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Hi Radix,

Yes that is what I assumed :)

squared is the inverse of square root.

16 squared = $$16^2=256$$

Square root of 16 = sqrt(16) = $$\sqrt{16}=4$$                

    43₆                         

     x 51₆                         

   --------

       43

     343

    ------

   3513₆

Please do not duble post like this Sabi92.

We do not hve time to answer the same post twice!

If you want to repost you are welcome to do so but please follow these instructions.

Hi anon,

Thankyou for trying to answer.

You can no longer post pictures unless you are a member.

Why don't you join up.  It is super easy, you just need to choose a username and a passsword :)

Thanks Heureka,

I would have done this as the sum of 2 cubes

$$\\\boxed{a^3+b^3=(a+b)(a^2-ab+b^2)}\\\\\\
(x(x+2))^3+8x^3=0\\\\
x^3((x+2)^3+8)=0\\\\
x=0\qquad or\\\\
(x+2)^3+2^3=0\\\\
((x+2)+2)((x+2)^2-2(x+2)+2^2)=0\\\\
(x+4)(x^2+4x+4-2x-4+4)=0\\\\$$

$$\\(x+4)(x^2+2x+4)=0\\\\
x=-4\qquad or\\\\
x^2+2x+4=0\\\\
$This has no real roots. \\\\So$\\\\
x=0 \;\;or\;\; x=-4\\\\$$

ACTUALLY, NOW I LOOK AT IT, HEUREKA'S ANSWER IS BETTER :)

We have two sums to consider.....

2*3^(-1) + 2*3^(-3) + 2*3^(-5)+ ....+2*3^-(2n-1)  =  (2/3) /(1 - 3^(-2)) = (2/3) / (1 - 1/9)  =

(2/3)/(8/9)  = (2/3)*(9/8)  = 18/24 = 3/4   ...... and...... 

3^(-2) + 3^(-4) + 3^(-6) + ....+ 3^-(2n)  =  (1/9) / (1 - 3^(-2))  = (1/9)/ ( 1 - 1/9)  = (1/9)/(8/9) =

(1/9) * (9/8)  =  9/72  = 1/8

So .......  3/4 + 1/8  =  6/8 + 1/8 =  7/8

Oh, ok now I see :))

I see we have interpreted the question differently Melody.

You have taken 242,797,824 as a list of single digits.  I've just assumed there are 242 million, 797 thousand, 8 hundred and 24 distinguishable items.

Who knows?!!

.

A:

$$\small{\text{$
\begin{array}{rcl}
\mathbf{
\left[~x \left( x+2 \right) \right~]^3 + 8 x^3 } &\mathbf{
=}& \mathbf{0} \\
x^3 \left( x+2 \right)^3 + 8 x^3 &=& 0\\
x^3 \left( x^3+3x^2\cdot 2 + 3x\cdot 2^2 + 2^3 \right) + 8 x^3 &=& 0\\
x^3 \left( x^3+6x^2 + 12x + 8 \right) + 8 x^3 &=& 0\\
x^3 \left( x^3+6x^2 + 12x + 8 +8\right) &=& 0\\
x^3 \left( x^3+6x^2 + 12x + 16 \right) &=& 0\\
\end{array}
$}}$$

I. First solution

$$x^3=0 \qquad \Rightarrow \qquad x=0$$

II. Second solution

$$\underbrace{x^3+6x^2 + 12x + 16}_{=(x+4)(x^2+2x+4)} = 0 \qquad \Rightarrow \qquad x = -4$$

III. No more solutions

$$x^2+2x+4 = 0$$

The parabola has no solutions, because the vertex of this parabola (-1 | 3) lies above the x-axis

B:

$$\small{\text{$
\begin{array}{rcl}
\mathbf{
\left[~x \left( x+2 \right) \right~]^3 + 8 x^3 } &\mathbf{
=}& \mathbf{0} \\
x^3 \left( x+2 \right)^3 + 8 x^3 &=& 0\\
x^3 \left[~~\left( x+2 \right)^3 + 8~~\right] &=& 0\\
\end{array}
$}}$$

I. First solution:

$$x^3=0 \qquad \Rightarrow \qquad x=0$$

II. Second solution:

$$\small{\text{$
\begin{array}{rcl}
\left( x+2 \right)^3 + 8 &=& 0\\
\left( x+2 \right)^3 &=& -8 \qquad | \qquad \sqrt[3]{}\\
x+2 &=& \sqrt[3]{-8}\\
x+2 &=& -2\\
x &=& -2-2\\
x &=& -4\\
\end{array}
$}}$$

Let's just divide, here......

5 / 1.5    = 10/3  batches of 15  =  50 cookies 

4 / 1.25  = 32/10 batches of 15  =  48 cookies

1.75 / .5 = 35/10  batches of 15  =  52.5 cookies

He could bake 48 cookies, at most........

