All Answers 358514 Answers

it is  [2nd]

Solve the epuation 4X2-24=8x for x by completing the spuare

$$\small{\text{
$
\begin{array}{rcrcr}
4x^2-24 &=& 8x & \quad | \quad & :4 \\
x^2-6 &=& 2x & \quad | \quad & -2x\\
x^2 -2x - 6 &=& 0 & \quad | \quad & +6\\
x^2 -2x &=& 6 & \\
\underbrace{(x-1)^2}_{=x^2-2x+1} -1 &=& 6 & \quad | \quad & +1\\\\
(x-1)^2 &=& 7 & \quad | \quad & \pm\sqrt{} \\
x-1 &=& \pm\sqrt{ 7 } & \quad | \quad & +1\\
x &=& 1\pm\sqrt{ 7 } \\ \\
x_1 &=& 1 + \sqrt{ 7 } \\
x_2 &=& 1 - \sqrt{ 7 } \\
\end{array}
$
}}$$

Maybe by the wingspan ?

Thanks Alan,

I think I am suffering from an embarrasing mental block.

C(10,0)     A(0,8)

equation AC:    

$$y=-0.8x+8$$

or

$$x= \frac{y-8}{-0.8}$$

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

For the SA of the curved surface why can't I just say

$$\\SA=\int_0^8\;\;2\pi x\;dy\\\\
SA=2\pi\int_0^8\;\;\frac{y-8}{-0.8}\;dy\\\\
SA=\frac{2\pi}{-0.8}\int_0^8\;\;y-8\;dy\\\\
SA=\frac{2\pi}{-0.8}\left[\;\;\frac{y^2}{2}-8y\right]_0^8\\\\\
SA=\frac{2\pi}{-0.8}\left[\frac{8^2}{2}-64\right]\\\\
SA=\frac{2\pi}{-0.8}\left[\frac{-62}{2}\right]\\\\
SA=80\pi\\\\$$

I know that this is wrong but I don't know why.

@@ End of Day Wrap   Mon 9/3/15   Sydney, Australia    Time 11:15 pm    ♪ ♫

Hi all,

Our great answers or the past 2 days were supplied by Rosala, Alan, Nauseated, Tetration, CPhill, Heureka, sasini, Geno3141, Hermzz, Bertie, InnovativeProgrammer and Silver27.   Thank you.   

Our little forum is continuing to mature.  I think that it is still a friendly place for the children although the number of purely social posts does seem to have reduced.      However, we are increasingly attracting higher level questions. I thank our regular high level answerers greatly for this.  Alan, Heureka, Bertie, CPhill, Geno3141 and occasionally DavidQD , and even Nauseated when he is behaving himself.   I cannot overstate how much I appreciate having you all here. 

I do not want Camelot to become just another high level forum, there are plenty of those, but I do want to cater for a very wide range of ability levels and I am very pleased with the present path that the forum is taking. 

Interest Posts:

A model of the eiffel tower that was purchased in a gift shop is 29.55 inches tall. the actual height of the eiffel tower is 985 feet, or 11,820 inches. what scale factor was used to make the model ?

$$\\\small{\text{
$
\dfrac{11820\ \mathrm{inches} } {29.55\ \mathrm{inches}} = 400
$
}}\\\\
\small{\text{
$
1:400
$
}}$$

Tues 10/3/15

What is the nth term of this sequence?               3, 9, 27, 81, 243

$$\\a_{n+1} =3\cdot a_n \\
a_n = 3\cdot a_{n-1}\\$$

(1+3+5+⋯+3983)/1992

$$\small{\text{
$
\dfrac{ 1+3+5+\cdots +3983} {1992} = \ ?
$
}}\\ \\
\small{\text{
The arithmetic Series $ 1+3+5+\cdots +3983 $ has $a_1 = 1$ and $a_n=3983$ and $d = 2$
}}\\
\small{\text{
The formula is $a_n = a_1+(n-1)\cdot d$ or
$ 3983 = 1 + (n-1)\cdot 2 $
}}\\
\small{\text{
So $n = 1992 $
}}\\
\small{\text{
The sum $ = 1+3+5+\cdots +3983 = n \left( \dfrac{a_1+a_n}{2} \right) = 1992\cdot \left( \dfrac{1+3983}{2} \right)
= 1992\cdot 1992
$
}}\\\\
\small{\text{
$
\dfrac{ 1+3+5+\cdots +3983} {1992} = \dfrac{1992\cdot 1992} {1992} = 1992$
}}\\ \\$$

Thanks Bertie,

Yes, so it is!

I am still slightly more likely to remember the derivation than i am to remember the formula.

I'm funny like that, I hardy remember any formulas.

[ I'll probably have to see it many more times before I really remember either.  :)  ]

I am very grateful to all three of you for your help, and I am very pleased that i actually do understand. 

120 ml per hour, how many hours for 1000 ml

$$\frac{1hour}{120ml}*\frac{1000ml}{1}=\frac{1000}{120}\;hours$$

$${\frac{{\mathtt{1\,000}}}{{\mathtt{120}}}} = {\frac{{\mathtt{25}}}{{\mathtt{3}}}} = {\mathtt{8.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

8 hours and 20 minutes

Do you mean

$$FV=PV*e^{kt}$$

Could you please give us a question to show you on?

