Facebook Page
Twitter
RSS
  • 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)
  • kaliprasad's Avatar
    Yesterday, 20:06
    you have done right except a small modification that you need to yo do $f(x) = A(x-6)(x^2+1)$ where A is a constant putting the values you get A...
    5 replies | 54 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)
  • kaliprasad's Avatar
    March 26th, 2017, 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 | 66 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)
  • 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)
  • 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 | 92 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 | 105 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 | 80 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 | 97 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 | 101 view(s)
More Activity

2 Visitor Messages

  1. Reply/View Conversation
    Hello Fallen Angel!

    As the moderator of the Challenge Problems sub-forum, I want to thank you for posting a challenge for our members to have fun solving with.

    Best,

    anemone
  2. Reply/View Conversation
    Hello and welcome to MHB, Fallen Angel!

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

    Best Regards,

    Mark.
Showing Visitor Messages 1 to 2 of 2
About Fallen Angel

Basic Information

Age
24
About Fallen Angel
Location:
Cantabria
Country Flag:
Spain

Signature


"Poetry is the art of giving different names to the same thing, Mathematics is the art of giving the same name to different things" H. Poincare

"Mathematics is the music of reason" J. J. Sylvester

Statistics


Total Posts
Total Posts
205
Posts Per Day
0.23
Thanks Data
Thanks Given
39
Thanks Received
342
Thanks Received Per Post
1.668
Visitor Messages
Total Messages
2
Most Recent Message
November 14th, 2014 12:12
General Information
Last Activity
December 21st, 2016 12:09
Last Visit
December 20th, 2016 at 16:02
Last Post
October 28th, 2016 at 03:18
Join Date
November 5th, 2014

6 Friends

  1. bwpbruce Offline

    MHB Apprentice

    bwpbruce
  2. Evobeus Offline

    MHB Apprentice

    Evobeus
  3. Fernando Revilla Offline

    MHB Journeyman

    Fernando Revilla
  4. kaliprasad Offline

    MHB Master

    kaliprasad
  5. MarkFL  Online

    Pessimist Singularitarian

    MarkFL
  6. Peter Offline

    MHB Master

    Peter
Showing Friends 1 to 6 of 6
No results to show...
Math Help Boards