All Answers 358087 Answers

They are good you know!

a notice to 7up:After making plans and thinking so much , I was struck by something, you said you also turn into a werewolf , well in that add you might be having wolf-y cells , and if I take them in would I turn into a female werewolf, and I can't affor to scone a wolf , it doesn't really look good wen girls become a wolf so ya know im quitting the idea of " tasting" you! If M44 wants , she can go ahead but this would be too much for me!

well only if humans would have created teeth that could filter blood then if only drink the human blood and leave the wolf-y one but unfortunately it doesn't exist! So I'll sadly have to spare you!

And anyway even if I turned into a wolf , I wouldn't be able to play with anyone, you'd be lost in studying and if I turn into a wolf, I straight away eat tin, so I'll prefer remaking a vampire! 

You can call me when you turn into a wolf, i can click a pic and let it spread and humans would catch you and cut you into pieces and maybe if people's luck you can be sent to Thailand or China to be seve at restaurants at the highest prices! Haha a wolf-curry! Lol 

so overall......beware....

$${\frac{\left({\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}\right)}{{\mathtt{2}}}}$$

= $${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{3}}}}\right)}{{\mathtt{2}}}}$$

= $${\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{3}}}}$$

Yes, definitely.   BUT please put a paper bag over your head so that yout brains will make less mess.

Make sure it is a paper bag because I would not want you to asphixiate BEFORE your brains explode.

That would not be fun at all !

If it was an episode of 'Criminal Minds' or better still 'Bones' then you would want your brains as well as other bits of your anatomy to be jetisoned as far as possible BUT if your mum has to clean it up then that would not be nice :/

7 . 685 517 241 379 310 3

Count the number of places behind the decimal point: 16

Rewrite the decimal portion as that number divided by 1 followed by as many zeros as there are decimal places:  685 517 241 379 310 3 / 1 000 000 000 000 000 0

So, the number becomes  7 + ( 6 855 172 413 793 103 / 10 000 000 000 000 000 ).

To turn it into one common fraction, place the whole number portion at the beginning of the numerator:  76 855 172 413 793 103 / 10 000 000 000 000 000 

@@ End of Day Wrap    Mon 17/8/15     Sydney, Australia Time   12:15am       ♪ ♫   

(Yes it is Tuesday already  )

Hello everyone,

Well our great answers were delivered by TitaniumRome, Alan, Radix, Heureka, Rubenhh, CPhill and jdh3010.    Thanks guys     

$$\\(5\frac{2}{5})^2 \\\\
= \left(\frac{27}{5}\right)^2\\\\
= \frac{27^2}{5^2}\\\\
= \frac{27^2}{5^2}\\\\
=\frac{729}{25}\\\\
=29\frac{4}{25}$$

June had twice as many apples as oranges. Ken had 1/4 as many oranges as apples. The number of oranges Ken had was 2/9 as many as the number of apples June had. Ken had 120 apples and oranges.

(a) express the number of oranges June had as a fraction of the number of apples Ken had. Give you answers in its SIMPLEST FORM.

$$\small{
\begin{array}{lrcl}
& O_1 &=& \text{ oranges June }\\
& A_1 &=& \text{ apples June }\\\\
& O_2 &=& \text{ oranges Ken }\\
& A_2 &=& \text{ apples Ken }\\\\
(1) & A_1 &=& 2\cdot O_1\\
(2) & O_2 &=& \frac14 \cdot A_2 \\\\
(3) & O_2 &=& \frac29\cdot A_1 \\\\
(4) & 120 &=& O_2 + A_2\\\\\\
(1), (2), (3) & \frac14 \cdot A_2 &=& \frac29 \cdot A_1 \quad |\quad A_1 = 2\cdot O_1 \\\\
& \frac14 \cdot A_2 &=& \frac29 \cdot 2 \cdot O_1 \\\\
& \frac14 \cdot A_2 &=& \frac49 \cdot O_1 \\\\
& O_1 &=& \frac14 \cdot \frac94 \cdot A_2 \\\\
& \mathbf{O_1} & \mathbf{=} & \mathbf{ \frac9{16} \cdot A_2 }\\\\
\end{array}
}$$

