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Excuse me young sir or young lady

camelot rules state you must register an account to post pictures now

Hold on Melody

if you want to reach your true potential, you cant cuta birthday cake with a knife

you cut it with:

an Ancient Japanese Samurai Katana, or A.J.S.K.

must use katana, no knife

do or do not, there is no try btw

you either can cut the cake with a katana, or not, no maybe

$${\frac{{\mathtt{4}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{3.14}}$$

We all know that 3.14 is equal to 22/7

so, we dont have to after all try to make 4/3 a decimal, cuz you can make 3.14 a fraction

$$\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{22}}}{{\mathtt{7}}}}\right)$$

no cross cancelling can be done sadly

so 4*22=88

3*7=21

now, its 88/21

if you however want to try to make me not be lazy and make me convert it to a mixed number fraction:

21 would go in 88 4 times

21*4=84 

4 additional pieces of 88 are included

overall answer:

4 4/21 or 88/21

.....so?... what do u want us to do?

To rosala-yup, we both got a matter with the british.

but 75 years before today, we unite to attack the third reich, or the n**i empire, the italians, and more importantly, the japanese.

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{y}}}}}{{solve}}{\left({\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{y}}{\mathtt{\,-\,}}{\mathtt{5}}\right)}{{\mathtt{6}}}}={\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{y}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}$$

$${\mathtt{y}} = -{\mathtt{15}}$$

$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{t}} = {\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{t}}$$

So 

4t-2=-7t+4

4t=-7t+6

11t=6

t=$${\frac{{\mathtt{6}}}{{\mathtt{11}}}}$$

1/2  x  ( 4.5  +  7.5  )  x  5.5  =   0.5  x  ( 12 )  x 5.5  =  6  x  5.5  =  33

$78,000 = 41607766.78 Cost Rica Colones according to the online currency converter website I just visited

I'm sure you could! I have done it; it's much easier than you think.

Very nice !!

$${{\mathtt{10}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{x}}}^{{\mathtt{21}}} = {\mathtt{0}}$$

$${{\mathtt{x}}}^{{\mathtt{21}}} = -{\mathtt{100}}$$

x = $${\sqrt[{{\mathtt{{\mathtt{21}}}}}]{-{\mathtt{100}}}} = -{\mathtt{1.245\: \!197\: \!084\: \!735\: \!032\: \!8}}$$

Gruß radix !

$${\frac{{\mathtt{11}}}{{\mathtt{15}}}}$$  

The simplest form   is   $${\frac{{\mathtt{11}}}{{\mathtt{15}}}}$$  !

$${\frac{{\mathtt{11}}}{{\mathtt{15}}}} = {\mathtt{0.733\: \!333\: \!333\: \!333\: \!333\: \!3}}$$  =  $$0.7\overline{3}$$

Gruß radix !

Haha! 

Dumb mistake meant 36. >.<

If this was made of mm^3 blocks (some would need to be cut up) How many cubes would be needed to cover the front face?     ???     Well there are 10 mm in a cm so you have to change the dimensions to mm and then work out the area of the front face.   I will leave you to do this.  :)

I knew that there was 10mm in a cm but the power bit puzzles me.

would it 36cm^2*10=

360cm^2

How many rows of block triangles would you need to make the whole shape?    15.7

if the blocks were 1cm^3 then you would be correct but they are 1mm^3  so you better answer again :)

Would it be the same by multiplying by 10.   15.7cm*10=

570mm^3

What is the volume of this shape? (Show working)

Formula = (L*H)/2*W

(8.2*6.4)/2=26.24cm^2

26.24*15.7= 411.968cm^2             No:  Volume is not measured in cm^2

411.968^3

Sorry Melody. But I'm really stuck. I haven't done this yet only one lesson in class about 2 months ago.

I haven't seen ^3 yet...

The only reason i know what I'm doing so far, is because of the video you gave me.

Me too! Only if admin made a Hindi forum  where we could speak Hindi with English alphabets and make me its mod then I can say id not miss a day..........but he hasn't excepted my deal........maybe he doesn't like being too popular ! Lol

Probably Rosala, you always know how to bring a party! :)

US kids on holiday, i guess that makes sence, as the admin of course you'd like it quiet

xD

Less problems to deal with, more time to chill. 

Not to mention more sleep!

I kind of like it too, but I'd like some questions to answer.

Happy Birthday !!!

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

$$\\\frac{((4x)^3-3^3)}{4x-3}\\\\
=\frac{((4x)-3)((4x)^2+3*(4x)+3^2}{4x-3}\\\\
=\frac{(4x-3)(16x^2+12x+9)}{4x-3}\\\\
=16x^2+12x+9\qquad $Where x cannot = 3/4$$$

Yes that is fine Fiora  

@@ End of Day Wrap  Sun 5/7/15   Sydney, Australia Time   11:45pm  ♪ ♫

Hi everyone,

Today our wonderful answerers were Alan, MathsGod1 Eloise1st, CiciPurple, CPhill, Radix, Badinage and Fiora.  Thank you