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It is 408.41 unit^2    Correct to 2 decimal place

Your teacher probably rounded pi off more than you did and that is why her/his answer is a little less acurate :)

9983576256787896*9999999999999999999999999

=99835762567878959999999990016423743212104

Nice one, Melody......!!!    [ I always like to try to figure these out  ]

if i have a cylinder with a radius of 5, and a height of 8, what is its surface area

if i have a cylinder with a radius of 5, and a height of 8, what is its surface area. p.s. i have already worked out the answer, but my teacher and i are having a debate on what the answer was. my answer was 408.41, and hers was 408.2. Which one is right ?

$$\small{\text{$
\begin{array}{rcl}
\mathrm{radius~}r=5\\
\mathrm{height~}h=8 \\\\
\hline
\\
\mathrm{surface~area~} O&=&\pi r^2+ \pi r^2 + 2\pi r*h \\
O&=&2\pi r ( r + h ) \\
O&=&2\pi 5 ( 5 + 8 ) \\
O&=&2\pi 5 \cdot 13 \\
O&=&2\cdot 5 \cdot 13 \pi \\
O&=&130 \pi \\
\end{array}
$}}$$

If you set $$\small{\text{$\pi=3.14$}}$$ you get $$\small{\text{$
O=130\cdot 3.14 = 408.2
$}}$$

But if you set $$\small{\text{$\pi=3.14159265359$}}$$  you get the better answer $$\small{\text{$
O=130\cdot 3.14159265359 = 408.407044967
$}}$$

Thanks Alan, that is really neat.     

I have been thinking about that problems since it was first posted. 

Cylinder:   r = 5     ;   h = 8

O=2*pi*r*(r+h)  = $${\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}\left({\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right) = {\mathtt{408.407\: \!044\: \!966\: \!673\: \!121}}$$

408.41  is right !

2x²-11x+12=0

(x-4)*(2x-3)=0    =>     x  =4      ;   2x-3=0    =>   x = 3/2= 1,5

Answer:  x1 = 4  and  x2 = 1.5

$${{\mathtt{4}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{16}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)} = {\frac{{{\mathtt{2}}}^{{\mathtt{3}}}}{{{\mathtt{2}}}^{{\mathtt{3}}}}}$$ =  1

$${{\mathtt{4}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)} = {{\left({{\mathtt{2}}}^{{\mathtt{2}}}\right)}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}$$ =  $${{\mathtt{2}}}^{{\mathtt{3}}}$$

$${{\mathtt{16}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)} = {\frac{{\mathtt{1}}}{\left({\left({{\mathtt{2}}}^{{\mathtt{4}}}\right)}^{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}\right)}}$$      = $${\frac{{\mathtt{1}}}{{{\mathtt{2}}}^{{\mathtt{3}}}}}$$

$${\frac{{{\mathtt{2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{1}}}{{{\mathtt{2}}}^{{\mathtt{3}}}}} = {\mathtt{1}}$$

That makes good sense, Thanks Bertie  

Here's a pictorial representation.  The dotted lines enclose the overlap of all four criteria and cover 10 people (from 25 to 35 on the diagram).

.

i had already worked out the answer, but my teacher and i are having a debate on what the answer was. my answer was 408.41, and hers was 408.2. Which one is right?

1,125,899,906,842,600 ;)

Hi Melody

As Alan has said $$r\angle \theta$$ is a shorthand notation used mainly by electrical engineers.

It's actually shorthand for the complex number $$r(\cos \theta + \imath \sin \theta).$$

Written out in full, we would have

$$r_{1}\angle \theta_{1}\times r_{2}\angle\theta_{2}\\=r_{1}r_{2}(\cos\theta_{1}+\imath\sin\theta_{1})(\cos\theta_{2}+\imath\sin\theta_{2})\\=r_{1}r_{2}(\cos\theta_{1}\cos\theta_{2}-\sin\theta_{1}\sin\theta_{2}+\imath(\sin\theta_{1}\cos\theta_{2}+\cos\theta_{1}\sin\theta_{2}))\\=r_{1}r_{2}(\cos(\theta_{1}+\theta_{2})+\imath\sin(\theta_{1}+\theta_{2}))\\=r_{1}r_{2}\angle(\theta_{1}+\theta_{2}).$$

@@ End of Day Wrap  Mon 15/6/15   Sydney, Australia Time   7:05am   ♪ ♫

Hello everyone,

It has been a very quiet two days.  I am told that is because the U.S students are mostly on a long summer break. 

Anyway, there have still been lots of questions and our wondeful answerers were: CPhill, Civonamzuk, Alan Radix, Pyramid, Fiora, MathsGod1 and BrittanyJ.  Thank you all  

Maybe from this you can work it out yourself :/

The circumference of the circle is the length of the rectangle.

I did say ' that you could argue that '... .

If the 'function' mentioned in the question were tan, then the discontinuity at pi/2 would make it unreasonable to talk about the limit without further information as to whether we were approaching from below or above.

However, the 'function' mentioned in the question is the inverse tangent arctan. This is multivalued and if you take usual view of working within the principal range, there is no discontinuity. Asking what happens to the inverse tangent as its argument tends to infinity is quite reasonable.

Tues 16/6/15

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' 

There are an infinite number of irrational numbers between 3 and 5

just the same as there is an infintie number of rational numbers between 3 and 5

pi is one of the irrational ones BUT remember that 3.14 IS NOT PI.    It is only an approximation for pi.

3.14 is a rational number.

If you will press the   tan -button a little bit longer, you get  tan^-1 .

Try it !

Thank you Heureka.   

Another Heureka special.            

