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I see that 210 = 2*3*5*7
And I see that 2310 = 2*3*5*7*11
But, for instance, the numbers in the list
that are being subtracted from 210 are
not all prime numbers. I feel that the numbers
in your list are presupposed, that you are
partially putting the cart before the horse
(or at least sideways to the cart).
Hi Miaflower;
I found a really great math site called...
You made a spelling error! I think it is spelled MathsisFun.
bobbyman,
you made a spelling error. It is "MathsIsFun."
The "i" is capitalized in the correct spelling.
Post Deleted
The question is to find the number of intersections in the x-y plane
of the functions 2[sup]x[/sup] and x[sup]12[/sup]. So I got;I don't think a calculator is allowed for this paper, but I was wondering if
this was the right method. I was going to rewrite ln(x) as a Taylor series
approximation, divide the whole thing by x and then see where that got me.
Is this the wrong way to do it?The possible answers are;
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4I'm pretty sure A is wrong, but I don't know about the rest.
If I had to guess I would say at least 2.
zetafunc, don't rely on the use of derivatives, because I don't know
if you're taking a math portion of the general GRE or of the
higher level math-specific GRE.
The exponential one is always above the x-axis, and approaches the
x-axis on the left side of the y-axis.
The polynomial touches at the origin.
Then there is one place of intersection to the left
of the y-axis.
The exponential one is above y = 1 on the right side
for all x-values, while the polynomial one is just rising
above the x-axis.
So there is an intersection there. The polynomial
one is steeper up to that point in that interval.
Then, because the exponential one becomes
steeper more to the right, it intersects once
more. And the the exponential one remains
above the polynomial one thereafter.
Look at how many intersections in total I have discussed.
2. Is sqrt(x) 7 a polynomial? Explain why or why not.
no sqrt(x) 7 is not a polynomial because
it has the variable inside a radical.
just checking to see if i did this right
the original problem says Divide (15x^7 45x^5)/(3x^4)
Divide [(3x 2)(x - 4) (x - 4)(6 5x)] /[(4 x)(8x 1)]
so i cross out the x-4 they all divide to give me oneleft with [(3x-2)-(6-5x)]/(8x-1)
so from [(3x 2)(x - 4) (x - 4)(6 5x)] /[(4 x)(8x 1)]
to [(3x-2)-(6 -5x)]/[-(x-4)(8x-1)] *to* (8x-8)/[(4-x)(8 x-1)]
zee-f, you must use grouping symbols, as in the above where I
have amended in these quote boxes (parentheses, brackets, or
braces) because of the Order of Operations.
Hi;
I did not mean you. !
Please do not try to read my mind. Give me the benefit of the doubt as you want it.
It's the same in return to you.
I didn't state it was me.
Please do not try to read my mind. Give me the benefit of the doubt.
Hmmmm, maybe later someone will raise a little controversy about this problem...
As a *moderator*, you should know better than to instigate
and/or make a dare against a user.
hi reconsideryouranswer
if this one isn't correct
The second problem:
hi reconsideryouranswer
for the
are you surethat in the second one x shouldn't go to 2?
ig it is then i
As far as the second one, x is supposed to approach 1, so that
it is of one of the indeterminate forms,
Another attempt for you, knowing this?
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As far as the first one, I got a different answer than you.
Pat and Chris played five games of checkers. Each won the same
number of games, but there were no draws (ties), nor were there
any unfinished games.
How is this possible?
User (moderator) gAr has been officially reported to the administration (contact form)
http://www.mathisfunforum.com/viewtopic.php?pid=185221#p185221
Note the first few lines of post #7 where he had/has zero business
challenging and disparaging about my educational teaching status.
This is certainly a kind of bashing (albeit it more subtle), as what has been
*alleged* by me.
And other users, including other moderators, will be considered for being
reported, depending on their future online behaviors in the future.
reconsideryouranswer wrote:The following list are examples, but it is not exhaustive, of what should
not be used as mathematical qualifiers:
easy, simple, hard, difficult, obvious, clear, trivial**.Except, possibly as in "trivial step" in a proof by induction
I believe that is the problem you are implying is arrogant.
Then I have already politely told you twice to stop with the
name calling. I do not consider his post arrogant or desperate.
Bullying is not allowed here!***
Don't misread what I have typed. Those words in my list *are* presumptuous.
What it is for you does not necessarily apply to me, and vice versa.
Typing those words do mean those qualities, as I am an expert in the
misuse/abuse of them. In all of the excitement of presenting a problem
with a subjective qualifier, this error is lost. You, as a moderator,
made up some statement in one of your posts that I haqd ignored you.
From there, you stated that you were closing that thread. I didn't ignore
what you typed, but in you closing that thread, it was easier to do that
than to engage me honestly.
At a different tutoring site a well-seasoned user got banned by the
administrator, in part, for challenging my education (schooling, teaching),
much as gAr did with me. gAr instigated his attack on me with what
is his equivalent name-calling, and is subject to being reported.
*** It is, and denying it is not currently being done by any of your
moderators, e.g. gAr, is [name calling removed by moderator]
The following list are examples, but it is not exhaustive, of what should
not be used as mathematical qualifiers:
easy, simple, hard, difficult, obvious, clear.
These give no objective information as to a math problem, proof,
step in a proof, etc. One of the ways around these, in certain
circumstances, is to use the word "relatively" with them.
Or, make it clear as an opinion, as in "I found it easy for me."
Don't presume to speak for others. The above list of words
smack of either 1) arrogance, 2) desperation for others to
read/attempt your problem, or both.
hi reconsideryouranswer the first one [post # 236]?
What idea(s) do you have to support this? As it is, it looks like "a stab in the dark."
You asked about "the first one" being the answer.
Suppose you were to look at it where n = 1:
This would eliminate your guess (my second case) and also my third case
out of the four cases.
hi guys
could you help me with:
i got a/b.is that correct?
this. Find the answer to this:
this.
Find the answer to this:
19)
Easy!
Problem poser, you have to state what restrictions there are
on the variable constant, a.
Unhide:
Did you observe anything by plugging in some large values for n?
I don't see (as in I am not seeing it at this moment) what your hint will do.
When k catches up to n, then I see the 'last' term as
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