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• Today, 13:44
MarkFL replied to a thread Product of the Roots in Pre-Calculus
"Ricky Recruit" was a nickname applied to recruits in their first week of boot camp...and "Ricky Raisin" was applied to those who didn't yet have...
17 replies | 132 view(s)
• Today, 12:41
MarkFL replied to a thread Product of the Roots in Pre-Calculus
Ah, I didn't catch that about your username. I got to enjoy Great Lakes RTC beginning in January...that was some brisk weather to be sure. (Rofl) I...
17 replies | 132 view(s)
• Today, 12:16
MarkFL replied to a thread Product of the Roots in Pre-Calculus
I'm 53, and yes life does seem to pass by more quickly the older I get. :D As a child and young adult, a year seemed to be an eternity...now a year...
17 replies | 132 view(s)
• Today, 11:51
MarkFL replied to a thread Product of the Roots in Pre-Calculus
Yes, I think it is a well-written textbook too. (Yes) When I was taking College Algebra, near the end of the semester I was approached by my...
17 replies | 132 view(s)
• Today, 11:31
I solved for $x$. :D But check your work...the areas aren't correct yet. ;)
13 replies | 107 view(s)
• Today, 11:25
MarkFL replied to a thread Product of the Roots in Pre-Calculus
When I took PreCalc back in 1991, it was David Cohen's textbook that we used...third edition. :D
17 replies | 132 view(s)
• Today, 01:11
MarkFL replied to a thread Variation Expression in Pre-Calculus
Let's let $F$ be the fixed costs and $M$ be the marginal cost (the cost to machine one part), and $x$ be the number of parts machined. Then the total...
3 replies | 46 view(s)
• Today, 00:39
If you have a piece of wire whose length is $a$, and we bend it into a square, then each side of the square will be: \frac{a}{4} And so what...
13 replies | 107 view(s)
• Today, 00:15
Let's go back to: f(x)=A(x-6)\left(x^2+1\right) Now, we set: f(2)=A(2-6)\left(2^2+1\right)=-20A=-10\implies A=\frac{1}{2} And so we have:
5 replies | 54 view(s)
• Yesterday, 23:27
MarkFL replied to a thread Sum of the Roots in Pre-Calculus
Here's another approach: Suppose we have: ax^2+bx+c=0 Them by the quadratic formula, we have that the sum $S$ of the roots is given by: ...
4 replies | 41 view(s)
• Yesterday, 23:22
MarkFL replied to a thread Product of the Roots in Pre-Calculus
Here's another approach: Suppose we have: ax^2+bx+c=0 Then by the quadratic formula, we know the product $P$ of the roots is: ...
17 replies | 132 view(s)
• Yesterday, 22:29
greg1313 replied to a thread Product of the Roots in Pre-Calculus
$$x^2-x(a+b)+ab=x^2+px+q\implies a+b=-p,ab=q$$
17 replies | 132 view(s)
• Yesterday, 17:36
greg1313 replied to a thread Product of the Roots in Pre-Calculus
Are you sure the product is p? It seems it would be q.
17 replies | 132 view(s)
• Yesterday, 17:34
Does the problem give a relationship between the two pieces of wire?
13 replies | 107 view(s)
• Yesterday, 17:25
greg1313 replied to a thread Sum of the Roots in Pre-Calculus
What do you get when you expand (x - a)(x - b)?
4 replies | 41 view(s)
• March 27th, 2017, 19:34
Any point in $\mathbb{R}^2$ can be described as the coordinates of the tip of a vector. As any vector can be the hypotenuse of the triangle formed by...
2 replies | 63 view(s)
• March 27th, 2017, 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 | 83 view(s)
• March 26th, 2017, 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...
5 replies | 88 view(s)
• March 26th, 2017, 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...
5 replies | 88 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 | 101 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 | 105 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 | 97 view(s)
• 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 | 65 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 | 105 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 | 97 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 | 80 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 | 105 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 | 97 view(s)
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