I don't think you're missing anything. My idea may work for finite sets, but I have one or more flaws with my demonstration
with these infinite sets. It does not work here.
Type: Posts; User: lookagain
I don't think you're missing anything. My idea may work for finite sets, but I have one or more flaws with my demonstration
with these infinite sets. It does not work here.
I'll use a different set-up (read: ordering), possibly as you suggested.
Every {2n}, a set which is an element of the powerset of A, gets paired to 2n, an even integer which
is an element of the...
If your "N" stands for the positive integers, then set A is the set of positive even integers.
Sets A and N have positive integers as their elements.
P(A), the powerset of A, has sets as its...
No,, I do not expect the first equation is correct.
Check the source (again?) to see if it should be "y = 3x - 7" instead.
Please do not give out a detailed solution without the student attempting and showing work here, as well as it has
only been the next day. Note how a prompt was given to the student just before...
You don't having matching parentheses. You don't need that extra parenthesis
in front. You do need grouping symbols around that fractional exponent:
(X/287,173.53)^(1/10) - 1 = .865
e^x \ge x + 1.
e^x = x + 1 \ \ only where x = 0.
y=22x/(2ln2) + C
The derivative of 22x \ \ne \ 2e2x
For s an integer, show that 360 divides \ s^2(s^2 - 1)(s^2 - 4).
Or:
For s an integer, show that \ s^2(s^2 - 1)(s^2 - 4) \ is divisible by 360.
Writing "no remainder" is redundant. ...
On the left-hand side of the equation, the "x" should not be as low as it is. It should on the same level as "log."
a^log_ax \ = \ x. \ \ \ Also, \ a \ne 1, as well as a > 0.
This is not a teaching site.
You are to learn that in the classroom. Tutors/helpers are to coach/reinforce what you have already gone over and have questions
over and/or shown some attempts here.
The natural log will take the absolute value bars in conjunction here:
(1/2)*ln|2x+1| + C
"Brute force" was stated, but it is not required.
Because it is a perfect square, it is of the form \ N^2.
Because it is also a perfect cube, it is also of the form \ M^3 .
x^6 \ = \...
Also, there is this:
.
And to think you were outstanding in your field.
Don't try to use the wording in either 1 or 4. See this related material:
...
You can't forget about the [cost of the] shared fence. The solutions don't involve pastures that are square-shaped.
That is, the pastures turn out not to be square-shaped.
"(Whatever) times greater/smaller than" is wrong/poor wording from the start. It should
never be used in writing. The problem should just have been phrased as:
"Find a double-digit number which...
.
The answers page is incorrect if you presented it fully. In the context of this specific problem,
it is not allowable to divide by the variable, because information is lost.
45x^6 = 9x^4
...
hamburgerman1212,
let d = the common difference.
3 - x + d = x
x + d = 1 - 3x
----------------
2x - d = 3
4x + d = 1
These particular denominators need grouping symbols:
{x(x - 7) + (6)(2)}/(2x) = 1/2
{x^2 - 7x + 12}/(2x) = 1/2
And this is how the truth value of the second step can be seen to be independent from the truth value of the first step.
Your first two sentences in that quote box where you address stapel make no sense,
because she never stated to what you objected/emphasized. In her question, she refers
to assuming for n = k, not...
Incorrect. Step 1 is independent from step 2. One must assume that the statement is true for n = k. And using that, one must show the statement is true for n = k + 1, which would be step 3.
...