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A librarian needed two different books.
The price of book (A) is $17.
The price of book (B) is $29.
How many books he bought if their total cost was $1000?
Kerim
Being of this world means that a human has no choice but to follow always the preprogrammed instructions embedded in his living flesh, known as instincts, as all other living things do. Some of the human’s preprogrammed instructions are the instincts of survival, superiority, selfishness and applying a certain justice on others.
Therefore, not being of this world means that a human can oppose and even defeat his natural robotic nature. Such a human loves even his enemies, is humble no matter how great he may be, sees other humans as an extension of his being and doesn't apply any justice on others.
Being and not being of this world were first known since about 2000 years ago.
Hi All,
A natural fact is that knowledge, scientific or else, has to be taught in steps according to age.
I chose ‘parallelism’ to explore it.
I was personally taught parallelism in 3 steps; 3 successive definitions while I grew up.
Definition 1:
If you remember well the first definition you heard of at school when you were little kids (first year of geometry), you can easily fill the gap in the following sentence:
Two straight lines are said parallel if...
I didn’t complete it because I like to know first if we all heard of the same definition at that age. Thank you.
For instance, the end result of this topic is more philosophical than mathematical. It is related to a natural truth that the multitudes around the world are not supposed to hear
But you will be aware of this hidden truth if you too were able to recall, as I do, the 3 definitions of parallelism we learnt while we grew up (I assume you are older than 20 years).
Kerim
To those who got the logical step and to complete the exercise, here is the mathematical step:
From the right triangle (SWY ), we can write:
[(S2*T1) + R]^2 = D^2 + (S1*T)^2 , //T1 = time from W to X//
(S2*T1)^2 + 2*S2*T1*R + R^2 = D^2 + (S1*T)^2
(S2*T1)^2 + 2*S2*T1*R + R^2 - D^2 - (S1*T)^2 = 0
If T2 = time from X to Y, T2=R/S3
T = T1 + T2
T1 = T – T2
T1 = T – R/S3
By replacing T1:
S2^2*(T – R/S3)^2 + 2*S2*(T – R/S3)*R + R^2 - D^2 - (S1*T)^2 = 0
S2^2*(T^2 – 2*R/S3*T + R^2/S3^2) + 2*S2*(T – R/S3)*R + R^2 - D^2 - (S1*T)^2 = 0
S2^2*T^2 – 2* S2^2*R/S3*T + S2^2*R^2/S3^2 + 2*S2*R*T – 2*S2*R^2/S3 + R^2 - D^2 - S1^2*T^2 = 0
(S2^2 - S1^2)*T^2 – 2* S2*R*(S2/S3 - 1)*T + S2^2*R^2/S3^2 – 2*S2*R^2/S3 + R^2 - D^2 = 0
(S2^2 - S1^2)*T^2 – 2* S2*R*(S2/S3 - 1)*T + R^2*(S2/S3 – 1)^2 - D^2 = 0
This is a quadratic equation whose general parameters (a), (b) and (c) are:
a = S2^2 - S1^2
b= –2* S2*R*(S2/S3 - 1)
c= R^2*(S2/S3 – 1)^2 - D^2
A numeric example:
D = 24.11 km
S1 = 30 km/h
S2 = 45 km/h
S3 = 150 km/h
R = 20 km
The mission time is T = 15 min.
Kerim
I will refer here to your original diagram (I guess you still have it).
I used to tell my students that, in general, finding how to solve a problem has two main directions, forwards and backwards.
To solve this exercise, walking backwards makes it a simple one.
Let us assume first that the full range (R) of the missile was used.
In this case, the warship had to be on the circle whose center is ‘Y’ and radius is (R).
If we draw this circle, it becomes clear that WX is the shortest distance from W to this circle.
Therefore, the mathematical solution of T=f(D,R,S1,S2,S3) could take advantage of Pythagoras theorem of the right triangle SWY.
One may think that it is not necessarily to use the full range of the missile, say (k*R) instead of (R), where 0<k<1.
The same reasoning above also applies in case (k*R). Only the radius of the circle becomes (k*R) instead of (R).
This means that T_new = f(D,k*R,S1,S2,S3). That is (R) is replaced with (k*R) in the final formula of (T).
And by analyzing the variation of the function (T_new) with the variable parameter (k), we will find out that T_new > T for all values of (k).
As the song of B.....s, "All I need is... pi=3.14" ![]()
Kerim
#8984.
