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#1 Re: Help Me ! » Simultaneous Equation? » 2024-05-06 03:03:50
Thanks, guys.
How do we know which is the height, and which is the base?
Good remark.
In general, a problem may have more than one solution (even more than two as it is the case here).
In real life, one usually chooses the solution which seems be the best one for his application.
For example, in this exercise, we use to see the base of a rectangle be longer than its height. So, our preference is likely to say that h=4.5 cm and b=12 cm. But this is valid as long no one complains (mainly a math teacher at one's school).
Also, although this answer comes from a quadratic equation and the quadratic formula can we get it from a simultaneous equation? I ask because someone in class said it was a sim.eq.
At the start we had indeed two simultaneous equations to solve. But, from them, we got a quadratic one to be solved.
I mean, let us consider the following two simultaneous equations.
3x + 2y - 7 = 0 [eq. 1]
x + 4y - 9 = 0 [eq. 2]
Multiplying [eq. 2] by 3
3x + 12y - 27 = 0 [eq. 3]
From [eq. 3] - [eq. 1], we get
0x - 10y - 20 = 0
10y - 20 = 0 [eq. 4]
Here, we derived a linear equation from the two simultaneous ones instead of a quadratic equation.
In both cases, we had two simultaneous equations to solve
Anyway, perhaps someone knows better what solving simultaneous equations does mean.
#2 Re: Help Me ! » Two Different Reasons. » 2024-05-05 21:41:21
And what about (a/0)/(b/0)? [∞/∞]?
#3 Re: Help Me ! » Simultaneous Equation? » 2024-05-05 11:29:16
I get b = 12 and b = 4.5.
Well done.
if Ax^2+Bx+C=0
x=[-B±√(B^2-4AC)]/2A
Here:
2b^2 - 33b + 108 = 0
A=2
B=-33
C=108
Therefore, if b=12 cm, h=54/12=4.5 cm
And, if b=4.5 cm, h=54/4.5=12 cm
#4 Re: Help Me ! » How Far Can A Person See? » 2024-05-02 19:03:35
I wonder if the required distance from the observation deck to the most distant seen point is in the air or on the earth surface.
If it is the straight distance (likely the case in this exercise), your equation is right.
#5 Re: Help Me ! » (Number)^0 = 1 » 2024-05-02 09:26:08
[a*a*a*a... m times / a*a*a*a... n times] = a^m/a^n = a^(m-n)
Similary, for any value of x we can write:
1 = a^x/a^x = a^(x-x) = a^0
#6 Re: Introductions » Hi from Brazil! » 2024-05-01 03:41:02
Hello!
(I might include a bible verse in my signature or elsewhere because I'm Christian, hope that doesn't bother anybody.)
Welcome to the forum, Gabriel.
And I hope you had the chance to become no more of the world.
I mean those who find themselves having no choice but to be guided by the natural preprogrammed instructions embedded in their living flesh (called instincts of survival, superiority, selfishness and imposing rules on others, to name a few, are of the world. And they are very important to its continuous progress, by building and destroying it whenever necessary, as all other living things are important.
The only real freedom, a human, unlike all other living things, may have, is having the choice to defeat or not his natural robotic nature; for example, by loving even his enemies, by being humble while he has all the means to impose his superiority on others, by seeing others as an extension of his existence and by not resisting evil in the name of any justice. But, by opposing/defeating the human robotic nature, a human cannot be seen of the world anymore and the world has no choice but to hate him in one way or another (up to kill him in some cases).
The good news is that no matter if someone is of the world or not, he is very welcomed in this scientific friendly forum.
Kerim (I am 75)
Sorry... It seems I was a philosopher since I was a kid playing with kids who used to tease me by shouting on the street, once a while, "Here is the modern philosopher" 
#7 Re: Puzzles and Games » Who took it? » 2024-04-25 02:20:52
In short, Lu took it how? Well he was the first to defend him self and remember only one tells the truth. And all the rest(liars) said Mu was lying. Am i right?
You are right.
#8 Re: Puzzles and Games » Who took it? » 2024-04-25 00:57:53
Fortunately, there were 4 students only and 3 of them are liars.
