; Taylor; INTLAB; dcc
Full Text:

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text:
# If I buy 2 lottery tickets do I double my chance of winning?

There's a lottery. There are 6 balls chosen randomly from 49 and you have to match all the balls to win.

I buy one ticket. If I buy two tickets with different numbers for the same draw, do I double my chance of winning the jackpot.

What's the correct formula here?

• my 2 cent: it is indeed doubled. there is the same chance to win the lottery each ticked. but I guess the confusion is because if you draw dice twice the chance to hit 6 is only $11/36$ which is less than $2*1/6$(which is 2 times the chance to hit 6 with one roll). anyway dice example is not valid here. – d_e Jul 22 '15 at 6:34
• You double the probability of having your combination drawn. Winning is more complicated, since in standard lotteries the jackpot is fairly often shared. (We are assuming that as in the standard 6/49, that possibly no one wins,) – André Nicolas Jul 22 '15 at 6:38
• Order does not matter in standard 6/49. – André Nicolas Jul 22 '15 at 6:45
• The reason why lotteries are successful is a lot simpler, the prize is less than the cost of tickets they sell. A simple corollary of this is that the expected value of a ticket is less than its price. – Jorge Fernández Hidalgo Jul 22 '15 at 7:49
• Though it does not answer the question, I have to remark that you double your expected loss by being two tickets (unless the organisers have made a massive mistake). – PJTraill Jul 22 '15 at 11:24

Yes, you do double your chance of winning. There are $\binom{49}{6}$ different possible draws. The probability of winning with one ticket is the chance that the 6 drawn balls are one specific combination, i.e. $\frac{1}{\binom{49}{6}}$. With two tickets, there are two winning combinations, so the probability is $\frac{2}{\binom{49}{6}}$

Basically, the reason it is doubled is because winning one ticket and winning the other ticket are mutually exclusive, i.e. you can't win both tickets (note the problem statement says "different tickets"). If there was a chance of winning both, the chance would be less than doubled.

• Which gives $$\frac{1}{{49\choose 6}}=0.00000007151123842...$$ I'd better get a better paid job. – Antinous Jul 22 '15 at 6:39
• @pbs Hey, I think it is at least a better chance than happen to discover the solution to one of the Milennium Problems and win $1M. – wythagoras Jul 22 '15 at 6:42 • In cases like this you can always logically check your results: If you would buy tickets for every combination, your chance of winning would be 100%, if you bought half of all possible tickets it is 50%, so it is proportional! The difference is when you play the lottery two times in a row! Then your chance is not doubled! (Again do the check, you could play the lottery trillions of times without ever winning 100%) – Falco Jul 22 '15 at 12:54 • @MihirSinghal I suspect R.. was referring to the error term when$n / { 49 \choose 6 }$is used as an estimate for the probability of winning at least once when$n$separate lotteries are played. (He was responding to Falco's comment). – anovstrup Jul 22 '15 at 19:21 • @wythagoras OTOH, if you do manage to solve a Millenium Problem, you'll benefit society more than you would by winning the lottery – Max Nanasy Jul 22 '15 at 20:05 It depends on the lottery. In your lottery, all bills are unique. So, the odds are indeed doubling as others have pointed out, and if you buy all different tickets, you have a 100% winning chance (though the cost of buying all those tickets is far greater than the lottery prize). Another lottery: Suppose that there is a lottery where all tickets are thrown in an pool. There is only one prize. The ticket that is randomly selected gets the prize. There are already 50 tickets sold. If you buy one ticket, then your chance of winning is$\frac{1}{51}$. If you buy two tickets, then your chance of winning is$\frac{2}{52}<\frac{2}{51}=2\cdot\frac{1}{51}$. So your odds are improved, but not doubled. I guess this is where the confusion on the internet is from. • It seems quite evident in the question that the winning combination is drawn independently of any tickets sold (and so quite possibly nobody gets the jackpot). So the "another lottery" stuff is just a needless diversion; don't add confusion that is absent in the question. – Marc van Leeuwen Jul 22 '15 at 13:46 "If I buy two tickets with different numbers for the same draw, do I double my chance of winning the jackpot." Does "winning the jackpot" mean exclusively? If so, then the answer is "no" because there is no mathematical formula that can predetermine how many tickets will be sold. If "winning the jackpot" means "picking the winning numbers," then yes, your odds of winning increase proportionally with the number of unique tickets you buy. So if there are 6 million combinations, your first ticket is 1 in 6 million, and it increases with each ticket you purchase: 1 = 1/6 million 2 = 1/3 million 3 = 1/2 million 4 = 1/1.5 million 5 = 1/1.2 million 6 = 1/1 million • Even if it does mean “win exclusively”, you double your chance by buying a second distinct ticket, if •(0) the draw is fair, •(1) there is only one winner, •(2) the rest of what happens is unaffected by your behaviour and •(3) the probabilities of other participants buying a ticket identical to your first or second ticket are the same. Under these assumptions, the probability of winning with either ticket is the same, and they are mutually exclusive, so may be added. Even if you drop (0, 2 & 3) but have no information about the true situation, your estimate of your chance should double. – PJTraill Jul 22 '15 at 13:00 It depends on how you choose the second ticket: If you choose the second ticket independently, you do not double the chance of winning, since there is a non-zero chance of getting a ticket with exactly the same numbers -- after all, in this kind of lottery you can choose the numbers, and you cannot prevent two persons choosing the same numbers. If you choose the tickets carefully such that you have two different number sequences, you indeed double the chance of winning. There are$\binom{49}{6}$possible numbers, and by choosing two you have a chance of winning of$\frac{2}{\binom{49}{6}}$as compared to$\frac{1}{\binom{49}{6}}$. • The problem states, "If I buy two tickets with different numbers" – msinghal Jul 22 '15 at 6:40 • Correct. I just wanted to clarify this explicitly, since this apparently causes the confusion in the internet the OP was writing about... – Bernhard Jul 22 '15 at 7:38 • So let me get this right. If I have a 1 in 14 million of chance of winning the lottery, if I buy a further ticket with a different sequence of numbers to the first one for the same draw my chance of winning is slashed to 1 in 7 million? – Rickie Jul 22 '15 at 8:16 • Yes, that is correct. – Bernhard Jul 22 '15 at 8:23 Well I wonder if I was rich enough to buy 14 million unique lottery tickets and the jackpot is a guaranteed 20 million for matching 6 balls im 100% chance of winning but it could be shared equally between other winners but let's say it's not so I win the 20 million plus all the 5 ball numbers and bonus ball plus all the 4 numbers all the 3 numbers etc is that the case or will it be 20 million and that's it The expected value of the win is typically doubled because expected value is additive. However, as you mention 6 out of 49: In this type of lottery the prize sum depends on the tickets submitted, not even this is true: If you have one ticket ans are the sole winner, you may win a million and if you are the only winner with two identical tickets each of these may win only half a million, so it may happen that there is no increase at all. If we model that each ticket has a certain independent winning probability$p$, then two tickets raise your winning probbaility not to$2p$, but to$1-(1-p)^2=2p-p^2\$, so slightly less than doubled.