The number of oranges June had is 9/16 of the number of apples Ken had.

$$\small{
\begin{array}{lrcl}
(A_2?) & 120 &=& O_2 + A_2 \qquad | \qquad O_2 = \frac14 \cdot A_2\\\\
& 120 &=& \frac14 \cdot A_2 + A_2\\\\
&120 &=& \frac54 \cdot A_2\\\\
& A_2 &=& 120 \cdot \frac45 \\\\
& \mathbf{A_2} & \mathbf{=} & \mathbf{96 }\\\\\\
(O_1?) &O_1 &=& \frac9{16}\cdot A_2 \\\\
& O_1 &=& \frac9{16} \cdot 96 \\\\
& \mathbf{O_1} & \mathbf{=} & \mathbf{54 }\\\\\\
(A_1?) &A_1 &=& 2\cdot O_1\\\\
& A_1 &=&2 \cdot 54\\\\
& \mathbf{A_1} & \mathbf{=} & \mathbf{ 108 }\\\\\\
(O_2?) &O_2 &=& \frac14\cdot A_2 \\\\
& O_2 &=& \frac14 \cdot 96\\\\
& \mathbf{O_2} & \mathbf{=} & \mathbf{ 24 }\\\\\\
\end{array}
}$$

$$\small{
\begin{array}{|r|r|r|}
\hline
\text {oranges} & \text{apples} & \\
\hline
54 & 108 & \text{June} \\
\hline
24 & 96 &\text{Ken}\\
\hline
\end{array}
}$$

Now to answer you real question.

Where did I go wrong!

I expect you know that   

$$\\3a+4a=(3+4)a\\\\
and\\\
xa+6a = (x+6)a\\\\
$say a was not $a$ but just some expression.$\\\\
3(x+2)+4(x+2\\=3 \; lots\; of\; (x+2)+4 \; lots\; of\; (x+2)\\= (3+4)\; lots\; of\; (x+2) \\= 7(x+2)\\\\
x(x+2)+6(x+2)\\=x \; lots\; of\; (x+2)+6 \; lots\; of\; (x+2)\\= (x+6)\; lots\; of\; (x+2) \\= (x+6)(x+2)\\\\$$

Now, this is what you wrote:

$$\\x(x-2)+6(x-1)\\\\(x+6)(2x-3)\\$$

You cannot do this because you have x lots of (x-2)  and 6 lots of (x-1)  

The brackets are not the same - they are unlike terms - you cannot add them like you have done :/

Hi Purpe,

Alan is right. Only expressions with rational roots can be done the 'factoring in pairs' way.

Yours is 

$$\\2(x^2+4x-6)=0\\\\
x^2+4x-6=0\\\\$$

Now you have to look for 2 numbers that multiply to 1*-6=-6

Since they multiply to a negative, one is pos and the other is neg

They have to add to +4

So the 'bigger' (absolute value) one is positive

There are no 2 rational numbers that meet this requirement so you cannot do it this way!

There are a number of methods that you can use.

Alan has shown you one of them.

I would just solve it using the quadratic formula :))

I want to try pepper and butter popcorn too MG.  It sounds pretty good :))

Would it be reasonable for me to do it like this?

A population of fruit flies starts with 6 flies. On Day 4, the population has grown to 94 flies. Write an exponential growth function to model the growth of the fly population.

$$\\P=6e^{k(t-1)}\\\\
94=6e^{3k}\\\\
15.\dot6 = e^{3k} \\\\
ln(15.\bar6 )= ln(e^{3k}) \\\\
ln(15.\bar6 )= 3k \\\\
k=ln(15.\bar6 )/3 \\\\
k\approx 0.91718\\\\
so\\\\
P=6e^{0.91718(t-1)}\\\\$$

Is that ok?

