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Yesterday, 21:17
I would let $a$ be the length (in cm) of the segments with 1 tick mark, and $b$ be the length (in cm) of the segments with 2 tick marks. And so,...
2 replies  21 view(s)

Yesterday, 19:30
Being that it is an optimization problem, on a function in two variables, the way I worked it using the calculus is the most straightforward way I...
3 replies  56 view(s)

Yesterday, 12:53
I think what I would do is orient a coordinate system such that south is up and west is to the right. At time $t=0$, have boat A at the origin, and...
3 replies  56 view(s)

Yesterday, 11:28
There are online calculators that will spit out the answer to this question, however if you want genuine help, we need to know what you've tried, or...
3 replies  84 view(s)

Yesterday, 03:19
Finding what to multiply together to get an expression. It is like "splitting" an expression into a multiplication of simpler expressions.
6 replies  55 view(s)

Yesterday, 02:50
I would say division is the inverse of multiplication, but factoring is certainly related to division. For example, we know:
...
6 replies  55 view(s)

Yesterday, 02:42
Looks good. (Yes)
2 replies  27 view(s)

Yesterday, 02:32
Perfect. (Yes)
6 replies  55 view(s)

Yesterday, 02:24
I was just rewriting that term as a perfect square:
9(ab+c)^2=3^2(ab+c)^2=(3(ab+c))^2
4 replies  37 view(s)

Yesterday, 02:16
I would first write the expression as:
(2ab)^2(3(ab+c))^2
Now, factor as the difference of squares, and simplify. :D
4 replies  37 view(s)

Yesterday, 00:48
As I said before, please post calculus questions in our "Calculus" forum, and please don't begin a new thread for the same question. I have moved and...
7 replies  79 view(s)

March 22nd, 2017, 20:04
True, the roots are no given, but you can compute them...we should begin by equating the given quadratic to zero:
2x^2+5x–k=0
Using the...
8 replies  95 view(s)

March 22nd, 2017, 19:49
You want to set the discriminant greater than or equal to zero, since you don't want it to be negative. If the discriminant is equal to zero, then...
6 replies  58 view(s)

March 22nd, 2017, 19:21
No, the roots themselves. Recall, if:
ax^2+bx+c=0
then:
x=\frac{b\pm\sqrt{b^24ac}}{2a}
What are the roots of the given quadratic?
8 replies  95 view(s)

March 22nd, 2017, 19:18
Yes, the discriminant is what's under the radical in the quadratic formula, and it cannot be negative if the roots ore real. :D
6 replies  58 view(s)

March 22nd, 2017, 14:21
Begin by stating the roots of the quadratic, using the quadratic formula...what do you have?
8 replies  95 view(s)

March 22nd, 2017, 14:17
Consider the general quadratic:
ax^2+bx+c
If this quadratic is to have real roots, then the discriminant has to be nonnegative, that is:
...
6 replies  58 view(s)

March 22nd, 2017, 13:51
Differentiation is linear, that is:
\frac{d}{dx}\left(f(x)+g(x)\right)=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)
This means that is a function is...
7 replies  79 view(s)

March 22nd, 2017, 13:33
In this example, you are given the closed form, rather than a recursion, so you simply need to use $n=10$, that is:
a_{10}=2(1)^{10}=?
In the...
3 replies  54 view(s)

March 22nd, 2017, 13:30
Okay, well let's see if you understand the rules of differentiation that you will need for this problem. Suppose we have:
y=f(x)+g(x)
What is...
7 replies  79 view(s)

March 22nd, 2017, 13:15
Hello, rahulk! (Wave)
We are not a "drop off your homework so we can do it for you" site. What good would that do for you, to have someone else do...
7 replies  79 view(s)

March 22nd, 2017, 11:41
No, I reversed them to get rid of the leading negative for that term. :D
6 replies  73 view(s)

March 22nd, 2017, 11:37
Please show what you have tried, and please post calculus problems in our Calculus forum. :D
3 replies  65 view(s)

March 22nd, 2017, 11:33
Let's first subtract \frac{1}{2} from both sides:
\frac{\left(U^WU^L\right)}{2(\alpha\beta)}<\frac{1}{2}
If $\alpha\beta>0$, then we may...
6 replies  73 view(s)

March 22nd, 2017, 03:47
\lim_{h\to0}\frac{\sin(h)}{\cos(h)+1}=\frac{\sin(0)}{\cos(0)+1}=\frac{0}{1+1}=0 :D
5 replies  74 view(s)

March 22nd, 2017, 03:34
Well, if $f'(c)=\cos(c)$ and c=\frac{\pi}{4}...then what do you get?
5 replies  74 view(s)

March 22nd, 2017, 02:59
Since you posted in the PreCalculus forum, I suppose you are to use the definition:
f'(x)\equiv\lim_{h\to0}\frac{f(x+h)f(x)}{h}
So, for...
5 replies  74 view(s)
 Date of Birth
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 BS Mathematics, East Tennessee State University
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He who knows not, and knows not that he knows not, is a fool, shun him; He who knows not, and knows that he knows not, is a child, teach him. He who knows, and knows not that he knows, is asleep, wake him. He who knows, and knows that he knows, is wise, follow him.
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