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• Yesterday, 22:32
That is the radius of the circle after it has been increased by $a$ units. If I tell you that my weight increased by 20 lbs., then you know my...
4 replies | 41 view(s)
• Yesterday, 22:19
Hi tmt, (Wave) For the quoted part in bold, should this read "If all cars are reserved for the day..."? I think there are two events here,...
1 replies | 38 view(s)
• Yesterday, 21:35
Let's let $0<a$ be the number units the radius must be increased. And so the change in area we can write as: \Delta A=\pi(r+a)^2-\pi r^2=b Now...
4 replies | 41 view(s)
• Yesterday, 20:41
Looks good. -Dan
2 replies | 28 view(s)
• Yesterday, 15:20
So, at a car rental company, 20% of car reservations are not claimed. There is a total of 22 cars and the manager takes 25 reservations a day. If...
1 replies | 38 view(s)
• Yesterday, 12:05
I like Serena replied to a thread ideal gas in Other Topics
Hi markosheehan, Yes, there are Van der Waals forces in all cases. However, Helium is the only one that is electrically neutral. That's because...
1 replies | 41 view(s)
• Yesterday, 11:52
I like Serena replied to a thread copper(2) oxide in Other Topics
Yes, there are exceptions. Most elements have a stable bonding with a full outer shell, but they typically also have alternative stable bondings....
5 replies | 69 view(s)
• Yesterday, 11:46
Rido12 replied to a thread copper(2) oxide in Other Topics
There are many exceptions. I remember back in my first year chemistry course, my professor criticized the textbook for providing incorrect...
5 replies | 69 view(s)
• Yesterday, 09:08
I like Serena replied to a thread copper(2) oxide in Other Topics
Hi markosheehan, I'm assuming EC stands for Electron Configuration? Before the bonding Copper has the configuration 2-8-18-1 (there are 18...
5 replies | 69 view(s)
• March 25th, 2017, 21:53
MarkFL replied to a thread The Distance Across in Geometry
\overline{MK}=\sqrt{(\sqrt{2}a)^2+(\sqrt{2}b)^2}=\sqrt{2\left(a^2+b^2\right)}=\sqrt{2(50)}=\sqrt{100}=10 :D
8 replies | 82 view(s)
• March 25th, 2017, 21:31
At a guess you are forgetting about the cross term. (a + b)^2 \neq a^2 + b^2. It is (a + b)^2 = a^2 + 2ab + b^2. You are going to end up having to...
4 replies | 54 view(s)
• March 25th, 2017, 21:07
MarkFL replied to a thread Lagrange Multipliers 2 in Calculus
I arbitrarily chose another point on the constraint, so that we could do a comparison like I mentioned just now in the other thread. :D
9 replies | 78 view(s)
• March 25th, 2017, 21:04
MarkFL replied to a thread Lagrange Multipliers in Calculus
I chose the point as it is on the constraint. Using that point, we can determine if our one critical point is a maximum or a minimum. If the...
9 replies | 67 view(s)
• March 25th, 2017, 20:02
Here is this week's POTW: ----- Let $f$ be a real function with a continuous third derivative such that $f(x),f'(x), f''(x), f'''(x)$ are...
0 replies | 23 view(s)
• March 25th, 2017, 20:00
We first note that \ Subtracting $S$ from this gives two sums, one of which is \ and the other of which sums to $xy^3/$. Therefore...
1 replies | 62 view(s)
• March 25th, 2017, 19:21
MarkFL replied to a thread Lagrange Multipliers 2 in Calculus
I agree that the point $(2,2)$, is the only one that meets all criteria. Now we need to compare the value of $f$ at another point on the constraint,...
9 replies | 78 view(s)
• March 25th, 2017, 19:08
MarkFL replied to a thread Lagrange Multipliers in Calculus
I agree that of the 3 critical points, $(1,1)$ is the only one in quadrant I. Now, we know this is either a maximum or a minimum, and to determine...
9 replies | 67 view(s)
• March 25th, 2017, 11:15
MarkFL replied to a thread Factoring...6 in Pre-Calculus
It might be more clear to state something like the following: The difference of cubes formula states: p^3-q^3=(p-q)\left(p^2+pq+q^2\right) ...
5 replies | 73 view(s)
• March 25th, 2017, 10:47
MarkFL replied to a thread Lagrange Multipliers 2 in Calculus
Consider: e^u=0 What do you get when solving for $u$? Okay, you correctly found $x^2=y^2$...what do you get when you substitute for...
9 replies | 78 view(s)
• March 25th, 2017, 10:39
MarkFL replied to a thread Lagrange Multipliers in Calculus
What I would do is use the constraint to determine $y=2-x$. Now substitute for $y$ in both equations you mentioned, and solve for $x$, then your...
9 replies | 67 view(s)
• March 25th, 2017, 06:59
Hey evinda!! (Smile) I haven't figured it out yet. :( However, I can see a couple of approaches... According to Euler we have: a^{\phi(pq)}...
1 replies | 57 view(s)
• March 25th, 2017, 02:30
This is a calculus question...please don't continue to post calculus questions in other forums. If given: ...
3 replies | 83 view(s)
• March 24th, 2017, 23:16
topsquark replied to a thread Factoring...7 in Pre-Calculus
I mentioned this in another thread. Don't set the LHS of "a = a + b" It's too confusing. -Dan
3 replies | 48 view(s)
• March 24th, 2017, 23:13
topsquark replied to a thread Factoring...6 in Pre-Calculus
You need to clarify your variables. "b = (a - b)" isn't right. Pick a variable name, say, p... Anything but that b on the LHS.... Then you have p...
5 replies | 73 view(s)
• March 24th, 2017, 23:12
MarkFL replied to a thread Lagrange Multipliers 2 in Calculus
If you solve both equations for $\lambda$ and then equate the results, you obtain: \frac{ye^{xy}}{2x}=\frac{xe^{xy}}{2y} Multiply through by 2:...
9 replies | 78 view(s)
• March 24th, 2017, 21:24
MarkFL replied to a thread Lagrange Multipliers in Calculus
Okay, so what this implies is: \frac{x}{\sqrt{6-x^2-y^2}}=\frac{y}{\sqrt{6-x^2-y^2}} Cross-multiply: x\sqrt{6-x^2-y^2}=y\sqrt{6-x^2-y^2} ...
9 replies | 67 view(s)
• March 24th, 2017, 21:05
topsquark replied to a thread Factoring...5 in Pre-Calculus
Yeppers! -Dan
2 replies | 33 view(s)
• March 24th, 2017, 17:47
Actually, f(n)=2n is not a bijection, since for instance 1 does not have an original. But it does free up all the odd rooms, so that a new arrival...
1 replies | 43 view(s)
• March 24th, 2017, 16:18
MarkFL replied to a thread The Distance Across in Geometry
Using the Pythagorean theorem, we find: \overline{MK}=\sqrt{(\sqrt{2}a)^2+(\sqrt{2}b)^2}=\sqrt{2\left(a^2+b^2\right)} Now, we know that...
8 replies | 82 view(s)
• March 24th, 2017, 06:10
MarkFL replied to a thread [SOLVED] Minimum of function under constraint in Calculus
I would use W|A: W|A - optimize 2x+y subject to xy=18
8 replies | 253 view(s)
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