• Yesterday, 20:06
MarkFL replied to a thread Related Rates in Calculus
Yes, the length of his shadow is increasing at a rate of 4/3 ft/s, and the tip of his shadow is moving at a rate of 10/3 ft/s. :D
8 replies | 96 view(s)
• Yesterday, 16:29
MarkFL replied to a thread Related Rates in Calculus
I got 4/3 ft/s where you got 3 ft/s. You have the line: \d{y}{t}=\frac{3}{2}\d{x}{t} Now, this implies: \d{x}{t}=\frac{2}{3}\d{y}{t}
8 replies | 96 view(s)
• Yesterday, 16:20
Nicely done, Chris! (Yes) This is the solution I found elsewhere: We are given to evaluate: \lim_{n\to\infty}\left(\sqrt{\int_0^1...
4 replies | 114 view(s)
• Yesterday, 15:15
MarkFL replied to a thread Related Rates in Calculus
Suppose we wanted to work the problem in general terms. Please refer to the following diagram: $P$ is the height of the pole, $M$ is the...
8 replies | 96 view(s)
• February 26th, 2017, 23:47
Compute the following limit: \lim_{n\to\infty}\left(\sqrt{\int_0^1 \left(1+x^n\right)^n\,dx}\right)
4 replies | 114 view(s)
• February 26th, 2017, 06:31
$m>r$ holds because by definition of $r$, $m=rd$ and by assumption $d>1$. But I don't see where $s>1$ is used in the proof. It can happen, for...
1 replies | 41 view(s)
• February 26th, 2017, 06:20
If $=$, then $x$ can be canceled and we have $i=j$.
1 replies | 42 view(s)
• February 26th, 2017, 02:44
I am reading Anderson and Feil - A First Course in Abstract Algebra. I am currently focused on Ch. 8: Integral Domains and Fields ... I need...
1 replies | 42 view(s)
• February 26th, 2017, 02:02
I am reading Anderson and Feil - A First Course in Abstract Algebra. I am currently focused on Ch. 8: Integral Domains and Fields ... I need...
1 replies | 41 view(s)
• February 26th, 2017, 01:48
We could likely model this with a function of the form (which comes from Newton's Law Of Cooling): f(t)=c_1e^{-kt}+350 Since we know $f(0)=70$,...
1 replies | 77 view(s)
• February 26th, 2017, 01:01
My solution: Consider the objective function: F(A,B,C)=\sin(A)+\sin(B)+\sin(C)-\cos(A)-\cos(B)-\cos(C) Subject to the constraint: ...
3 replies | 114 view(s)
• February 25th, 2017, 23:35
Let's look at the behavior of the solutions near the equilibrium solution $y=3$. Suppose we have: y(0)=2.9 Then we see: y'<0 And for:
4 replies | 78 view(s)
• February 24th, 2017, 11:34
3 replies | 94 view(s)
• February 24th, 2017, 11:28
There's no need to link to the commercial site you used. Thanks! :D
3 replies | 99 view(s)
• February 24th, 2017, 00:26
Let's find the $p$th term. I think what I would do is set: \frac{n(n+1)}{2}=p Now solve for $n$... n(n+1)=2p
1 replies | 58 view(s)
• February 23rd, 2017, 22:15
Perhaps you meant how can we derive this formula. Let's begin with the definition of average acceleration: \overline{a}=\frac{\Delta v}{\Delta...
6 replies | 103 view(s)
• February 23rd, 2017, 20:43
I made a mistake in my above post...the IVP should be: \displaystyle \d{F}{t}=2-\frac{5F}{4(50-t)} where $F(0)=0$. So, this leads to: ...
7 replies | 125 view(s)
• February 23rd, 2017, 16:33
I would let $F(t)$ be the amount of fertilizer in the tank at time $t$. From the problem statement, we get the following IVP: ...
7 replies | 125 view(s)
• February 23rd, 2017, 14:31
I don't follow...it's already in the form we want to give us the acceleration. :D
6 replies | 103 view(s)
• February 23rd, 2017, 14:28
The kinematic formula you want is: \overline{a}=\frac{v_f^2-v_i^2}{2\Delta x}
6 replies | 103 view(s)
• February 23rd, 2017, 02:40
Our goal is to make MHB stand out as much as we can from our competitors by offering features you just don't see anywhere else. If we were to package...
3 replies | 128 view(s)
• February 23rd, 2017, 02:35
MarkFL replied to a thread Tapatalk... in Chat Room
Personally, I don't think Tapatalk is ready for vBulletin, much less sites using $\LaTeX$. From what I've gathered, many vBulletin admins refuse to...
4 replies | 94 view(s)
• February 23rd, 2017, 01:29
MarkFL replied to a thread Tapatalk... in Chat Room
I'd be happy if Tapatalk didn't lazily bypass most vBulletin PHP plugin hooks. (Giggle)
4 replies | 94 view(s)
• February 22nd, 2017, 20:01
Well! ... that is a thoroughly convincing proof!!! Thanks HallsofIvy ... really helpful ... Peter
5 replies | 119 view(s)
• February 22nd, 2017, 12:54
The scale along the $x$ line is not clear from the picture, so it is impossible to say whether this is the graph of $y=x/2+3$.
5 replies | 154 view(s)
• February 22nd, 2017, 04:08
Your solution to the given IVP isn't correct...we are given: \d{y}{x}=9x^8e^{-x^9} where y(0)=8 \int_{y(0)}^{y(x)}\,du=\int_0^x...
1 replies | 66 view(s)
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