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  • Euge's Avatar
    Yesterday, 22:53
    Euge replied to a thread Real Analysis in Analysis
    No. Use the conditions $f \ge 0$ and $\int_a^b f = 0$, along with basic properties of the integral to prove that result.
    12 replies | 154 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 | 48 view(s)
  • Jameson's Avatar
    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 | 45 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 | 48 view(s)
  • Euge's Avatar
    Yesterday, 21:25
    Euge replied to a thread Ball Application in Pre-Calculus
    Yes. Yes.
    2 replies | 28 view(s)
  • Euge's Avatar
    Yesterday, 21:19
    Euge replied to a thread Find All Real Solutions in Pre-Calculus
    Yes, it's correct.
    2 replies | 37 view(s)
  • Euge's Avatar
    Yesterday, 21:15
    Euge replied to a thread Real Analysis in Analysis
    Almost. You have a typo in the last line: the fraction $\frac{x + h - h}{h}$ should be $$\frac{x + h - \color{red}{x}}{h}$$
    12 replies | 154 view(s)
  • topsquark's Avatar
    Yesterday, 20:41
    Looks good. -Dan
    2 replies | 29 view(s)
  • Euge's Avatar
    Yesterday, 12:39
    Euge replied to a thread Real Analysis in Analysis
    To prove problem 1 directly, fix $x\in $ and show that $$f(x) = \lim_{h\to 0^+} \frac{1}{h}\int_x^{x+h} f(t)\, dt$$ Then show that for all...
    12 replies | 154 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 | 45 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 | 74 view(s)
  • Rido12's Avatar
    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 | 74 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 | 74 view(s)
  • Euge's Avatar
    Yesterday, 02:05
    Euge replied to a thread Real Analysis in Analysis
    Well, the difference between the Riemann sum and integral is made less than $\epsilon$ in magnitude when $n\ge N$, but since $\epsilon$ was...
    12 replies | 154 view(s)
  • Euge's Avatar
    Yesterday, 01:53
    Euge replied to a thread Real Analysis in Analysis
    Not quite. Fix $k$. If $n \ge N$ and $x\in \left$, then $\lvert \frac{k}{n} - x\rvert \le \frac{1}{n} \le \frac{1}{N} < \delta$. Thus, for all $n \ge...
    12 replies | 154 view(s)
  • Euge's Avatar
    Yesterday, 00:35
    Euge replied to a thread Real Analysis in Analysis
    Hi joypav, You're not bothering us with your questions, so feel free to ask whenever you have trouble. :)
    12 replies | 154 view(s)
  • Euge's Avatar
    March 25th, 2017, 23:17
    Yes. Let $h > 0$ such that $c + h\in (a,b)$. Then $$F(c + h) - F(c) - f(c+)h = \int_c^{c+h} \, dt $$ Let $\epsilon > 0$. There exists a $\delta...
    7 replies | 165 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 | 83 view(s)
  • topsquark's Avatar
    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 | 56 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 | 82 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 | 70 view(s)
  • greg1313's Avatar
    March 25th, 2017, 20:52
    Rewrite as $\sqrt{2t+5}+\sqrt{2t+8}=\sqrt{8t+25}$. What do you get when you square both sides?
    4 replies | 56 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 | 82 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 | 70 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 | 82 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 | 70 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)
  • greg1313's Avatar
    March 24th, 2017, 23:17
    greg1313 replied to a thread Factoring...8 in Pre-Calculus
    1. Correct. 2. Correct. 3. Yes, apply the sum of cubes formula to $8a^3+27b^3$. Be careful with your variable names. $a$ does not necessarily...
    2 replies | 39 view(s)
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    Happy Birthday, Rido!
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    *rido
    oops lol
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    Thanks ride hope everything is good for you too
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    Happy Birthday Rido12! I wish you another healthy year with loads of fun, excitement, beautiful memories and all your wishes & dreams come true!
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    Happy Birthday, Rido12!!!
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    wb when you get the chance
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    you never clear you inbox i left something for you on the wb
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    Thanks Rido , sorry for late reply. I am busy with exams these days.
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About Rido12

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