Which as you said gives:

Your two roots are given by

and

Therefore

Finally:

Nice answer, Melody.....

BTW....I think Melody meant to write.......  3(81) + 5    =  243 + 5  = 248

Thanks Chris :)

That's a nice approach and nice thinking, Melody....

6 = 6 + x     =>    x = 0     divide by  3,14    =>   0

Thank you Melody,

I thought  squared =  sqrt   = $${\sqrt{{\mathtt{1}}}} = {\mathtt{1}}$$

Hi Radix

They probably mean

$${{\mathtt{58}}}^{{\mathtt{2}}} = {\mathtt{3\,364}}$$

Only one other integer requires nine cubes of positive integers to represent its value as a sum of cubes. This other integer is between 200 and 300. Can you find it ?

$$239=5^3+3^3+3^3+3^3 +2^3+2^3+2^3+2^3+1^3$$

Do you mean this ?

$${\sqrt{{\mathtt{58}}}} = {\mathtt{7.615\: \!773\: \!105\: \!863\: \!908\: \!3}}$$

$${\mathtt{125}}{\mathtt{\,\times\,}}{\mathtt{336}} = {\mathtt{42\,000}}$$

$${\frac{{\mathtt{1}}}{{\mathtt{20}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{10}}}} = {\frac{{\mathtt{1}}}{{\mathtt{20}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}}{{\mathtt{20}}}} = {\frac{{\mathtt{3}}}{{\mathtt{20}}}}$$

$${\frac{{\mathtt{1}}}{{\mathtt{20}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{10}}}} = {\frac{{\mathtt{3}}}{{\mathtt{20}}}} = {\mathtt{0.15}}$$

Find the equation of the straight line which is parallel to the line whose equation is 3x + 4y = 0, and which passes through the point of intersection of the lines x - 2y = a and x + 3y = 2a. Express the answer in the form Ax + By = C

I. The slope of the new straight line:

$$\small{\text{$
3x + 4y = 0 \rm{~~or~~}y = -\frac34 \cdot x \rm{~~so~the ~ slope ~is ~~} -\frac34
$.}}
\small{\text{$
\rm{~The ~ slope ~of~a ~parallel ~line ~is ~ also ~} -\frac34
$}}$$

II. intersection of the lines x - 2y = a and x + 3y = 2a:

$$\small{\text{$
\begin{array}{lrcl}
(1): & x - 2y &=& a \\
(2): & x+3y &=& 2a\\
\hline
(2)-(1): & x-x + 3y-(-2y) &=& 2a-a \\
& 3y+2y &=& a \\
& 5y &=& a \\
& \textcolor[rgb]{1,0,0}{y} & \textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{\frac15 a}\\
\hline
(1): & x - 2y &=& a \\
& x - 2y &=& a \\
& x &=& a + 2y \quad | \quad y = \frac15 a \\
& x &=& a + 2(\frac15 a) \\
& \textcolor[rgb]{1,0,0}{x} & \textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{\frac75 a} \\
\end{array}
$}}$$

The intersection is $$(~\frac75 a ~|~ \frac15 a~)$$

III. Find the equation of the straight line:

$$\small{\text{$
\begin{array}{rcl}
y &=&mx+b \qquad | \qquad m = -\frac34 \\
y&=&-\frac34 \cdot x+b
\end{array}
$}}$$

find b:

$$\small{\text{$
\begin{array}{rcl}
y_s&=&-\frac34 \cdot x_s+b \qquad | \qquad x_s = \frac75 a \rm{~~and~~} y_s = \frac15 a\\\\
\frac15 a &=&-\frac34 \cdot \frac75 a+b \\\\
b&=& \frac15 a+\frac34 \cdot \frac75 a \\\\
b&=& \frac{25}{20} a \\\\
\textcolor[rgb]{1,0,0}{b}&\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{\frac{5}{4} a} \\\\
\end{array}
$}}$$

The equation of the straight line is:

$$\small{\text{$
\begin{array}{lrcl}
&y &=& mx+b \qquad | \qquad m = -\frac34 \rm{~~and~~} b = \frac54 a\\\\
&\mathbf{y} &\mathbf{=}& \mathbf{-\frac34 \cdot x + \frac{5}{4} a}\\\\
\rm{or~~} \\\\
&\mathbf{\frac34 \cdot x } + 1\cdot \mathbf{y} &\mathbf{=}& \mathbf{ \frac{5}{4} a }
\end{array}
$}}$$

@@ End of Day Wrap  Mon 8/6/15   Sydney, Australia Time   5:50pm   ♪ ♫

Good afternoon / morning everyone,  

It is not often that I can leave out 'night', I am writing this wrap at a respectable time today.  

Great answers today from Geno3141, Eloise1st, CPhill, Dbg4thebest, Heureka, Alan and radix.  A big thank you to each of you