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Yes.
Bob wrote:Both look good to me, well done.
Bob
Well, well. . .it's about time. I finally do something right, huh?
By the way, it is always about time for me too. Some problems that need innovative ways to solve took me many years to finish properly.
I mean if one needs 'really' to solve something, he will do it 'for sure' sooner or later.
Inversely, even the greatest genius won't solve a problem in which he is not interested.
KerimF wrote:In general, to solve f(x)/g(x) when their ratio becomes 0/0 for x=a, we calculate f'(a)/g'(a).
The rule you mention here is L'Hopital's rule, which does NOT calculate "f(x)/g(x) when their ratio becomes 0/0 for x=a"
but rather calculates the LIMIT of f(x)/g(x) as x approaches a where f(a)/g(a) is indeterminate.That limit may well exist, even if f(a)/g(a) is undefined.
E.g., The graph may be smooth but with a hole at f(a)/g(a).
The graph of (x^2 -3x + 2)/(x-2)
looks like the straight line y = x-1, but (e.g., in Desmos) hover over the line at x=2 and it will tell you (2,undefined).This is an example of two functions equal at all but a finite number of points.
You are right.
But, to me in the least (in practice), if I have f(x)= (x^2 - 3x + 2) / (x - 2) and x -> 2, f(x) -> 1. And the answer '1' is the required result.
For instance, my interest in math when I was at school (then universities) was to use it (after graduation) in designing real things; I ended up using it in electronics, hardware and software.
For example, math says that the voltage (as 5V) of a charged capacitor (C) while being discharged by a resistor (R) reaches 0V at infinity. In practice (not math), we can say that, in general, it could be seen as being 0V after 10*R*C sec only, instead of infinity.
But I also understand that, at school, students (including the one I was for about 20 years) have to follow the various definitions and rules that are approved by their teachers (who give them the grades).
My teachers now are those who buy my designed products
KerimF wrote:What is the standard form of the equation of the circle whose radius is sqrt(5) and its center (0,0)?
It is:
x^2 + y^2 = 5And this is your answer here though the center of your given circle is at M(2, 2) which you did well in finding it.
Is it right or wrong? It's either yes or no....
Ok, it is wrong.
The center of your circle is at (0, 0). It should be at M(2, 2) as you found out.
KerimF wrote:You did well for r by applying the right equation of a circle. Then... I didn't see it.
I guess you know that:
x^2 + y^2 = r^2
means that the circle center is at (0, 0).Is my answer right?
But... your exercise is to find the standard form of the equation of the circle with center (-3, 1) ... not (0,0)
The standard form of an equation of a circle with radius r and center (h, k) is given by:
(x - h)^2 + (y - k)^2 = r^2.
What is the standard form of the equation of the circle whose radius is sqrt(5) and its center (0,0)?
It is:
x^2 + y^2 = 5
And this is your answer here though the center of your given circle is at M(2, 2) which you did well in finding it.
You did well for r by applying the right equation of a circle. Then... I didn't see it.
I guess you know that:
x^2 + y^2 = r^2
means that the circle center is at (0, 0).
All the factors are:
1, 2, 3, 4, 6, 12, 1x, 2x, 3x, 4x, 6x, 12x, 1x^2, 2x^2, 3x^2, 4x^2, 6x^2, 12x^2
If I understood well the exercise, amnkb gave the right answer.
For example. let us assume x=5
12x^2 = 300
So, its factors are:
1, 2, 3, 4, 6, 12, 5, 10, 15, 20, 30, 60, 25, 50, 75, 100, 150, 300
----------------- ----------------------- -----------------------------
What I know is that 2a/a becomes 0/0 if a=0. In this case 0/0 = 2
In other words, to know the value of 0/0, its origin needs to be known first.
In practice, 0/0 (as ∞/∞) comes (be the result) from an algebraic expression under certain conditions, otherwise it is just three consecutive letters 0, / and 0
In general, to solve f(x)/g(x) when their ratio becomes 0/0 for x=a, we calculate f'(a)/g'(a). If f'(a)/g'(a) becomes 0/0 too, we calculate f''(x)/g''x)... and so on.
My first attempt at every post is usually full of typos. After I've read it through and corrected a few times I think I've got an error free post. Let me know if you discover this statement is false.
Bob
Same is here. I never write something new without editing it several times.
