• Yesterday, 22:28
I am answering my own question aincve I have now noticed that regarding my question on the first semtence of the proof of Theorem 45.6 ... the...
1 replies | 24 view(s)
• Yesterday, 22:03
MarkFL replied to a thread Two Pieces of Wire in Pre-Calculus
We require both: 0<x 0<12-x Combined, this gives us: 0<x<12
18 replies | 128 view(s)
• Yesterday, 21:41
For better forum organization, I have merged the 4 threads pertaining to Stanley A. Ellisen quotes into one thread. This way they can all be...
4 replies | 37 view(s)
• Yesterday, 21:12
MarkFL replied to a thread Two Pieces of Wire in Pre-Calculus
x and x - 12 represent lengths (assumed to be in cm)...and so must be positive. Lengths cannot be negative, and if we are to actually have two pieces...
18 replies | 128 view(s)
• Yesterday, 19:13
MarkFL replied to a thread Two Pieces of Wire in Pre-Calculus
At this point you can multiply through by 8 to get: x^2-12x+32>0 Then factor: (x-4)(x-8)>0 Now solve this inequality, subject to:
18 replies | 128 view(s)
• Yesterday, 11:04
When I was a student, I found one of the best ways to understand things is to look at them in general terms. For example, in this problem, rather...
21 replies | 175 view(s)
• Yesterday, 10:49
What we see is that we must have: L=W And so, what we have found is that for a rectangle of a given area, the perimeter is the smallest it can...
21 replies | 175 view(s)
• Yesterday, 10:34
Your next step should be: (L-W)^2=0 So, how must L and W be related in order for this to be true?
21 replies | 175 view(s)
• Yesterday, 10:28
MarkFL replied to a thread Two Pieces of Wire in Pre-Calculus
No, what we want is: A\left(W_1\right)+A\left(W_2\right)>5 A(x)+A(12-x)>5 Now, the arguments for the area function represent wire lengths,...
18 replies | 128 view(s)
• Yesterday, 10:10
MarkFL replied to a thread Two Pieces of Wire in Pre-Calculus
Yes, now we need to answer the question: For which values of x will the combined areas of the squares exceed 5 cm^2? We have two pieces of wire...
18 replies | 128 view(s)
• Yesterday, 10:00
MarkFL replied to a thread Two Pieces of Wire in Pre-Calculus
Okay, you have a piece of wire whose length is let's say W...you bend it into a square...what is the perimeter of the square? Then using your formula...
18 replies | 128 view(s)
• Yesterday, 09:55
Yes, once you simplify, you will find the implication. :D
21 replies | 175 view(s)
• Yesterday, 09:53
Yes, now combine terms on the LHS...:D
5 replies | 47 view(s)
• Yesterday, 09:44
MarkFL replied to a thread Two Pieces of Wire in Pre-Calculus
Let's use x for the length of the sides of the square, so we have: P=4x A=x^2 Now, we want to replace x in the formula for the area with an...
18 replies | 128 view(s)
• Yesterday, 09:39
You're just seeing what the equation implies about the relationship between L and W. :D
21 replies | 175 view(s)
• June 23rd, 2017, 23:01
I am reading Anderson and Feil - A First Course in Abstract Algebra. I am currently focused on Ch. 45: The Splitting Field ... ... I need...
1 replies | 24 view(s)
• June 23rd, 2017, 22:50
We have: \sqrt{LW}\le\frac{L+W}{2} When I ask "when does equality occur," I am essentially asking you to solve: \sqrt{LW}=\frac{L+W}{2} ...
21 replies | 175 view(s)
• June 23rd, 2017, 21:56
I would look at a substitution like: u=\sqrt{x^2+4x+3}\implies dx=\frac{u}{x+2}\,du And the integral then becomes: I=\int\frac{1}{u^2+1}\,du
1 replies | 27 view(s)
• June 23rd, 2017, 19:33
MarkFL replied to a thread Two Pieces of Wire in Pre-Calculus
What is the area $A$ of a square whose perimeter is $P$?
18 replies | 128 view(s)
• June 23rd, 2017, 19:25
You don't want to multiply an inequality by an expression whose sign is unknown...arrange everything to one side and then get your critical values...
5 replies | 47 view(s)
• June 23rd, 2017, 19:21
You are assuming x is positive when you do that. A better step is to subtract x/2 from both sides: \frac{2}{x}-\frac{x}{2}<0 Combine terms: ...
2 replies | 35 view(s)
• June 23rd, 2017, 15:00
MarkFL replied to a thread Quadratic Inequality in Pre-Calculus
Yes, we have a parabola opening upwards, and given that it has two real roots, we should expect to find it to be negative in between its roots. :D
7 replies | 73 view(s)
• June 23rd, 2017, 14:44
Okay, I have several issues here: 1.) We ask that you post no more than 2 questions per thread: 2.) You've shown no effort:
1 replies | 64 view(s)
• June 23rd, 2017, 13:38
In this problem, you have a function representing a parabola that opens upwards. We should then expect, given that it has two real roots, that the...
7 replies | 76 view(s)
• June 22nd, 2017, 22:28
Thanks Euge ... that helped in a big way ... Most grateful for your help and support ... Peter
2 replies | 62 view(s)
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