Let us assume that Ku took it:
Ku=Mu took it=wrong
Lu=I did not take it=right
Mu=Lu is lying=wrong
Nu=Mu is lying=right
Let us assume that Lu took it:
Ku=Mu took it=wrong
Lu=I did not take it=wrong
Mu=Lu is lying=wrong
Nu=Mu is lying=right
Let us assume that Mu took it:
Ku=Mu took it=right
Lu=I did not take it=right
Mu=Lu is lying=wrong
Nu=Mu is lying=right
Let us assume that Nu took it:
Ku=Mu took it=wrong
Lu=I did not take it=right
Mu=Lu is lying=wrong
Nu=Mu is lying=right
Therefore, the answer is...
Added:
In case, there is one liar only among them, the answer would be...
#9 Re: Science HQ » Outer Space » 2024-04-24 08:11:15
I mean, see what happened to Galileo for example. He chose to live hard days for presenting openly his discoveries which were not supposed to be heard by the world's multitudes of his time.
#10 Re: Science HQ » Outer Space » 2024-04-24 02:09:52
I wonder what kind of difference believing (or not believing) the big bang theory, for example, may do in a human life.
But, in general, a wise human pretends believing what most people around him are supposed to believe in order to be on the safe side.
#11 Re: Exercises » solve » 2024-04-14 00:13:40
From the original equation:
[1] x-1801+sqrt[(y-1860)*(x-1801)] = 2*sqrt(x-1801) - 1
[2] sqrt[(y-1860)*(x-1801)] = -(x-1801)+ 2*sqrt(x-1801) - 1
[3] sqrt[(y-1860)*(x-1801)] = -[sqrt(x-1801) - 1]^2
Also from the original equation:
x-1801 > 0
x > 1801
From the equation [3] above:
The right side is always negative.
And for the sqrt on left side to exist, the right side should be 0.
[4] [sqrt(x-1801) - 1] = 0
[5] sqrt(x-1801) = 1
[5] (x-1801) = 1
[6] x = 1801 + 1
[7] x = 1802
Also the left side of [3] should be zero.
[8] sqrt[(y-1860)*(x-1801)] = 0
[9] (y-1860)*(x-1801) = 0
By replacing x = 1802
[10] (y-1860)* 1 = 0
[11] (y-1860) = 0
[12] y = 1860
Therefore, there is one solution only:
x = 1802
y = 1860
#12 Re: Exercises » solve » 2024-04-13 22:34:53
One solution is:
x=1802
y=1860
#13 Re: Help Me ! » Words To Algebraic Expression » 2024-04-01 02:42:37
What do you say? I say it has everything to do with understanding what is going on in the problem.
By the way, I am not talking about simple problems...
Indeed, this is the first step in designing every part of a new application.
#14 Re: Help Me ! » Why Does y = f(x)? » 2024-03-29 21:22:04
If y=x+2 we can write y=f(x)=x+2
If y=h+2 we can write y=f(h)=h+2
If z=w+2 we can write z=f(w)=w+2
...
In brief, the notation f(a) means that we have a function (its name could be any symbol) in which the 'independent' variable is 'a'.
For example:
If y=3*x+2*a +1 and we read y=f(a), it means that 'x' here is not the independent variable of 'y', it is just added aa a parameter. In this case, we plot 'y' versus 'a', for each value of the parameter 'x' of interest. But saying y=f(x), 'x' is the independent variable and 'a' is the parameter in 'y'.
#15 Re: Help Me ! » Vertices of Equilateral Triangle » 2024-03-29 06:23:23
The main, if not crucial, role of the human scientific brain is to find out the formula, the function or the equation which can emulate/reflect, as possible, a real problem that needs to be solved.
Therefore, the purpose of the various math's exercises is to gain, also as possible, the logical reasoning on how to get the required/needed results from analyzing a formula/function of interest or solving a well-defined equation(s). Fortunately, there are also many ready-made tools to assist someone in doing this to save time (after he learns how to use them).
#16 Re: Help Me ! » Use Graph to Find Domain and Range » 2024-03-19 17:02:54
KerimF wrote:[img]https://i.imgur.com/a/IbFoc46.jpeg[/img]https://i.imgur.com/a/IbFoc46.jpeg
This time it didn't work on my side.That's the link to the webpage containing the image, not to the image on it. It shouldn't have the .jpeg extension.
Here's an image of that webpage. It's just a small version of it, done by using the letter 't' (for small) size modifier before the '.jpeg' extension:
https://i.imgur.com/XGtFi7kt.jpgThe link to the image on that webpage is this:
[img]https://i.imgur.com/xx9dW9G.jpeg[/img]And here's the image:
https://i.imgur.com/xx9dW9G.jpegl
Thank you.