Someone said (not me, to avoid getting another big zero
):
"The most precious thing that a human can give to another is sincerity".
If an object (A) is moving and another object (B) must aim to travel straight towards A then B's path is continuously changing direction. This is illustrated by the Guardian's Four Dogs Puzzle:
This may be applied in real life. But, as you know, the situation here is just a math exercise.
For instance, when I saw your first diagram, I thought you got quickly the right logical answer of the situation related to the 2 directions of W and the missile.
Unfortunately, you changed your mind ![]()
As we will see, proving that your first diagram is right, and it gives indeed the minimum time of the mission is somehow simple if one discovers how to start it.
I believe you can prove it, right?
I've realised that WX and XY should not be part of the same straight line.
I couldn't get well the relation between this sentence and your diagram.
On your diagram, they do have the same direction, unless 'same straight line' and 'same direction' have different meaning.
hi KerimF
I have drawn a diagram. The question says "(while moving on a straight line towards the side of the other ship)."
That seems contradictory. I'm marking the two vessel's positions with S for ship and W for warship. If W heads directly towards S while S is moving forwards then W must continuously change the direction of travel so that it is once more heading at S. If W doesn't do this it will end up where S was rather than where S is. But that means W's movement is not in a straight line.
Sorry for the confusion. I thought that the meaning of 'towards the side of the other ship' is totally different from "towards the other ship'.
Let us assume that the two parallel straights of S and W are drawn horizontally.
The ordinary ship, S, was on the northern line and the warship, W, was on the southern one. Both were moving from west to east.
At t=0, the captain would have to change the direction of his warship, W, towards north-east. The angle formed by the new direction (straight line) of W and the horizontal line could be called θ for example. In theory, the angle θ could be any value between 0 and 90 degrees but the captain had to decide its exact value to achieve the minimum time for his mission. The trajectory of the missile is also a straight line (between the position of the warship, W, when the missile was launched and the position of the other ship, S, when it was hit by the missile).
Please don't hesitate to notify me if I missed something to clarify about the exercise.
Kerim
I liked to thank you for your great and continuous efforts in providing this collection of interesting info.
I wonder if you had the idea to make an index list of all your entries in this thread. Obviously, it will need to be updated continuously since it is related to a growing book (not a fixed one).
Best Regards,
Kerim
An ordinary ship and a warship were moving, side by side, on two parallel straight lines. The distance between them was (D). Their speed was constant and equal to (S1).
At time zero, t=0, the captain of the warship was instructed to shut down the other ship.
Let us assume that the new constant speed of the warship is (S2), S2>S1 (while moving on a straight line towards the side of the other ship).
The captain can launch a missile whose constant speed is (S3) which is much higher than (S2). The maximum range of the missile is (R), R<D.
Find the minimum time (T) to hit the other ship, that is:
t_min = T = f(D,R,S1,S2,S3)
Kerim
Note1:
I heard of this exercise when I was in the final year of high school (Baccalauréat). At that time, math teachers used to skip it. So, it was out of question to me not to solve it, and naturally I did it (because I was very interested in doing it). Later, I heard that it was removed in the printings of newer math books of the final year.
Note2:
I am afraid that its solution needs to be done in two steps:
(1) Logical
(2) Mathematical
If the logical step is skipped, the mathematical solution will be somehow undetermined, instead of being obvious.
It is clear, to me in the least, that you both, amnkb and you (Bob), are very good teachers.
so i did it wrong b/c im not a teacher?
huh
twi
But, thanks to your first good work and my reply to you, our friend Bob was able to notice. more easily. the second right answer.
So, I guess, if Bob presented his solution before you did, you will likely be the teacher
by noticing the second option as he did.
Anyway, this situation may look like a humble example of the saying:
"But many that are first shall be last; and the last shall be first."
So, as possible, I try always to be the last ![]()
Very nice quotes, thank you.
This reminds me what happened to me about quoting when I was at school.
My French teacher used to ask us to write always some quotes which could be related to the topic of homework. My problem was (still is) that I used to be interested solely in math and applied science more than anything else (I seldom read books offered by famous writers). Once, I wrote a sentence very clearly and carefully to reflect my idea about the topic of interest. I followed it with ‘Kerim said’ ![]()
On the next day, my grade was ‘a Big Zero’ followed with ‘Your teacher, Joseph, said’ ![]()
Bob
ps. Aren't we all still students?
