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• 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)
• 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) • 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) • 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) • 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) • Yesterday, 20:41 Looks good. -Dan 2 replies | 32 view(s) • 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) • 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) • 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) • 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 | 76 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 | 76 view(s) • 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) • 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) • 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) • 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) • 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) • 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) • 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) • 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 | 84 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 | 72 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 | 84 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 | 84 view(s)
• March 24th, 2017, 22:18
Yes. Then in case of doubt you after factoring can multiply and see the result
5 replies | 73 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 | 72 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 | 46 view(s)
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