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Today, 02:28
'Factor:
(x^29)(x^216)\le0
Continue factoring...:D
1 replies  16 view(s)

Today, 02:25
You want to determine the roots of both the numerator and denominator as your critical numbers (i.e., values for which the expression can change...
1 replies  22 view(s)

Today, 02:02
Move everything to the LHS, and then combine all terms into a rational expression...what do you then have?
1 replies  20 view(s)

Yesterday, 22:03
We require both:
0<x
0<12x
Combined, this gives us:
0<x<12
18 replies  135 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  47 view(s)

Yesterday, 21:12
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  135 view(s)

Yesterday, 19:13
At this point you can multiply through by 8 to get:
x^212x+32>0
Then factor:
(x4)(x8)>0
Now solve this inequality, subject to:
18 replies  135 view(s)

Yesterday, 18:54
https://www.google.com/search?safe=off&site=&source=hp&q=rational+equations+worksheet&oq=rational+equations+&gs_l=hp.1.1.0l10.1587.6218.0.9144.19.19.0...
6 replies  62 view(s)

Yesterday, 11:23
yes
looks like one to me ...
4 replies  49 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  180 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  180 view(s)

Yesterday, 10:34
Your next step should be:
(LW)^2=0
So, how must L and W be related in order for this to be true?
21 replies  180 view(s)

Yesterday, 10:28
No, what we want is:
A\left(W_1\right)+A\left(W_2\right)>5
A(x)+A(12x)>5
Now, the arguments for the area function represent wire lengths,...
18 replies  135 view(s)

Yesterday, 10:10
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  135 view(s)

Yesterday, 10:00
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  135 view(s)

Yesterday, 09:55
Yes, once you simplify, you will find the implication. :D
21 replies  180 view(s)

Yesterday, 09:53
Yes, now combine terms on the LHS...:D
5 replies  49 view(s)

Yesterday, 09:44
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  135 view(s)

Yesterday, 09:39
You're just seeing what the equation implies about the relationship between L and W. :D
21 replies  180 view(s)

Yesterday, 08:31
How are the the functions $f$ and $g$ related?
1 replies  31 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  180 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  29 view(s)

June 23rd, 2017, 20:42
$\dfrac{ax+b}{\sqrt{x}} > 2\sqrt{ab}$
since $\sqrt{x} > 0$ ...
$ax+b > 2\sqrt{abx}$
$(ax)^2 + 2abx + b^2 > 4abx$
$(ax)^2  2abx + b^2 >...
4 replies  49 view(s)

June 23rd, 2017, 19:33
What is the area $A$ of a square whose perimeter is $P$?
18 replies  135 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  49 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  38 view(s)

June 23rd, 2017, 15:48
Sketching the graph of a function helps one visualize the domain & range.
2 replies  60 view(s)

June 23rd, 2017, 15:00
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  74 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  69 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  78 view(s)
 Date of Birth
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 BS Mathematics, East Tennessee State University
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