What is 9 to the power of 999 ?

change of basis: 9 to 10

$$b_1^{ e_1 } = b_2^{ e_2 } \quad | \quad \ln() \\
e_1 \cdot \ln{(b_1)} = e_2 \cdot \ln{(b_2)} \\\\
e_2 = e_1 \cdot \dfrac{ \ln{(b_1)} } { \ln{(b_2)} } \\
\boxed{
b_1^{ e_1 } = b_2^{ e_1
\left(
\cdot \dfrac{ \ln{(b_1)} } { \ln{(b_2)} }
\right)
}}\\\\
\small{\text{
$
b_1 = 9 \qquad e_ 1 = 999 \qquad b_2 = 10
$
}} \\\\
\small{\text{
$
9^{999} = 10^{999
\left(
\cdot \dfrac{ \ln{(9)} } { \ln{(10)} }
\right)
}
$
}} \\\\
\small{\text{
$ 9^{999} = 10^{999 \cdot( 0.95424250944 ) } $ }} \\
\small{\text{
$ 9^{999} = 10^{953.288266930} $
}} \\
\small{\text{
$ 9^{999} = 10^{0.288266930}\cdot 10^{953} $
}} \\
\small{\text{
$ 9^{999} = 1.94207916858
\cdot 10^{953} $
}}$$

Well that's the question I was struggling with but then my teacher told me that in the question it would be given that the sum has to be compounded annually or half yearly or by using simple interest ! It would all be given like how much will $45 be after two years at the rate of 5% compounded annually? So like that it would be given and you just have to use the formula with a sprinkle of your brain!

Hi MathsGod,

No,

9! means 9*8*7*6*5*4*3*2*1

Say you had the letters  A,B,C,D,E,F,G,H,I

There are 9 letters.

So there are 9 letters to chose from for the first letter.

then there are 8 letters to choose from next

then 7 and so on untill there is just 1 then 0 letters left

so the number of ways that they can be ordered is

9*8*7.....*1     as I wrote above.

9!/(2!3!) =   $$\frac{9!}{2!3!}=\frac{1*2*3*4*5*6*7*8*9}{1*2\;\;*\;\;1*2*3}$$                      $${\frac{{\mathtt{9}}{!}}{\left({\mathtt{2}}{!}{\mathtt{\,\times\,}}{\mathtt{3}}{!}\right)}} = {\mathtt{30\,240}}$$

that page of questions that you found are far to hard for you, that is why you cannot understand my explanations.

If you would like to extend yourself that is excellent but you need to find something just a little different from what you are  doing in class.  Not something that is years above you.  :)

How about if I find something for you.  

Maybe you will enjoy this video clip.

Actually Melody, my first method is exactly equivalent to Alan's method.

The equation ds/dt = (D/s)(dD/dt) can be written as (s/D)(ds/dt) = dD/dt, which, in Alan's notation is

v.cos(theta)=95, (with or without the negative sign)

Thanks Bertie.  

That really did help.  I especially like your first method but I can see that both would work fine.  :)

Alan, if I can remember yours I can see it is easier but i am not good at remembering things that I do not understand.  

Thanks Chris,

I am only just getting the hang of these.    

It is a tricky way of finding high powers that the calc cannot find on its own. 

If you can do a calc that the calc cannot do on its own does that kind of make you = 42?

I interpreted the question as

4 (tan 3 + sin√5)

If we are dealing with rads, the answer becomes .....  2.576810353892

{Who knows ???}

This question illustrates the need for people to put brackets in their questions so that the intended question is actually answered!

Hi Alan,

You have not answered the same question as I did, and neither of us answered the question that was actually asked.  I think that this is quite funny.

I'm going to try and answer the original question.

acot(sec(acsc -2sqrt 3/2))

I think technically this should be interpreted as;

$$\\acot(sec(\frac{acsc (-2) *\sqrt 3)}{2}))\\\\
=acot(sec(\frac{asin (\frac{1}{-2}) *\sqrt 3)}{2}))\\\\
=acot(sec(\frac{\frac{-\pi}{6} *\sqrt 3)}{2}))\\\\
=acot(sec(\frac{-\sqrt 3\pi}{12} ))\\\\$$

$${sec}{\left({\mathtt{\,-\,}}{\frac{\left({\frac{{\mathtt{180}}}{{\mathtt{\pi}}}}\right){\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{12}}}}\right)} = {\mathtt{1.112\: \!419\: \!829\: \!676\: \!128\: \!1}}$$

$${acot}{\left({sec}{\left({\mathtt{\,-\,}}{\frac{\left({\frac{{\mathtt{180}}}{{\mathtt{\pi}}}}\right){\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{12}}}}\right)}\right)} = {\mathtt{41.953\: \!677\: \!923\: \!85^{\circ}}}$$

Check with Wolfram|Alpha

We can evaluate   999log(9)  = 953.288

This equals about

 (10⁹⁵³)*10^.288 =

10⁹⁵³ * 1.942 =

1.942 x 10⁹⁵³

All in all......a really big number.....1942 followed by 950 zeros.......!!! {more or less}

$${\mathtt{4}}{\mathtt{\,\times\,}}\left({\mathtt{tan3}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{sin5}}}^{{\mathtt{0.5}}}\right) = {\mathtt{1.390\: \!517\: \!175\: \!974\: \!257\: \!7}}$$

ok!So just enter 272/165 in the on-site calculator on the homepage and hit enter and youll get your answer , and add a cm to it too!