All Answers 358115 Answers

you can place 5 into 10 twice

why?

cuz, theyre both equal

nauseated, i dont think youd want to steal Spider man's quote

if anyone ever gets mad at me in this situation just becuz i prove a point out, theres always one thing to do: RUN!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

its  true what civonamzuk said

4/6 would equal:

.66666666666666666666666

7/9 would equal:

.77777777777777777777777

overall, you were right about 7/9 being larger than 4/6. your answer was correct. your work though is when things went a bit, eh, lets just say a bit sloppy

Another approach:

$$\\
\text{Let }\theta=\cos^{-1}x\\\\
x=\cos\theta\quad\text{so }dx=-\sin\theta.d\theta\\\\
\sqrt{1-x^2}=\sqrt{1-\cos^2\theta}=\sin\theta\\\\
\text{So }\\\\
\int\frac{(\cos^{-1}x)^2}{\sqrt{1-x^2}}dx=-\int\frac{\theta^2.\sin\theta}{\sin\theta}d\theta=-\int\theta^2d\theta=-\frac{\theta^3}{3}=-\frac{(\cos^{-1}x)^3}{3}$$

There is a constant to be added of course.

.

$${\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{3.5}}\right)}{\left({\mathtt{2.44}}{\mathtt{\,\times\,}}{\mathtt{1.22}}\right)}} = {\mathtt{5.878\: \!796\: \!022\: \!574\: \!576\: \!7}}$$

minimum of 6 

$$\\250(1/2)^{10/2}\\\\
=250*(\frac{1}{2})^5\\\\
=250*\frac{1}{32}\\\\
=\frac{125}{16}\\\\
=7\frac{13}{16}\\\\$$

Integrate (cos^-1x)^2 /(1-x^2)^1/2

$$\small{\text{$
\begin{array}{lrcl}
& \int{
\dfrac{[ \cos^{-1}{(x)} ]^2 }
{ (1-x^2)^{\frac{1}{2}} } \ dx }
= \mathbf{ \int{
[\arccos{(x)}]^2 \cdot
\dfrac{ 1 }{ \sqrt{ (1-x^2) } } \ dx } }\\\\
\end{array}
$}}\\\\$$

$$\small{\text{$
\boxed{
\begin{array}{lrcl}
\mathrm{Formula:~}
& y &=& \dfrac{ f(x)^{n+1} } {n+1} \\\\
& y' &=& \left(\dfrac{n+1}{n+1}\right) \cdot f(x)^{n+1-1}\cdot f'(x) \\\\
& y' &=& f(x)^{n}\cdot f'(x) \\\\
&\mathbf{
\int{ f(x)^{n}\cdot f'(x) \ dx }
} & \mathbf{=} & \mathbf{ \dfrac{ f(x)^{n+1} } {n+1} }
\end{array}}
$}}$$

$$\small{\text{$
\begin{array}{lrcl}
&\mathbf{
\int{ f(x)^{n}\cdot f'(x) \ dx }
} & \mathbf{=} & \mathbf{ \dfrac{ f(x)^{n+1} } {n+1} }
\qquad f(x) = \arccos{(x)}
\qquad f'(x)= -\dfrac{ 1 }{ \sqrt{ (1-x^2) } }\\\\
& \int{ \left[ \arccos{(x)} \right]^2 \cdot
\left( -\dfrac{ 1 }{ \sqrt{ (1-x^2) } }
\right) \ dx }
& = &
\dfrac{ \left[ \arccos{(x)} \right]^{3} } {3} \\\\
& -\int{ \left[ \arccos{(x)} \right]^2 \cdot
\left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } }
\right) \ dx }
& = &
\dfrac{ \left[ \arccos{(x)} \right]^{3} } {3} \\\\
& \int{ \left[ \arccos{(x)} \right]^2 \cdot
\left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } }
\right) \ dx }
& = &
-\dfrac{ \left[ \arccos{(x)} \right]^{3} } {3} \\\\
& \int{ \left[ \arccos{(x)} \right] ^2 \cdot
\left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } }
\right) \ dx }
& = &
-\dfrac{1}{3} \cdot \left[ \arccos{(x)} \right]^{3} \\\\
& \int{ \left[ \arccos{(x)} \right]^2 \cdot
\left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } }
\right) \ dx }
& = &
-\dfrac{1}{3} \cdot \left[\cos^{-1}{(x)}\right]^3 \\\\
\end{array}
$}}$$

