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  • MarkFL's Avatar
    Today, 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 | 63 view(s)
  • lfdahl's Avatar
    Today, 11:09
    The function, $f$, is defined on the interval $$, and satisfies the following conditions: (a). $f(0) = f(1) = 0$. (b). For any $a,b \in $:...
    0 replies | 17 view(s)
  • MarkFL's Avatar
    Yesterday, 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 | 58 view(s)
  • MarkFL's Avatar
    Yesterday, 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 | 58 view(s)
  • lfdahl's Avatar
    Yesterday, 15:14
    I am so sorry, that I have posted a challenge, the solution of which, I am not certain. My problem is the use of the Rearrangement Inequality in the...
    1 replies | 91 view(s)
  • I like Serena's Avatar
    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 | 46 view(s)
  • I like Serena's Avatar
    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 | 76 view(s)
  • I like Serena's Avatar
    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 | 76 view(s)
  • kaliprasad's Avatar
    Yesterday, 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 | 60 view(s)
  • MarkFL's Avatar
    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 | 84 view(s)
  • MarkFL's Avatar
    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 | 84 view(s)
  • MarkFL's Avatar
    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 | 72 view(s)
  • MarkFL's Avatar
    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 | 84 view(s)
  • MarkFL's Avatar
    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 | 72 view(s)
  • MarkFL's Avatar
    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)
  • MarkFL's Avatar
    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 | 84 view(s)
  • MarkFL's Avatar
    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 | 72 view(s)
  • I like Serena's Avatar
    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)
  • MarkFL's Avatar
    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 | 84 view(s)
  • MarkFL's Avatar
    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 | 84 view(s)
  • kaliprasad's Avatar
    March 24th, 2017, 22:18
    kaliprasad replied to a thread Factoring...6 in Pre-Calculus
    Yes. Then in case of doubt you after factoring can multiply and see the result
    5 replies | 73 view(s)
  • MarkFL's Avatar
    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 | 72 view(s)
  • I like Serena's Avatar
    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 | 46 view(s)
  • MarkFL's Avatar
    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 | 84 view(s)
  • MarkFL's Avatar
    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 | 254 view(s)
  • MarkFL's Avatar
    March 24th, 2017, 04:29
    The objective function is linear, so it describes a plane, and so I don't believe there will be any saddle points, or in fact any critical points...
    4 replies | 62 view(s)
  • MarkFL's Avatar
    March 24th, 2017, 04:04
    MarkFL replied to a thread Boat direction in Calculus
    Is this a calculus-based physics course?
    6 replies | 92 view(s)
  • MarkFL's Avatar
    March 24th, 2017, 01:37
    What I would do is observe that we have cyclical symmetry, that is we may interchange $x_1$ and $x_2$ with no change in either the objective function...
    4 replies | 62 view(s)
  • MarkFL's Avatar
    March 24th, 2017, 01:25
    MarkFL replied to a thread The Distance Across in Geometry
    Not quite...it would be \sqrt{2}a and \sqrt{2}b...so what would the diagonal of the rectangle be?
    8 replies | 84 view(s)
  • MarkFL's Avatar
    March 23rd, 2017, 21:17
    MarkFL replied to a thread The Distance Across in Geometry
    I would let $a$ be the length (in cm) of the segments with 1 tick mark, and $b$ be the length (in cm) of the segments with 2 tick marks. And so,...
    8 replies | 84 view(s)
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7 Visitor Messages

  1. Reply/View Conversation
    Welcome back to MHB, friend! I'm so glad to see your return to our forum again!
  2. Reply/View Conversation
    Sorry... There seems to be an echo in here [my fault! ]
  3. Reply/View Conversation
    Hello K!

    I just wanted to say, i think you do an absolutely sterling job posting all sorts of brain-teasers on the Puzzles board. Keep up the great work!

    Gethin
  4. Reply/View Conversation
    Hello K!

    I just wanted to say, i think you do an absolutely sterling job posting all sorts of brain-teasers on the Puzzles board. Keep up the great work!

    Gethin
  5. Reply/View Conversation
    Hi there,

    I didn't realize Mark has already taught you how to type square root in LaTeX... and what I mentioned in the challenge thread that we both participated is essentially the same as Mark's ...
  6. Reply/View Conversation
    Hey Kali,

    To create a square root using $\LaTeX$, use the command \sqrt{} where the radicand goes in the braces. For example:

    \sqrt{x^2+y^2} gives you $\sqrt{x^2+y^2}$

    To create other roots, use the command \sqrt[]{} where the degree of the roots goes in the square brackets. For example:

    \sqrt[n]{a^m}=a^{\frac{m}{n}} gives you $\sqrt[n]{a^m}=a^{\frac{m}{n}}$

    Hope this helps, and please do not hesitate to ask me if you need to know any other $\LaTeX$ commands.

    Best Regards,

    Mark.
  7. Reply/View Conversation
    Hello and welcome to MHB, kaliprasad!

    If you have any questions or comments, pleases feel free to address them to me or another staff member. We are happy to help, and we look forward to your participation here!

    Best Regards,

    Mark.
Showing Visitor Messages 1 to 7 of 7
About kaliprasad

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Name: Secondary School/High School POTW Award (2016)
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