• Today, 17:48
Hey RTCNTC, Apparently it is not given how the piece of wire is cut into pieces. And I assume the total length of the wire is $L$? So let's...
2 replies | 19 view(s)
• Today, 13:26
Just for illustration purposes. (Smile) \begin{tikzpicture} \newcommand\Square{+(-#1,-#1) rectangle +(#1,#1)} \draw foreach \r in {0,...,16} {...
3 replies | 72 view(s)
• Yesterday, 16:27
Hey FallArk! (Smile) Let's start with the proper definition of the derivative. That is: $$f(x)=\begin{cases}\tan x &\text{if } -\frac\pi 2 \le x... 3 replies | 69 view(s) • Yesterday, 16:26 Putting a=0 and b=1, you see that f\bigl(\frac12\bigr) \leqslant 0+0 = 0. Now suppose that f\bigl(\frac12\bigr) = k < 0. Then (putting... 2 replies | 74 view(s) • Yesterday, 12:24 Just to follow up, here is the completed table: Sum S Probability of S: P(S) Net Gain/Loss (in dollars) G Product G\cdot P(S) 2 replies | 79 view(s) • March 26th, 2017, 22:32 MarkFL replied to a thread Radius of Circle in Pre-Calculus 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 | 84 view(s) • March 26th, 2017, 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 | 49 view(s) • March 26th, 2017, 21:35 MarkFL replied to a thread Radius of Circle in Pre-Calculus 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 | 84 view(s) • March 26th, 2017, 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 | 48 view(s) • March 26th, 2017, 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 | 78 view(s) • March 26th, 2017, 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 | 78 view(s) • March 26th, 2017, 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 | 78 view(s) • March 26th, 2017, 02:53 you can apply either way. but next steps become simpler if you apply difference of square x^6-y^6= (x^3+y^3)(x^3-y^3) 1st term is sum of cubes... 4 replies | 64 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 | 91 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 | 90 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 | 75 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 | 90 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 | 75 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 | 77 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 | 90 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 | 75 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 | 63 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 | 87 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 | 90 view(s)
• March 24th, 2017, 22:18
Yes. Then in case of doubt you after factoring can multiply and see the result
5 replies | 77 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 | 75 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 | 52 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 | 91 view(s)
• March 24th, 2017, 11:35
We have that two right triangles and from the Pythagorean Theorem for those two we get: $$464^2+e^2=a^2$$ and $$464^2+(1218-e)^2=b^2$$ From the...
2 replies | 73 view(s)
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