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  • MarkFL's Avatar
    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)
  • MarkFL's Avatar
    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)
  • MarkFL's Avatar
    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)
  • MarkFL's Avatar
    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)
  • MarkFL's Avatar
    Today, 11:31
    I solved for $x$. :D But check your work...the areas aren't correct yet. ;)
    13 replies | 107 view(s)
  • MarkFL's Avatar
    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)
  • MarkFL's Avatar
    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)
  • MarkFL's Avatar
    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)
  • MarkFL's Avatar
    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)
  • MarkFL's Avatar
    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)
  • MarkFL's Avatar
    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)
  • greg1313's Avatar
    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)
  • greg1313's Avatar
    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)
  • greg1313's Avatar
    Yesterday, 17:34
    Does the problem give a relationship between the two pieces of wire?
    13 replies | 107 view(s)
  • greg1313's Avatar
    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)
  • greg1313's Avatar
    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)
  • MarkFL's Avatar
    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)
  • MarkFL's Avatar
    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...
    5 replies | 88 view(s)
  • MarkFL's Avatar
    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...
    5 replies | 88 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 | 101 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 | 105 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 | 97 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 | 65 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 | 105 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 | 97 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 | 80 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 | 105 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 | 97 view(s)
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