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Looks good to me.

Here's the derivative for the function of power, P=Av^2+\frac{B}{v} (\text{Where }A\text{ and }B\text{ are positive constants)} that they give in the introduction to the problem: P'(v)=2Av-\frac{B}{v^2} To answer \text{(a) What speed }vP\text{ minimizes power?} we set this expression...

That depends on what a, b and c are.

yeah should have zoomed the page... lol... :)

In terms of getting useful replies, it would probably be best if you showed us where in the two problems you are having difficulty and how we can help. If you need to refer directly to the math in either of the two problems, please be quote it directly in your post. (clicking on the .pdf link...

\[ \lim_{N\rightarrow\infty}\frac{x}{2^n}\sin^2\left(\frac{x}{2^n}\right)\rightarrow0\times0^2\rightarrow0 \] but as the limit is taken over positive $x$ the limit tends to infinity.

There are a lot of online tutoring sites and which ones you may be interested in are are likely to be the ones concerning mathematics you want to study. https://khanacademy.org is one for elementary topics up to college/university topics.. https://desmos.com has a good online graphing calculator.

You need the roots $a,b$ of the equation $3\sqrt{x}-4=3x\sqrt{5}-\frac{8}{5}$ Once you have established these roots use them as endpoints in $\int_{a}^{b}\left(3\sqrt{x}-4-3x\sqrt{5}+\frac{8}{5}\right)dx$ The result is the area $A$ of $R$.

In general, periodic functions are of interest due to their frequent occurrence in natural phenomenon. As speculation, this particular function may be of interest due the times and places it occurs.

Note the concavity of $f(x)$. What does that tell you about lowering the graph of $f(x)$? (the line $y=0$ does not concern us presently).

Note the range of $f(x)$. A and B are vertical shifts, C and D are horizontal shifts. Which one of the given shifts would result in the graph of $f(x)$ not crossing the $x-\text{axis}$?

What do you get when you expand the LHS and equate the resulting terms with the corresponding terms in the RHS of the given equation

First, we need to establish $\sin\theta$ and $\cos\theta$. $9^2+(-5)^2=106$ (Pythagorean theorem) so $\sin\theta$ is $\sqrt{\frac{|-5|}{106}}, \text{that is}, \left(\frac{opp}{hyp}\right)$ and $\cos\theta$ is $\frac{3}{\sqrt{106}}, \text{that is}, \left(\frac{adj}{hyp}\right)$ (recall that...

Country Boy Is correct in stating that the slope of the tangent line is 2. So the tangent line equation is y = 2x + 3 and the equation of the normal is then y = -x/2 + 11/2

Haha!

Yes, I agree. My error was missing i = 30 and assuming i = 1 .

Oh! Those definitions are implied... :poop: f(i)=1.8. and g(i)=1.2 both suffice as definitions for f,\,g if I am not mistaken... After a few basic calculations we may arrive at: \frac{420-90+600}{2}=465. Do you see that too? Hint: use the fact that summation is associative and sum each addend...

Do you have definitions for f and g?

Actually, I've never seen it before...and not everyone here is necessarily as "smart" as you might be assuming they are. I've edited my initial post in this thread. Please review it and post accordingly, if you will. :)

Hi riffwraith. Please use accurate terminology in your thread titles. The use of "ftg" is ambiguous as, as far as I know, it is not standard terminology. In fact, please try to avoid abbreviations altogether. I've edited the title for you. .

[8, 72]

You need to look at left-hand and right-hand limits. I don't see where "$\pm$" comes into it.

\begin{align*}2(\sin2x+\sin2y+\sin2z)&=\sin x\cos(y-z)+\sin y\cos(x-z)+\sin z\cos(x-y) \\ &=6\sin x\sin y\sin z+\sin x\cos y\cos z+\sin y\cos x\cos z+\sin z\cos x\cos y \\ &=6\sin x\sin y\sin z+\sin x\cos x+\sin y\cos y+\sin z\cos z \\ \frac32(\sin2x+\sin2y+\sin2z)&=6\sin x\sin y\sin z \\...

If $x+y+z=\pi$ then \sin2x+\sin2y+\sin2z=4\sin x\sin y\sin z

In his 1859 paper entitled "On the Number of Primes Less than a Given Magnitude", Riemann gives as his point of departure the equation \prod\frac{1}{1-\frac{1}{p^s}}=\sum\frac{1}{n^s} where $p$ is all primes and $n$ is all natural numbers. The function of the complex variable $s$, wherever...