#17 Re: Help Me ! » Cross-sectional Area » 2024-03-19 16:45:52
The calculus way is based on the knowledge of the derivatives of functions.
#18 Re: Help Me ! » Find Domain & Range » 2024-03-18 00:59:14
Just a note.
Here, it is clear that:
Domain is for x (supposed being an independent variable) and Range is for y (supposed being a dependent variable, function of x).
#19 Re: Help Me ! » Sequences; finding the formula that gives us the nth term » 2024-03-18 00:46:13
Thank you, Bob.
Your steps are much better than mine.
#20 Re: Help Me ! » Find Domain & Range » 2024-03-18 00:27:25
Here is the image:
http://i.imgur.com/aflDtIp.gif
I don't know why it won't display as the syntax looks correct to me.
Bob
It works here too. Perhaps your image was uploaded in a different way.
#21 Re: Help Me ! » Find Domain & Range » 2024-03-17 22:26:03
It seems you did it.
But on my side, I just see the file name of a hidden image which is 'IbFoc46.jpeg'.
#22 Re: Help Me ! » Use Graph to Find Domain and Range » 2024-03-17 22:19:13
[img]https://i.imgur.com/a/IbFoc46.jpeg[/img]
This time it didn't work on my side.
But the following one is ok.
[img]https://i.imgur.com/7NtStGDl.jpeg[/img]
#23 Re: Help Me ! » Sequences; finding the formula that gives us the nth term » 2024-03-17 22:10:33
KerimF wrote:How did you come up with the formula?
I did it by following primitive steps since I forgot, at age 75, the advanced ones.
f(n) = f(n-1)*2+1
f(n-1) = f(n-2)*2+1
f(n) = (f(n-2)*2+1)*2+1
f(n-2) = f(n-3)*2+1
f(n) = ((f(n-3)*2+1)*2+1)*2+1
f(n-3) = f(n-4)*2+1
f(n) = (((f(n-4)*2+1)*2+1)*2+1)*2+1
f(n-4) = f(n-5)*2+1
f(n) = ((((f(n-5)*2+1)*2+1)*2+1)*2+1)*2+1
f(n) = ((((f(n-5)*2+1)*2+1)*2+1)*2+1)*2+1
f(n) = (((f(n-5)*2+1)*2+1)*2+1)*2*2+2+1
f(n) = (((f(n-5)*2+1)*2+1)*2*2*2+2*2+2+1
f(n) = (((f(n-5)*2+1)*2*2*2*2+2*2*2+2*2+2+1
f(n) = f(n-5)*2*2*2*2*2+2*2*2*2+2*2*2+2*2+2+1
f(n) = f(n-a)*2^a… +2^(a-1)+2^(a-2)+2^(a-3)+2^(a-4)+1
Um=k*r^(m-1)
In the following series
… +2^(a-1)+2^(a-2)+2^(a-3)+2^(a-4)+1
we have
k=1
m=a
f(n) = f(n-a)*2^a+2^(a)-1
f(n) = 2^a*[f(n-a)+1]-1
Let us assume:
n-a=1
n=a+1
f(a+1) = 2^a*[f(1)+1]-1
But
f(1)= 2
Therefore
f(a+1) = 2^a*3-1
Again, let us assume:
a+1=n
a=n-1
f(n) = 2^(n-1)*3-1
f(n) = 3*2^(n-1)-1 #24 Re: Help Me ! » Find Domain of A » 2024-03-17 22:02:17
KerimF wrote:What confuses me is how one gives a result as 'a range' and as 'a domain'!
I don't understand your confusion.
Find the domain is not the same as find the range.
I type all questions as stated in the textbooks.
Sorry if I gave you the impressing that I am arguing with you or else. Please note that I believe whatever you say.
I just knew things in math that are somehow different from what I see here sometimes. That is all.
So, when in doubt, I believe that Bob, for example, can help clarifying the math definitions of interest, as they are followed on his side and yours.
On my side, I have no reason not to accept anything you say. After all, at work, I don't follow any definition. I just solve math problems and get numerical/practical results.
#25 Re: Help Me ! » Find Domain of A » 2024-03-17 12:22:50
What confuses me is how one gives a result as 'a range' and as 'a domain'!