Lets see

I will type  5^x=25 into the input box

Output:

$${{\mathtt{5}}}^{{\mathtt{x}}} = {\mathtt{25}} \Rightarrow {\mathtt{x}} = {\frac{{ln}{\left({\mathtt{25}}\right)}}{{ln}{\left({\mathtt{5}}\right)}}} \Rightarrow {\mathtt{x}} = {\mathtt{2}}$$

There you go - that worked   :))

infinity

But this is a concept - not a number :)

I think you should join up and post a pic because I do not know what you are talking about.  Sorry :/

tan(angle) = opposite/adjacent, so opposite = adjacent*tan(angle)

$${\mathtt{opposite}} = {\mathtt{12}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{36}}^\circ\right)} \Rightarrow {\mathtt{opposite}} = {\mathtt{8.718\: \!510\: \!336\: \!06}}$$

.

.

.

Two different answers above!  Here's my version:

.

Amy      :   x $

Tim        :  x + 39,35

( x + 39,35 ) *7 / 9 = 114.45

( x + 39,35) =  $${\frac{{\mathtt{114.45}}{\mathtt{\,\times\,}}{\mathtt{9}}}{{\mathtt{7}}}} = {\frac{{\mathtt{2\,943}}}{{\mathtt{20}}}} = {\mathtt{147.15}}$$

x = $${\mathtt{147.15}}{\mathtt{\,-\,}}{\mathtt{39.35}} = {\frac{{\mathtt{539}}}{{\mathtt{5}}}} = {\mathtt{107.8}}$$

Amy  had  107.80 $  . She spend  3/11   =>

  $${\frac{{\mathtt{107.8}}{\mathtt{\,\times\,}}{\mathtt{11}}}{{\mathtt{3}}}} = {\frac{{\mathtt{5\,929}}}{{\mathtt{15}}}} = {\mathtt{395.266\: \!666\: \!666\: \!666\: \!666\: \!7}}$$      =>   395.27 $

Amy spend on her clothes 395.27 $ .

Gruß radix !

Try this :  click twice upon    =      !!

4 pi = w - 6pi

w = 4 pi + 6 pi            =>  w  = 10 pi  = $${\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{\pi}} = {\mathtt{31.415\: \!926\: \!535\: \!897\: \!932\: \!4}}$$

$${\mathtt{V}} = {\frac{{\mathtt{1.2}}{\mathtt{\,\times\,}}{\mathtt{0.4}}{\mathtt{\,\times\,}}{\mathtt{0.2}}{\mathtt{\,\times\,}}{\mathtt{4}}}{{\mathtt{5}}}} \Rightarrow {\mathtt{V}} = {\mathtt{0.076\: \!8}}$$     m³  are in the tank  = 76,8 litre

$${\frac{{\mathtt{76.8}}}{{\mathtt{1.3}}}} = {\frac{{\mathtt{768}}}{{\mathtt{13}}}} = {\mathtt{59.076\: \!923\: \!076\: \!923\: \!076\: \!9}}$$     You can fill  59   1.3 litre-pails .

$${\mathtt{76.8}}{\mathtt{\,-\,}}{\mathtt{59}}{\mathtt{\,\times\,}}{\mathtt{1.3}} = {\frac{{\mathtt{1}}}{{\mathtt{10}}}} = {\mathtt{0.1}}$$        0.1 litre = 100 ml = 100 cm³  is left in the tank.

Gruß radix !

Lynn has  x $  and spents   1/4 * x

Stacy spents   60 $          ;    together  450 $

$${\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{60}} = {\mathtt{450}}$$       =>      $${\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{390}}$$      =>  x = 1560  $

Lynn had 1560 $

$${\frac{{\mathtt{1\,560}}}{{\mathtt{4}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{60}} = {\mathtt{450}}$$

Gruß radix !

$${\mathtt{48}}\% = {\frac{{\mathtt{48}}}{{\mathtt{100}}}} = {\frac{{\mathtt{24}}}{{\mathtt{50}}}} = {\frac{{\mathtt{12}}}{{\mathtt{25}}}}$$