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by Luke53
Tue Feb 11, 2014 8:20 am
Forum: Calculus
Topic: The stiffness of a rectangular beam varies as its breadth...
Replies: 2
Views: 1450

Re: The stiffness of a rectangular beam varies as its breadt

Thank you for putting me on the right track. Solving the derivative of: (144-h^2)^0.5 * h^3 = 0, this gives: 3 h^2(144-h^2)^0.5 - h^4(144-h^2)^-0.5 = 0 (by the poduct rule). Solving this eqn: h = -6*sqrt(3) and h = 6*sqrt(3), (only the positive value is valid). So the height should be 10.4 inches an...
by Luke53
Sun Feb 09, 2014 10:04 am
Forum: Calculus
Topic: The stiffness of a rectangular beam varies as its breadth...
Replies: 2
Views: 1450

The stiffness of a rectangular beam varies as its breadth...

Hi, can someone help me getting started with this problem?

The stiffness of a rectangular beam varies as its breadth and as the cube of its hight. Find the dimensions of the stiffest beam which can be cut from a circular log 12 inches in diameter.

Thanks.
Luke.
by Luke53
Fri Nov 29, 2013 9:07 am
Forum: Advanced Algebra ("pre-calculus")
Topic: Finding the graph of an equation
Replies: 6
Views: 2005

Re: Finding the graph of an equation

Thanks for your hint and showing me the solution on the Wolfram website.
Part of the implicit plot is shown there.
This plot is what I was after.
How can I get this same plot on my graphic calculator (if this is possible)?
Thanks for the answer.
Luke.
by Luke53
Thu Nov 28, 2013 7:50 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: Finding the graph of an equation
Replies: 6
Views: 2005

Re: Finding the graph of an equation

If I could graph this equation, I would be able to see what points (x,y) are on this graph, so then I could derive the slopes of the tangent lines at several of these points. One solution is (2,1) so this surely is one point on the graph and the slope of the tangent line here is -(11/10) (after impl...
by Luke53
Thu Nov 28, 2013 2:57 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: Finding the graph of an equation
Replies: 6
Views: 2005

Finding the graph of an equation

How can I graph the following equation?
x^3 * y^2 = x * y^3 + 6
I don't know how to isolate y (to get y as a function of x).
Thanks.
Luke.
by Luke53
Fri Jun 07, 2013 2:14 pm
Forum: Intermediate Algebra
Topic: Exponential system of eqn's.
Replies: 4
Views: 2040

Re: Exponential system of eqn's.

OK, the log rules helped me out, I found the solutions now, thanks for your help.

Greetings.

Luke.
by Luke53
Thu Jun 06, 2013 1:53 pm
Forum: Intermediate Algebra
Topic: Exponential system of eqn's.
Replies: 4
Views: 2040

Re: Exponential system of eqn's.

Sorry, but I still can't solve these eqn's so please help. This is what I have till now: ln (4 * e^(.2/y))= ln e ^(1/y)) => (1.386 * .2)/y ln e = (1/y) ln e ( with ln e = 1) So (1.386 * .2)/y = 1/y . But these are clearly not equal to each other, so I must have done something wrong, but I can't figu...
by Luke53
Tue Jun 04, 2013 5:48 pm
Forum: Intermediate Algebra
Topic: Exponential system of eqn's.
Replies: 4
Views: 2040

Exponential system of eqn's.

How can I solve this system? (e = natural log)

x * e ^ -(0.2/y) = 4 (1)

x * e ^ -(1/y) = 1 (2)

I know that x= 5.65 and y= .577 but how do I get these values?

From the second eqn. I get:

ln x = 1/y, but how do I solve x = 4 * e ^(.2/y) ?

Thanks.
by Luke53
Tue Sep 18, 2012 7:14 pm
Forum: Intermediate Algebra
Topic: Solving a system of two eqn's.
Replies: 3
Views: 1393

Solving a system of two eqn's.

How to solve the following system of two equations?
(13/x) - (2/y) = - 0.3
(14.6/x) - (0.4/y) = 0.04
Thanks.
by Luke53
Wed Mar 14, 2012 6:46 pm
Forum: Intermediate Algebra
Topic: solve irrational eqn.: (x-9)^(1/3) - (x+26)^(1/3) + 5 = 0
Replies: 3
Views: 1081

Re: solve irrational eqn.: (x-9)^(1/3) - (x+26)^(1/3) + 5 =

Hi, I checked the equation and there are no errors, when I solve this eqn. with my graphic calculator (TI89) it gives me two roots 1 and -18, but sadly this thing only gives me the results and not how to get to them. According to my textbook it should be done like you mentioned by cubing both member...

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