• It's not independent though, the events are mutually exclusive. You can't win both tickets at once the way the problem is stated. – msinghal Jul 22 '15 at 6:39

Currently for the mega millions there are 302,575,350 different combinations of numbers. So your probability of winning on your first ticket is 1 / 302,575,350, which equals a percentage of 0.00000033049618880057500%. After you play your first ticket you have 302,575,349 different combinations left and a probability of winning on your second ticket of 1 / 302,575,349, which equals 0.00000033049618989285200%. The latter odds is twice that as the first. So ALMOST YES, mathematically your chances of winning improve by just under 100%. However, you'll notice that your ACTUAL ODDS do not improve that much.

Understanding lottery. Let us say there is a lottery where there is 100 tickets but there will be only 1 prize. This is a game of chance, nothing to do with probability. Buy 1 ticket and your chance of winning the prize is 1 in 100. Buy 2 tickers and your chance of winning is 1 in 99 and not 1 in 50 as most people think.

The rational is as follows: of the 2 tickets 1 has to be a looser but you do not know which. Therefore it is unmistakable that you have a looser. It follows then that you may have the winner, so mentally you discard 1 as a loser and maybe the other is the winner reducing your chance to only 1 in 99 and not 1 in 50 as many go on believing.

Think about it - you have to be bright enough to have posed the question.

## protected by Community♦Nov 22 '15 at 0:23

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text:
# area

0th

Percentile

##### Surface covered by a LAS object or by a LAScatalog.

The area is computed with the convex hull for LAS objects or x,y coordinates. If the data is not convex, the resulting area is only an approximation. The area is computed as the sum of the extents of each file for a LAScatalog.

##### Usage
area(x, y)
##### Arguments
x

An object of the class LAS or LAScatalog or numeric

y

If x is numeric, then provide also y.

##### Value

numeric. The area of the object computed in the same units as the coordinate reference system

• area
##### Examples
# NOT RUN {
LASfile <- system.file("extdata", "Megaplot.laz", package="lidR")