There are 20 terms in an arithmetic sequence. The sum of the terms with odd numbers is equal to 220. The sum of the terms with the even numbers is equal to 250. Find the two middle terms of the sequence

Mmm, the 2 middle terms are the 10th and the 11th.

I am going to split ths into 2 sequences.

odd terms

a,   a+2d,   a+4d,      ....    a+(n-1)*2d  ........  a+9*2d  

 sum=10/2(a+a+18d)=5(2a+18d)

$$\\5(2a+18d)=220\\
10(a+9d)=220\\
a+9d=22\qquad (1)\\$$

even terms

a+d,   a+3d,    a+5d,   .....    (a+d)+(n-1)*2d   ......   (a+d)+9*2d  

 sum=10/2(a+d+a+d+18d=5(2a+20d)

$$\\5(2a+20d)=250\\
10(a+10d)=250\\
a+10d=25\qquad (2)\\$$

(2) - (1)

d=3,    substitute     a= -5

----------------

check   the fist 20 terms should add up to 470

$$\\s_{n}=(n/2)[2a+(n-1)d \\\\
s_{20}= 10[-10+19*3] = 10[-10+57]=470 \qquad correct$$

-------------------

$$\\t_{10}=-5+9*3 = -5+27 = 22\\\\
t_{11}=22+3=25$$

The 2 middle terms are 22 and 25

12 719 726 = 1 271 972 600 % = 1 271 972 600/100 = 12 719 726

how do you solve 2sin(theta) = cos(theta/3)  ?

$$\small{\text{$
\boxed{
~~ 2\sin{(\theta)} = \cos \left( \frac{ \theta}{3}\right) \quad \mathrm{~we~set~} \theta = 3\alpha \quad
2\sin(3\alpha ) = \cos{( \alpha )}
~~}
$}}$$

$$\small{\text{$
\begin{array}{rcl|rrcl|l}
&&&&&&& \mathrm{Formula:}\\
2\sin(3\alpha ) &=& \cos{( \alpha )}
&(1)&\sin{(3\alpha)}&=&
\sin{(\alpha+2\alpha)}=
\sin{(\alpha)}\cos{(2\alpha)} +\cos{(\alpha)}\sin{(2\alpha)}
&\cos{2\alpha} = 1-2\sin^2{(\alpha)} \\
&&&&\sin{(3\alpha)}&=&
\sin{(\alpha)}(1-2\sin^2{(\alpha)}) +2\sin{(\alpha)}\cos^2{(\alpha)}
&\sin{2\alpha} = 2\sin{(\alpha)}\cos{(\alpha)} \\
&&&&\sin{(3\alpha)}&=&
\sin{(\alpha)}(1-2\sin^2{(\alpha)}) +2\sin{(\alpha)}( 1-\sin^2{(\alpha)} )
&\cos^2{\alpha} = 1-\sin^2{(\alpha)} \\
&&&&& \cdots && \\
&&&&\sin{(3\alpha)}&=&
\sin{(\alpha)} \left[(3-4\sin^2{(\alpha)} \right]
&\\
2\sin{(\alpha)} \left[(3-4\sin^2{(\alpha)} \right] &=& \cos{( \alpha )}
&&&&&\\
2\left[(3-4\sin^2{(\alpha)} \right] &=& \cot{( \alpha )}
&&&&&\\
& \cdots &
&&&&&\\
8\sin^2{(\alpha)} &=& 6-\cot{( \alpha )}
&&&&& \frac{1}{ \sin^2{(\alpha)} } = 1+\cot^2{(\alpha)}\\
8 &=& \left[ (6-\cot{( \alpha )} \right] \left[1+\cot^2{(\alpha)} \right]
&&&&&\\
& \cdots &&&&&&\\
\cot^3{(\alpha)}-6\cot^2{(\alpha)}+\cot{(\alpha)}+2 &=& 0 &&&&&\\
& \alpha=\frac{\theta}{3} &&&&&&\\
\cot^3{(\frac{\theta}{3})}-6\cot^2{(\frac{\theta}{3})}+\cot{(\frac{\theta}{3})}+2 &=& 0 &&&&&\\
\end{array}
$}}$$

$$\small{\text{$
\boxed{~~
\cot^3{\left(\frac{\theta}{3}\right)}
-6\cot^2{\left(\frac{\theta}{3}\right)}+\cot{\left(\frac{\theta}{3}\right)}+2 = 0
~~}
$}}\\\\\\
\small{\text{$
\mathrm{substitute:~~}
\boxed{~~u = \cot{ \left(\frac{\theta}{3} \right) }=\frac{1}{\tan{ \left(\frac{\theta}{3} \right) }}
\qquad \theta = 3\arctan{\left(~\frac{1}{u}~\right)} \pm 3\pi \cdot k \quad k=0,1,2,3\cdots ~~ }
$}}\\ \\
\small{\text{$
\boxed{~~u^3 - 6u^2 + u +2 = 0 ~~}
$}}\\\\
\small{\text{$
\begin{array}{rclcc}
u_1 &=& 5.766 435 484
& \theta_1=3\arctan{(\frac{1}{u_1})}
& \theta_1=0.515 128 919 \pm3\pi\cdot k\\\\
u_2 &=& -0.483 611 621
& \theta_2=3\arctan{(\frac{1}{u_2})}
& \theta_2=-3.361 035 503 \pm3\pi\cdot k\\\\
u_3 &=& 0.717 176 136
& \theta_3=3\arctan{(\frac{1}{u_3})}
& \theta_3=2.845 906 583 \pm3\pi\cdot k\\\\
\end{array}
$}}$$