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@anonimnystefy: Nope.
@Agnishom: Nope.
Try again.
If you get the hang of it, you can try the next one.
#2. [1 − cos[sup]2[/sup]()][sup]1/2[/sup]
Nope. You were actually closer the first time.
The plus/minus sign followed by a times sign? Doesn't make much sense...
You are very, very, VERY close to the answer.
debjit625, I'm sorry for my bad-tempered post above. I suppose I had had a bad day and somehow let it spill over onto the forum. It was totally wrong of me and I apologize.
In this particular question, the statement is "
() has roots in the ratio ". In order for this to be true, we must have . This is the condition that has to be fulfilled in order for the given statement to be true.What the bleep is wrong? The question asks for the condition under which certain things hold. I worked out and gave the condition. Just what are you not happy with?
Ok I think you guys didn't understood what I was asking for...
I want to know what the question is asking for ,what condition ?
Look. You asked a question. I took the time to help you with it. It was late, I could have gone to bed, but I stayed up to help you. Is this the gratitude I get in return?
Maybe what you don't understand is not the question but the English language?
Then I make it the speed of the person when the distance between the tree and the car is
is .Try
No, it doesn't work.
Eliminating
givesas the required condition.
Test:
have roots in the ratio .You also need to know the diamter of the tree trunk.
Oops, pardon me.
The number of positive divisors of
(where the
are distinct primes) is .It cannot be prpved. It is an axiom of natural numbers and of higher number sets, too.
It depends on how you are defining your sets of numbers. If you are defining them by axioms, then there is nothing to prove since the axioms will include commutativity and associativity of addition. But if you are constructing them from smaller sets of numbers, then commutativity and associativity need to be proved.
After checking that the operation is well defined, one can proceed to verify that
is commutative and associative.Moreover, zero
in is the equivalence class and multiplication in is defined as .Similarly
can be constructed from as equivalence classes of the equivalence relation on defined by iff . Letting the equivalence class containing be denoted addition in is definied asI think (c) is unneccessary there, and the rest just represents the axiom of induction.
The sequence is absolutely convergent iff both and converge.
It is conditionally convergent iff converges while diverges.
Examples.
is absolutely convergent. We have and .
is conditionally convergent. We have while is divergent.
A new prime number has been discovered this week by GIMPS:
It is now the largest known prime, a whopping 17-million-digit Mersenne prime.
So where are the imaginary numbers?
Hi. Consider the following problem:
[align=center]
[/align]This is my solution:
LHS is
for all x, y.RHS is
for all y since .Hence LHS = RHS if and only if
and
The second equation gives
; substituting into the first equation givesHence the solution set is
.So my question is: Is my proof okay? Is everyone happy with my proof?
Hi Al-Allo. Your equations are wrong. They should be as follows
Aaron reads 3 times faster than Patrick and Shad reads 4 times faster than Aaron. If Patrick reads 4 pages in 1 hour, how many pages reads each of his two comrades? I know how to resolve the problem, but I just wanna be sure I resolved it in the right way.
X: Number of pages Shad read
X/4: Number of pages Aaron read. (4 times slower than the numbers of pages Shad read.)
X/12: Number of pages Patrick read. (12 times slower than the numbers of pages Shad read.)
Now, I resolve : X/12=4 (pages read by Patrick)
p and q are both less than 0
Do you mean more than 0?
Now how can we manipulate/transform/ get conditions from these two inequalities which must hold: i.e. q+2p>1 and 2p+4p>2??