$p'(t)=v(t),\quad v'(t)=a(t)$

Hmmm... $p(t)=3t^2-\frac{t^3}{3}+C$ (Wondering)

(a-b)^3=a^3-3a^2b+3ab^2-b^3 a^3-b^3=3a^2b-3ab^2+(a-b)^3 a^3-b^3=3ab(a-b)+(a-b)^3 a^3-b^3=(a-b)(3ab+(a-b)^2) a^3-b^3=(a-b)(3ab+a^2-2ab+b^2) a^3-b^3=(a-b)(a^2+ab+b^2)

\begin{align*}x^3-7x^2+14x-8&=x^3-8-7x(x-2) \\ &=(x-2)(x^2+2x+4)-7x(x-2) \\ &=(x-2)(x^2+2x+4-7x) \\ &=(x-2)(x^2-5x+4) \\ &=(x-2)(x-1)(x-4)\end{align*}

Observing that $y=e^{\tan x} - 2$ has a root in [0, 1] at $\text{atan}(\log2)$, we need to evaluate $2\sec^2(\text{atan}(\log2))$. That is approximately 2.961, hence choice D is correct.

What angle are you referring to? What do you mean by an "exact" proof?

A link to an interesting article I found is below: https://www.quantamagazine.org/neutrinos-lead-to-unexpected-discovery-in-basic-math-20191113/

Re: 231 value of second dirivative See https://mathhelpboards.com/calculus-10/297-ap-calculus-exam-2nd-derivative-26690.html.

u^2=25-x^2 -u\,du=x\,dx \int_0^5u^2\,du=\frac{125}{3}-0=\frac{125}{3}

(x+2y)\frac{dy}{dx}=2x-y x\frac{dy}{dx}+2y\frac{dy}{dx}=2x-y \frac{dy}{dx}+x\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+2y\frac{d^2y}{dx^2}=2-\frac{dy}{dx} 2+3\frac{d^2y}{dx^2}+8=0 \frac{d^2y}{dx^2}=-\frac{10}{3}

Maybe the Dirac delta function?

y=(x - 2)(x - 4)=x^2-6x+8

Hey, it's not an equation it's an expression! Here's what an online CAS thinks.

The limit is with respect to $x$, so $a$ and $b$ are treated as constants. Getting things into a form where we can apply L'Hopital's rule, \]\exp\left(\frac{\log\left(1+\frac ax\right)}{\frac{1}{bx}}\right) Now *differentiate and simplify* inside the brackets; you'll end up with $e^{ab}$.

\sin^2x+\cos^2x=1 1+\cot^2x=\csc^2x \cot^2x-\csc^2x=-1

u=e^{\sqrt x} 2\,du=\frac{e^{\sqrt x}}{\sqrt x}\,dx 2\int_e^{e^2}\,du=2e(e-1)

3x = 15 x = 15/3 x = 5

W=\frac AL \frac{5A}{L}+2L=550 5A+2L^2=550L A=110L-\frac{2L^2}{5} $A$ has a maximum at the vertex of this inverted parabola, so $L=\frac{275}{2}$. Finding $A$ and $W$ from here should be straightforward.

Here's a start: Let $W$ be width, $L$ be length an $A$ be the desired area. Then, 5W+2L=550 LW=A Can you make any progress from there?

\frac{4x}{4x}\cdot\frac{\tan4x}{6x}=\frac{4x}{4x}\cdot\frac{\sin4x}{6x\cdot\cos{4x}}=\frac{4x}{6x}\cdot\frac{\sin4x}{\cos4x\cdot4x} \frac23\lim_{x\to0}\frac{\sin4x}{\cos4x\cdot4x}=\frac23

Hello Mooija and welcome to MHB. :) √20 = √4√5 = 2√5 10/√5 = 10√5/5 = 2√5

See stuff in red - otherwise looks good. :) Use ... tags.

g(x)=\log(2x)=\log(2)+\log(x),\quad g'(x)=\frac1x or, by the chain rule, g'(x)=\frac{2}{2x}=\frac1x

(-1-h)^2+(3-k)^2=(6-h)^2+(2-k)^2=(-2-h)^2+(-4-k)^2 (-1-h)^2-(6-h)^2+(3-k)^2-(2-k)^2=0 -7(5-2h)+(5-2k)=0 -35+14h+5-2k=0 -30+14h-2k=0\quad[1] (-1-h)^2-(-2-h)^2+(3-k)^2-(-4-k)^2=0 (-3-2h)+7(-1-2k)=0 -10-2h-14k=0\quad[2] [1]+7*[2]\implies k=-1\implies h=2 (x-2)^2+(y+1)^2=25