• Yesterday, 06:59
Hello and welcome to MHB! (Wave) We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions....
1 replies | 33 view(s)
• Yesterday, 01:55
MarkFL replied to a thread Domain...2 in Pre-Calculus
No factor in the denominator can be zero, and nothing under a square root radical can be negative...so this leads to: t-1\ne0 0<t What do...
1 replies | 27 view(s)
• Yesterday, 01:47
I would write: y=\frac{2x}{x-1}=\frac{2x-2+2}{x-1}=\frac{2(x-1)+2}{x-1}=2+\frac{2}{x-1} We see this will have a horizontal asymptote at $y=2$,...
1 replies | 29 view(s)
• Yesterday, 01:31
Let's look at a definition: |u|=\begin{cases}u, & 0\le u \\ -u, & u<0 \\ \end{cases} Can you see that we must have: 0\le|u| ?
2 replies | 46 view(s)
• March 19th, 2018, 22:30
Yes, and as Wilmer was pointing out, you would likely want to use the binomial theorem. If you raise both sides to the 5th power, you get: ...
10 replies | 106 view(s)
• March 19th, 2018, 21:56
You're being asked to find an area, and in essence, you're doing so by adding up a bunch of vertical lines, the length of which are determined by the...
5 replies | 54 view(s)
• March 19th, 2018, 21:48
To follow up, we get: \d{x}{t}=\frac{\d{y}{t}\left(x-12y^2\right)}{9x^2-y} Plugging in the given values, we find: ...
4 replies | 52 view(s)
• March 19th, 2018, 21:07
That's already included in the "top curve minus the bottom curve." :)
5 replies | 54 view(s)
• March 19th, 2018, 09:46
Let's first look at the bounded region: And so the area is: A=\int_0^3 (-x^2+6x)-(x^2-2x)\,dx=2\int_0^3 -x^2+4x\,dx=2\left_0^3=2(18-9)=18 ...
5 replies | 54 view(s)
• March 19th, 2018, 08:38
To follow up, the washer method gives us: V=\pi\int_{x_1}^{x^2} R^2-r^2\,dx The volume of an arbitrary washer is: ...
3 replies | 82 view(s)
• March 19th, 2018, 02:24
I would first observe that $1+\sqrt{5}$ is a root of: x^2-2x-4=0 And so, the coefficients of the expansion: (1+\sqrt{5})^n Can be found...
10 replies | 106 view(s)
• March 19th, 2018, 02:02
2 replies | 36 view(s)
• March 19th, 2018, 02:00
2 replies | 42 view(s)
• March 18th, 2018, 23:25
1 replies | 105 view(s)
• March 18th, 2018, 23:03
We consider that both $x$ and $y$ are functions of time $t$, and so beginning with: 3x^3+4y^3=xy We apply on the left the power and chain...
4 replies | 52 view(s)
• March 18th, 2018, 22:27
Hello, and welcome to MHB! (Wave) We are given the curve: 3x^2+4y^3=xy In order to answer the given question, it will be helpful to...
4 replies | 52 view(s)
• March 18th, 2018, 07:58
MarkFL replied to a thread More integral substitution in Calculus
Is it possible it wants a decimal approximation there? Of course, then the issue is how many digits of accuracy? Gotta love these automated systems,...
12 replies | 104 view(s)
• March 18th, 2018, 07:44
MarkFL replied to a thread More integral substitution in Calculus
Okay, proceeding to part (b), I get (leaving off the constant of integration): \int e^{4x}\left(3+e^{4x}\right)^3\,dx=\frac{1}{4}\int...
12 replies | 104 view(s)
• March 18th, 2018, 06:57
MarkFL replied to a thread More integral substitution in Calculus
Let's step through the substitution a bit more slowly... \int e^{4x}\left(3+e^{4x}\right)^3\,dx=\int...
12 replies | 104 view(s)
• March 18th, 2018, 06:29
It is a related rate problem involving differential calculus (so that's why I moved it). Many places teach elementary calculus in high school. :)
6 replies | 58 view(s)
• March 18th, 2018, 06:26
MarkFL replied to a thread More integral substitution in Calculus
Where did the $u-3$ in the denominator of the integrand come from? It should just be: \frac{u^3}{4} If we pull the constant inside.
12 replies | 104 view(s)
• March 18th, 2018, 06:20
For a sphere of radius $r$, the volume is: V=\frac{4}{3}\pi r^3 Differentiating w.r.t time $t$, we have: \d{V}{t}=4\pi r^2\d{r}{t}\implies...
6 replies | 58 view(s)
• March 18th, 2018, 05:49
MarkFL replied to a thread More integral substitution in Calculus
Okay, if we make the suggested substitution, this means: u=3+e^{4x}\implies du=4e^{4x}\,dx And so for part (a) we may write: \int...
12 replies | 104 view(s)
• March 18th, 2018, 02:04
MarkFL replied to a thread More integral substitution in Calculus
I would try the substitution: u=2x Using this substitution, what do you get?
12 replies | 104 view(s)
• March 16th, 2018, 13:24
MarkFL replied to a thread Horizontal Asymptote in Pre-Calculus
Observe that: \frac{x^2+3}{x^2+5}=\frac{x^2+5-2}{x^2+5}=1-\frac{2}{x^2+5} As $x\to\pm\infty$, the rational term vanishes (gets smaller and...
5 replies | 42 view(s)
• March 16th, 2018, 07:37
Yes, that would be acceptable, except in the case where you are instructed to rationalize the denominator where you could write: ...
6 replies | 89 view(s)
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