$${\mathtt{250}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{10}}}{{\mathtt{2}}}}\right)} = {\frac{{\mathtt{125}}}{{\mathtt{16}}}} = {\mathtt{7.812\: \!5}}$$

@@ End of Day Wrap  Fri 19/6/15   Sydney, Australia Time   9:35pm     ♪ ♫

Hi all,

It was very quiet until a short time ago when some fun hard ones came along.  Anyway, there were some really good questions and answers.  Today our honorable answerers were CPhill, Civonamzuk, Syllogist, Headingnorth, Alan, Radix and Heureka.  Thanks all  

Let me think about this

What is the lowest common multiple of 30 and 95?

The prime factors of 30 = 3*2*5

The prime factors of 95 are  5*19

so the lowest common multiple is  3*2*5*19

$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{19}} = {\mathtt{570}}$$

Syllogist is right :))

$$\\X^6-X^3=2\\\\
Let \;Y=X^3\\\\
Y^2-Y=2\\\\
$Completing the square$\\\\
Y^2-Y+\frac{1}{4}=2+\frac{1}{4}\\\\
(Y-\frac{1}{2})^2=\frac{9}{4}\\\\
Y-\frac{1}{2}=\pm\frac{3}{2}\\\\
Y-\frac{1}{2}=\frac{3}{2}\qquad OR \qquad Y-\frac{1}{2}=-\frac{3}{2}\\\\
Y=2\qquad \qquad OR \qquad \qquad Y=-1\\\\
X^3=2\qquad \qquad OR \qquad \qquad X^3=-1\\\\
X=\sqrt[3]{2}\qquad \qquad OR \qquad \qquad X=-1\\\\$$

These are just the real solutions :)

Let me think about the complex solutions

I think they are

$$\\x=\sqrt[3]{2}\;\;e^{\frac{2\pi i}{3}},
\qquad x=\sqrt[3]{2}\;\;e^{\frac{-2\pi i}{3}}, \qquad
x=e^{\frac{-\pi i}{3}}, \quad and \quad x= e^{\frac{\pi i}{3}}$$

I hope I didn't s***w that one up :/ 

Equation  X^6-X^3=2

$$\small{\text{$
\begin{array}{rcl}
x^6-x^3&=&2 \qquad \mathrm{we ~substitute~and ~set~} z = x^3 \\
\mathrm{then~we~have~} \quad z^2-z &=& 2 \\
z^2-z -2 &=& 0\\
z_{1,2} &=& \dfrac{1\pm \sqrt{1-4\cdot(-2)}}{2} \\
z_{1,2} &=& \dfrac{1\pm \sqrt{9}}{2} \\\\
z_{1,2} &=& \dfrac{1\pm 3 }{2} \\\\
z_1 &=& \frac{1 + 3 }{2} = 2 \\\\
z_2 &=& \frac{1 - 3 }{2} = - 1\\\\\\
x_1 & =& \sqrt[3]{z_1}= \sqrt[3]{2}= 1.25992104989\\\\
\mathbf{x_1} & \mathbf{=}& \mathbf{1.25992104989}\\\\\\
x_2 &=& \sqrt[3]{z_2}= \sqrt[3]{-1}= -1\\\\
\mathbf{x_2} &\mathbf{=}& \mathbf{-1}\\\\
\end{array}
$}}$$