So, at work, I see myself having a good luck if I can end up writing an error free code (for MCU) after correcting 50 bugs, not 500's, if not more, which happened sometimes
hi KerimF
Didn't spot you were on line too. Looks like we both saw the same thing
Bob
Off topic, after I realized, at school, that I can't avoid doing typo mistakes (not only in math), I used after finishing an exam (or the like) to revise what I did but as if I were a bad rival of Kerim who insisted to show him (Kerim) the many mistakes he did. By doing this seriously, it was possible for me to get the highest grade in most exams.
Kerim
x^2 - 2x + 1 + y^2 + 4y + 4 - 9 = 0
x^2 + y^2 - 2x + 4y + 1 + 5 - 9 = 0
x^2 - 2x + 1 + y^2 + 4y + 4 - 9 = 0
to
x^2 + y^2 - 2x + 4y + 1 + 5 - 9 = 0
It is actually:
x^2 + y^2 - 2x + 4y + 1 + 4 - 9 = 0
Your typo was 5 instead of 4.
I recall how, me too, I did this type of mistake in some math exams (when I was a student many decades ago). I even failed in one of them because of a silly typo I did.
(x - 1)^2 + (y + 2)^2 = 9
...
x^2 + y^2 - 2x + 4y - 3 = 0
(x - 1)^2 + (y + 2)^2 = 9
Let us assume x=1:
(1 - 1)^2 + (y + 2)^2 = 9
(y + 2)^2 = 9
y + 2 = 3
y = 3 - 2
y = 1
Now,
x^2 + y^2 - 2x + 4y - 3 = 0
Let us verify if this final form is also satisfied for x=1 and y=1
1^2 + 1^2 - 2*1 + 4*1 - 3 = 0
1 + 1 - 2 + 4 - 3 = 1 ≠ 0
I am afraid that you did a typo somewhere.
In that case, the equation should be
x^2 + y^2 + 10x + 4y - 20 = 0
Yes?
Yes.
The general form of Q1 is:
x^2 + y^2 - 9 = 0
Perhaps he meant the surface area of a sphere which is 4*pi*R^2.
As I hear it, "Express F in terms of d" is equivalent to "Express F as a function of d". And this is what you did.
a) 4
b) 10/4 = 2.5
c) 2.5 + 4 = 6.5
d) 6.5 / 2 = 3.25
a) 3.25
b) 10/3.25 = 3.077
c) 3.077 + 3.25 = 6.327
d) 6.327 / 2 = 3.163
...
You can get around this problem by restricting the domain of the inverse so that only one input exists.
Thank you.
In other words, in the full definition of a function, the domain in which it is valid is also specified.
I wonder how we can see this case:
An event occurs whenever Time = arcsine (0.5) is satisfied.
Perhaps we need to differentiate between a function and its inverse.
In this exercise, the words 'a number of boxes' is equivalent to 'a natural number'.
So, as you found out already, we can't have small boxes or medium ones whose number is not natural, even if the ratio 2:5 is satisfied.
Just a thought:
The matter in our universe (level 0) is formed by the universe of atoms (level -1).
This could lead us to also say that the matter in the atomic universe (level -1) is formed by the universe of sub-atoms (level -2).
And the matter in the sub-atomic universe (level -2) is formed by the universe of sub-sub-atoms (level -3).
etc.
In the opposite direction, our huge universe (level 0) could be seen by a much bigger universe (level 1) as a small piece of matter in it... the same view applies between (level 1) and (level 2)... etc.
Long ago, humans were aware of Earth as being the center of Existence.
Then, humans became aware of the huge universe (level 0)
Lately, humans became aware of the atomic universe (level -1)
Will humans exist long enough to be aware of the upper universe (level 1) and/or the next lower one (level -2)?
After all, it is just a thought.
Kinetic energy = K.E. = m*v^2/2
Potential gravitational energy = P.E. = (m+M)*g*h
Based on the energy conservation, we can write (since the exercise is about an ideal situation):
m*v^2/2 = (m+M)*g*h
therefore:
h = m*v^2/(m+M)/g/2
I have usually insisted to find the key to a locked door which looks related to my life and whose key is assumed out of reach. Then, I simply opened it and looked for another one to open.
Unfortunately, at age 74, I can't find more locked doors (unanswered crucial questions related to my existence and the real world) to also find their hidden keys
Being rational, math's door was among the easiest ones to open and gave me the chance to enjoy playing on its yard since I was teen till now.