Hi Bob,
Yes, we are all still students in life, since learning/discovering has no limit.
But, here, only a teacher may notice that 'k' is given twice in (2) and (5), though presented in different forms.
Following the norm, a student expects here that the right answer is (or has to be) given once on the list. So, when he finds the right answer, he just checks (x) on the expression (2) or (5), the one he got at his last step. But, by doing this, he also says, though without his knowledge, that none of the remaining 4 expressions is equal to the right value of 'k' which is wrong. ![]()
Kerim
Hi amnkb,
Your work is right, and you got k = 9 / [4*sqrt(3)], thank you.
But the question of this specific exercise was: "Find the right value of ‘k’ in the 5 expressions below:"
I am afraid, you didn't specify exactly where you find it in these 5 expressions. You just said k = 9 / [4*sqrt(3)].
Please remember, it is an 'unusual' tricky exercise to all students, but not to skilled teachers.
Have fun.
Kerim
The perimeter of a square (Ps) is equal to the perimeter of an equilateral triangle (Pe).
And the area of the square (As) is ‘k’ times equal to the area of the equilateral triangle (Ae), that is As = k*Ae.
Find the right value of ‘k’ in the 5 expressions below:
(1) 6 / [3*sqrt(3)]
(2) 9 / [4*sqrt(3)]
(3) 2*sqrt(3) / 3
(4) 6 / [4*sqrt(3)]
(5) 3*sqrt(3) / 4
As we will see, the way this exercise is stated lets it be tricky to all students but not necessarily to skilled teachers.
Have fun.
Kerim
I thought that such exercise is for students, not teachers ![]()
Anyway, I guess you meant by your last line that the condition to be satisfied by the parameters A, B, C and D for f(x) doesn’t have local high and low limits is:
B^2 - 3*A*C ≤ 0
And the parameter (D) may have any value.
Thank you.
The way I, a realistic rational person, see it is that every neural network is created by the human's brain.
Therefore, these neural networks cannot be made cleverer than their human creators though some creators try their best to give the world the impression they are, at least in some fields (while in reality they are not). They are just advanced tools which speed up the world's progress for the good and bad as well.
Anyway, if it is possible someday to make something more intelligent than the human's brain then the human up-to-date profession in every field, not only of photographer, will disappear altogether ![]()
For instance, the profession of photographer has now a chance to achieve new things by the use of new advanced tools; things which were impossible to make/create in the past.
We have f(x) = A*x^3 + B*x^2 + C*x + D
What is the condition to be satisfied by the parameters A, B, C and D for f(x) doesn’t have local high and low limits?
Actually, you have the right idea, it's just that the ± symbol needs to come slightly earlier. Instead of:
you want to say:
And what amnkb is saying is that .
Just to give an idea about my nature, I respect always other's views and knowledge no matter if they sound right or wrong to me.
And when I contact others, I simply say what I know while I understand fully that they may have their good reasons not to accept everything I say/know.
Perhaps you were expecting a solution like this:
The linear formula is of the form:
F = aC + b
where,
F= Fahrenheit (°F) degrees
C= Celsius (°C) degrees
'a' and 'b' are the unknown parameters to be found.
If C=0, F=32, then
32 = a*0 + b
b=32 , the intercept with the F, up, axis.
if C=100, F=212
212 = a*100 + 32 [was b]
212 - 32 = a*100
180 = a*100
a = 180/100 = 1.8 , the slope of the line (usually called m)
Note:
The slope m (or 'a' here) could be calculated directly if we know two points (x1,y1) and (x2,y2).
Here, x1=C1=0 and y1=F1=32, also x2=C2=100 and y2=F2=212
m= (y2-y1)/(x2-x1) = (F2-F1)/(C2-C1) = (212-32)/(100-0) = 180/100 = 1.8
C=1.8*F + 32
Sorry, I wrote the lines above while I am sure that you know them already.
hi mrpace,
I draw an 18x18 table on Excel. I filled its diagonal with 18 zeros.
I let its below part be a replica of its upper one, since (P1,P2) = (P2,P1) for example.
By filling the rows of upper cells with 8 one's for each player, I got the total of matches equal to 144 (8*18).
And to check that every player plays 8 matches, I summed the cells of each row/player (horizontally) and of each column/player (vertically). It was possible to let every sum be 8.
Kerim