Melody

Not sure what the question is asking bu you could take away 124, 117, 98  or 61 little cubes.

1^3=1                125-1=124

2^3=8                 125-8=117

3^3=27               125-27=98

4^3=64                125-64 =61

5^3=125

Rainbow squirrel you are spot on.  Great answer.

The easiest way to look at this is that you pick up a ball and put it in any box.

probabiliy will be 1 since there will be a ball in a box and it does not matter which box.

How you have a second ball and a choice of 2 places on where to put it.

Prob that it goes into the empty box is 1/2

1  *  1/2   =  1/2

-12, -11, -10 -9, -8 ..... 8

adding them

-12+-11+-10+-9  and the rest cancel out

-12-11-10-9 = ?? 

Put the five identical balls in a line use a line to sepearte them into 2 lines.

Assuming that all the balls can go into one box is an option then the problems is very easy

6C1=1

(x-(1+sqrt2))(x-(1-sqr2)(x+a)

\((x-(1+\sqrt2))(x-(1-\sqrt2)(x+a)=0\\ (x-1-\sqrt2)(x-1+\sqrt2)(x+a)=0\\ ((x-1)^2-2))(x+a)=0\\ (x^2-2x-1)(x+a)=0\\ \)

Yoou can check and finish it.

Note:

You are NOT finding the value at x=2, you are finding the value as x approaches 2 (from below)

So it is not   0/0

and 

infinity is not a number, it is a concept.

Does this help?

\(\displaystyle \lim_{x\rightarrow2^-}\;\frac{x}{x^2-4}\)

If x is between 0 and 2 (not including 2)  then this will be  pos divided by neg  which is negative

that is why this limit HAS to be negative

\(\displaystyle \lim_{x\rightarrow2^-}\;\frac{x}{x^2-4}\\~\\ =​​\displaystyle \lim_{x\rightarrow2^-}\;\frac{x/x^2}{(x^2-4)/x^2}\\~\\ =​​\displaystyle \lim_{x\rightarrow2^-}\;\frac{1/x}{1-4/x^2}\\~\\ ==> \frac{approaching \;0.5}{appraoching\; 0\; from\; negative \;side}\\~\\ =-\infty\)

\(f(x) = \left\{ \begin{array}{cl} 3-x & \text{ if } x \leq 3, \\ -x^3+2x^2+3x & \text{ if } x>3. \end{array} \right.\)

consider 

y=3-x     

when x=3 y= 0

when x<=3    y>=3

since y=3-x

x=3-y

and the inverse becomes    y=3-x   where  x>=0

so

\(f^{-1}(x)=3-x\qquad if \qquad x\ge0\\ f^{-1}(0)=3\\ f^{-1}(6)=3-6=-3\\ \\~\\f^{-1}(0)+f^{-1}(6)=3-3=0 \)

\(1_6 + 2_6 + 3_6 + \cdots + 45_6 + 50_6 + 51_6 \)

Probably the most straight forward way to takle this problem is to work out what 51 base 6 is in base 10,

do the sum and then convert back.

51base6 = 5*6+1 = 31 base 10

1+2+...+31 = 31/2(2+30)= 31*16 = 496

496/6 = 82R4

82/6=13R4

13/6=2R1

2/6=0R2

496 base 10 = 2144 base6

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

I just want to play with another method  (all working is in base 6)

(0+ 1+2+3+4+5)   +   (  10+ .... 15)     +  (  20 .....+25 )      + ( 30 ....+35 )    + ( 40+....45 )+  50 +51

=   5(0+1+2+3+4+5) +0+1   +     (10*10)   + (20*10)  +(30*10)  + (40*10) +50*2

=    5* (10+10+3)  +1            +       100     +      200  +   300  +   400  + 140

=      5* 23  +1                       +      1540

=          204  +1540

=   2144 base 6

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

This is how I would do it.

And yes our answers do agree.  Thanks guest.

\(\frac{1}{\sqrt{2} + \sqrt{5} + \sqrt{7}}\\ \frac{1}{(\sqrt{2} + \sqrt{5}) + \sqrt{7}}*\frac{(\sqrt{2} + \sqrt{5}) - \sqrt{7}}{(\sqrt{2} + \sqrt{5}) - \sqrt{7}}\\ =\frac{(\sqrt{2} + \sqrt{5} - \sqrt{7})}{(2+5+2\sqrt{10}) -7}\\ =\frac{(\sqrt{2} + \sqrt{5} - \sqrt{7})}{(2\sqrt{10}) }\\ =\frac{(\sqrt{2} + \sqrt{5} - \sqrt{7})}{(2\sqrt{10}) }*\frac{\sqrt{10}}{\sqrt{10}}\\ =\frac{\sqrt{10}(\sqrt{2} + \sqrt{5} - \sqrt{7})}{20}\\ etc \)

You need to check my working for careless errors