Thanks Syllogist.  You're right of course, I just want to try and explain.  

$$\\\int\;\frac{(cos^{-1}x)^2 }{(1-x^2)^{1/2}}\;dx\\\\
$What I notice is that $\;\; \frac{d}{dx}cos^{-1}x=\frac{-1}{\sqrt{1-x^2}}\\\\
$So I think the answer will be $ some constant* (cos^{-1}x)^3 \;\;+c\\\\
$Now if i differentiated this (without the constant multiple) I get$\\\\
\frac{d}{dx}(cos^{-1})^3x=3(cos^{-1})^2x\times \frac{-1}{\sqrt{1-x^2}}\\\\
$I need to get rid of the - and the 3 so I divide by -3 (that will be the constant multiple$\\\\
\int\;\frac{(cos^{-1}x)^2 }{(1-x^2)^{1/2}}\;dx=\frac{-1}{3}cos^{-1}x+c$$

Just like Syllogist said:))

NOTE: I can never remember that inverse trig differential, I have to work it out from scratch every time.  

(It doesn't take long to work out)

lcm(30, 95) = 570

$${\mathtt{y}} = {{\mathtt{x}}}^{{\mathtt{3}}}$$

$${{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{y}} = {\mathtt{2}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{2}}\\
{\mathtt{y}} = -{\mathtt{1}}\\
\end{array} \right\}$$

$${{\mathtt{x}}}^{{\mathtt{3}}} = {\mathtt{2}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\frac{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\mathtt{0.629\: \!960\: \!524\: \!947\: \!436\: \!6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.091\: \!123\: \!635\: \!972\: \!428\: \!7}}{i}\\
{\mathtt{x}} = {\mathtt{\,-\,}}\left({\mathtt{0.629\: \!960\: \!524\: \!947\: \!436\: \!6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.091\: \!123\: \!635\: \!972\: \!428\: \!7}}{i}\right)\\
{\mathtt{x}} = {\mathtt{1.259\: \!921\: \!049\: \!894\: \!873\: \!2}}\\
\end{array} \right\}$$

$${{\mathtt{x}}}^{{\mathtt{3}}} = -{\mathtt{1}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = -{\mathtt{1}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.866\: \!025\: \!403\: \!785}}{i}\right)\\
{\mathtt{x}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.866\: \!025\: \!403\: \!785}}{i}\\
{\mathtt{x}} = -{\mathtt{1}}\\
\end{array} \right\}$$

So the real solutions are:

• $${\mathtt{x}} = {{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$
• $${\mathtt{x}} = -{\mathtt{1}}$$

3/4 of an hour = 45 minutes

-1/3 cos^(-1)(x)^3 + constant

Thanks Chris

I had to think quite hard about this - maybe I shouldn't admit that   :))

He has a maximum of   $100,000 for fuel

Fuel costs $400 per gallon so he has enough money for

100,000 / 400 = 250 gallons

Now the distance is 600 miles so it will be alright so long has he gets at least 

$$\frac{600\;miles}{250\;gallons}=\frac{2400\;miles}{1000\;gallons}=2.4miles/gallon$$

Just like Chris said   

Sat 20/6/15

If you would like to comment on other site issues please do so on the Lantern Thread.  Thank you.    

Interest Posts: 

FTJ means 'For the juniors' 

9-10=-1

because

9-10 = 9-9-1 = 0-1 = -1

The area for ANY triangle is half the base x the perpendicular height.

There you go - I was not even in the right country.  Thanks Alan.

I like the way that you have done this anon, I just want to have a play with it :)

$$\\3y^{1/3} -2=4\\\\
3y^{1/3} =6\\\\
y^{1/3} =2\\\\
(y^{1/3})^3 =2^3\\\\
y=8$$

Mmm so you didn't really need to use logs but both our answers are good :)

You might find this helpful

The question does not really make much sense to me although anon's interpretation is alright :)

Anon, your working is good :)