All Answers 358087 Answers

$$\\(-6)^{-1}=\frac{1}{(-6)^1}=\frac{1}{-6}=-\frac{1}{6}\\\\\\
-\left(\frac{1}{2}\right)^{-2}
=-\left(\frac{2}{1}\right)^{+2}
=-\frac{2^2}{1^2}
=-4$$

So they are both negative.  Does that help or do you want to quiz me about it?

Added later:  Oh i will do the heading one too

$$\left(\frac{-1}{2}\right)^{-2}
=\left(\frac{2}{-1}\right)^{+2}
=\frac{2^2}{(-1)^2}
=\frac{2*2}{-1*-1}
=\frac{4}{1}
=4$$

Remember that :   a^-n = 1/aⁿ

So (-6)^-1  =  1/-6 = -(1/6)

And (-1/2)^-2  =  1 / (-1/2)²  =   1/ [ (-1/2) * (-1/2)] = 1 / (1/4)  = 4

So isn't there enough blocks to make a bigger cube  

I think that there are  :/

The cube is exactly the same size as the original - constructed of 488 cubes  with 512 cubes removed......

Huh?

How big is your new cube and how many blocks are left over?

Here's my take on this one:

If we reconstructed the cube, we would have 1000 smaller cubes. Removing a "central" cube of 8 x 8 x 8 would leave us with the same size cube as before, but with 512 of the cubes removed. This would leave us with 488 cubes, which would be composed of two layers of 100 cubes on the "top" and "bottom" and 8 layers in between. And each of these layers will have (100- 64) cubes.

So   100 + 100 + 8(100- 64)   leaves 488 cubes. And the 512 we removed would total to 1000.

Well there will be 1000 cm cubes

Let the side length of the new cube be x cm

The front and back will have    $$x^2$$     cubes each

The top and bottom will have a further   x(x-2)   cubes each

and the sides will have a further   (x-2)(x-2) cubes each

So

$$\\2[x^2+x(x-2)+(x-2)^2]\le1000\\\\
2[x^2+x^2-2x+x^2-4x+4]\le1000\\\\
x^2+x^2-2x+x^2-4x+4\le500\\\\
3x^2-6x+4\le500\\\\
3x^2-6x-496\le0\\\\$$

$${\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{496}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{1\,497}}}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{3}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{1\,497}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{{\mathtt{3}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{11.897\: \!028\: \!081\: \!435\: \!402\: \!3}}\\
{\mathtt{x}} = {\mathtt{13.897\: \!028\: \!081\: \!435\: \!402\: \!3}}\\
\end{array} \right\}$$

This is a concave up parabola.

It will be < 0 between the two roots.

So  the highest integer x value will be 13.

Number of cubes used will be

$$\\f(x)=2[x^2+x(x-2)+(x-2)^2]\\\\
f(13)=2[13^2+13(13-2)+(13-2)^2]\\\\
f(13)=2[169+143+121]\\\\
f(13)=2[169+143+121]=866\\\\
$Left over blocks $=1000-866 =\; 134 \;blocks$$

I think that method is correct     

WOW!!!  ......WolframAlpha's interpretation of that is odd!!!

That is really interesting radix.

Sometimes Wolfram|Alpha does misinterprete the meaning of words in a question and I believe that this is one of those times.    

SORRY !

Gruß radix !

Hello anonymous,

here is a calculator for  quadratic equation:

Isn't it awful trying to work out the nuances of another language :))

I admire your efforts radix :)

It would be completely beyond me   

Rats....Melody beat me to it !!!!

Radix....I think they wanted to divide 18004 by 8

18004/8 = 2250.5

Hi Radix,

I think it is meant to be the other way around

8 into 18004   means      $$18004 \div 8$$

$${\frac{{\mathtt{18\,004}}}{{\mathtt{8}}}} = {\frac{{\mathtt{4\,501}}}{{\mathtt{2}}}} = {\mathtt{2\,250.5}}$$

I had an easier answer

cos(6x - x)  = cos6xcosx + sin6xsinx 

so

sin6xsinx = cos5x  - cos6xcosx

That is a difference   

8  into  18004

$${\frac{{\mathtt{8}}}{{\mathtt{18\,004}}}} = {\frac{{\mathtt{2}}}{{\mathtt{4\,501}}}}$$

$${\frac{{\mathtt{8}}}{{\mathtt{18\,004}}}} = {\frac{{\mathtt{2}}}{{\mathtt{4\,501}}}} = {\mathtt{0.000\: \!444\: \!345\: \!700\: \!955\: \!3}}$$

3x - 9 = 0      add 9 to both sides

3x = 9       divide both sides by 3

x = 3

@@ End of Day Wrap   Wed 26/8/15     Sydney, Australia Time   7:01pm       ♪ ♫   

Good evening everyone,

There are a llot of unanswered questions on the forum tonight, most of them are fairly easy so if anywone would like to answer a few that would be good.

We did, however, have some great answers from asinus, geno3141, CPhill, Douglas, Dragonlance and BlackskirtHollywood.  Thank you.   

cos(6x - x)  = cos6xcosx + sin6xsinx   (1)

cos(6x + x) = cos6xcosx - sin6xsinx     (2)

Subtract   (2) from (1)   and we have  :

cos(6x - x) - cos (6x + x) = sin6xsinx + sin6xsinx   .....  simplify

cos(5x) - cos(7x)  = 2sin6xsinx

2sin6xsinx = cos(5x) - cos(7x)

sin6xsinx  = (1/2)[cos(5x) -  cos(7x) ]

 [Thanks to Melody for spotting an earlier couple of typos  !!! ]

a flow of lava is traveling at the rate of 3m per day. how far will the lava have traveled at the end of 30 days

3m the fiurst day

3*2m at the end of the second day

3*3m at the end of the 3rd day

3*30m  at the end of the 30th day    

If a*b=0 then either a or b or both must equal zeros    SO

if    -2x * (3x-4)  = 0    then   either    

-2x=0       or    3x-4 = 0

x=0           or       3x=4

x=0           or         x=4/3

Input like this

sqrt(3/4)

this is the output 

$${\sqrt{{\frac{{\mathtt{3}}}{{\mathtt{4}}}}}} = {\mathtt{0.866\: \!025\: \!403\: \!784\: \!438\: \!6}}$$

1060

=10*106 = 10*53*2 = 2*5*53*2 = 2*2*5*53

$$\sqrt{1060}=\sqrt{2*2*5*53}=2\sqrt{5*53}=2\sqrt{265}$$